Solutions of Maxima and Minima – Previous Years Board Questions (2000 to Sem1-2021) – ISC – Class 12 – Mathematics Maxima and Minima (2000 to Sem 1 – 2021) . Click Here to open Solutions. 2 thoughts on “ Maxima and Minima. Previous Years Board Questions (2000 to Sem 1-2021) with Solutions of ISC Class 12 Mathematics. ”- Pingback: Contents - Tapati's Classes | Tapati's Classes
- Pingback: Integration Previous Years Board's Questions (1990 - 2011) with Solutions of ISC Class 12 Mathematics | Tapati's Classes
Leave a Reply Cancel replyYour email address will not be published. Required fields are marked * - 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 MinimaRD 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 18.5carouselExampleControls112 Previous Next Access other exercises of RD Sharma Solutions For Class 12 Chapter 18 – Maxima and MinimaExercise 18.1 Solutions Exercise 18.2 Solutions Exercise 18.3 Solutions Exercise 18.4 Solutions Access answers to Maths RD Sharma Solutions For Class 12 Chapter 18 – Maxima and Minima Exercise 18.5Exercise 18.5 page no: 18.72. 1. Determine two positive numbers whose sum is 15 and the sum of whose squares is minimum. 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 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? 1 + 3(½ – a) 2 (-1) = 0 1 – 3(½ – a) 2 = 0 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) 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 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. 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 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. Hence, the wire of length 20 m should be cut into two pieces of lengths 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 10. Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long. 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 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 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 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? 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) (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 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) 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. 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 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) 17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. 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. 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 Let us say the area of the rectangle = A = xy 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. 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. 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 21. Prove that the semi – vertical angle of the right circular cone of given volume and least curved surface is cot -1 √2 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. 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 23. Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6√3r. 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) ⇒ PER = 2(√3r) + 4(√3r) ⇒ PER = 6√3r ∴ Thus proved Leave a Comment Cancel replyYour Mobile number and Email id will not be published. Required fields are marked * Request OTP on Voice Call Post My Comment Register with BYJU'S & Download Free PDFsRegister with byju's & watch live videos. CBSE Class 12th- UP Board 10th
- UP Board 12th
- Bihar Board 10th
- Bihar Board 12th
- Top Schools
- Top Schools in India
- Top Schools in Delhi
- Top Schools in Mumbai
- Top Schools in Chennai
- Top Schools in Hyderabad
- Top Schools in Kolkata
- Top Schools in Pune
- Top Schools in Bangalore
Products & Resources- JEE Main Knockout April
- Free Sample Papers
- Free Ebooks
- NCERT Notes
NCERT Syllabus- NCERT Books
- RD Sharma Solutions
- Navodaya Vidyalaya Admission 2024-25
NCERT Solutions- NCERT Solutions for Class 12
- NCERT Solutions for Class 11
- NCERT solutions for Class 10
- NCERT solutions for Class 9
- NCERT solutions for Class 8
- NCERT Solutions for Class 7
- JEE Main Exam
- JEE Advanced Exam
- BITSAT Exam
- View All Engineering Exams
- Colleges Accepting B.Tech Applications
