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Class 12 Maths Case Study Questions

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Class 12 Maths question paper will have 1-2 Case Study Questions. These questions will carry 5 MCQs and students will attempt any four of them. As all of these are only MCQs, it is easy to score good marks with a little practice. Class 12 Maths Case Study Questions are available on the myCBSEguide App and Student Dashboard .

Why Case Studies in CBSE Syllabus?

CBSE has introduced case study questions in the CBSE curriculum recently. The purpose was to make students ready to face real-life challenges with the knowledge acquired in their classrooms. It means, there was a need to connect theories with practicals. Whatsoever the students are learning, they must know how to apply it in their day-to-day life. That’s why CBSE is emphasizing case studies and competency-based education .

Case Study Questions in Maths

Let’s have a look over the class 12 Mathematics sample question paper issued by CBSE, New Delhi. Question numbers 17 and 18 are case study questions.

Focus on concepts

If you go through each MCQ there, you will find that the theme/case study is common but the questions are based on different concepts related to the theme. It means, that if you have done ample practice on the various concepts, you can solve all these MCQs in minutes.

Easy Questions with a Practical Approach

The difficulty level of the questions is average or say easy in some cases. On the other hand, you get four options to choose from. So, you get two levels of support to get full marks with very little effort.

Practice Questions Regularly

Most of the time we feel that it’s easy and neglect it. But in the end, we have to pay for this negligence. This may happen here too. Although it’s easy to score good marks on the case study questions if you don’t practice such questions, you may lose your marks. So, we suggest students should practice at least 30-40 such questions before writing the board exam.

12 Maths Case-Based Questions

We are giving you some examples of case study questions here. We have arranged hundreds of such questions chapter-wise on the myCBSEguide App. It is the complete guide for CBSE students. You can download the myCBSEguide App and get more case study questions there.

Case Study Question – 1

  • A is a diagonal matrix
  • A is a scalar matrix
  • A is a zero matrix
  • A is a square matrix
  • If A and B are two matrices such that AB = B and BA = A, then B 2 is equal to

Case Study Question – 2

  • 4(x 3  – 24x 2   + 144x)
  • 4(x 3 – 34x 2   + 244x)
  • x 3  – 24x 2   + 144x
  • 4x 3  – 24x 2   + 144x
  • Local maxima at x = c 1
  • Local minima at x = c 1
  • Neither maxima nor minima at x = c 1
  • None of these

Case Study Questions Matrices -1

Answer Key:

Case Study Questions Matrices – 2

Read the case study carefully and answer any four out of the following questions: Once a mathematics teacher drew a triangle ABC on the blackboard. Now he asked Jose,” If I increase AB by 11 cm and decrease the side BC by 11 cm, then what type of triangle it would be?” Jose said, “It will become an equilateral triangle.”

Again teacher asked Suraj,” If I multiply the side AB by 4 then what will be the relation of this with side AC?” Suraj said it will be 10 cm more than the three times AC.

Find the sides of the triangle using the matrix method and  answer the following questions:

  • (a) 3  ×  3

Case Study Questions Determinants – 01

DETERMINANTS:  A determinant is a square array of numbers (written within a pair of vertical lines) that represents a certain sum of products. We can solve a system of equations using determinants, but it becomes very tedious for large systems. We will only do 2 × 2 and 3 × 3 systems using determinants. Using the properties of determinants solve the problem given below and answer the questions that follow:

Three shopkeepers Ram Lal, Shyam Lal, and Ghansham are using polythene bags, handmade bags (prepared by prisoners), and newspaper envelopes as carrying bags. It is found that the shopkeepers Ram Lal, Shyam Lal, and Ghansham are using (20,30,40), (30,40,20), and (40,20,30) polythene bags, handmade bags, and newspapers envelopes respectively. The shopkeeper’s Ram Lal, Shyam Lal, and Ghansham spent ₹250, ₹270, and ₹200 on these carry bags respectively.

  • (b) Shyam Lal
  • (a) Ram Lal

Case Study Questions Determinants – 02

Case study questions application of derivatives.

  • R(x) = -x 2  + 200x + 150000
  • R(x) = x 2  – 200x – 140000
  • R(x) = 200x 2  + x + 150000
  • R(x) = -x 2  + 100 x + 100000
  • R'(x) > 0
  • R'(x) < 0
  • R”(x) = 0
  • (a) -x 2  + 200x + 150000
  • (a) R'(x) = 0
  • (c) 257, -63

Case Study Questions Vector Algebra

  • tan−1⁡(5/12)
  • tan−1⁡(12/3)
  • (b) 130 m/s
  • (a)  tan−1⁡(5/12)
  • (b) 170 m/s

More Case Study Questions

These are only some samples. If you wish to get more case study questions for CBSE class 12 maths, install the myCBSEguide App. It has class 12 Maths chapter-wise case studies with solutions.

12 Maths Exam pattern

Question Paper Design of CBSE class 12 maths is as below. It clearly shows that 20% weightage will be given to HOTS questions. Whereas 55% of questions will be easy to solve.

1.  Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.
 Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas
4455
2.  Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.2025
3.
Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations
1620

Present and defend opinions by making judgments about information, the validity of ideas, or quality of work based on a set of criteria.

Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions
80100
  • No. chapter-wise weightage. Care to be taken to cover all the chapters
  • Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections

Periodic Tests ( Best 2 out of 3 tests conducted)10 Marks
Mathematics Activities10 Marks

12 Maths Prescribed Books

  • Mathematics Part I – Textbook for Class XII, NCERT Publication
  • Mathematics Part II – Textbook for Class XII, NCERT Publication
  • Mathematics Exemplar Problem for Class XII, Published by NCERT
  • Mathematics Lab Manual class XII, published by NCERT

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Class 12 Maths: Case Study Based Questions PDF Download

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You must practice some good Case Study questions of Class 12 Maths to boost your preparation to score 95+% on Boards. In this post, you will get Case Study Questions of All Chapters which will come in CBSE Class 12 Maths Board Exams.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

Download Books for Boards

We have provided here Case Study questions for the Class 12 Maths exams. You can read these chapter-wise Case Study questions. Prepared by subject experts and experienced teachers. The answer key is also provided so that you can check the correct answer for each question. Practice these questions to score well in your Board Final exams.

We are providing Case Study questions for class 12 Biology based on the latest syllabi. There is a total of 13 chapters included in CBSE class 12 Maths exams. Students can practice these questions for concept clarity and score better marks in their exams.

Table of Contents

CBSE Class 12th – MATHS : Chapterwise Case Study Question & Solution

CBSE will ask two Case Study Questions in the CBSE class 12 maths questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them.

Case Study Based Questions for Class 12 Maths

Class 12 Physics Case Study Questions Class 12 Chemistry Case Study Questions Class 12 Biology Case Study Questions Class 12 Maths Case Study Questions

Books for Class 12 Maths

Strictly as per the new term-wise syllabus for Board Examinations to be held in the academic session 2022-23 for class 12 Multiple Choice Questions based on new typologies introduced by the board- Stand-Alone MCQs, MCQs based on Assertion-Reason Case-based MCQs. Include Questions from CBSE official Question Bank released in April 2022 Answer key with Explanations What are the updates in the book: Strictly as per the Term wise syllabus for Board Examinations to be held in the academic session 2022-23. Chapter-wise -Topic-wise Multiple choice questions based on the special scheme of assessment for Board Examination for Class 12th.

case study questions on maxima and minima class 12

Class 12 Maths Syllabus 2022-23

Unit-I: Relations and Functions

1. Relations and Functions (15 Periods)

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions.

2. Inverse Trigonometric Functions (15 Periods)

Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit-II: Algebra

1. Matrices (25 Periods)

Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operation on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Oncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants 25 Periods

Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit-III: Calculus

1. Continuity and Differentiability (20 Periods)

Continuity and differentiability, chain rule, derivative of inverse trigonometric functions, 𝑙𝑖𝑘𝑒 sin −1  𝑥 , cos −1  𝑥 and tan −1  𝑥, derivative of implicit functions. Concept of exponential and logarithmic functions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.

2. Applications of Derivatives (10 Periods)

Applications of derivatives: rate of change of bodies, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as reallife situations).

3. Integrals (20 Periods)

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

jagran josh

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals (15 Periods)

Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only)

5. Differential Equations (15 Periods)

Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

jagran josh

Unit-IV: Vectors and Three-Dimensional Geometry

1. Vectors (15 Periods)

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors.

2. Three – dimensional Geometry (15 Periods)

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines.

Unit-V: Linear Programming

1. Linear Programming (20 Periods)

Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Unit-VI: Probability

1. Probability 30 (Periods)

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

case study questions on maxima and minima class 12

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RD Sharma solutions for Class 12 Maths chapter 18 - Maxima and Minima [Latest edition]

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RD Sharma solutions for Class 12 Maths chapter 18 - Maxima and Minima - Shaalaa.com

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Solutions for chapter 18: maxima and minima.

Below listed, you can find solutions for Chapter 18 of CBSE, Karnataka Board PUC RD Sharma for Class 12 Maths.

RD Sharma solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.1 [Page 7]

f(x) = 4x 2 + 4 on R .

f(x) = - (x-1) 2 +2 on R ?

f(x)=| x+2 | on R .

f(x)=sin 2x+5 on R .

f(x) = | sin 4x+3 | on R ?

f(x)=2x 3   +5 on R .

f (x) = \[-\] | x + 1 | + 3 on R .

f(x) = 16x 2 \[-\] 16x + 28 on R ?

f(x) = x 3  \[-\] 1 on R .

