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

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) (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 (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=22500 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0, 150. Here's your graph:

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

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