- Top Engineering Colleges in India
- Engineering Colleges in India
- Engineering Colleges in Tamil Nadu
- Engineering Colleges Accepting JEE Main
- Top IITs in India
- Top NITs in India
- Top IIITs in India
- JEE Main College Predictor
- JEE Main Rank Predictor
- MHT CET College Predictor
- AP EAMCET College Predictor
- GATE College Predictor
- KCET College Predictor
- JEE Advanced College Predictor
- View All College Predictors
- JEE Advanced Cutoff
- JEE Main Cutoff
- GATE Registration 2025
- JEE Main Syllabus 2025
- Download E-Books and Sample Papers
- Compare Colleges
- B.Tech College Applications
- JEE Main Question Papers
- View All Management Exams
Colleges & Courses- Top MBA Colleges in India
- MBA College Admissions
- MBA Colleges in India
- Top IIMs Colleges in India
- Top Online MBA Colleges in India
- MBA Colleges Accepting XAT Score
- BBA Colleges in India
- XAT College Predictor 2025
- SNAP College Predictor
- NMAT College Predictor
- MAT College Predictor 2024
- CMAT College Predictor 2024
- CAT Percentile Predictor 2024
- CAT 2024 College Predictor
- Top MBA Entrance Exams 2024
- AP ICET Counselling 2024
- GD Topics for MBA
- CAT Exam Date 2024
- Download Helpful Ebooks
- List of Popular Branches
- QnA - Get answers to your doubts
- IIM Fees Structure
- AIIMS Nursing
- Top Medical Colleges in India
- Top Medical Colleges in India accepting NEET Score
- Medical Colleges accepting NEET
- List of Medical Colleges in India
- List of AIIMS Colleges In India
- Medical Colleges in Maharashtra
- Medical Colleges in India Accepting NEET PG
- NEET College Predictor
- NEET PG College Predictor
- NEET MDS College Predictor
- NEET Rank Predictor
- DNB PDCET College Predictor
- NEET Result 2024
- NEET Asnwer Key 2024
- NEET Cut off
- NEET Online Preparation
- Download Helpful E-books
- Colleges Accepting Admissions
- Top Law Colleges in India
- Law College Accepting CLAT Score
- List of Law Colleges in India
- Top Law Colleges in Delhi
- Top NLUs Colleges in India
- Top Law Colleges in Chandigarh
- Top Law Collages in Lucknow
Predictors & E-Books- CLAT College Predictor
- MHCET Law ( 5 Year L.L.B) College Predictor
- AILET College Predictor
- Sample Papers
- Compare Law Collages
- Careers360 Youtube Channel
- CLAT Syllabus 2025
- CLAT Previous Year Question Paper
- NID DAT Exam
- Pearl Academy Exam
Predictors & Articles- NIFT College Predictor
- UCEED College Predictor
- NID DAT College Predictor
- NID DAT 2025
- NID DAT Syllabus 2025
- Design Colleges in India
- Top NIFT Colleges in India
- Fashion Design Colleges in India
- Top Interior Design Colleges in India
- Top Graphic Designing Colleges in India
- Fashion Design Colleges in Delhi
- Fashion Design Colleges in Mumbai
- Top Interior Design Colleges in Bangalore
- NIFT Cutoff
- NIFT Fees Structure
- NIFT Syllabus 2025
- Free Design E-books
- List of Branches
- Careers360 Youtube channel
- IPU CET BJMC 2024
- JMI Mass Communication Entrance Exam 2024
- IIMC Entrance Exam 2024
- Media & Journalism colleges in Delhi
- Media & Journalism colleges in Bangalore
- Media & Journalism colleges in Mumbai
- List of Media & Journalism Colleges in India
- CA Intermediate
- CA Foundation
- CS Executive
- CS Professional
- Difference between CA and CS
- Difference between CA and CMA
- CA Full form
- CMA Full form
- CS Full form
- CA Salary In India
Top Courses & Careers- Bachelor of Commerce (B.Com)
- Master of Commerce (M.Com)
- Company Secretary
- Cost Accountant
- Charted Accountant
- Credit Manager
- Financial Advisor
- Top Commerce Colleges in India
- Top Government Commerce Colleges in India
- Top Private Commerce Colleges in India
- Top M.Com Colleges in Mumbai
- Top B.Com Colleges in India
- IT Colleges in Tamil Nadu
- IT Colleges in Uttar Pradesh
- MCA Colleges in India
- BCA Colleges in India
Quick Links- Information Technology Courses
- Programming Courses
- Web Development Courses
- Data Analytics Courses
- Big Data Analytics Courses
- RUHS Pharmacy Admission Test
- Top Pharmacy Colleges in India
- Pharmacy Colleges in Pune
- Pharmacy Colleges in Mumbai
- Colleges Accepting GPAT Score
- Pharmacy Colleges in Lucknow
- List of Pharmacy Colleges in Nagpur
