Question

The campus of a college has plans to construct a rectangular parking lot on land bordered on one side by a highway. There are 680 ft of fencing available to fence the other three sides. Let x represent the length of each of the two parallel sides of fencing. A rectangle has width x. x x
​(a) Express the length of the remaining side to be fenced in terms of x.
​(b) What are the restrictions on​ x? ​
(c) Determine a function A that represents the area of the parking lot in terms of x. ​
(d) Determine the values of x that will give an area between 20 comma 000 and 50 comma 000 ftsquared.
​(e) What dimensions will give a maximum​ area, and what will this area​ be?

Answer 1

Answer:

a). (680 - 2x) ft

b). 0 < x < 340

c). A(x) = x(680 - 2x)

d). 32.52 < x < 107.55 or 232.45 < x < 302.48

e). 170 ft, 170 ft, 340 ft.

Step-by-step explanation:

(a). Remaining side of the triangle will be = (680 - 2x) ft

(b). Since 680 - 2x > 0

2x < 680

x < 340

and x > 0

So 0< x < 340 will be the restriction for x

(c) Let the area of the parking lot is given by A(x) then

A(x) = x(680 - 2x) will be the area of the rectangular park.

(d) If area is between 20000 and 50000 square feet then

20000 < x(680 - 2x) < 50000

and we have to determine the values of x.

Since 20000 < 680x - 2x² < 50000

10000 < 340x - x² < 25000 [by deviding the inequality by 2]

Now we can break this inequality t solve for the value of x.

10000 < 340x - x²  and 340x - x² > 25000

-x² + 340x > 10000

or x² - 340x + 10000 < 0

Now we convert this inequality into a whole square form

x² - 2(170x) + 10000 + 18900 < 18900

(x - 170)² < 18900

170 - √18900 < x < 170 + √18900

32.52 < x < 302.48 -----------(1)

and -x² + 340x < 25000

x² - 340x + 25000 > 0

(x - 170)² > 3900

So (x - 170) < -(√3900) or (x - 170) > √3900

x < 107.55 or x > 232.45 -------(2)

From inequalities 1 and 2

32.52 < x < 107.55 or 232.45 < x < 302.48

(e) Since area of the rectangular parking lot is A(x) = 680x - 2x²

To calculate the maximum area of the parking we will find the derivative of the area and equate it to zero.

then $\frac{dA}{dx}=\frac{d(680x-2x^{2})}{dx}$ = 0

680 - 4x = 0

4x = 680

x = 170

So dimensions of the parking will be 170 ft, 170 ft, 340 ft.