Question

The manager of a department store wants to build a 600 square-foot rect- angular enclosure in the stores parking lot to display some equipment. Three sides of the enclosure will be built of redwood fencing at a cost of $14 per running foot. The forth side will be built with cement blocks, at a cost of $28 per running foot. Find the dimensions of the enclosure that will minimize the total cost of the building materials. [Hint: You’ll need to find an equation for the total cost of the materials in terms of only one variable.]

Answer 1

1. The total area in sq. ft. of the store parking lot in terms of dimensions x and y is:
xy = 600

2. The perimeter of the the lot in terms of x and y is:
P(x,y) = 3x + y

3. The total cost of the lot in terms of x and y is:
C(x,y) = 42x + 28y

4. The total cost of the lot in terms of x only is:
C(x) = 42x + 28(600/x)
C(x) = 42x + 16800/x
C(x) = 42x^2/x + 16800/x
C(x) = (42x^2 + 16800)/x

5. The derivative of C'(x) is:
C'(x) = 42(x^2 - 400)/x^2

6. Set C'(x) = 0
0 = 42(x^2 - 400)/x^2
0 = 42(x^2 - 400)
0 = x^2 - 400
400 = x^2
x = 20
y = 30

7. Thus the dimensions, (x,y), of the enclosure that will minimize the total cost of building materials is (20,30) in feet.