- GPAT Result
- GPAT 2024 Admit Card
- GPAT Question Papers
- NCHMCT JEE 2024
- Mah BHMCT CET
- Top Hotel Management Colleges in Delhi
- Top Hotel Management Colleges in Hyderabad
- Top Hotel Management Colleges in Mumbai
- Top Hotel Management Colleges in Tamil Nadu
- Top Hotel Management Colleges in Maharashtra
- B.Sc Hotel Management
- Hotel Management
- Diploma in Hotel Management and Catering Technology
Diploma Colleges- Top Diploma Colleges in Maharashtra
- UPSC IAS 2024
- SSC CGL 2024
- IBPS RRB 2024
- Previous Year Sample Papers
- Free Competition E-books
- Sarkari Result
- QnA- Get your doubts answered
- UPSC Previous Year Sample Papers
- CTET Previous Year Sample Papers
- SBI Clerk Previous Year Sample Papers
- NDA Previous Year Sample Papers
Upcoming Events- NDA 2 Admit card 2024
- SSC CGL Admit card 2024
- CDS 2 Admit card 2024
- UGC NET Admit card 2024
- HP TET Result 2024
- SSC CHSL Result 2024
- UPTET Notification 2024
- SBI PO Notification 2024
Other Exams- SSC CHSL 2024
- UP PCS 2024
- UGC NET 2024
- RRB NTPC 2024
- IBPS PO 2024
- IBPS Clerk 2024
- IBPS SO 2024
- Top University in USA
- Top University in Canada
- Top University in Ireland
- Top Universities in UK
- Top Universities in Australia
- Best MBA Colleges in Abroad
- Business Management Studies Colleges
Top Countries- Study in USA
- Study in UK
- Study in Canada
- Study in Australia
- Study in Ireland
- Study in Germany
- Study in China
- Study in Europe
Student Visas- Student Visa Canada
- Student Visa UK
- Student Visa USA
- Student Visa Australia
- Student Visa Germany
- Student Visa New Zealand
- Student Visa Ireland
- CUET PG 2025
- DU Admission 2024
- UP B.Ed JEE 2024
- LPU NEST 2024
- IIT JAM 2025
- AP OAMDC 2024
- Universities in India
- Top Universities in India 2024
- Top Colleges in India
- Top Universities in Uttar Pradesh 2024
- Top Universities in Bihar
- Top Universities in Madhya Pradesh 2024
- Top Universities in Tamil Nadu 2024
- Central Universities in India
- CUET DU Cut off 2024
- IGNOU Date Sheet 2024
- CUET DU CSAS Portal 2024
- CUET 2025 Syllabus
- CUET PG Syllabus 2025
- CUET Participating Universities 2024
- CUET Previous Year Question Paper
- IGNOU Result 2024
- E-Books and Sample Papers
- CUET College Predictor 2024
- CUET Exam Date 2024
- CUET Cut Off 2024
- NIRF Ranking 2024
- IGNOU Exam Form 2024
- CUET PG Counselling 2024
- CUET Counselling 2024
Engineering Preparation- Knockout JEE Main 2024
- Test Series JEE Main 2024
- JEE Main 2024 Rank Booster
Medical Preparation- Knockout NEET 2024
- Test Series NEET 2024
- Rank Booster NEET 2024
Online Courses- JEE Main One Month Course
- NEET One Month Course
- IBSAT Free Mock Tests
- IIT JEE Foundation Course
- Knockout BITSAT 2024
- Career Guidance Tool
Top Streams- IT & Software Certification Courses
- Engineering and Architecture Certification Courses
- Programming And Development Certification Courses
- Business and Management Certification Courses
- Marketing Certification Courses
- Health and Fitness Certification Courses
- Design Certification Courses
Specializations- Digital Marketing Certification Courses
- Cyber Security Certification Courses
- Artificial Intelligence Certification Courses
- Business Analytics Certification Courses
- Data Science Certification Courses
- Cloud Computing Certification Courses
- Machine Learning Certification Courses
- View All Certification Courses
- UG Degree Courses
- PG Degree Courses
- Short Term Courses
- Free Courses
- Online Degrees and Diplomas
- Compare Courses
Top Providers- Coursera Courses
- Udemy Courses
- Edx Courses
- Swayam Courses
- upGrad Courses
- Simplilearn Courses
- Great Learning Courses
