Answer to Question #280008 in Calculus for Prioty

Question

Interior and exterior walls of a rectangular 80,000 square foot building cost 90$ per running foot. The building is to be divided into 10 rooms by four interior walls in the x direction and one interior wall in the y direction. What should be the building dimensions if the wall cost is to be minimum.

Answer 1

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

$xy=80000$

Now,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 wall is

$12x+6y$

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

First, let's express the number of linear feet of walls as a function of single variable as follows;

$y={80000\over x}$

$L(x)=12x+6.({80000\over x})=12x+{480000\over x}$

Secondly, let's find where the derivative of the function is equal to zero

${dL\over dx}=12-{480000\over x^2}=0$

$12={480000\over x^2}$

$12x^2=480000$

$x^2=40000$

$x=200$

The total number of linear feet,and therefore the total cost of the walls, is minimum when then width is

$x=200$ feet and the length is ${80000\over x}={80000\over 200}=400$ feet