Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #59

A coaxial right circular cylinder rests on the base of a cone of height h>0 and base radius r>0, with its upper rim touching the cone. Prove the cylinder of greatest volume has height h/3 and volume 4/9 of the cone volume.

✅ Step-by-Step Solution

Final Answer: Height h/3; radius 2r/3; volume 4πr²h/27=(4/9)Vcone.

Let the cylinder height be H and radius R, with 0≤H≤h. Similar triangles give R/r=(h−H)/h.
Therefore V(H)=πr²H(h−H)²/h².
V′(H)=(πr²/h²)(h−H)(h−3H), whose interior zero is H=h/3.
The derivative is positive on (0,h/3) and negative on (h/3,h); endpoint volumes are 0, so this is the global maximum.
At H=h/3, R=2r/3 and Vmax=4πr²h/27=(4/9)(πr²h/3).

📋 CBSE Stepwise Marking Scheme

[1 mark] Let the cylinder height be H and radius R, with 0≤H≤h. Similar triangles give R/r=(h−H)/h.
[1 mark] Therefore V(H)=πr²H(h−H)²/h².
[1 mark] V′(H)=(πr²/h²)(h−H)(h−3H), whose interior zero is H=h/3.
[1 mark] The derivative is positive on (0,h/3) and negative on (h/3,h); endpoint volumes are 0, so this is the global maximum.
[1 mark] At H=h/3, R=2r/3 and Vmax=4πr²h/27=(4/9)(πr²h/3).

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Class 12 Maths · Applications Of Derivatives · Very Long Answer #59

✅ Step-by-Step Solution

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Master Chapter 6: Applications Of Derivatives

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