Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #69

Prove that among closed cuboids with square base and fixed positive volume V, the least surface area occurs for a cube.

✅ Step-by-Step Solution

Final Answer: Cube of side V^(1/3); Smin=6V^(2/3).

Let the square-base side be x>0 and height y>0. The volume constraint gives y=V/x².
The closed surface area is S=2x²+4xy=2x²+4V/x.
S′=4x−4V/x²=0 gives x³=V.
S′<0 before V^(1/3) and S′>0 afterwards (also S″=4+8V/x³>0), proving the global minimum.
At x=V^(1/3), y=V/x²=x, so the cuboid is a cube and Smin=6V^(2/3).

[1 mark] Let the square-base side be x>0 and height y>0. The volume constraint gives y=V/x².
[1 mark] The closed surface area is S=2x²+4xy=2x²+4V/x.
[1 mark] S′=4x−4V/x²=0 gives x³=V.
[1 mark] S′<0 before V^(1/3) and S′>0 afterwards (also S″=4+8V/x³>0), proving the global minimum.
[1 mark] At x=V^(1/3), y=V/x²=x, so the cuboid is a cube and Smin=6V^(2/3).

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