SOLUTION: John has 300 feet of lumber to frame a rectangular patio. (the perimeter of a rectangle is 2 times length plus two times width). He wants to maximize the area of his patio ( area

Algebra ->  Linear-equations -> SOLUTION: John has 300 feet of lumber to frame a rectangular patio. (the perimeter of a rectangle is 2 times length plus two times width). He wants to maximize the area of his patio ( area       Log On


   

Question 23299: John has 300 feet of lumber to frame a rectangular patio. (the perimeter of a rectangle is 2 times length plus two times width). He wants to maximize the area of his patio ( area of a rectangle is length times width). What should the dimensions of the patio be?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Perimeter = 2l+2w = 300
l+w =150
w = 150-l
Area =lw
=l(150-l)
=150l-l^2
This is a quadratic in "l" where
a=-1, b=150, c=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -1x%5E2%2B150x%2B0+=+0) has the following solutons:

x%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=%28150%29%5E2-4%2A-1%2A0=22500.

Discriminant d=22500 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-150%2B-sqrt%28+22500+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28150%29%2Bsqrt%28+22500+%29%29%2F2%5C-1+=+0
x%5B2%5D+=+%28-%28150%29-sqrt%28+22500+%29%29%2F2%5C-1+=+150

Quadratic expression -1x%5E2%2B150x%2B0 can be factored:
-1x%5E2%2B150x%2B0+=+-1%28x-0%29%2A%28x-150%29
Again, the answer is: 0, 150. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-1%2Ax%5E2%2B150%2Ax%2B0+%29

The maximum is at l=-b/2a = 150/2=75
Maximum area occurs when l=75 and w=75
Cheers,
stan H.