RD Sharma solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.2 [Page 16]

f(x) = (x \[-\] 5) 4 .

f(x) = x 3  \[-\] 3x .

f(x) = x 3   (x \[-\] 1) 2  .

f(x) =  (x \[-\] 1) (x+2) 2 . 

f(x) = \[\frac{1}{x^2 + 2}\] .

f(x) =  x 3  \[-\] 6x 2  + 9x + 15 . 

f(x) = sin 2x, 0 < x < \[\pi\] .

f(x) =  sin x \[-\] cos x, 0 < x < 2\[\pi\] .

f(x) =  cos x, 0 < x < \[\pi\] .

`f(x)=sin2x-x, -pi/2<=x<=pi/2`

`f(x)=2sinx-x, -pi/2<=x<=pi/2`

f(x) =\[x\sqrt{1 - x} , x > 0\].

f(x) = x 3  (2x \[-\] 1) 3 .

f(x) =\[\frac{x}{2} + \frac{2}{x} , x > 0\] .

RD Sharma solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.3 [Page 31]

f(x) = x 4 \[-\] 62x 2  + 120x + 9.

f(x) = x 3 \[-\] 6x 2  + 9x + 15

f(x) = (x - 1) (x + 2) 2 .

`f(x) = 2/x - 2/x^2,  x>0`

f(x) = xe x .

`f(x) = x/2+2/x, x>0 `.

`f(x) = (x+1) (x+2)^(1/3), x>=-2` .

`f(x)=xsqrt(32-x^2),  -5<=x<=5` .

f(x) = \[x^3 - 2a x^2 + a^2 x, a > 0, x \in R\] .

f(x) = \[x + \frac{a2}{x}, a > 0,\] , x ≠ 0 .

f(x) = \[x\sqrt{2 - x^2} - \sqrt{2} \leq x \leq \sqrt{2}\] .

f(x) = \[x + \sqrt{1 - x}, x \leq 1\] .

f(x) = (x \[-\] 1) (x \[-\] 2) 2 .

`f(x)=xsqrt(1-x),  x<=1` .

f(x) = \[- (x - 1 )^3 (x + 1 )^2\] .

The function y = a log x+bx 2  + x has extreme values at x=1 and x=2. Find a and b ?

Show that \[\frac{\log x}{x}\] has a maximum value at x = e ?

Find the maximum and minimum values of the function f(x) = \[\frac{4}{x + 2} + x .\]

Find the maximum and minimum values of y = tan \[x - 2x\] .

If f(x) = x 3  + ax 2  + bx + c has a maximum at x = \[-\] 1 and minimum at x = 3. Determine a, b and c ?

Prove that f(x) = sinx + \[\sqrt{3}\] cosx has maximum value at x = \[\frac{\pi}{6}\] ?

RD Sharma solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.4 [Page 37]

f(x) = 4x \[-\] \[\frac{x^2}{2}\] in [ \[-\] 2,4,5] .

f(x) = (x \[-\] 1) 2  + 3 in [ \[-\] 3,1] ?

`f(x) = 3x^4 - 8x^3 + 12x^2- 48x + 25 " in "[0,3]` .

f(x) = (x \[-\] 2) \[\sqrt{x - 1} \text { in  }[1, 9]\] .

Find the maximum value of 2x 3 \[-\] 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [ \[-\] 3, \[-\] 1].

Find the absolute maximum and minimum values of the function of given by \[f(x) = \cos^2 x + \sin x, x \in [0, \pi]\] .

Find the absolute maximum and minimum values of a function f given by `f(x) = 12 x^(4/3) - 6 x^(1/3) , x in [ - 1, 1]` .

Find the absolute maximum and minimum values of a function f given by \[f(x) = 2 x^3 - 15 x^2 + 36x + 1 \text { on the interval }  [1, 5]\] ?

RD Sharma solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5 [Pages 72 - 74]

Determine two positive numbers whose sum is 15 and the sum of whose squares is maximum.

Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

How should we choose two numbers, each greater than or equal to `-2, `whose sum______________ so that the sum of the first and the cube of the second is minimum?

Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm 3 , which has the minimum surface area?

A beam is supported at the two end and is uniformly loaded. The bending moment M at a distance x from one end is given by \[M = \frac{WL}{2}x - \frac{W}{2} x^2\] .

Find the point at which M is maximum in a given case.

A beam is supported at the two end and is uniformly loaded. The bending moment M at a distance x from one end is given by \[M = \frac{Wx}{3}x - \frac{W}{3}\frac{x^3}{L^2}\] .

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?

A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle is minimum?

Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle.

Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.   

Two sides of a triangle have lengths 'a' and 'b' and the angle between them is \[\theta\]. What value of \[\theta\] will maximize the area of the triangle? Find the maximum area of the triangle also.  

A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Find this maximum volume.

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, in cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

A tank with rectangular base and rectangular sides, open at the top, is to the constructed so that its depth is 2 m and volume is 8 m 3 . If building of tank cost 70 per square metre for the base and Rs 45 per square metre for sides, what is the cost of least expensive tank?

A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening.

A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window.

Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius R is \[\frac{2R}{\sqrt{3}} .\]

A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area ?

Prove that a conical tent of given capacity will require the least amount of  canavas when the height is \[\sqrt{2}\] times the radius of the base.

Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to \[ \frac{2}{3} \] of the diameter of the sphere.

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is \[\cot^{- 1} \left( \sqrt{2} \right)\] .

An isosceles triangle of vertical angle 2 \[\theta\] is inscribed in a circle of radius  a . Show that the area of the triangle is maximum when \[\theta\] = \[\frac{\pi}{6}\] .

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is \[6\sqrt{3}\]r. 

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides ?

Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm ?

A closed cylinder has volume 2156 cm 3 . What will be the radius of its base so that its total surface area is minimum ?

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius \[5\sqrt{3 cm} \text { is }500 \pi  {cm}^3 .\]

Show that among all positive numbers x and y with x 2 + y 2 =r 2 , the sum x+y is largest when x=y=r \[\sqrt{2}\] .

Determine the points on the curve x2 = 4y which are nearest to the point (0,5) ?

Find the point on the curve y 2 = 4x which is nearest to the point (2,\[-\] 8).

Find the point on the curve x 2 = 8y which is nearest to the point (2, 4) ?

Find the point on the parabolas x 2  = 2y which is closest to the point (0,5) ?

Find the coordinates of a point on the parabola y=x 2 +7x + 2 which is closest to the strainght line y = 3x \[-\] 3 ?

Find the point on the curvey y 2 = 2x which is at a minimum distance from the point (1, 4).

Find the maximum slope of the curve y = \[- x^3 + 3 x^2 + 2x - 27 .\]

The total cost of producing x radio sets per  day is Rs \[\left( \frac{x^2}{4} + 35x + 25 \right)\] and the price per set  at which they may be sold is Rs. \[\left( 50 - \frac{x}{2} \right) .\] Find the daily output to maximum the total profit.

Manufacturer can sell x items at a price of rupees \[\left( 5 - \frac{x}{100} \right)\] each. The cost price is Rs  \[\left( \frac{x}{5} + 500 \right) .\] Find the number of items he should sell to earn maximum profit.

An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.

A box of constant volume c is to be twice as long as it is wide. The material on the top and four sides cost three times as much per square metre as that in the bottom. What are the most economic dimensions?

The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its semi-circular ends is \[\pi : (\pi + 2)\].

The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius a ?

A straight line is drawn through a given point P(1,4). Determine the least value of the sum of the intercepts on the coordinate axes ?

The total area of a page is 150 cm 2 . The combined width of the margin at the top and bottom is 3 cm and the side 2 cm. What must be the dimensions of the page in order that the area of the printed matter may be maximum?

The space s described in time  t  by a particle moving in a straight line is given by S = \[t5 - 40 t^3 + 30 t^2 + 80t - 250 .\] Find the minimum value of acceleration.

A particle is moving in a straight line such that its distance at any time  t  is given by  S = \[\frac{t^4}{4} - 2 t^3 + 4 t^2 - 7 .\]  Find when its velocity is maximum and acceleration minimum.

RD Sharma solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.6 [Page 80]

Write necessary condition for a point x = c to be an extreme point of the function f(x).

Write sufficient conditions for a point x = c to be a point of local maximum.

If f(x) attains a local minimum at x = c, then write the values of `f' (c)` and `f'' (c)`.

Write the minimum value of f(x) = \[x + \frac{1}{x}, x > 0 .\]

Write the maximum value of f(x) = \[x + \frac{1}{x}, x > 0 .\] 

Write the point where f(x) = x log, x attains minimum value.

Find the least value of f(x) = \[ax + \frac{b}{x}\], where a > 0, b > 0 and x > 0 .

Write the minimum value of f(x) = x x  .

Write the maximum value of f(x) = x 1 /x .

Write the maximum value of f(x) = \[\frac{\log x}{x}\], if it exists .

RD Sharma solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.7 [Pages 80 - 82]

The maximum value of x 1 /x , x > 0 is __________ .

none of these

If \[ax + \frac{b}{x} \frac{>}{} c\] for all positive x where a,b,>0, then _______________ .

`ab<c^2/4`

`ab>=c^2/4`

`ab>=c/4`

The minimum value of \[\frac{x}{\log_e x}\] is _____________ .

For the function f(x) = \[x + \frac{1}{x}\]

x = 1 is a point of maximum

x = \[-\] 1 is a point of minimum

maximum value > minimum value

maximum value < minimum value

Let f(x) = x 3 +3x 2  \[-\] 9x+2. Then, f(x) has _________________ .

a maximum at x = 1

a minimum at x = 1

neither a maximum nor a minimum at x = - 3

The minimum value of f(x) = \[x4 - x2 - 2x + 6\] is _____________ .