RD Sharma Solutions Class 12 Mathematics Chapter 17 MCQClass 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. Latest: JEE Main: high scoring chapters | Past 10 year's papers Don't Miss: Most scoring concepts for NEET | NEET papers with solutions New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now RD Sharma Class 12 Solutions Chapter 17 MCQ Maxima and Minima - Other ExerciseMaxima 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 Maxima and Minima exercise Multiple choice question, question 2 Maxima and Minima exercise MCQ question 3 Maxima and Minima exercise MCQ question 4 Maxima and Minima exercise MCQ question 5 Answer: Option (b) a minimum at x = 1 Maxima and Minima exercise MCQ question 6 Answer: option (b) 4 Maxima and Minima exercise MCQ question 7 Maxima and Minima exercise MCQ question 9. Maxima and Minima exercise MCQ question 10. Maxima and Minima exercise MCQ question 11 Maxima and Minima exercise MCQ question 12 Answer: option(c) x =1 Maxima and Minima exercise MCQ question 13. Answer: option(d) 0 Maxima and Minima exercise MCQ question 14. Maxima and Minima exercise MCQ question 15 Maxima and Minima exercise MCQ question 16. Maxima and Minima exercise MCQ question 17. Maxima and Minima exercise MCQ question 18. Answer: option(c)Maxima and Minima exercise MCQ question 19 Maxima and Minima exercise MCQ question 20 Maxima and Minima exercise MCQ question 21. Maxima and Minima exercise MCQ question 22 Maxima and Minima exercise MCQ question 23 Maxima and Minima exercise MCQ question 24 Maxima and Minima exercise MCQ question 25 Maxima and Minima exercise MCQ question 26 Maxima and Minima exercise MCQ question 27 Maxima and Minima exercise MCQ question 28 Maxima and Minima exercise MCQ question 29 Maxima and Minima exercise MCQ question 30 Answer: option(d) f (x) is an increasing function.Maxima and Minima exercise MCQ question 31 Maxima and Minima exercise MCQ question 33 Maxima and Minima exercise MCQ question 34Maxima and Minima exercise MCQ question 35 Answer: (b) 12 Maxima and Minima exercise MCQ question 36 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 - Latest Articles
- Popular Articles
Upcoming School ExamsNational means cum-merit scholarship. Application Date : 01 August,2024 - 16 September,2024 International General Knowledge OlympiadExam Date : 19 September,2024 - 19 September,2024 Unified Cyber OlympiadExam Date : 20 September,2024 - 20 September,2024 International English OlympiadExam Date : 26 September,2024 - 26 September,2024 International Olympiad of MathematicsApplication Date : 30 September,2024 - 30 September,2024 Applications for Admissions are open.Tallentex 2025 - ALLEN's Talent Encouragement ExamRegister for Tallentex '25 - One of The Biggest Talent Encouragement Exam JEE Main Important Physics formulasAs per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters JEE Main Important Chemistry formulasAs per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters TOEFL ® Registrations 2024Accepted by more than 11,000 universities in over 150 countries worldwide PTE Registrations 2024Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer! JEE Main high scoring chapters and topicsAs per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE Explore on Careers360- Board Exams
- Navodaya Vidyalaya
- NCERT Solutions for Class 10
- NCERT Solutions for Class 9
- NCERT Solutions for Class 8
- NCERT Solutions for Class 6
NCERT Exemplars- NCERT Exemplar
- NCERT Exemplar Class 9 solutions
- NCERT Exemplar Class 10 solutions
- NCERT Exemplar Class 11 Solutions
- NCERT Exemplar Class 12 Solutions
- NCERT Books for class 6
- NCERT Books for class 7
- NCERT Books for class 8
- NCERT Books for class 9
- NCERT Books for Class 10
- NCERT Books for Class 11
- NCERT Books for Class 12
- NCERT Notes for Class 9
- NCERT Notes for Class 10
- NCERT Notes for Class 11
- NCERT Notes for Class 12
- NCERT Syllabus for Class 6
- NCERT Syllabus for Class 7
- NCERT Syllabus for class 8
- NCERT Syllabus for class 9
- NCERT Syllabus for Class 10
- NCERT Syllabus for Class 11
- NCERT Syllabus for Class 12
- CBSE Date Sheet
- CBSE Syllabus
- CBSE Admit Card
- CBSE Result
- CBSE Result Name and State Wise
- CBSE Passing Marks
CBSE Class 10- CBSE Board Class 10th
- CBSE Class 10 Date Sheet
- CBSE Class 10 Syllabus
- CBSE 10th Exam Pattern
- CBSE Class 10 Answer Key
- CBSE 10th Admit Card
- CBSE 10th Result
- CBSE 10th Toppers
- CBSE Board Class 12th
- CBSE Class 12 Date Sheet
- CBSE Class 12 Admit Card
- CBSE Class 12 Syllabus
- CBSE Class 12 Exam Pattern
- CBSE Class 12 Answer Key
- CBSE 12th Result
- CBSE Class 12 Toppers
CISCE Board 10th- ICSE 10th time table
