Question #280008

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.


1
Expert's answer
2021-12-15T17:24:25-0500

Let xx be the width (in feet) and yy be the length. Then we have

xy=80000xy=80000

Now,counting interior and exterior walls,there are 12 walls in the xdirectionx-direction and 6 in the ydirectiony-direction .the number of linear feet of the wall is

12x+6y12x+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=80000xy={80000\over x}


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

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

dLdx=12480000x2=0{dL\over dx}=12-{480000\over x^2}=0

12=480000x212={480000\over x^2}

12x2=48000012x^2=480000

x2=40000x^2=40000

x=200x=200

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

x=200x=200 feet and the length is 80000x=80000200=400{80000\over x}={80000\over 200}=400 feet


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