The number which exceeds its square by the greatest possible quantity is _________________ .

\[\frac{1}{2}\]

\[\frac{1}{4}\]

\[\frac{3}{4}\]

Let f(x) = (x \[-\] a) 2  + (x \[-\] b) 2  + (x \[-\] c) 2 . Then, f(x) has a minimum at x = _____________ .

\[\frac{a + b + c}{3}\]

\[\sqrt[3]{abc}\]

\[\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\]

The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is ______________ .

\[\frac{1}{8}\]

The function f(x) = \[\sum^5_{r = 1}\] (x \[-\] r) 2 assumes minimum value at x = ______________ .

At x= \[\frac{5\pi}{6}\] f(x) = 2 sin 3x + 3 cos 3x is ______________ .

If x lies in the interval [0,1], then the least value of x 2 + x + 1 is _______________ .

The least value of the function f(x) = \[x3 - 18x2 + 96x\] in the interval [0,9] is _____________ .

The maximum value of f(x) = \[\frac{x}{4 - x + x^2}\] on [ \[-\] 1, 1] is _______________ .

\[ \frac{1}{4}\]

\[- \frac{1}{3}\]

\[\frac{1}{6}\]

\[\frac{1}{5}\]

The point on the curve y 2 = 4x which is nearest to, the point (2,1) is _______________ .

\[1, 2\sqrt{2}\]

If x+y=8, then the maximum value of xy is ____________ .

The least and greatest values of f(x) = x 3 \[-\] 6x 2 +9x in [0,6], are ___________ .

f(x) = \[\sin + \sqrt{3} \cos x\] is maximum when x = ___________ .

\[\frac{\pi}{3}\]

\[\frac{\pi}{4}\]

\[\frac{\pi}{6}\]

If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is ______________ .

\[\frac{1}{3}\]

\[\frac{2}{3}\]

The minimum value of \[\left( x^2 + \frac{250}{x} \right)\] is __________ .

If(x) = x+\[\frac{1}{x}\],x > 0, then its greatest value is _______________ .

If(x) = \[\frac{1}{4x^2 + 2x + 1}\] then its maximum value is _________________ .

\[\frac{4}{3}\]

Let x, y be two variables and x>0, xy=1, then minimum value of x+y is _______________ .

\[2\frac{1}{2}\]

\[3\frac{1}{3}\]

f(x) = 1+2 sin x+3 cos 2 x, `0<=x<=(2pi)/3` is ________________ .

Minimum at x =\[\frac{\pi}{2}\]

Maximum at x = sin \[- 1\] ( \[\frac{1}{\sqrt{3}}\])

Minimum at x = \[\frac{\pi}{6}\]

Maximum at `sin^-1(1/6)`

The function f(x) = \[2 x^3 - 15 x^2 + 36x + 4\] is maximum at x = ________________ .

The maximum value of f(x) = \[\frac{x}{4 + x + x^2}\] on [ \[-\] 1,1] is ___________________ .

\[- \frac{1}{4}\]

Let f(x) = 2x 3 \[-\] 3x 2 \[-\] 12x + 5 on [ 2, 4]. The relative maximum occurs at x = ______________ .

The minimum value of x log e x is equal to ____________ .

The minimum value of the function `f(x)=2x^3-21x^2+36x-20` is ______________ .

RD Sharma solutions for Class 12 Maths chapter 18 - Maxima and Minima

Shaalaa.com has the CBSE, Karnataka Board PUC Mathematics Class 12 Maths CBSE, Karnataka Board PUC solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. RD Sharma solutions for Mathematics Class 12 Maths CBSE, Karnataka Board PUC 18 (Maxima and Minima) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Class 12 Maths chapter 18 Maxima and Minima are Maximum and Minimum Values of a Function in a Closed Interval, Maxima and Minima, Simple Problems on Applications of Derivatives, Graph of Maxima and Minima, Approximations, Tangents and Normals, Increasing and Decreasing Functions, Rate of Change of Bodies or Quantities, Introduction to Applications of Derivatives.

Using RD Sharma Class 12 Maths solutions Maxima and Minima exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in RD Sharma Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board PUC Class 12 Maths students prefer RD Sharma Textbook Solutions to score more in exams.

Get the free view of Chapter 18, Maxima and Minima Class 12 Maths additional questions for Mathematics Class 12 Maths CBSE, Karnataka Board PUC, and you can use Shaalaa.com to keep it handy for your exam preparation.

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Chapter 18: Maxima and Minima

Maxima and minima.

One of the major applications of derivatives is determination of the maximum and minimum values of a function. We know that the derivative provides the information about the gradient of the graph of a function which can be used to identify the points where the gradient is zero. These points are usually related to the largest and the smallest values of the function. For more details use the links given below:

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Chapter 18: Maxima and Minima – Exercise...

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Class 12-science RD SHARMA Solutions Maths Chapter 18 - Maxima and Minima

Maxima and minima exercise mcq, solution 30.

case study questions on maxima and minima class 12

f'(x) = 2x - 8

Take f'(x) = 0, we get

Again differentiating, we get

f"(x) = 2 > 0 for all real x

Therefore, x = 4 is the point of minima.

The minimum value of f(x) is

f(4) = 16 - 32 + 17 = 1

Solution 32

The maximum value of f(x) is

Solution 33

Taking log on both the sides, we get

From equation (ii), we get

Solution 34

Solution 35

Again differentiating w.r.t x, we get

So, the slope is maximum at x = 1

Solution 36

Now, let's find f(x) at x = 2 or -1

Therefore, x = -1 is the point of local maxima and the maximum value is 11.

Whereas, x = 2 is the point of local minima and the minimum value is -16.

Hence, f(x) has one maximum and one minimum.

case study questions on maxima and minima class 12

Solution 10

case study questions on maxima and minima class 12

Solution 11

case study questions on maxima and minima class 12

Solution 12

case study questions on maxima and minima class 12

Solution 13

case study questions on maxima and minima class 12

Solution 14

case study questions on maxima and minima class 12

Solution 15

case study questions on maxima and minima class 12

Solution 16

case study questions on maxima and minima class 12

Solution 17

case study questions on maxima and minima class 12

Solution 18

case study questions on maxima and minima class 12

Solution 19

case study questions on maxima and minima class 12

Solution 20

case study questions on maxima and minima class 12

Solution 21

case study questions on maxima and minima class 12

Solution 22

case study questions on maxima and minima class 12

Solution 23

case study questions on maxima and minima class 12

Solution 24

case study questions on maxima and minima class 12

Solution 25

case study questions on maxima and minima class 12

Solution 26

case study questions on maxima and minima class 12

Solution 27

case study questions on maxima and minima class 12

Solution 28

case study questions on maxima and minima class 12

Solution 29

case study questions on maxima and minima class 12

The idea behind introducing this question is to identify whether the student has understood the logical sequences that should be performed while performing any type of calculation. These are of very basic level questions where the focus is to understand the importance of mathematical operator, based on its position. If a , — x, v sign comes before a number then it has entirety different significance fromwhat it gives when comes after the number. This is where the practice of BODIVIAS is required.

Maxima and Minima Exercise Ex. 18.1

case study questions on maxima and minima class 12

Maxima and Minima Exercise Ex. 18.2

case study questions on maxima and minima class 12

Maxima and Minima Exercise Ex. 18.3

Solution 1(i).

case study questions on maxima and minima class 12

Solution 1(ii)

case study questions on maxima and minima class 12

Solution 1(iii)

case study questions on maxima and minima class 12

Solution 1(iv)

case study questions on maxima and minima class 12

Solution 1(v)

case study questions on maxima and minima class 12

Solution 1(vi)

case study questions on maxima and minima class 12

Solution 1(vii)

case study questions on maxima and minima class 12

Solution 1(viii)

case study questions on maxima and minima class 12

Solution 1(ix)

case study questions on maxima and minima class 12

Solution 1(x)

case study questions on maxima and minima class 12

Solution 1(xi)

case study questions on maxima and minima class 12

Solution 1(xii)

case study questions on maxima and minima class 12

Solution 2(i)

case study questions on maxima and minima class 12

Solution 2(ii)

case study questions on maxima and minima class 12

Solution 2(iii)

case study questions on maxima and minima class 12

Take f'(x) = 0

Differentiating f'(x) w.r.t x, we get

Maxima and Minima Exercise Ex. 18.4

case study questions on maxima and minima class 12

Maxima and Minima Exercise Ex. 18.5

case study questions on maxima and minima class 12

Solution 37

Solution 38.

case study questions on maxima and minima class 12

Solution 39

case study questions on maxima and minima class 12

Solution 40

case study questions on maxima and minima class 12

Solution 41

Solution 42

case study questions on maxima and minima class 12

Solution 43

case study questions on maxima and minima class 12

Solution 44

case study questions on maxima and minima class 12

Solution 45

case study questions on maxima and minima class 12

Let r and h be the radius and height of the cylinder.

From (i), we get

Taking O as the centre of the circle, join OE, OF and OD such that

OE = OF = OD = r (radius)

Similarly, AF = r cot x

P = AB + BC + CA

  = AE + EC + BD + DC + AF + BF

Taking second derivative of P, we get

Maxima and Minima Exercise Ex. 18VSAQ

case study questions on maxima and minima class 12

CBSE Class 10 Maths MIQs, Subjective Questions & More

Once you complete your vast syllabus, you can continue to revise the concepts learned in your class. Once you complete your vast syllabus, you can continue to revise the concepts learned in your class. Once you complete your vast syllabus, you can continue to revise the concepts learned in your class. Once you complete your vast syllabus, you can continue to revise the concepts learned in your class.

