fencing 9.5 18

Question

ccu College is planning to construct a rectangular parking lot on land bordered on one side by a highway. The plan is to use
560
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?


The width is
_____feet. The length is
_______ feet.

Answer 1

Let l = length parallel to the highway and w the other dimension of the lot.

 

Also, l + 2w = 560  and A = area = l*w.

 

Since l = 560 - 2w, then A = l*w = (560 - 2w) * w = 560w - 2w²

 

Non Calculus

 

A = 560w - 2w² = -2(w² - 280w)

 

Complete the square: A = -2(w² - 280w +140²) + 2*140²

 

                                      = 2*140² - 2(w - 140)²

 

When w = 140, the area is a maximum otherwise the area becomes smaller since a number would be subtracted from the maximum value.

 

Calculus

 

dA/dx = 560 - 4w = 0 => w = 140

 

from d²A/dx² = -4 < 0, this would be a maximum area.

 

In either case the length is l = 560 - 2w = 280.

 

 

Answer 2

the maximum area occurs when you have a square

560/4=140

since one side is the highway, add that side's length of 140 to the opposite side

140 by 280 by 140 by "the highway"

algebraically...

P=2l+2w

P=2l+w because one side is the highway

560=2l+w

560-2l=w

A=lw

A=l(560-2l)

0=560l-2l^2

this is a parabola that opens downward so the vertex, (h,k) is the maximum point

h=-b/2a from the general form of the equation for a parabola

-b/2a=-560/(2)(-2)=-560/-4=140

again, since one side is the highway, add 140 to the opposite side

140 feet by 280 feet by 140 feet by "the building"

the length is 280 feet and the width is 140 feet