SOLUTION: The perimeter of a rectangle is 32", and the area is 60 square inches. Find the length and width of the rectangle.

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Question 71147: The perimeter of a rectangle is 32", and the area is 60 square inches. Find the length and width of the rectangle.
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Let L=length and h=height. In this case the equations for perimeter and area are:
2L+2h=32 and L*h=60
L=60/h
Now plug in L=60/h for L in the previous equation

Multiply 2nd term on left side to reach common denominator
Multiply both sides by h
Get everything to one side by subtraction
Now we have a quadratic to solve. Use the quadratic equation to solve.
Solved by pluggable solver: SOLVE quadratic equation with variable
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=64 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 10, 6. Here's your graph:

So the height is either 6 or 10. Lets solve for L.


If h=10 then L=6


If h=6 then L=10, see a pattern?
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