Once you complete your vast syllabus, you can continue to revise the concepts learned in your class.

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  • Maxima and Minima

As the name suggests, this topic is devoted to the method of finding the maximum and the minimum values of a function in a given domain. It finds application in almost every field of work, and in every subject. Let’s find out more about the maxima and minima in this topic.

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Some day-to-day applications are described below:

  • To an engineer – The maximum and the minimum values of a function can be used to determine its boundaries in real-life. For example, if you can find a suitable function for the speed of a train; then determining the maximum possible speed of the train can help you choose the materials that would be strong enough to withstand the pressure due to such high speeds, and can be used to manufacture the brakes and the rails etc. for the train to run smoothly.
  • To an economist – The maximum and the minimum values of the total profit function can be used to get an idea of the limits the company must put on the salaries of the employees, so as to not go in loss.
  • To a doctor – The maximum and the minimum values of the function describing the total thyroid level in the bloodstream can be used to determine the dosage the doctor needs to prescribe to different patients to bring their thyroid levels to normal.

Types of Maxima and Minima

The maxima or minima can also be called an extremum i.e. an extreme value of the function. Let us have a function y = f(x) defined on a known domain of x. Based on the interval of x, on which the function attains an extremum, the extremum can be termed as a ‘local’ or a ‘global’ extremum. Let’s understand it better in the case of maxima.

Browse more Topics under Application Of Derivatives

  • Rate of Change of Quantities
  • Approximations
  • Increasing and Decreasing Functions
  • Tangents and Normals
  • Local Maxima

A point is known as a Local Maxima of a function when there may be some other point in the domain of the function for which the value of the function is more than the value of the local maxima, but such a point doesn’t exist in the vicinity or neighborhood of the local maxima. You can also understand it as a maximum value with respect to the points nearby it.

Global Maxima

A point is known as a Global Maxima of a function when there is no other point in the domain of the function for which the value of the function is more than the value of the global maxima. Types of Global Maxima:

  • Global maxima may satisfy all the conditions of local maxima. You can also understand it as the Local Maxima with the maximum value in this case.
  • Alternately, the global maxima for an increasing function could be the endpoint in its domain ; as it would obviously have the maximum value. In this case, it isn’t a local maximum for the function.

Similarly, the local and the global minima can be defined. Look at the graph below to identify the different types of maxima and minima.

Maxima and Minima

Stationary Points

A stationary point on a curve is defined as one at which the derivative vanishes i.e. a point (x 0 , f(x 0 )) is a stationary point of f(x) if \({[\frac{df}{dx}]_{x = x_0} = 0}\). Types of stationary points:

  • Local Minimas
  • Inflection Points

We won’t discuss inflection points here. As of now though, you must note that all the points of extremum are stationary points.

Proof:  I’ll prove the above statement for the case of a Local Maxima. Others will simply follow from this. Let us have a function y = f(x) that attains a Local Maximum at point x = x 0 . Near the extremum point, the curve will look something like this:

Maxima and Minima

Clearly, the derivative of the function has to go to 0 at the point of Local Maximum; otherwise, it would never attain a maximum value with respect to its neighbors.

The Second Derivative Test

This test is used to determine whether a stationary point is a Local Maxima or a Local Minima. Whether it is a global maxima/global minima can be determined by comparing its value with other local maxima/minima. Let us have a function y = f(x) with x = x 0 as a stationary point. Then the test says:

  • If \({[\frac{d^2f}{d^2x}]_{x = x_0} < 0}\), then x = x 0 is a point of Local Maxima.
  • If \({[\frac{d^2f}{d^2x}]_{x = x_0} > 0}\), then x = x 0 is a point of Local Minima.
  • If for x > x 0 , \({[\frac{df}{dx}]_{x = x_0} < 0 }\) and for x < x 0 , \({[\frac{df}{dx}]_{x = x_0} > 0}\) i.e. the function is increasing for x < x 0 and decreasing for x > x 0 ; we can conclude that x = x 0 is a point of Local Maxima.
  • Similarly, if for x > x 0 , \({[\frac{df}{dx}]_{x = x_0} > 0 }\) and for x < x 0 , \({[\frac{df}{dx}]_{x = x_0} < 0}\) i.e. the function is decreasing for x < x 0 and increasing for x > x 0 ; we can conclude that x = x 0 is a point of Local Minima.

The proof of the third case can be understood by looking at Fig 1. above for local maxima. Similarly, for local minima, one can get:

Maxima and Minima

Proof of the Second Derivative Test

We’ll prove the test for the case of a Local Minima. The proof for a Local Maxima will follow in a similar fashion. Take a look at the Fig 2. above. One can see that the slope of the tangent drawn at any point on the curve i.e. \({\frac{dt}{dx}}\) changes from a negative value to 0 to a positive value, near the point of local minima. T

his means that the function that is represented by(say)  \({f(x) = \frac{dt}{dx}}\) behaves like an increasing function. The condition for a function to be increasing is: $$ {\frac{df}{dx} > 0}{\text{ i.e.} \frac{d^2y}{d^2x} > 0} $$ This confirms that the function will have a local minima if the first derivative is 0, and the second derivative is positive at that point.

Solved Examples for You on Maxima and Minima

Question 1 : Find the local maxima and minima for the function y = x 3 – 3x + 2.

Answer : We’ll need to find the stationary points for this function, for which we need to calculate \({\frac{df}{dx}}\). We’ll proceed as follows:

$$ { y = x^3 – 3x +2} $$ $${\frac{dy}{dx} = 3x^2 – 3} $$

At stationary points, \({\frac{dy}{dx} = 0}\). Thus, we have;

$$ {3x^2 – 3 = 0} $$ $$ {3(x^2 – 1) = 0} $$ $$ {(x – 1)(x + 1) = 0} $$ $$ { \text{ x = 1 / x = -1}} $$

Now we have to determine whether any of these stationary points are extremum points. We’ll use the second derivative test for this:

$${\frac{dy}{dx} = 3x^2 – 3} $$ $${\frac{d^2y}{d^2x} = 6x } $$

  • For x = 1; \( {\frac{d^2y}{d^2x} = 6/times{1} = 6}\), which is positive. Thus the point (1, y(x = 1)) is a point of Local Minima.
  • For x = -1; \( {\frac{d^2y}{d^2x} = 6/times{-1} = -6}\), which is positive. Thus the point (-1, y(x = -1)) is a point of Local Maxima.

We can see from the graph below to verify our calculations:

Maxima and Minima

This concludes our discussion on this topic of maxima and minima.

Question 2: What are relative maxima and relative minima?

Answer: Finding out the relative maxima and minima for a function can be done by observing the graph of that function. A relative maxima is the greater point than the points directly beside it at both sides. Whereas, a relative minimum is any point which is lesser than the points directly beside it at both sides.

Question 3: How to find out the absolute maxima of a function?

Answer: Finding the absolute maxima:

Firstly, find out all the critical numbers of the function within the interval [a, b].

Then, plug in every single critical number from the first step into the function i.e. f(x).

Plugin the ending points that are (a) and (b) into the function f(x).

Finally, the biggest value is the absolute maxima and the lowest value is the absolute minima.

Question 4: What is the absolute maxima?

Answer: The biggest value that a mathematical function can consume over its whole curve. The absolute maxima on the graph takes place at x = d, and the absolute minima of that graph takes place at x = a.

Question 5: What are the local and global maxima and minima?

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  • RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima (Ex 18.3) Exercise 18.3
  • RD Sharma Solutions

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RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima (Ex 18.3) Exercise 18.3 - Free PDF

Free PDF download of RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima Exercise 18.3 solved by Expert Mathematics Teachers on Vedantu. All Chapter 18 - Maxima and Minima Ex 18.3 Questions with Solutions for RD Sharma Class 12 Maths to help you to revise the complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other engineering entrance exams.

Topics included in Class 12 RD Sharma Solutions for Chapter 18, that is, Maxima and Minima

Class 12 Chapter 18, that is, Maxima and Minima of RD Sharma include topics like maximum and minimum values of a function, definition and meaning of maximum and minimum, Local Maxima and Minima, first derivative test for Local Maxima and Minima, higher-order derivative test, theorem and algorithm based on the higher derivative test, point of inflexion and inflexion, properties of Maxima and Minima, maximum and minimum values in a closed interval and applied problems on Maxima and Minima.

What are the features of RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima?

The main features of RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima are that RD Sharma Class 12 contains detailed theory and illustrations, an algorithmic approach, a large number of graded illustrative examples and exercises and a summary of concepts and formulae. These features make RD Sharma one of the best books to study for Class 12 students. They act as revision notes for the students because students get concept clarity, example problems and formulae all in one single place.

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FAQs on RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima (Ex 18.3) Exercise 18.3

1. What topics are covered in Exercise 18.3 of RD Sharma Class 12 Solutions Chapter 18 Maxima and Minima?

Chapter 18, that is, Maxima and Minima of RD Sharma Class 12 is very interesting and easy to understand the chapter. The exercise is concerned with some major parts of the chapter. The major topics covered in Chapter 18, that is, Maxima and Minima are higher-order derivative tests, theorems, and algorithms based on the higher derivative tests, points of inflexion, points of inflexion, and properties of Maxima and Minima. The questions from these topics are asked in exams as well, therefore, the students must prepare this chapter well.

2. From where can I download the offline version of RD Sharma Class 12 Solutions for Chapter 18, that is, Maxima and Minima?

RD Sharma Class 12 Solutions for Chapter 18, that is, Maxima and Minima, exercise 18.3 can be downloaded as a PDF file from the official website of Vedantu. The students will register themselves at the official website of Vedantu through their email address and they can then download the PDF file of the solutions available on the website. This PDF file will be available with the students and they can assess these files in an offline mode as well.

