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 680 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

Answer:

The maximum length is 170 feet and maximum width is 340 feet.

Step-by-step explanation:

Let the length of the parking = x feet and width = y feet.

Since, the fencing is to be applied on three sides of the lot.

So,we get, $2x+y=680$

i.e. $y=680-2x$

Since, the area of the rectangular lot = Length × Width

So, area of the lot = $x \times y$

i.e. area of the lot = $x \times (680-2x)$

i.e. area of the lot = $680x-2x^2$

Now, we know the maximum value of 'x' in $ax^2+bx+c$ is $\frac{-b}{2a}$.

So, the maximum length of the lot is x= $\frac{-680}{2\times -2}$ = $\frac{680}{4}$ = 170 feet.

Then, $y=680-2x$ i,plies $y=680-2\times 170$ implies $y=680-340$ i.e. y= 340 feet.

So, the maximum length is 170 feet and maximum width is 340 feet.