- ICSE 10th Syllabus
- ICSE 10th exam pattern
- ICSE 10th Question Papers
- ICSE 10th Result
- ICSE 10th Toppers
- ISC 12th Board
- ISC 12th Time Table
- ISC Syllabus
- ISC 12th Question Papers
- ISC 12th Result
- IMO Syllabus
- IMO Sample Papers
- IMO Answer Key
- IEO Syllabus
- IEO Answer Key
- NSO Syllabus
- NSO Sample Papers
- NSO Answer Key
- NMMS Application form
- NMMS Scholarship
- NMMS Eligibility
- NMMS Exam Pattern
- NMMS Admit Card
- NMMS Question Paper
- NMMS Answer Key
- NMMS Syllabus
- NMMS Result
- NTSE Application Form
- NTSE Eligibility Criteria
- NTSE Exam Pattern
- NTSE Admit Card
- NTSE Syllabus
- NTSE Question Papers
- NTSE Answer Key
- NTSE Cutoff
- NTSE Result
Schools By Medium- Malayalam Medium Schools in India
- Urdu Medium Schools in India
- Telugu Medium Schools in India
- Karnataka Board PUE Schools in India
- Bengali Medium Schools in India
- Marathi Medium Schools in India
By Ownership- Central Government Schools in India
- Private Schools in India
- Schools in Delhi
- Schools in Lucknow
- Schools in Kolkata
- Schools in Pune
- Schools in Bangalore
- Schools in Chennai
- Schools in Mumbai
- Schools in Hyderabad
- Schools in Gurgaon
- Schools in Ahmedabad
- Schools in Uttar Pradesh
- Schools in Maharashtra
- Schools in Karnataka
- Schools in Haryana
- Schools in Punjab
- Schools in Andhra Pradesh
- Schools in Madhya Pradesh
- Schools in Rajasthan
- Schools in Tamil Nadu
- NVS Admit Card
- Navodaya Result
- Navodaya Exam Date
- Navodaya Vidyalaya Admission Class 6
- JNVST admit card for class 6
- JNVST class 6 answer key
- JNVST class 6 Result
- JNVST Class 6 Exam Pattern
- Navodaya Vidyalaya Admission
- JNVST class 9 exam pattern
- JNVST class 9 answer key
- JNVST class 9 Result
Download Careers360 App'sRegular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile Certifications We Appeared in- Application of derivatives
- Maxima and Minima
- Active page
Application of derivatives of Class 12From the figure for the function y = f(x) (iii) Boundary points - Tangent and Normal
- Angle Between Two Curves
- Monotonicity
- Working Rule For Finding Maxima And Minima
- Rolle’s Theorem
- Lagrange’s Mean value Theorem (LMVT)
Talk to Our counsellor- Andhra Pradesh
- Chhattisgarh
- West Bengal
- Madhya Pradesh
- Maharashtra
- Jammu & Kashmir
- NCERT Books 2022-23
- NCERT Solutions
- NCERT Notes
- NCERT Exemplar Books
- NCERT Exemplar Solution
- States UT Book
- School Kits & Lab Manual
- NCERT Books 2021-22
- NCERT Books 2020-21
- NCERT Book 2019-2020
- NCERT Book 2015-2016
- RD Sharma Solution
- TS Grewal Solution
- TR Jain Solution
- Selina Solution
- Frank Solution
- Lakhmir Singh and Manjit Kaur Solution
- I.E.Irodov solutions
- ICSE - Goyal Brothers Park
- ICSE - Dorothy M. Noronhe
- Micheal Vaz Solution
- S.S. Krotov Solution
- Evergreen Science
- KC Sinha Solution
- ICSE - ISC Jayanti Sengupta, Oxford
- ICSE Focus on History
- ICSE GeoGraphy Voyage
- ICSE Hindi Solution
- ICSE Treasure Trove Solution
- Thomas & Finney Solution
- SL Loney Solution
- SB Mathur Solution
- P Bahadur Solution
- Narendra Awasthi Solution
- MS Chauhan Solution
- LA Sena Solution
- Integral Calculus Amit Agarwal Solution
- IA Maron Solution
- Hall & Knight Solution
- Errorless Solution
- Pradeep's KL Gogia Solution
- OP Tandon Solutions
- Sample Papers
- Previous Year Question Paper
- Important Question
- Value Based Questions
- CBSE Syllabus
- CBSE MCQs PDF
- Assertion & Reason
- New Revision Notes
- Revision Notes
- Question Bank
- Marks Wise Question
- Toppers Answer Sheets
- Exam Paper Aalysis
- Concept Map
- CBSE Text Book
- Additional Practice Questions
- Vocational Book
- CBSE - Concept
- KVS NCERT CBSE Worksheets
- Formula Class Wise
- Formula Chapter Wise
- Toppers Notes
- Most Repeated Question
- Diagram Based Question
- Study Planner
- JEE Previous Year Paper
- JEE Mock Test
- JEE Crash Course
- JEE Sample Papers
- Important Info
- SRM-JEEE Previous Year Paper
- SRM-JEEE Mock Test
- VITEEE Previous Year Paper
- VITEEE Mock Test
- BITSAT Previous Year Paper
- BITSAT Mock Test
- Manipal Previous Year Paper
- Manipal Engineering Mock Test
- AP EAMCET Previous Year Paper