3. How can RD Sharma Class 12 Solutions for Chapter 18 help me in my exams?

RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima are very important for the students. The RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima is a book written in a simple language and it provides students with an easy way to prepare for the exams. The RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima involve chapter-wise solutions to each question. The questions available in RD Sharma are also asked in final exams, therefore, students can score better in exams by practising these questions.

4. Why should I prefer Vedantu for downloading RD Sharma Class 12 Solutions for Chapter 18, that is, Maxima and Minima?

Students should download RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima from Vedantu because of some added benefits that the students may get:

The PDF file of RD Sharma Class 12 Solutions for Chapter 18, that is, Maxima and Minima are free of cost at the official website of Vedantu.

The RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima contain important questions which may be asked in the examination.

The RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima covers all the important concepts and therefore, students will be able to revise the full course in a very short period.

5. How many exercises are present in RD Sharma Class 12 Solutions for Chapter 18 Maxima and Minima?

The RD Sharma Class 12 Solutions for Chapter 18, that is, Maxima and Minima contains a total of 5 exercises. The first exercise of Chapter 18 covers questions wherein we need to find the maximum and minimum values. The second exercise of Chapter 18 covers the questions to calculate the points of Local Maxima and Minima of the given functions. The third exercise prepares students to calculate Local Maxima and Minima, corresponding Local Maxima and Minima, and Local extremism. The fourth exercise allows students to find the absolute maximum and minimum values of the given function. The last exercise of the chapter allows students to apply the concept of integration. 

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  • RD Sharma Solutions
  • Chapter 18 Maxima And Minima
  • Exercise 18.5

RD Sharma Solutions for Class 12 Maths Exercise 18.5 Chapter 18 Maxima and Minima

RD Sharma Solutions for Class 12 Maths Exercise 18.5 Chapter 18 Maxima and Minima is available here. The solutions are designed by subject matter experts based on the grasping abilities of students. Students can make use of the solutions PDF while solving exercise-wise problems, also understanding the other possible ways of obtaining answers easily.

The PDF of RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5 is available here. Students who aim to clear the exams with good marks can download the PDF from the given links.

RD Sharma Solutions for Class 12 Chapter 18 – Maxima and Minima Exercise 18.5

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Access answers to Maths RD Sharma Solutions For Class 12 Chapter 18 – Maxima and Minima Exercise 18.5

Exercise 18.5 page no: 18.72.

1. Determine two positive numbers whose sum is 15 and the sum of whose squares is minimum.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 14

Which implies S is minimum when a = 15/2 and b = 15/2.

2. Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Let the two positive numbers be a and b.

Given a + b = 64 … (1)

We have, a 3  + b 3  is minima

Assume, S = a 3  + b 3

(From equation 1)

S = a 3  + (64 – a) 3

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 15

Hence, the two number will be 32 and 32.

3. How should we choose two numbers, each greater than or equal to –2, whose sum is ½ so that the sum of the first and the cube of the second is minimum?

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 16

1 + 3(½ – a) 2 (-1) = 0

1 – 3(½ – a) 2 = 0

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 17

4. Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Let the given two numbers be x and y. Then,

x + y = 15 ….. (1)

y = (15 – x)

Now we have z = x 2 y 3

z = x 2 (15 – x) 3 (from equation 1)

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 18

5. Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm 3 , which has the minimum surface area?

Let r and h be the radius and height of the cylinder, respectively. Then,

Volume (V) of the cylinder = πr 2 h

⟹ 100 = πr 2 h

⟹ h = 100/ πr 2

Surface area (S) of the cylinder = 2 πr 2 + 2 πr h = 2 πr 2 + 2 πr × 100/ πr 2

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 19

6. A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by

Find the point at which M is maximum in each case.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 23

7. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?

Suppose the given wire, which is to be made into a square and a circle, is cut into two pieces of length x and y m respectively. Then,

x + y = 28 ⇒ y = (28 – x)

We know that the perimeter of a square, 4 (side) = x

Area of square = (x/4) 2 = x 2 /16

Circumference of a circle, 2 π r = y

r = y/ 2 π

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 26

8. A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?

Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length x and y respectively. Then,

x + y = 20 ⇒ y = (20 – x) …… (1)

Again we know that the perimeter of a triangle, 3 (side) = y.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 30

Hence, the wire of length 20 m should be cut into two pieces of lengths

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 33a

9. Given the sum of the perimeters of a square and a circle, show that the sum of their areas is least when one side of the square is equal to diameter of the circle.

Let us say the sum of the perimeter of the square and circumference of the circle be L

Given the sum of the perimeters of a square and a circle.

Assuming the side of the square = a and the radius of the circle = r

Then, L = 4a + 2πr ⇒ a = (L – 2πr)/4… (1)

Let the sum of the area of the square and circle be S

So, S = a 2  + πr 2

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 34

10. Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 36

11. Two sides of a triangle have lengths ‘a’ and ‘b’ and the angle between them is θ. What value of θ will maximize the area of the triangle? Find the maximum area of the triangle also.

It is given that two sides of a triangle have lengths a and b, and the angle between them is θ.

Let the area of the triangle be A

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 41

12. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 42

Given side length of big square is 18 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 18 – 2a

Breadth, B = 18 – 2a and

Height, H = a

Assuming, volume of box, V = LBH = a (18 – 2a) 2

Condition for maxima and minima is

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 43

13. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 45

Given the length of the rectangle sheet = 45 cm

Breath of rectangle sheet = 24 cm

Length, L = 45 – 2a

Breadth, B = 24 – 2a and

Assuming, volume of box, V = LBH = (45 – 2a)(24 – 2a)(a)

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 46

(45 – 2a) (24 – 2a) + (- 2) (24 – 2a) (a) + (45 – 2a) (- 2)(a) = 0

4a 2  – 138a + 1080 + 4a 2  – 48a + 4a 2  – 90a = 0

12a 2  – 276a + 1080= 0

a 2  – 23a + 90= 0

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 47

14. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m 3 . If building of tank cost Rs 70 per square metre for the base and Rs 45 per square metre for sides, what is the cost of least expensive tank?

Let the length, breadth and height of the tank be l, b and h, respectively.

Also, assume the volume of the tank as V

h = 2 m (given)

2lb = 8 (given)

b = 4/l … (1)

Cost for building base = Rs 70/m 2

Cost for building sides = Rs 45/m 2

Cost for building the tank, C = Cost for base + cost for sides

C = lb × 70 + 2(l + b) h × 45

C = l(4/l) × 70 + 2(l + 4/l) (2) × 45 [Using (1)]

= 280 + 180 (l + 4/l)

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 48

C = Rs 1000

15. A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimensions of the rectangular part of the window to admit maximum light through the whole opening.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 49

Let the radius of the semicircle, length and breadth of the rectangle be r, x and y respectively

AB = x = 2r (semicircle is mounted over rectangle) …1

Given the Perimeter of the window = 10 m

x + 2y + πr = 10

2r + 2y + πr = 10

2y = 10 – (π + 2).r

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 50

16. A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle that will produce the largest area of the window.

Let the dimensions of the rectangle be x and y.

Therefore, the perimeter of window = x + y + x + x + y = 12

3x + 2y = 12

y = (12 – 3x)/2 …. (1)

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 52

17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 55

18. A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimensions of the rectangle so that its area is maximum. Find also the area.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 58

Let the length and breadth of rectangle ABCD be 2x and y, respectively

The radius of semicircle = r (given)

In triangle OBA, where is the centre of the circle and mid-point of the side AC

r 2  = x 2  + y 2  (Pythagoras theorem)

y 2  = r 2  – x 2

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 59

Let us say the area of the rectangle = A = xy

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 60

19. Prove that a conical tent of given capacity will require the least amount of canvas when the height is √2 times the radius of the base.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 63

20. Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to 2/3 of the diameter of the sphere.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 67

Let the radius and height of the cone be r and h, respectively

The radius of the sphere = R

R 2  = r 2  + (h – R) 2

R 2  = r 2  + h 2  + R 2  – 2hR

r 2  = 2hR – h 2  … (1)

Assuming the volume of the cone to be V

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 68

21. Prove that the semi – vertical angle of the right circular cone of given volume and least curved surface is cot -1 √2

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 70

22. An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when θ = π/6.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 77

Δ ABC is an isosceles triangle such that AB = AC.

The vertical angle BAC = 2θ

Triangle is inscribed in the circle with centre O and radius a.

Draw AM perpendicular to BC.

Since Δ ABC is an isosceles triangle, the circumcenter of the circle will lie on the perpendicular from A to BC.

Let O be the circumcenter.

BOC = 2 × 2θ = 4θ (Using central angle theorem)

COM = 2θ (Since, Δ OMB and Δ OMC are congruent triangles)

OA = OB = OC = a (radius of the circle)

In Δ OMC,

CM = asin2θ

OM = acos2θ

BC = 2CM (Perpendicular from the center bisects the chord)

BC = 2asin2θ

Height of Δ ABC = AM = AO + OM

AM = a + acos2θ

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 78

23. Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6√3r.

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 81

QR at X and PR at Z.

OZ, OX, OY are perpendicular to the sides PR, QR, PQ.

