Value determining for cuboid

Question

A manufacturer wishes to make a container in the form of a closed box with thin sheet metal. The box is in the shape of a cuboid with a rectangular base whose length is twice the width of x metres and has height , h metres.

The volume, V of the box is 72m3.

Determine the values of x and h, for which the surface area of the box, A is a minimum value.

Answer 1

Volume, V = 72 m³ = x(2x)h= 2x²h --> h = 72/(2x²) = 36/x²

Surface area, A = 2(2x²) + 2(xh) + 2(2xh) = 4x²+6xh = 4x² +6x(36/x²) = 4x² + 216/x

To find the minimum, take the derivative of A(x) and then set to zero and solve for x, then h.

A'(x) = 8x -216/x² = (8x³-216)/x²

Only need to set the numerator to zero.

8x³-216 = 0

x = (216/8)^1/3 = 27^1/3 = 3 m

h = 36/3² = 4 m

Answer 2

V = 2x²h = 72; h = 36/x².

Surface area S = 4x² + 2xh + 4xh = 4x² + 6xh = 4x² + 6x·36/x² = 4x² + 216/x;

S' = 8x - 216/x² = 0; 8x³ - 216 = 0; x³ = 27; x = 3.

S'' = 8 + 432/x³; S''(3) = 8 + 432/27 = 24 > 0. So, we have minimum.

Answer: x = 3 m and h = 36/9 = 4 m