Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #64

Show that a right circular cylinder of greatest volume inscribed in a sphere of radius R>0 has height 2R/√3, and find that volume.

✅ Step-by-Step Solution

Final Answer: Height 2R/√3; Vmax=4πR³/(3√3).

Let y be half the cylinder height and ρ the base radius. Its axial rectangle satisfies ρ²+y²=R², 0≤y≤R.
V(y)=2πy(R²−y²).
V′=2π(R²−3y²)=0 gives y=R/√3.
The derivative is positive before this point and negative after it, while endpoint volumes are zero. Thus this is the global maximum.
The height is 2R/√3 and Vmax=2π(R/√3)(2R²/3)=4πR³/(3√3).

[1 mark] Let y be half the cylinder height and ρ the base radius. Its axial rectangle satisfies ρ²+y²=R², 0≤y≤R.
[1 mark] V(y)=2πy(R²−y²).
[1 mark] V′=2π(R²−3y²)=0 gives y=R/√3.
[1 mark] The derivative is positive before this point and negative after it, while endpoint volumes are zero. Thus this is the global maximum.
[1 mark] The height is 2R/√3 and Vmax=2π(R/√3)(2R²/3)=4πR³/(3√3).

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