Question 234475: How do you find the dimensions of a rectangle whose perimeter is inches and whose area is square inches?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! P = 2L + 2W
A = LW
Solve for L in terms of W in the second equation.
Use that value of L in the first equation to solve for W.
Example:
2L + 2W = 100
LW = 400
Solve for L in second equation to get L = 400/W
Use that value in the first equation to get 2(400/W) + 2W = 100
Once you find W, then you go back to the first equation and solve for L
I'll do this one for you.
Equations are:
2L + 2W = 100
LW = 400
I use second equation to solve for L in terms of W to get L = W/400
I replace L with 400/W in the first equation to get:
2*(400/W) + 2*W = 100
I multiply both sides of this equation by W to get:
2*400 + 2*W*W = 100*W which becomes:
800 + 2W^2 = 100W
I subtract 100W from both sides of this equation and reorder the terms to get:
2W^2 - 100W + 800 = 0
I divide both sides by 2 to get:
W^2 - 50W + 400 = 0
I factor this quadratic equation to get:
(W-10) * (W-40) = 0
The result is that:
W = 10 or:
W = 40
I then go back to the first equation and use these values of W to solve for L.
I get:
L = 10 or:
L = 40
When L = 10, W = 40
When L = 40, W = 10
I confirm these values by substituting in the original equations of:
2L + 2W = 100
LW = 400
They confirm so the answers are good.
I get:
L = 10 or 40
W = 10 or 40
When L = 10, W = 40
When L = 40, W = 10
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