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Let's define some variables. We'll call the length L, and the width W. Then we can express the above algebraically:

300 (perimeter) = 2L + 2W

3600 (area) = L x W

Now let's find one variable in terms of the other:

300 = 2L + 2W

300 - 2W = 2L

(300 - 2W)/2 = L

Since we now have the length in terms of width, we can substitute for L in the second equation:

3600 = L x W

3600 = [(300 - 2W)/2] x W

3600 = (300W - 2W2)/2

7200 = 300W - 2W2

---- That was part one of the solution. Now we have to reformat the equation and use the quadratic formula (for the formula and syntax, visit the Wikianswers link below):

-2W2 + 300W - 7,200 = 0

W = [-300+or-(3002 - 4(-2)(-7,200)1/2]/[2(-2)]

W = [-300+or-(90,000 - 57,600)1/2]/-4

W = [-300+or-(32,400)1/2]/-4

W = (-300 + or - 180)/-4

Quadratic equations usually have two answers, since they are parabolas and will most likely cross the x-axis twice.

The first answer is:

w = (-300 + (-180))/-4

W = -480/-4

W = 120

(L = 30)

Or the second answer:

W = (-300 - (-180))/-4

W = (-120)/-4

W = 30

(L = 120)

Either way, the dimensions of the plot must be 120 ft by 30 ft

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Q: A farmer has 300 feet of fencing and wants to enclose a rectangular area of 3600 square feet What should the dimensions be?
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