Question 76574: I don't know exactly how to go about solving this one problem. Here's is the question I have to answer. Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to maximize the area of her patio ( area of a rectangle is length times width). What should the dimensions of the patio be, and show how maximum area of the patio is calculated from the algebraic equation. Use the vertex form to find the maximum area. I can only get so far and don't know how to go about solving it.
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! First, you know that the perimeter is given as P = 400 ft. and you have the formula for the perimeter of a rectangle: P = 2L + 2W, so you can write:
 Solve this for L
 Divide both sides by 2.
 You also know that the area of a rectangle is given by:
 Now substitute the L=200-W from above into this equation and solve for W.
 Simplify this.
 You can rewrite this as:
Ok, now you have a quadratic equation which, when graphed, will show a parabola that opens downward (because of the negative coefficient).
This means that the vertex of this parabola will be a maximum and it represents the maximum value of A, the area. This is exactly what you are looking for, right?
So, let's find the value of W that will give us the maximum value of A.
Using the relationship:  where: W is the width, as it was defined above. The a is the coefficient of the  term, that's (-1), and the b is the coefficient of the  term, and that's 200, so:
Now, when W = 100, then L = 200-100 = 100
So, to obtain the maximum area from the given perimeter of 400 ft, the width must be 100 ft and the length must be 100 ft.
In othe words, the patio will be a 100 by 100 ft square.
When you graph the quadratic equation, you will see the following parabola:
This shows you the relationship between the width, W, (the horizontal axis) and the area, A, (the vertical axis).
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