- AP EAMCET Mock Test
- COMEDK Previous Year Paper
- COMEDK Mock Test
- GUJCET Previous Year Paper
- GUJCET Mock Test
- KCET Previous Year Paper
- KCET Mock Test
- KEAM Previous Year Paper
- KEAM Mock Test
- MHT CET Previous Year Paper
- MHT CET Mock Test
- TS EAMCET Previous Year Paper
- TS EAMCET Mock Test
- WBJEE Previous Year Paper
- WBJEE Mock Test
- AMU Previous Year Paper
- AMU Mock Test
- CUSAT Previous Year Paper
- CUSAT Mock Test
- AEEE Previous Year Paper
- AEEE Mock Test
- UPSEE Previous Year Paper
- UPSEE Mock Test
- CGPET Previous Year Paper
- BCECE Previous Year Paper
- JCECE Previous Year Paper
- Crash Course
- Previous Year Paper
- NCERT Based Short Notes
- NCERT Based Tests
- NEET Sample Paper
- Previous Year Papers
- Quantitative Aptitude
- Numerical Aptitude Data Interpretation
- General Knowledge
- Mathematics
- Agriculture
- Accountancy
- Business Studies
- Political science
- Enviromental Studies
- Mass Media Communication
- Teaching Aptitude
- Verbal Ability & Reading Comprehension
- Logical Reasoning & Data Interpretation
- CAT Mock Test
- CAT Important Question
- CAT Vocabulary
- CAT English Grammar
- MBA General Knowledge
- CAT Mind Map
- CAT Study Planner
- CMAT Mock Test
- SRCC GBO Mock Test
- SRCC GBO PYQs
- XAT Mock Test
- SNAP Mock Test
- IIFT Mock Test
- MAT Mock Test
- CUET PG Mock Test
- CUET PG PYQs
- MAH CET Mock Test
- MAH CET PYQs
- NAVODAYA VIDYALAYA
- SAINIK SCHOOL (AISSEE)
- Mechanical Engineering
- Electrical Engineering
- Electronics & Communication Engineering
- Civil Engineering
- Computer Science Engineering
- CBSE Board News
- Scholarship Olympiad
- School Admissions
- Entrance Exams
- All Board Updates
- Miscellaneous
- State Wise Books
- Engineering Exam
RD Sharma Class 12 Solutions Maxima and Minima PDF DownloadAll 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. How to Download the RD Sharma Class 12 Solutions Maxima and Minima from the official website of Selfstudys?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.
- 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 ’.
- 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.
Why Should Students Go Through the RD Sharma Class 12 Solutions Maxima and Minima? 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.
- The RD Sharma Class 12 Solutions Maxima and Minima are created by the subject matter experts at selfstudys which have made sure to develop them as per the latest curriculum. The students get an in-depth knowledge of all the topics as all the methods used in these solutions are well-detailed and easy. This makes students interested in Maths subjects rather than most of the students who fear it.
- The students should regularly practise the RD Sharma Class 12 Solutions Maxima and Minima as it helps in increasing the accuracy of all the students. Apart from this, these solutions help the students in remembering the topics which they have previously learned and also helps to stick these topics in their minds so that they can remember them for a longer period of time. These Maxima and Minima solutions also give an idea to the students about their progress in exam preparation and also the strong areas and weaker areas which makes students aware about the areas in Maxima and Minima where they need to put more effort.
What Are the Tips to Study From the RD Sharma Class 12 Solutions Maxima and Minima?There are many tips to study from the RD Sharma Class 12 Solutions Maxima and Minima. All the students are advised to complete their textbook first and then they should go for the RD Sharma solutions Class 12 Maxima and Minima. Some of the most important tips are: - Go through the RD Sharma solutions Class 12 Maxima and Minima strategically: It is advisable for all the students to go through these notes strategically to ensure that they are understanding every topic deeply and clearly.
- Brush up on the basic mathematical concepts: Maths is a difficult subject if you are not aware of the basic concepts So it is always advisable for all the students to brush up on the basic mathematical concepts when going through the RD Sharma Class 12 Solutions Maxima and Minima.
