Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #62

Prove that a right circular cone of greatest volume inscribed in a sphere of radius r>0 has height 4r/3. Find that volume.

✅ Step-by-Step Solution

Final Answer: Height 4r/3; Vmax=32πr³/81.

Let H be the cone height and ρ its base radius. A central axial section gives ρ²=r²−(H−r)²=2rH−H², 0≤H≤2r.
V(H)=πρ²H/3=π(2rH²−H³)/3.
V′=πH(4r−3H)/3 has its only interior zero at H=4r/3.
V′ is positive before and negative after that point; both endpoint volumes are zero. This proves the global maximum.
Vmax=(π/3)[2r(4r/3)²−(4r/3)³]=32πr³/81.

[1 mark] Let H be the cone height and ρ its base radius. A central axial section gives ρ²=r²−(H−r)²=2rH−H², 0≤H≤2r.
[1 mark] V(H)=πρ²H/3=π(2rH²−H³)/3.
[1 mark] V′=πH(4r−3H)/3 has its only interior zero at H=4r/3.
[1 mark] V′ is positive before and negative after that point; both endpoint volumes are zero. This proves the global maximum.
[1 mark] Vmax=(π/3)[2r(4r/3)²−(4r/3)³]=32πr³/81.

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