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put this solution on YOUR website!OK. So the playground will be bound by the barn's wall on one side, and 80 feet of fencing will make up the other three sides. Let's call the barn side the length. There would then be 2 widths and 1 length for the 80 meters of fencing. So far we know that 2W + L = 80.
The thing is, there are many ways to split up the 80 meters into three parts where two are going to be the width and the other (leftover) one to be the length. You can have a width of 5 meters and a length of 70 meters, making a really skinny playground (because 2*5 + 70 = 80 ), or one that has a width of 20 meters and a length of 40 meters (still rectangular but not as skinny). So you'll have different length and width pairs so that 2W + L = 80, but which pair of length and width gives you the largest W*L (or largest area)?
You know that the area of a rectangle is A = L*W. We already agreed that 2W + L = 80. We need to take the 2W + L = 80 and solve for a variable so that we can plug it into A = L*W so that A = L*W will only be in one variable instead of two. So I'd solve for L because it's easier.
------->
<---- substituted 80 - 2W for the L.
<----- used distributive property.
Now, we have an expression for the area. If we choose any width W, the L should be calculated automatically (because we did the substitution). Now, we have to find what width gives us the maximum area.
(BTW, in a lot of area problems, they already give you the area and you have to find the dimensions to make up the area. In here, it's like we have to keep on plugging in multiple values for W until we're convinced that we have found the single W value that maximizes A, the area. But we don't need to go through trial and error.)
One way to attack this is to graph it and look at the maximum point. The y-value of the maximum point will be the largest area, and the x-value of the maximum point will be the measure of the width that caused the maximum area. But just plotting and looking at a graph is inaccurate! Let's just do it algebraically.
We're going to use completing the square here. We're going to rearrange our area equation so that we can pick out the coordinates of the maximum point easily.
<---- Start here
<---- just switched the position of the terms.
<---- factored out the -2
Now we're going to force the (W^2 - 40W) to be a perfect trinomial square. First, we must be assured that a(b) = a(b + c) - ac. What we did here was to force-add a c to the b inside the parentheses. Once we do this, we throw off the equation's balance. That's why we undo (subtract) the c we added in. BUT we multiply the "undo" c with the a because the "throw-off" c was also multiplied by the a. That way, our expression just changed appearance but preserved its value. OK. Let's move on.
<---- We left off here.
<----- This is the step that forced the expression inside the parentheses to be a perfect trinomial square. What we did was we took the -40 and divided it by 2 (you ALWAYS divide by 2 in this step, for ANY quadratic equation) to get -20. This -20 comes into play later on. We took that -20 and squared it to get the 400. Now, our equation is thrown off balance when we added the 400.
<----- So to bring back the balance, we undo the addition of 400 by subtracting it. Since the added 400 is inside the parentheses that's multiplied by the -2, the subtracted 400 must also be multiplied by -2. That's why we had to make sure that a(b) = a(b + c) - ac.
<----- We said earlier that the -20 will have a part. When we unfoil the
, we get
. Doing the step "half the -40 and square the result" guarantees this.
Now, just by looking at the transformed equation, the largest possible area you can have is 800 square meters. A width of 20 meters caused the largest area of 800 square meters. If 20 meters is the width, then the length would have to be 40 meters. Since 2*20 + 40 = 80.
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