- Do Group Study: Doing group study with friends who are also going through the RD Sharma solutions Class 12 Maxima and Minima can prove to be a great advantage to the student. It also helps a student to identify their strong areas and weak areas and work on the weaker areas to get good marks.
- Stay Focused: As Maxima and Minima is a hard topic and requires constant focus and dedication, all the students must stay focused and motivated and also should regularly go through the RD Sharma solutions Class 12 Maxima and Minima.
- Make a routine: All the students are advised to make a routine and go through the RD Sharma Class 12 Solutions Maxima and Minima daily at least for half an hour. This will ensure that the students will remember the information for a longer period of time.
What are the features of the RD Sharma Class 12 Solutions Maxima and Minima? There are various features of the RD Sharma solutions Class 12 Maxima and Minima. Some of the most important of them are: - In-Depth Knowledge: One of the most important benefits of the RD Sharma solutions Class 12 Maxima and Minima is that the students get an in-depth knowledge of the concepts as each and every problem is well-explained.
- Various Important Questions: In the RD Sharma Class 12 Solutions Maxima and Minima, there are different questions including Multiple Choice Questions, Fill ups, true or false which can help the students create a strong base in their minds for all the mathematical concepts. Not only this, but there are also HOTS (High order thinking skills) questions which forces the students to think on an advanced level. Also, there are solutions of the previous year question papers in the RD Sharma solutions Class 12 Maxima and Minima which gives the students a brief idea about how to answer the questions in the exam.
- Brief Summary: In the end of each chapter of the RD Sharma Class 12 Solutions Maxima and Minima, a brief summary is also provided which gives a theoretical idea to all the students about the important concepts and key points.
- Examples are also included: A lot of different examples are also included in the RD Sharma solutions Class 12 Maxima and Minima which helps the students in getting a clear understanding of all the concepts.
What are the Benefits of Using RD Sharma Class 12 Solutions Maxima and Minima? There are various benefits of using the RD Sharma Class 12 Solutions Maxima and Minima. Some of the most important benefits of them are: - Helps in Solving Tricky Questions: One of the biggest benefits of the RD Sharma solutions Class 12 Maxima and Minima is that it helps the students in solving the tricky and complex questions.
- Increases Speed: The RD Sharma Class 12 Solutions Maxima and Minima are explained in detail to all the students which help them increase the speed of their work.
- Helps the students to boost their conceptual knowledge: The RD Sharma solutions Class 12 Maxima and Minima helps in boosting the conceptual knowledge of the students and also helps them to prepare well for their final examinations.
- A great revision tool: After completing your exam preparation, every student needs to know their progress of preparation. Also, to stick all the important concepts and topics of Maxima and Minima in their minds, students need to revise each and every topic which includes important key points, questions, formulas etc. The RD Sharma solutions Class 12 Maxima and Minima acts as a great tool for revision which can enhance the preparation of all the students.
- Learn Various Approaches: All the students regularly going through the RD Sharma Class 12 Solutions Maxima and Minima can learn different approaches to solve the problems. This not only boosts their knowledge but also increases their chances of scoring well in their examinations.
How is the RD Sharma Class 12 Solutions Maxima and Minima at Selfstudys is Different From Other Solutions?There are various reasons why the RD Sharma solutions Class 12 Maxima and Minima at Selfstudys are different from other solutions. Some of the most important of them are: - All the solutions are exercise wise: All the solutions in the RD Sharma Class 12 Solutions Maxima and Minima are exercise-wise which helps the students to study in an organised way and also prevents them from being confused.
- Solutions are explained with easy tricks: All the solutions in the RD Sharma solutions Class 12 Maxima and Minima are explained with easy tricks to ensure that the student does not get confused and also grasps the concepts easily.
- All the solutions are solved by subject matter experts: All the solutions in the RD Sharma Class 12 Solutions Maxima and Minima are solved by the subject matter experts which means that they are accurate.
