SOLUTION: A rope 100 feet long is stretched around four posts set at the corners of a rectangle whose area is 576 square feet. Find the dimensions of the rectangle.

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Question 171632: A rope 100 feet long is stretched around four posts set at the corners of a rectangle whose area is 576 square feet. Find the dimensions of the rectangle.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A rope 100 feet long is stretched around four posts set at the corners of a rectangle whose area is 576 square feet. Find the dimensions of the rectangle.
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Perimeter = 2L + 2W = 100
Area = L*W = 576
2L + 2W = 100
L+W = 50
L = 50-W
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A = L*W = (50-W)*W
576 = 50W - W^2
W^2 - 50W + 576 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-50x%2B576+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-50%29%5E2-4%2A1%2A576=196.

Discriminant d=196 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--50%2B-sqrt%28+196+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-50%29%2Bsqrt%28+196+%29%29%2F2%5C1+=+32
x%5B2%5D+=+%28-%28-50%29-sqrt%28+196+%29%29%2F2%5C1+=+18

Quadratic expression 1x%5E2%2B-50x%2B576 can be factored:
1x%5E2%2B-50x%2B576+=+%28x-32%29%2A%28x-18%29
Again, the answer is: 32, 18. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-50%2Ax%2B576+%29

W = 18, L = 32