SOLUTION: Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.

Question 159731: Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.
Found 2 solutions by KnightOwlTutor, jojo14344:
Answer by KnightOwlTutor(293) (Show Source): You can put this solution on YOUR website!
We know that the perimeter is equal to the length of all the sides
X=width
Y=length
The perimeter of the rectangle=X+X+Y+Y=26
Simplify to 2X+2Y=26
we know that the area is the width multplied by the length
XY=40
We isolate X by dividing Y on both sides X=40/Y
Replace 40/Y for X in the perimeter equation
2(40/Y) +2Y=26
Multiply each term on both sides by Y
80+2Y^2=26Y
subtract 26Y from both sides
80-26Y+2Y^2=0
This is a quadratic equation

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=36 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 8, 5. Here's your graph:

Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!

First
------>

, EQN 1
And,
, substitute EQN 1

, perfect square, can be factor out

2 values, either,
, or
Get via EQN 1:
, and
Go back Perimeter and Area to check: use either or
1)

2)

Thank you,
Jojo
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