a) --- up to you
let the length of the side which is parallel to the highway be x , x < 640
let the length of the other two equal sides by y each
x + 2y = 640 ---> y = (640-x)/2 = 320 - x/2
area = xy = x(320-x/2)
= 320x - (1/2)x^2
so we need the vertex of this downwards opening parabola
the x of the vertex = -b/(2a) = -320/(-1) = 320
if x = 320, then y = (640-320)/2 = 160
and the maximum area is (160)(320) = 51200 ft^2
for part d)
30000 = 320x - x^2/2
x^2 - 640x + 60000 = 0
x = (640 ± √169600)/2 = 114.1 ft or 529 ft
if x = 114.1 then y = 262.95 and area = 30003
if x = 529 ft then y = negative, no way
40,000 = 320x - x^2/2
x^2 -640x + 80000 = 0
x = 170.33 , then y = 234.83 ft, area = 39998.6
I will leave it up to you to split up my solutions into the individual parts b), c) etc
One campus of HCC has plans to construct rectangular parking lot on land bordered on one side by a highway. There are 640 ft of fencing available to fence the other sides. Let x represent the length of each of the two parallel sides of fencing.
a. draw the illustration of the problem
b.express length of remaining side to be fenced in terms of x.
c. what're the restrictions on x?
d. Determine values of x that'll give area between 30,000 and 40,000 ft squared.
e. what dimensions will give a max area, and what will this area be?
f. Determine a function A that represents the area of the parking lot in terms of x.
1 answer
2 answers