Katie R.

asked • 10/14/17

# Word Problem Help!

A fourth-grade class decides to enclose a rectangular garden, using the side of the school as one side of the rectangle. What is the max area that the class can enclose using 32 ft of fence? What should the dimension of the garden be in order to yeild this area?

## 1 Expert Answer

By:

Arthur D. answered • 10/14/17

Effective Mathematics Tutor

Katie R.

First off thank you for answering. Though I'm still confused as to how you got some of your numbers. So I understand that P=l +2w and that 32 = l + 2w. But I don't understand how and why you used l = 32-2w. Then when I get to the area it makes even less sense because you have (32w-2w)(w). I don't get how you got that (w).
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10/14/17

Arthur D.

tutor
you asked for the area, A=l*w for a rectangle
Area=length times width  (l*w)
the length comes from P=l+2w where P (perimeter) is the amount of fence 32 feet, thus 32=l+2w
32=l+2w means l=32-2w
the length is 32-2w (not 32w-2w) and the width is w
A=length* width
A=(32-2w)*w
use the distributive property
A=32w-2w2
0=32w-2w2
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10/15/17

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