Question 155818
For security reasons a company will enclose a rectangular area of 11200ft^2 in the rear of its plant.  One side will be bounded by the building and the other three sides by fencing.  If 300 ft of fencing will be used, what will be the dimensions of the rectangular area?
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Let x = length of rectangular area
and y = width of rectangular area
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Since we have two variables, we'll need to find two equations.
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Our first equation is derived from the perimeter of a rectangle.
Because we have 300 ft of fencing available and we're using the building as one side, the perimeter of fence:
2y+x = 300  (equation 1)
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Our second equation is derived from the area of a rectangle.
The problem tells us that the area is 11200 ft^2
xy = 11200 (equation 2)
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Let's start with equation 2 and solve for y:
xy = 11200
y = 11200/x
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Subsitute the above into equation 1 and solve for x:
2y+x = 300
2(11200/x)+x = 300
Multiplying both sides by x:
2(11200) + x^2 = 300x
22400 + x^2 = 300x
x^2 - 300x + 22400 = 0
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Factoring the left side:
(x-160)(x-140) = 0
Solutions: x={160,140}
Note: you can always use the "quadratic equation" if you can't factor
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Conclusion: area of rectangle is 160 ft by 140 ft
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