Here PQR is an isosceles triangle with sides PQ = PR, and also from the figure,

⇒ PY = PZ = x

⇒ YQ = QX = XR = RZ = y

From the figure we can see that,

⇒ Area (ΔPQR) = Area (ΔPOR) + Area (ΔPOQ) + Area (ΔQOR)

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 82

⇒ PER = 2(√3r) + 4(√3r)

⇒ PER = 6√3r

∴ Thus proved

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RD Sharma Solutions Class 12 Mathematics Chapter 17 MCQ

Class 12 RD Sharma chapter 17 exercise MCQ solution is a sought-after book that is used by almost all students in class 12. The RD Sharma class 12th exercise MCQ covers questions from the entire NCERT maths book and provides intensive knowledge and essential information for all concepts and chapters. RD Sharma Solutions It is designed to aid students to perform well in any school and entrance exam and expand their knowledge on the maths subject.

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RD Sharma Class 12 Solutions Chapter 17 MCQ Maxima and Minima - Other Exercise

Maxima and minima excercise: mcq, rd sharma chapter-wise solutions.

  • Chapter 17 - Maxima and Minima - Ex 17.1
  • Chapter 17 - Maxima and Minima - Ex 17.2
  • Chapter 17 - Maxima and Minima - Ex 17.3
  • Chapter 17 - Maxima and Minima - Ex 17.4
  • Chapter 17 - Maxima and Minima - Ex 17.5
  • Chapter 17 - Maxima and Minima - Ex FBQ
  • Chapter 17 - Maxima and Minima - Ex CSBQ
  • Chapter 17 - Maxima and Minima- Ex VSA

Maxima and Minima exercise MCQ question 1

e^{\frac{1}{e}}

Maxima and Minima exercise Multiple choice question, question 2

ab\geq \frac{c^2}{4}

Maxima and Minima exercise MCQ question 3

f(x)=\frac{x}{\log _e x}

Maxima and Minima exercise MCQ question 4

f(x)=x+\frac{1}{x}

Maxima and Minima exercise MCQ question 5 Answer: Option (b) a minimum at x = 1

f(x)=x^3+3x^2-9x+2

Maxima and Minima exercise MCQ question 6 Answer: option (b) 4

f(x)=x^4-x^2-2x+6

Maxima and Minima exercise MCQ question 7

\frac{1}{2}

Maxima and Minima exercise MCQ question 9.

f(x)=\frac{1}{x}+\frac{1}{y}

Maxima and Minima exercise MCQ question 10.

f(x)=\sum_{r-1}^{5}(x-r)^2

Maxima and Minima exercise MCQ question 11

f(x)=2\sin 3x+3\cos 3x

Maxima and Minima exercise MCQ question 12

Answer: option(c) x =1

f(x)=x^2+x+1

Maxima and Minima exercise MCQ question 13.

Answer: option(d) 0

f(x)=x^3-18x^2+96x

Maxima and Minima exercise MCQ question 14.

\frac{1}{4}

Maxima and Minima exercise MCQ question 15

y^2=2x

Maxima and Minima exercise MCQ question 16.

x+y =8\Rightarrow y=8-x

Maxima and Minima exercise MCQ question 17.

f(x)=x^3-6x^2+9x

Maxima and Minima exercise MCQ question 18.

Answer: option(c)

f(x)=\sin x+\sqrt{3}\cos x

Maxima and Minima exercise MCQ question 19

\frac{2}{3}

Maxima and Minima exercise MCQ question 20

f(x)=x^2+\frac{250}{x}

Maxima and Minima exercise MCQ question 21.

\Rightarrow x= 1(x>0)

Maxima and Minima exercise MCQ question 22

\frac{4}{3}

Maxima and Minima exercise MCQ question 23

\Rightarrow y=\frac{1}{x}

Maxima and Minima exercise MCQ question 24

x=\frac{\pi}{2}

Maxima and Minima exercise MCQ question 25

f(x)=2x^3-15x^2+36x+4

Maxima and Minima exercise MCQ question 26

\frac{1}{6}

Maxima and Minima exercise MCQ question 27

f(x)=2x^3-3x^2-12x+5

Maxima and Minima exercise MCQ question 28

\frac{-1}{e}

Maxima and Minima exercise MCQ question 29

f(x)=2x^3-21x^2+36x-20

Maxima and Minima exercise MCQ question 30

Answer: option(d) f (x) is an increasing function.

f(x)=2x+\cos x

Maxima and Minima exercise MCQ question 31

y=x^2-8x+17

Maxima and Minima exercise MCQ question 33

f(x)=\left (\frac{1}{x} \right )^x

Maxima and Minima exercise MCQ question 34

\frac{1}{e}

Maxima and Minima exercise MCQ question 35

Answer: (b) 12

y=-x^3+3x^2+9x-27

Maxima and Minima exercise MCQ question 36

f(x)=2x^3-3x^2-12x+4

RD Sharma class 12th exercise MCQ comprises 36 MCQs that incorporates the entire syllabus of chapter 17. The solutions are elaborate yet easy to understand. These solutions will work to solve all your queries and doubts. It include the following concepts:

Maximum and minimum values

Local maxima

Local minima

First derivative test for local maxima and minima

Higher-order derivative test

Theorem and algorithm based on higher derivative test

Application based problems on maxima and minima

The RD sharma class 12 solution of Maxima and Minima exercise MCQ is the multiple choice questions, it gives you options to select the correct one which makes it quite easy for you to find accurate answers. The questions provided in the RD Sharma class 12th exercise MCQ are hand-picked by maths experts from across the country and their exceptional tips gives you the benefit and skill of solving the questions in less time.

The RD Sharma class 12 solutions chapter 17 exercise MCQ is particularly praised by students as the questions answered in these texts often appear in school and board exams. Therefore, these solutions are trusted by all the students for their prominence in the maths field. RD Sharma class 12th exercise MCQ has helped countless students with it's basic, simple and straightforward concepts.

The RD Sharma class 12th exercise MCQ can easily be studied by downloading it from the Career360 website and can be used offline from any device. The online PDFs of the RD Sharma class 12th exercise MCQ is available free of cost on the Career360 website. So don't wait anymore and get yourself the solutions without any hustle.

  • Chapter 1 - Relations
  • Chapter 2 - Functions
  • Chapter 3 - Inverse Trigonometric Functions
  • Chapter 4 - Algebra of Matrices
  • Chapter 5 - Determinants
  • Chapter 6 - Adjoint and Inverse of a Matrix
  • Chapter 7 - Solution of Simultaneous Linear Equations
  • Chapter 8 - Continuity
  • Chapter 9 - Differentiability
  • Chapter 10 - Differentiation
  • Chapter 11 - Higher Order Derivatives
  • Chapter 12 - Derivative as a Rate Measurer
  • Chapter 13 - Differentials, Errors and Approximations
  • Chapter 14 - Mean Value Theorems
  • Chapter 15 - Tangents and Normals
  • Chapter 16 - Increasing and Decreasing Functions
  • Chapter 17 - Maxima and Minima
  • Chapter 18 - Indefinite Integrals
  • Chapter 19 - Definite Integrals
  • Chapter 20 - Areas of Bounded Regions
  • Chapter 21 - Differential Equations
  • Chapter 22 - Algebra of Vectors
  • Chapter 23 - Scalar Or Dot Product
  • Chapter 24 - Vector or Cross Product
  • Chapter 25 - Scalar Triple Product
  • Chapter 26 - Direction Cosines and Direction Ratios
  • Chapter 27 - Straight Line in Space
  • Chapter 28 - The Plane
  • Chapter 29 - Linear programming
  • Chapter 30- Probability
  • Chapter 31 - Mean and Variance of a Random Variable

Frequently Asked Question (FAQs)

  RD Sharma Solutions can be extremely useful for students preparing for their upcoming board exams. Using these answers, students can keep track of their knowledge and performance at home.

Careers360 will provide free copies of the Class 12 RD Sharma Chapter 17  MCQs pdf to all students. It can be downloaded for free from any device.

  RD Sharma Class 12 Solutions from Careers360 is one of the best reference materials for their Class 12 RD Sharma Chapter 17 MCQs solution that can be used by CBSE students. The solutions follow an updated syllabus and provide answers to all questions in NCERT Books.

There are 36 questions in Class 12 RD Sharma Chapter 17  MCQ

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Application of derivatives of Class 12

From the figure for the function y = f(x)

Maxima and Minima

(iii) Boundary points

Maxima and Minima

  • Tangent and Normal
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RD Sharma Class 12 Solutions Maxima and Minima PDF Download

All the students must solve the RD Sharma Class 12 Solutions Maxima and Minima who are preparing for their final examinations and want to score well in their examinations. RD Sharma Solutions is one the best Maths solutions and is very common among the students of class 12. All types of questions are included in this book, from easy to medium to difficult which helps the students to get an idea about the type of questions of Maxima and Minima which can come in the exam. 

The RD Sharma solutions Class 12 Maxima and Minima helps all the students in doing the exam preparation for final exams. Our highly qualified subject matter experts who have years of experience in the education industry have created the Maxima and Minima solutions in a stepwise manner to ensure that the students go through them and grasp all the important concepts. Not only for studying but these solutions can also be a great tool for revision. 

The students can download these RD Sharma Class 12 Solutions Maxima and Minima from the official website of selfstudys i.e. selfstudys.com. Below we will discuss how the students can download them in just 2 minutes. 

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Downloading the RD Sharma Class 12 Solutions Maxima and Minima is very easy for all the students and takes only 2 minutes. Below are the steps to download these solutions:

  • The first step is to open the official website of selfstudys i.e. selfstudys.com. 

case study questions on maxima and minima class 12

  • Once the website is completely loaded, you need to click on the three lines which you will find on the upper left side. After clicking, select the option ‘ Books and Solutions ’. After clicking, you will see a couple of books and you have to select the ‘ RD Sharma Solutions ’. 

