SOLUTION: 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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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?








Found 2 solutions by oscargut, nerdybill:
Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
length=L
width=W
LW=720000
2L+W=2500
then
L(2500-2L)=720000
-2L^2+2500L-720000=0
2L^2-2500L+720000=0
L+=+%282500+%2B-+sqrt%28+2500%5E2-4%2A2%2A720000+%29%29%2F%282%2A2%29+=
L+=+%282500+%2B-+sqrt%28490000+%29%29%2F%284%29+=
L+=+%282500+%2B-+700%29%2F%284%29+=
L=800 or L=450
then L=800 and W=900
or
L=450 and W=1600


Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
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?
.
Let L = length of rectangular field
and W = width of rectangular field
.
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
.
From the given area (720000 sq meters) we can get our second equation (area):
LW = 720000
.
Solving equation 1 for L we get:
2W+L = 2500
L = 2500-2W
.
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
.
If W=800m then
2W+L = 2500
2(800)+L = 2500
1600+L = 2500
L = 2500-1600
L = 900
.
If W=450m then
2W+L = 2500
2(450)+L = 2500
900+L = 2500
L = 2500-900
L = 1600
.
So, the dimensions of the field can be:
900m by 800m
or
1600m by 450m
.
The quadratic solution follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1250x%2B360000+=+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=%28-1250%29%5E2-4%2A1%2A360000=122500.

Discriminant d=122500 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1250%2B-sqrt%28+122500+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1250%29%2Bsqrt%28+122500+%29%29%2F2%5C1+=+800
x%5B2%5D+=+%28-%28-1250%29-sqrt%28+122500+%29%29%2F2%5C1+=+450

Quadratic expression 1x%5E2%2B-1250x%2B360000 can be factored:
1x%5E2%2B-1250x%2B360000+=+1%28x-800%29%2A%28x-450%29
Again, the answer is: 800, 450. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1250%2Ax%2B360000+%29