John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus?

Question

Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained. Use the vertex formula to find the maximum area.

Answer 1

I have two answers depending on whether your patio is free standing and the boards go all around the outside, or if the patio is attached to a house and the boards go around three sides.

Please clarify.

CASE 1 - maximum area is a square.
Let L be the length
Let W be the width
We know the perimeter is:
2(L + W) = 300
Solving for L:
L + W = 150
L = 150 - W

Now write a formula for the area:
Area = L * W
Area = (150 - W) * W
Area = -W² + 150W

This is the formula for a parabola:
f(x) = -x² + 150x

But it isn't in vertex form:
f(x) = a(x - h) + k

To get it in that form, first pull out the -1 to get x²:
f(x) = -1(x² - 150x)

Now take half the coefficient on the x term (-150), take half of it (-75) and square it (5625). Add and subtract this:
f(x) = -1(x² - 150x + 5625 - 5625)
Now you can write this as a square:
f(x) = -1[ (x - 75)(x - 75) - 5625)
Simplify:
f(x) = -1(x - 75)² + 5625

You have a downward facing parabola, with a maximum vertex at the point (h, k) or (75, 5625)

So the maximum is when the width is 75. When you solve for length you get that the length is also 75. So the maximum area for a free standing patio is a square with sides of length 75 feet.

Answer 2

perimeter=2(length + width)
300= 2(l+w)
150= l +w