SOLUTION: How do you find the dimensions of a rectangle whose perimeter is inches and whose area is square inches?

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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