Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #80

Prove that the greatest-volume right circular cone inscribed in a sphere of radius a>0 occupies 8/27 of the sphere volume.

✅ Step-by-Step Solution

Final Answer: Vmax=(8/27)Vsphere=32πa³/81.

For cone height H and base radius ρ, an axial section gives ρ²=a²−(H−a)²=2aH−H², 0≤H≤2a.
The volume is V=π(2aH²−H³)/3.
V′=πH(4a−3H)/3=0 gives the unique interior critical height H=4a/3.
V′ changes from positive to negative there and endpoint volumes are zero, proving the global maximum.
Vmax=32πa³/81=(8/27)(4πa³/3)=(8/27)Vsphere.

[1 mark] For cone height H and base radius ρ, an axial section gives ρ²=a²−(H−a)²=2aH−H², 0≤H≤2a.
[1 mark] The volume is V=π(2aH²−H³)/3.
[1 mark] V′=πH(4a−3H)/3=0 gives the unique interior critical height H=4a/3.
[1 mark] V′ changes from positive to negative there and endpoint volumes are zero, proving the global maximum.
[1 mark] Vmax=32πa³/81=(8/27)(4πa³/3)=(8/27)Vsphere.

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