If the box is square with side length x, the surface area is

a = xy+2yz+2xz

= x^2 + 2z(2x)

= x^2 + 4x(32/(x^2))

= x^2 + 128/x

This will be minimized when the derivative with respect to x is zero.

2x - 128/x^2 = 0

2x^3 - 128 = 0

x^3 = 64

x = 4

A square box that is

Its area will be 16 ft^2 + 32 ft^2 = 48 ft^2.

If the box is 4.1 ft by 3.9 ft by 2.00125 ft, the area will be ≈ 48.01 ft^2. Thus, it looks like the box should be square for minimum surface area.

a = xy+2yz+2xz

= x^2 + 2z(2x)

= x^2 + 4x(32/(x^2))

= x^2 + 128/x

This will be minimized when the derivative with respect to x is zero.

2x - 128/x^2 = 0

2x^3 - 128 = 0

x^3 = 64

x = 4

A square box that is

**4 ft by 4 ft and 2 ft deep**will have minimum surface area.Its area will be 16 ft^2 + 32 ft^2 = 48 ft^2.

If the box is 4.1 ft by 3.9 ft by 2.00125 ft, the area will be ≈ 48.01 ft^2. Thus, it looks like the box should be square for minimum surface area.