Question 148073
A rectangular field is fenced in by using a river as one side . If 2500 m
of fencing are used for the 720,000-m2 field , what are its dimensions? 
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Let L = length of rectangular field
and W = width of rectangular field
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Since we have two variables we'll need two equations.
From the fact that there is 2500 m of fencing, we can get our first equation (perimeter):
2W+L = 2500
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From the given area (720000 sq meters) we can get our second equation (area):
LW = 720000
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Solving equation 1 for L we get:
2W+L = 2500
L = 2500-2W
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Plug the above into equation 2 and solve for L:
LW = 720000
(2500-2W)W = 720000
2500W-2W^2 = 720000
0 = 2W^2 - 2500W + 720000
dividing through by 2 we get:
0 = W^2 - 1250W + 360000
Using the quadratic equation we get two solutions for W:
800m, 450m
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If W=800m then
2W+L = 2500
2(800)+L = 2500
1600+L = 2500
L = 2500-1600
L = 900
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If W=450m then
2W+L = 2500
2(450)+L = 2500
900+L = 2500
L = 2500-900
L = 1600
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So, the dimensions of the field can be:
900m by 800m
or
1600m by 450m
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The quadratic solution follows:
*[invoke quadratic "x", 1, -1250, 360000 ]