# Misc 17

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2 3 . Also find the maximum volume. Let R be the radius of sphere Let h be the height & be the diameter of cylinder In Using Pythagoras theorem 2 + 2 = 2 h2 + 2 = + 2 h2 + 2 = 2 2 h2 + 2 = 4R2 2 = 4R2 h2 We need to find maximum volume of cylinder Let V be the volume of cylinder V = 2 V = 2 2 V = 2 4 V = 4 2 2 4 V = 4 2 4 3 4 V = hR2 3 4 Different w.r.t = 2 3 /4 = R2 4 3 = R2 4 3 2 = R2 3 4 h2 Putting =0 R2 3 4 2 =0 3 4 2 = 2 h2 = 2 3 4 h2 = 4 2 3 h = 4 2 3 h = 2 3 Finding 2 2 = 2 3 4 2 Different w.r.t h 2 2 = 2 3 4 2 2 2 = 0 3 4 2 2 2 = 3 4 Putting h = 2 3 2 2 = 3 2 2 3 = 3 < 0 h = 2 3 is point of maxima So, volume is maximum when h = 2 3 From (1) 2 = 4R2 h2 2 = 4R2 2 3 2 2 = 4R2 4 2 3 2 = 12 2 4 2 3 2 = 8 3 2 Maximum value of volume is V = 2 2 V = 4 2 Putting value of 2 & h V = 4 8 2 3 2 3 V = 16 3 12 3 V = cubic unit

Chapter 6 Class 12 Application of Derivatives

Ex 6.3,5
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Ex 6.3,7 Important

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Ex 6.3,15 Important

Ex 6.3,26 Important

Example 35 Important

Example 37 Important

Example 38 Important

Example 40 Important

Ex 6.5,1 Important

Ex 6.5,5 Important

Ex 6.5,7 Important

Ex 6.5,11 Important

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Example 46 Important

Example 47 Important

Misc 6 Important

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Misc 17 Important You are here

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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.