Minimizing Packaging Cost

Question

A rectangular box is to have a square base and a volume of 54 ft³. If the material for the base costs $0.33/ft², the material for the sides costs $0.12/ft², and the material for the top costs $0.15/ft², determine the dimensions (in ft) of the box that can be constructed at minimum cost. (Refer to the figure below.)

Answer 1

Box with square bottom with sides x, and height y.

Volume can be expressed as x²y = 54

Cost can be expressed as

bottom: x²(0.33)

4 sides: 4*xy*(0.12)

top: x²(0.15)

Cost = x²(0.33) + 4xy(0.12) + x²(0.15) = x²(0.48) + xy(0.48) = 0.48(x²+xy)

To do min/max problems we will do the derivate for the cost equation,but first we need to substitute out y so that its only in terms of x. We can do that using our volume expression. x²y = 54

y = 54/x²

Cost = 0.48(x²+54/x)

dC/dx = 0.48(2x -54/x²)

We now set the derivate equal to 0 and solve for x

0.48(2x - 54/x²) = 0

2x - 54/x² =

2x = 54/x²

2x³= 54

x³ = 27

x = 3.

Remember y = 54/x²

y = 54/9 = 6.

Dimensions of box = 3x3x6.