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November 22, 2014

November 22, 2014

Posted by **Andy G** on Tuesday, January 14, 2014 at 9:12pm.

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.

- Pre-Calc -
**Reiny**, Tuesday, January 14, 2014 at 10:54pma) --- 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

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