Minimizing Costs

Question

 A pencil cup with a capacity of 16 in.³ is to be constructed in the shape of a rectangular box with a square base and an open top. If the material for the sides costs 12¢/in.² and the material for the base costs 48¢/in.², what should the dimensions of the cup be to minimize the construction cost?

h=

l=

w=

Answer 1

Let the square base have side length x inches and height h inches.

Volume:

x²h = 16

So:

h = 16/x²

The four sides have total area:

4xh

At 12¢/in², side cost is:

12(4xh) = 48xh

The base has area x² and costs 48¢/in²:

48x²

Therefore, total cost is:

C = 48xh + 48x²

Substitute h = 16/x²:

C(x) = 48x(16/x²) + 48x²

C(x) = 768/x + 48x²

Differentiate:

C'(x) = -768/x² + 96x

Set C'(x) = 0:

-768/x² + 96x = 0

96x³ = 768

x³ = 8

x = 2

Since the base is square:

l = 2 in

w = 2 in

Now find the height:

h = 16/(2²)

h = 4 in

Final answers:

h = 4 in

l = 2 in

w = 2 in