RD Sharma Class 12 Solutions Maxima and Minima, RD Sharma Class 12 Solutions Maxima and Minima PDF, RD Sharma Class 12 Maxima and Minima Solutions, RD Sharma Maxima and Minima Solutions, RD Sharma Solutions Maxima and Minima PDF, How to Download the RD Sharma Class 12 Solutions Maxima and Minima

  • After clicking on the ‘RD Sharma Solutions’, a new page will appear.
  • Now you can select the class for which you need the solutions, in this case, you need to select and click on the RD Sharma solutions Class 12 Maxima and Minima. 

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There are various reasons why students should go through the RD Sharma solutions Class 12 Maxima and Minima. Some of the most important of them are: 

  • All the students are advised to go through the RD Sharma solutions Class 12 Maxima and Minima if they want to score good marks in their board examinations as they are explained step wise. It helps in building a strong base of all the topics. These solutions also help in boosting the logical-thinking skills of the students as all the key points are included in the Maxima and Minima.
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Class 12 RD Sharma Solutions – Chapter 18 Maxima and Minima – Exercise 18.5 | Set 1

Question 1. determine two positive numbers whose sum is 15 and the sum of whose squares is minimum..

Let us assume the two positive numbers are x and y,  And it is given that x + y = 15         …..(i) So, let P = x 2 + y 2                     …..(ii) From eq (i) and (ii), we get P = x 2 + (15 – x) 2 On differentiating w.r.t. x, we get dP/dx = 2x + 2(15 – x)(-1) = 2x -30 +2x = 4x -30 For maxima and minima. Put dP/dx = 0 ⇒ 4x – 30 = 0 ⇒ x = 15/2 Since, d 2 P/dx 2 = 4 > 0 So, x = 15/2 is the point of local minima, From eq(i), we get  y = 15 – 15/2 = 15/2 So, the two positive numbers are 15/2, 15/2.

Question 2. Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Let us assume 64 is divide into two parts that is x and y  So, x + y = 64                    …..(i) Let P = x 3 + y 3               ………(ii) From eq(i) and (ii), we get P = x 3 + (64 – x) 3 On differentiating w.r.t. x, we get dP/dx = 3x 2 + 3(64 – x) 2 × (-1) = 3x 2 – 3(4096 – 128x + x 2 ) = -3 (4096 – 128x) For maxima and minima. Put dP/dx = 0  ⇒ -3(4096 – 128x) = 0 ⇒ x = 32 Now, d 2 s/dx 2 = 384 > 0 So, x=32 is the point of local maxima. Hence, the 64 is divide into two equal parts that is (32, 32)

Question 3. How should we choose two numbers, each greater than or equal to -2, whose sum is 1/2 so that the sum of the first and the cube of the second is minimum?

Let us assume x and y be the two numbers, such that x, y ≥ -2 and x + y = 1/2             ……(i) So, let P = x + y 3             …….(ii) From eq(i) and (ii), we get P = x + (1/2 – x) 3 On differentiating w.r.t. x, we get dP/dx = 1 + 3(1/2 – x) 2 × (-1) = 1 – 3(1/4 – x + x 2 ) = 1/4 +3x -3x 2 For maximum and minimum, Put dP/dx = 0 ⇒ 1/4 + 3x – 3x 2 = 0 ⇒ 1 + 12x – 12x 2 = 0 ⇒ 12x 2 – 12x – 1 = 0 ⇒  ⇒ x = 1/2 ± (8√3/24) ⇒ x = 1/2 ± (1/√3) ⇒ x = {1/2 – (1/√3)}, {1/2 + (1/√3)} Now, d 2 P/dx 2 = 3 – 6x So, at x =1/2 – (1/√3), d 2 P/dx 2 = 3(1 – 2(1/2 – 1/√3)) = 3(+2/√3) = 2√3 > 0 Hence, x = 1/2 – 1/√3 is point of local minima From eq(i), we get y = 1/2 – (1/2 – 1/√3) = 1/√3 So, the numbers are (1/2 – 1/√3) and 1/√3

Question 4. Divide 15 into two parts such that the square of one multiplied with the cube of the other minimum.

Let us assume 15 is divide into two parts that is x and y  So, x + y = 15 Also, P = x 2 y 3 From eq(i) and (ii), we get P = x 2 (15 – x) 3 On differentiating w.r.t. x, we get dP/dx = 2x(15 – x) 3 – 3x 2 (15 – x) 2 = (15 – x) 2 [30x – 2x 2 – 3x 2 ] = 5x(15 – x) 2 (6 – x) For maxima and minima, Put dP/dx = 0 ⇒ 15(15 – x) 2 (6 – x) = 0 ⇒ x = 0, 15, 6 Now, So, d 2 P/dx 2 = 5(15 – x) 2 (6 – x) – 5x × 2(15 – x)(6 – x) –  5x(15 – x) 2 At x = 0, d 2 P/dx 2 = 1125 > 0 So, x = 0 is point of local minima At x = 15, d 2 P/dx 2 = 0 So, x = 15 is an inflection point. At x = 6, d 2 P/dx 2 = -2430 < 0 So, x = 6 is the point of local maxima So, the 15 is divide into two parts that are 6 and 9.

Question 5. Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm 3 which has the minimum surface area?

Let us assume r be the radius of the cylinder and h be the height of the cylinder So, the volume of the cylinder is 100 cm 3 i.e., V = πr 2 h = 100 h = 100/πr 2 ……(i) Now we find the surface area of the cylinder is A = 2πr 2 + 2πrh  = 2πr 2 + 200/r On differentiating w.r.t. r, we get dA/dr = 4πr – 200/r 2 ,  ⇒ d 2 A/dr 2 = 4π + 400/r 3 For maxima and minima, dA/dr = 0  ⇒ 4πr = 200/r 2 ⇒ r 3 = 200/4π = 50/π r = (50/π) 1/3 So, when r = (50/π) 1/3 , d 2 s/dr 2 > 0 Hence, from the second derivative test, the surface area is the minimum  when the radius of the cylinder is (50/π) 1/3 cm Now put the value of r in eq(i), we get h = 100 / π(50/π) 1/3 = (2×50)/(50 2/3 π 1-2/3 ) = 2(50/π) 1/3

Question 6. A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by

(1) m = (wl/2)x – (w/2)x 2, (ii) m = wx/3 –  (w/3) (x 3 /l 2 ), find the point at which m is maximum in each case..

(i) M = (WL/2)x – (w/2)x 2 On differentiating w.r.t. x, we get dM/dx = WL/2 – Wx For maxima and minima, Put dM/dx = 0  WL/2 – Wx = 0    x = L/2 Now, d 2 M/dx 2 = -W < 0 So, x = L/2 is point of local maxima. Hence, M is maximum when x = L/2 (ii) M = Wx/3 – (W/3)(x 3 /L 2 ) On differentiating w.r.t. x, we get dM/dx = W/3 – Wx 2 /L 2 For maxima and minima, Put dM/dx = 0  W/3 – Wx 2 /L 2 = 0   x = ± L/√3 Now, d 2 M/dx 2 = – 2xW/L 2 So, at x = L/√3, ⇒ d 2 M/dx 2 =-2W/√3L < 0 (for max value) at x = -L/√3, ⇒ d 2 M/dx 2 = 2W/√3L > 0 (for min value) Hence, M is maximum when x = L/√3

Question 7. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made square and the other into a circle. What should be the lengths of the two pieces so that combined area of the circle and the square is minimum?  

Let us assume l m be the piece of length cut from the given wire to make a square. and  the other piece of wire that is used to create a circle is of length (28-l) m. So, the side of square = l/4 Now, let us considered the radius of the circle is r.  Then, 2πr = 28 – l ⇒ r = (1/2π)(28 – l) Now we find the combined area of square and circle A = l 2 /16 + π[(1/2π)(28 – l)] 2 = l 2 /16 + 1/4π (28 – l) 2 On differentiating w.r.t. l, we get dA/dl = 2l/16 + (2/4π)(28 – l)(-1) = l/8 – (1/2π)(28 – l) d 2 A/dl 2 = l/8 + (1/2π) > 0 For maxima and minima, Put dA/dl = 0  ⇒ l/8 – (1/2π)(28 – l) = 0 ⇒ {πl – 4(28 – l)}8π = 0 ⇒ (π + 4)l – 112 = 0 ⇒ l = 112/(π + 4) So, at l = 112/(π + 4), d 2 A/dl 2 > 0 Hence, using second derivative test, the area is the minimum when l = 112/(π + 4) So, the length of the two pieces of wire are 112/(π + 4) and 28π/(π + 4) cm.

Question 8. A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?  

According to the question The length of the wire is 20 m and the wire cut into two pieces x and y. So, x length wire is used to make a square and y length wire is used to make triangle. Now. x + y = 20           …..(i) x = 4l and y = 3a So, A = sum of area of square and triangle A = l 2 + √3/4a 2         ……(ii) We have, 4l + 3a  =20 4l = 20 – 3a l = (20 – 3a) / 4 From eq(i), we have, A = (20 – 3a) 2 /4 + √3/4a 2 On differentiating w.r.t. a, we get dA/da = 2{(20 – 3a)/4}(-3/4) + 2a × √3/4 For maxima and minima, Put dA/da = 0 ⇒ 2 {(20 – 3a)/4}(-3/4) + 2a × √3/4 = 0 ⇒ -3(20 – 3a) + 4a√3 = 0 ⇒ -60 + 9a + 4a√3 = 0 ⇒ 9a + 4a√3 = 60 ⇒ a(9 + 4√3) = 60 ⇒ a = 60/(9 + 4√3) Again differentiating w.r.t. a, we get d 2 s/da 2 = (9 + 4√3)/8 > 0 So, the sum of the areas of the square and triangle is minimum when a = 60 / (9 + 4√3) So l = (20 – 3a)/4 ⇒ l =  ⇒ l = (180 + 80√3 – 180)/{4(9 + 4√3)} ⇒ l = 20√3/(9 + 4√3)

Question 9. Given the sum of the perimeters of a square and a circle show that the sum of their areas is least when one side of the square is equal to the diameter of the circle.

