Will give 12 points

community recreational center has 500 ft of fencing with which to enclose a rectangular parking lot in the back of the property. the building itself will be used as one of the sides of the enclosed area. what is the maximum area that can be enclosed by the fencing? enter your answer in the box. (box right here) ft²

Let b be the dimension opposite the building and s be the dimension of the sides, and m be the amount of material used...

m=b+2s

b=m-2s

The area of this parking lot is:

A=bs, using b found above you get:

A=(m-2s)s

A=ms-2s^2

dA/ds=m-4s and d2A/ds2=-4

Since acceleration is a constant negative, when the velocity equals zero, it is at the point where A(s) is the absolute maximum.

dA/dx=0 only when m=4s, so s=m/4

We are told that m=500, so s=125

A=ms-2s^2  so the maximum area is:

A(125)=500(125)-2(125^2)

A(125)=31250 ft^2