# Example 38 - Chapter 6 Class 12 Application of Derivatives

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.

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25

Example 26

Example 27

Example 28

Example 29

Example 30

Example 31

Example 32

Example 33

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important You are here

Example 39

Example 40 Important

Example 41

Example 42

Example 43

Example 44

Example 45

Example 46 Important

Example 47 Important

Example 48

Example 49

Example 50

Example 51

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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.