SOLUTION: Interior and exterior walls of 80,000 square foot rectangular warehouse cost $90 per running foot. The warehouse is to be divided into 10 rooms by four interior walls running in th

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: Interior and exterior walls of 80,000 square foot rectangular warehouse cost $90 per running foot. The warehouse is to be divided into 10 rooms by four interior walls running in th      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1145205: Interior and exterior walls of 80,000 square foot rectangular warehouse cost $90 per running foot. The warehouse is to be divided into 10 rooms by four interior walls running in the x direction and one running in the y direction.
a) What dimension will lead to minimal total wall cost?
b) What is this minimal cost?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let x be the width (in feet) and y the length. Then

xy+=+80000

Counting interior and exterior walls, there are 12 walls in the x direction and 6 in the y direction. The number of linear feet of the walls is

12x%2B6y

We need to minimize the cost of the walls, which means minimizing the number of linear feet of the walls.

(1) Express the number of linear feet of walls as a function of a single variable:

y+=+80000%2Fx
L%28x%29+=+12x%2B6%2880000%2Fx%29+=+12x%2B480000%2Fx

(2) Find where the derivative of the function is equal to zero.

dL%2Fdx+=+12-480000%2Fx%5E2+=+0
12+=+480000%2Fx%5E2
12x%5E2+=+480000
x%5E2+=+40000
x+=+200

The total number of linear feet, and therefore the total cost of the walls, is minimum when the width is x=200 feet and the length is 80,000/x = 400 feet.

ANSWERS:
(a) width 200 feet, length 400 feet
(b) (12(200)+6(400))*$90 = (4800)*$90 = $432,000