# Example 38

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 38 Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. Let OC = r be the radius of cone & OA = h be height of cone & OAQ = be the semi vertical angle of cone Let OE = x be the radius of cylinder Height of cylinder = OO From (1) & (2) = = = = = Now, Curved Surface Area Of cylinder = 2 Radius of cylinder Height of cylinder S = 2 S = 2 S = 2 2 S = 2 We need to minimize S, So, finding S (x) S = 2 S = 2 S = 2 Putting S = 0 0 = 2 2 = 0 = 2 Now, Finding S (x) at x = 2 S = 2 S = 2 S = 0 2 S = 2 So, S = 2 Hence, at = 2 = 2 <0 = is maxima of S. Hence, radius of cylinder with greatest curved surface area which can be inscribed in a given cone is half of that cone.

Chapter 6 Class 12 Application of Derivatives

Ex 6.3,5
Important

Ex 6.3,7 Important

Ex 6.3,12 Important

Ex 6.3,15 Important

Ex 6.3,26 Important

Example 35 Important

Example 37 Important

Example 38 Important You are here

Example 40 Important

Ex 6.5,1 Important

Ex 6.5,5 Important

Ex 6.5,7 Important

Ex 6.5,11 Important

Ex 6.5,18 Important

Ex 6.5,20 Important

Ex 6.5,23 Important

Ex 6.5,26 Important

Ex 6.5,28 Important

Example 46 Important

Example 47 Important

Misc 6 Important

Misc 11 Important

Misc 13 Important

Misc 17 Important

Misc 22 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.