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

To maximize the enclosed area of the rectangular parking lot, we can use calculus. By setting up an optimization problem, we can find the dimensions of the parking lot that will result in the maximum area. In this case, the dimensions should be 110 feet by 220 feet.

Step-by-step explanation:

To find the dimensions of the rectangular parking lot that will maximize the enclosed area, we need to solve the optimization problem using calculus. Let's assume the width of the parking lot is w feet. Since the fencing is only required on the three sides other than the highway side, the length of the two equal sides will be w/2 feet each. The total length of these two sides will be 2(w/2) = w feet.

Now, the length of the highway side is unknown and can be expressed as 440 - 2w feet, where 440 feet is the total length of the fencing available. The enclosed area of the parking lot is given by the product of its length and width, which is Area = (w)(440 - 2w) = 440w - 2w².

To maximize the enclosed area, we need to find the critical points of the area function by taking its derivative and setting it to zero. The derivative of the area function is d(Area)/dw = 440 - 4w. Setting it to zero and solving for w, we get w = 110. Substituting this value back into the area function, we find the maximum enclosed area to be Area = (110)(440 - 2(110)) = 24200 square feet.

Therefore, to maximize the enclosed area, the parking lot should have dimensions of 110 feet by 220 feet.

Answer 2

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)=(d)/(dy)(440y)-(d)/(dy)(2y^2)`

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

Let us find critical point by equating derivative to 0.

`440-4y=0`

`440=4y`

`(440)/(4)=(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.