Maximum Area Word Problem College Algebra

Question

Jim wants to build a rectangular parking lot along a busy street but only has 2900 feet of fencing available. If not fencing is required along the street, find the maximum area of the parking lot.

Answer 1

Let x = length

Let y = width

 

 

The perimeter is 2900 ft.

 

2(x + y) = 2900                

 

x + y = 1450                     eq1

 

Area = xy                          eq2

 

 

We need to substitute eq1 into eq2 so that we have a function for area in terms of x.

 

Area = x(1450 - x)

 

Area = -x² + 1450x

 

 

We end up with a quadratic function for the area.  To find the maximum area, we need to find the vertex of the function.

 

Vertex has coordinate (h, k).

 

h = -b / (2a)

 

where:

a = -1

b = 1450

 

 

Plug in these values to find h. 

 

h = -1450 / (2*-1)

h = 725

 

This is x coordinate of the vertex.  Now, we evaluate the area when x=725.

 

Area = -(725)² + 1450(725)

 

Area = 525625

 

 

The maximum area is 525625 square feet.

Answer 2

You can solve this with simple arithmetic or use algebra.

The maximum area is always a square.

2900/4=725

You want a square 725 by 725 but fencing is not needed along the street.

Add 725 feet to the side opposite the street to make that side 725+725=1450 feet long.

You have a rectangle 725 by 1450 by 725 by "the street".

The maximum area is 725*1450=1,051,250 square feet.

Using algebra...

P=2l+2w

2900=2l+2w

1450=l+w

l=1450-w

A=l*w

A=(1450-w)*w

A=1450w-w²

This is a parabola that opens downward with vertex (h,k) where h is the "maximizing width" and k is the area.

use -b/2a

-1450/-2=725 which we got before; now use the same reasoning as above

also notice that the 725 by 1450 rectangle can be made into 2 squares 725 by 725 each