Acommunity 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?

This is an optimization problem. This means that you have to find the equation that models the situation, find the first derivative and make it equal to zero.

The condition is that the fence will form three sides of the parking lot, because the other side of the rectangle will be a wall of the building.

Call x the length of the side on the rectangle that is parallel to the building.

Call y to the other side.

The lengths of the three sides that the fence will cover is x + y + y = 500 ft

Equation (1) x + 2y = 500.

The area enclosed by the fence will be x * y

Equation (2) Area = x * y.

Now you want that the deriviative of the area = 0. First, put the area as a function of one variable (x or y).

I will use y.

From equation (1) x + 2y = 500 => x = 500 - 2y

Then area = (500 - 2y) * y = 500y - 2y^2

=> (area)' = 500 - 4y = 0 => y = 500/4 = 125 ft.

Thex x = 500 - 2*(125) = 250 ft

And the area is: area = 250 ft * 125 ft =31,250 ft^2

31,250 ft^2