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

Question

John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 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, and show how the maximum area of the patio is calculated from the algebraic equation.

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.

please show work so i can understand ty

Answer 1

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

You can see this is the formula for a parabola. I'll simplify things by expressing it in terms of x:
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 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 the patio is a square with sides of length 75 feet. The total area will be 5,625 sq. feet, a maximum.

Answer 2

Lets call the width x. Then the length must be 150-x, as width + length = half of total perimeter.
So the area is x(150-x) = 150x - x^2.
Now, the next step can be done in several ways. If you know any calculus, you can differentiate to get 150 - 2x = 0, ie x = 75.
However, from the sounds of your question you want a more algebraic way. In that case, lets try completing the square:
A = 150x - x^2
x^2 - 150x + A = 0
(x^2 - 150x + 75^2) + A = 75*75
(x-75)^2 + A = 75^2
A = 75^2 - (x-75)^2.
So A is definitely less than 75^2, so the maximum must occur when the second bit is 0, ie x = 75. Then the area is 75^2.