How do you prove that a square-based cuboid takes the largest possible volume when it is a cube considering the point that it has surface area 520cm2?

Answer 1

Let the side of the square base of the cuboid be s > 0;

Let the height of the cuboid be h > 0;

Let the surface area be A > 0;

Then:

A = 2 × (s × s + s × h + h × s) = 2(s² + 2sh) = 2s² + 4sh

→ 4sh = A - 2s²

→ h = (A - 2s²)/4s

V = s × s × h

= s²h

= s²(A - 2s²)/4s

= s(A - 2s²)/4

= sA/4 - s³/2

This has a maximum value when dV/ds = 0

dV/ds = A/4 - 3s²/2

→ 3s²/2 = A/4

→ s² = A/6

→ s = √(A/6)

now:

h = (A - 2s²)/4s

= s × (A - 2s²)/4s²

= s × (A - 2 × A/6)/(4 × A/6)

= s × (A - A/3)/(2A/3)

= s × (2A/3)/(2A/3)

= s × 1

= s

→ The maximum volume (V) for a cuboid for a given surface area (A) is when the cuboid is a cube.