Question

A college is planning to construct a rectangular parking lot on land bordered on one side by a highway. The plan is to use 440 feet of fencing to fence off the other three sides. What dimensions should the lot have if the enclosed area is to be a​ maximum?

Answer 1

Let x represent side opposite to highway and y represent other two opposite sides as shown in the diagram.

The perimeter of the parking lot will be sum of 3 sides that is $x+y+y=x+2y$.

We have been given that the plan is to use 440 feet of fencing to fence off the other three sides. This means that perimeter of 3 sides is 440.

$x+2y=440$

$x=440-2y$

We know that area of rectangle is length times width, so area of parking lot will be $A=x\cdot y$

Upon substituting value of x, we will get:

$A(y)=(440-2y)\cdot y$

$A(y)=440y-2y^2$

Now we will find the derivative of area function as:

$A'(y)=\frac{d}{dy}(440y)-\frac{d}{dy}(2y^2)$

$A'(y)=440-4y$

Let us find critical point by equating derivative to 0.

$440-4y=0$

$440=4y$

$\frac{440}{4}=\frac{4y}{4}$

$110=y$

Now we will substitute this value is equation $x=440-2y$ to solve for x as:

$x=440-2(110)$

$x=440-220=220$

Therefore, the dimensions of 220 feet by 110 feet will enclose the maximum area.