Let us assume the radius of the circle is r  We have, 2πr + 4a = k (k is constant) a = (k – 2πr)/4 Now we find the sum of the areas of the circle and the square: A = πr 2 + a 2 = πr 2 + (k – 2πr) 2 /16 On differentiating w.r.t. r, we get dA/dr = 2πr + 2(k – 2πr)(-2π)/16  = 2πr- π(k – 2πr)/4 For maxima and minima, Put, dA/dr = 0 2πr = π(k – 2πr)/4 8r = k – 2πr r = k / (8 + 2π)= k / 2(4 + π) Now, d 2 A/dr 2 = 2π + π 2 /2 > 0 So, at r = k / 2(4 + π), d 2 A/dr 2 > 0 Hence, the sum of the areas minimum when r = k / 2(4 + π) So, a =  = 2r  Hence Proved

Question 10. Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm long.

Let us assume PQR is a right-angled triangle,  So, the hypotenuse h = PR = 5 cm. Now, le us assume a and b be the remaining sides of the triangle.  So, a 2 + b 2 = 25         ……(i) Now we find the area of PQR = 1/2 QR × PQ A = 1/2 ab             ……(ii) From eq(i) and (ii), we get ⇒ A = 1/2 x √(25 – a 2 ) On differentiating w.r.t. a, we get dA/da =  =  =  For maxima and minima, Put dA/da = 0 ⇒  ⇒ a = 5/√2 Now, d 2 A/d 2 a =  At a = 5√2, d 2 s/d 2 =  = – 5/2 < 0 So, x = 5/√2 is a point local maxima, Hence, the largest possible area of the triangle = 1/2 × (5/√2) × (5/√2) = 25/4 square units

Question 11. Two sides of a triangle have lengths ‘a’ and ‘b’ and the angle between them is θ. What value of θ will maximize the area of the triangle? Find the maximum area of the triangle also.

Let us assume ABC is a triangle such that AB = a, BC = b and ∠ABC = θ and AD in perpendicular to BC. BD = asinθ So, the area of △ABC = 1/2 × BC × AD ⇒ A = 1/2 × b × a × sinθ On differentiating w.r.t. θ, we get dA/dθ = 1/2 × abcosθ For maxima and minima, Put dA/dθ = 0 ⇒ 1/2 × abcosθ = 0 ⇒ cosθ = 0 ⇒ θ = π/2 Now, d 2 A/dθ 2 = -1/2 ab sinθ At θ = π/2, d 2 A/dθ 2 = -1/2ab < 0 So, θ = π/2, is point of local maxima Hence, the maximum area of the ABC triangle is 1/2 × absin(π/2) = 1/2 ab.

Question 12. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each comer and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume.  

Let us assume that x cm be the side of the square to be cut off. Now, the length and the breadth of the box will be (18 – 2x) cm each and  the x cm be the height of the box. So, the volume of the box is  V (x) = x(18 – 2x) 2 On differentiating w.r.t. x, we get V'(x) = (18 – 2x) 2 – 4x(18-2x) = (18 – 2x)[18 – 2x – 4x] = (18 – 2x)(18 – 6x) = 6 × 2(9 – x)(3 – x) = 12(9 – x)(3 – x) Again on differentiating w.r.t. x, we get V”(x) = 12 [-(9 – x) – (3 – x)] = -12 (9 – x + 3 – x) = -12 (12 – 2x) = -24 (6 – x) For maxima and minima, Put V'(x) = 0 12(9 – x)(3 – x) = 0 x = 9, 3 When x = 9 length and breadth of the box become zero. So, x ≠ 9 When x = 3, V”(x) = -24 (6 – x) = -72 < 0 So, x = 3 is the point of maxima Hence, the maximum volume is V x = 3 = 3(18 – 2 × 3) 2 ⇒ V = 3 × 12 2 ⇒ V = 3 × 144 ⇒ V = 432 cm 3  

Question 13. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Let us assume that x cm be the side of the square to be cut off. So, the height of the box = x,  the length of the box = 45 – 2x,  and the breadth of the box = 24 – 2x. So, the volume of the box is  V(x) = x (45 – 2x)(24 – 2x) = x (1080 – 90x – 48x + 4x 2 ) = 4x 3 – 138x 2 + 1080x On differentiating w.r.t. x, we get V ‘(x)= 12x 2 – 276x + 1080 = 12(x 2 – 23x + 90) = 12(x – 18) (x – 5) Again on differentiating w.r.t. x, we get V ”(x) = 24x – 276 = 12 (2x – 23) For maxima and minima, Put V'(x) = 0 4x 3 – 138x 2 + 1080x = 0 x(x – 18) – 5(x – 18) = 0 (x – 5)(x – 18) = 0 So, x = 18 and x = 5 when x = 18 it is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet.  So, x ≠ 18 When x = 5, V ”(5) = 12 (10 – 23) = 12(-13) = -156 < 0 So, x = 5 is the point of maxima. Hence, the volume of the box is maximum when x = 5.

Question 14. A tank with rectangular base and rectangular sides open at the top is to be constructed so that its depth is 2 m and volume is 8m 3 . If building of tank costs ₹ 70 per square meter for the base and ₹ 45 per square meter for sides, what is the cost of least expensive tank?

Let us considered the length, breadth and height of the tank be l, b, and h According to the question The height of the tank is 2 and the volume is 8m 3 So, the volume of the tank is   V = l × b × h  8 = l × b × 2 lb = 4   ⇒ b = 4/l    ….(i) Now, we find the area of the base = lb = 4 and the area of the four walls (A) = 2h (l + b) A = 4 (l + l/4) On differentiating w.r.t. l, we get ⇒ dA/dl = 4 (l – 4/l 2 ) Again differentiating w.r.t. l, we get  d 2 A/dl 2 = 32/l 3 For maxima and minima, Put dA/dl = 0 ⇒ l – 4/l 2 = 0  ⇒ l 2 = 4 ⇒ l = ±2 As we know that the length cannot be negative. So, l ≠ 2 When l = 2, d 2 A/dl 2 = 32/8 = 4 > 0 So, l = 2 is the point of minima. Now put the value of l = 2 cm in the eq(i) b = 4/l = 4/2 = 2 So, l = b = h = 2 Hence, the area is the minimum when l = 2. So, the cost of building the base = 70 × (lb) = 70 × 4 = 280 Cost of building the walls = 2h (l + b) × 45 = 90 × 2 × (2 + 2) = 8 × 90 = 720 Hence, the total cost = 280 + 720 = 1000

Question 15. A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimensions of the rectangular part of the window to admit maximum light through the whole opening.

Let us assume x and y be the length and breadth of the rectangle.  So the radius of the semicircular opening = x/2 From the question it is given that the perimeter of the window is 10 m. So, x + 2y + πx/2 = 10 ⇒ x(1 + π/2) + 2y = 10 ⇒ 2y = 10 – x(1 + π/2) ⇒ y = 5 – x(1/2 + π/4) Now, the area of the window is A = xy + (π/2)(x/2) 2 = x [5 – x(1/2 + π/4)] + (π/8)x 2 = 5x – x 2 (1/2 + π/4) + (π/8)x 2 On differentiating w.r.t. x, we get dA/dx = 5 – 2x(1/2 + π/4) + (π/4)x = 5 – x(1 + π/2) + (π/4)x Again differentiating w.r.t. x, we get d 2 A/dx 2 = -(1 + π/2) + π/4 = -1 – π/4 For maxima and minima, Put dA/dx = 0 ⇒ 5 – x(1 + π/2) + (π/4)x = 0 ⇒ 5 – x – (π/4)x = 0 ⇒ x (1 + π/4) = 5 ⇒ x = 5/(1 + π/4) = 20/(π + 4) So, when x = 20/(π + 4), d 2 A/dx 2 < 0 So, the area is the maximum when length (x) = 20/(π + 4) Now, y = 5 – 20/(π + 4){(2 + π)/4} = 5 – 5(2 + π)/(π + 4) = 10/(π + 4) m  Hence, the length of the rectangle is 20/(π + 4) m and the breadth is 10/(π + 4) m.

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    Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others. Question of Class 12-Maxima and Minima : From the figure for the function y = f (x) We find that the function gets local maximum and local minimum at the points.

  21. RD Sharma Class 12 Solutions Maxima and Minima PDF Download

    The RD Sharma solutions Class 12 Maxima and Minima helps all the students in doing the exam preparation for final exams. Our highly qualified subject matter experts who have years of experience in the education industry have created the Maxima and Minima solutions in a stepwise manner to ensure that the students go through them and grasp all the important concepts. Not only for studying but ...

  22. Important Questions for CBSE Class 12 Maths Maxima and Minima

    6 Marks Questions. Filed Under: CBSE Tagged With: Class 12 Maths, Maths Maxima and Minima. Free Resources. RD Sharma Class 12 Solutions. RD Sharma Class 11. RD Sharma Class 10. RD Sharma Class 9.

  23. Class 12 RD Sharma Solutions

    Class 12 RD Sharma Solutions - Chapter 18 Maxima and Minima - Exercise 18.5 | Set 1. Last Updated : 08 Dec, 2022. Question 1. Determine two positive numbers whose sum is 15 and the sum of whose squares is minimum. Solution: Let us assume the two positive numbers are x and y, And it is given that x + y = 15 …..