- NCERT Solutions for Class 12 Maths
- NCERT Solutions for Class 10 Maths
- CBSE Syllabus 2023-24
- Social Media Channels
- Login Customize Your Notification Preferences
One Last Step...- Second click on the toggle icon
Provide prime members with unlimited access to all study materials in PDF format. Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding. Provide prime users with access to exclusive PDF study materials that are not available to regular users. NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12 Important Questions for CBSE Class 12 Maths Maxima and MinimaNovember 17, 2015 by Sastry CBSE Application of Derivatives Important Questions for CBSE Class 12 Maths Maxima and MinimaPrevious Year Examination Questions4 marks questions. 6 Marks QuestionsImportant Questions for Class 12 Maths Class 12 Maths NCERT Solutions Home Page Free ResourcesNCERT SolutionsQuick Resources- School Guide
- Mathematics
- Number System and Arithmetic
- Trigonometry
- Probability
- Mensuration
- Maths Formulas
- Class 8 Maths Notes
- Class 9 Maths Notes
- Class 10 Maths Notes
- Class 11 Maths Notes
- Class 12 Maths Notes
Class 12 RD Sharma Solutions – Chapter 18 Maxima and Minima – Exercise 18.5 | Set 1Question 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. Please Login to comment...Similar reads. - RD Sharma Solutions
- School Learning
- Maths-Class-12
- RD Sharma Class-12
- Top Android Apps for 2024
- Top Cell Phone Signal Boosters in 2024
- Best Travel Apps (Paid & Free) in 2024
- The Best Smart Home Devices for 2024
- 15 Most Important Aptitude Topics For Placements [2024]
Improve your Coding Skills with PracticeWhat kind of Experience do you want to share? |
IMAGES
VIDEO
COMMENTS
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.
CBSE Class 12 Maths Case Study Questions With Solutions CBSE Class 12 Mathematics Case Study Questions are introduced this year in the updated CBSE Board Exam Pattern. According to that this year the candidates need to prepare for the case study problems along with the questions that are legacy of CBSE Board.
We have provided here Case Study questions for the Class 12 Maths term 2 exams. You can read these chapter-wise Case Study questions for your term 2 science paper. These questions are prepared by subject experts and experienced teachers. Answer key is also provided so that you can check the correct answer for each question. Practice these questions to score well in your term 2 exams.
Students can get free Updated 2023-24 RD Sharma Class 12 Solutions Chapter 18 Maxima and Minima here. Visit BYJU'S to download the PDF version of RD Sharma Solutions Class 12.
Get free RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima solved by experts. Available here are Chapter 18 - Maxima and Minima Exercises Questions with Solutions and detail explanation for your practice before the examination
In RD Sharma Solutions for Class 12th Maths Chapter 18 maxima and minima, students first obtain knowledge on how to describe maximum and minimum values. Maximum is a point at which the value of a function is greatest, whereas minimum in maths is a point at which the value of a function is less than or equal to the value at any nearby point (local point) or an absolute minimum. Along with this ...
Free PDF download of RD Sharma Solutions for Class 12 Maths Chapter 18 - Maxima and Minima solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 18 - Maxima and Minima Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced), NEET, Engineering and Medical entrance exams.
Free PDF download of RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima Exercise 18.2 solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 18 - Maxima and Minima Ex 18.2 Questions with Solutions for RD Sharma Class 12 Maths to help you to revise complete Syllabus and Score More marks.
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 ...
RD Sharma Solutions for Class 12-science Maths CBSE Chapter 18: Get free access to Maxima and Minima Class 12-science Solutions which includes all the exercises with solved solutions. Visit TopperLearning now!
Download free pdf of RD Sharma Class 12 Maths chapter Maxima and Minima CSBQ solutions to see shortcut methods and simple formulas to get more marks in your exams.
Get Free RD Sharma Class 12 Solutions Chapter 18 Ex 18.1. Maxima and Minima Class 12 Maths RD Sharma Solutions are extremely helpful while doing your homwork or while preparing for the exam.
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.
Free PDF download of RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima Exercise 18.1 solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 18 - Maxima and Minima Ex 18.1 Questions with Solutions for RD Sharma Class 12 Maths to help you to revise complete Syllabus and Score More marks.
Download free pdf of RD Sharma 12th maths chapter 17 Maxima and Minima exercise 17.5 solutions to see shortcut methods and simple formulas to get more marks in your exams.
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.
Solutions of Maxima and Minima - Previous Years Board Questions - (2000 to 2020) - ISC - Class 12 - Mathematics Maxima and Minima. Click here to open solutions Scroll Down for solutions of Sem1-2021.
Updated 2023-24 RD Sharma Solutions for Class 12 Maths Chapter 18 - Maxima and Minima Exercise 18.5 covers the topic of applied problems on maxima and minima of a given function, with a descriptive solution.
Download free pdf of RD Sharma 12th maths chapter 17 Maxima and Minima MCQ solutions to see shortcut methods and simple formulas to get more marks in your exams.
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.
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 ...
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.
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 …..