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) About Me  (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
2%2860%2Fh%29%2B2h=32
120%2Fh%2B2h%5E2%2Fh=32Multiply 2nd term on left side to reach common denominator
%28120%2B2h%5E2%29%2Fh=32Multiply both sides by h
%28120%2B2h%5E2%29=32hGet everything to one side by subtraction
2h%5E2-32h%2B120=0Now we have a quadratic to solve. Use the quadratic equation to solve.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ah%5E2%2Bbh%2Bc=0 (in our case 2h%5E2%2B-32h%2B120+=+0) has the following solutons:

h%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=%28-32%29%5E2-4%2A2%2A120=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--32%2B-sqrt%28+64+%29%29%2F2%5Ca.

h%5B1%5D+=+%28-%28-32%29%2Bsqrt%28+64+%29%29%2F2%5C2+=+10
h%5B2%5D+=+%28-%28-32%29-sqrt%28+64+%29%29%2F2%5C2+=+6

Quadratic expression 2h%5E2%2B-32h%2B120 can be factored:
2h%5E2%2B-32h%2B120+=+2%28h-10%29%2A%28h-6%29
Again, the answer is: 10, 6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-32%2Ax%2B120+%29

So the height is either 6 or 10. Lets solve for L.
10L=60
L=60%2F10
L=6If h=10 then L=6
6L=60
L=60%2F6
L=6If h=6 then L=10, see a pattern?