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I am having trouble solving a math problem for a test review. Here it is:

What are the dimensions of the largest area of a rectangle with a semicircle on a short side given 80ft of fencing?

I am assuming the figure looks like this:

+-------------+ + 
+             +  +
+             +    +
+             +  +
+-------------+ + 

               ^^^^^
               a semi circle

Can someone please tell me how to solve this problem step by step?

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Let the side of the rectangle with the semicircle attached and its opposite side be $a$ and let the other two sides be $b$. Can you see why $Area = ab + \frac{1}{2} \pi (\frac{a}{2})^2$ and $a + 2b + \pi \frac{a}{2} = 80$? After that, can you express the area in terms of just $a$ and find the maximum of the quadratic? –  Sp3000 Nov 16 '11 at 5:16
    
is this your figure ? –  pedja Nov 16 '11 at 5:18
1  
You were very impatient for someone to answer. Now someone has answered - and you have nothing to say? –  Gerry Myerson Nov 16 '11 at 12:11
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1 Answer

We have two variables, the height and width of the rectangle (note that the height of the rectangle determines the radius of the circle). We can write two expressions: one relating these variables to the perimeter of the region, and another relating to the area of the region. Let's call the height $h$ and the width $w$; then the radius of the circle is $h/2$. Then our expressions are $$ \text{Perimeter: } 80 = 2w + h + \frac{\pi h}{2} $$ $$ \text{Area: } A = wh + \frac{\pi h^2}{8} $$ We want to maximize $A$ using these equations. We can do this by solving the perimeter equation for one of the variables, and then substituting into the area equation; we will end up with area as a function of one variable, which we can then maximize. Solving for w, $$ w = 40 - h(\frac{1}{2} + \frac{\pi}{4}) $$ and substituting, $$ A = -h^2(\frac{\pi + 4}{8}) + 40h $$ Since the problem specifies that the semicircle must be on the short side, we can't simply maximize this function over its whole domain; we need to find a constraint based on this requirement. We can do this by setting $w=h$ and solving for $h$; this will give us the maximum value for $h$ (i.e., making $h \le w$ ensures that $h$ is the short side). We find that this constraint is $h\le \frac{160}{\pi+6}$.

Now we can maximize our function for $A$; since it is a parabola, this is relatively simple. The vertex is at $-\frac{b}{2a} = \frac{160}{\pi+4}$, and since our constraint is less than this value, the maximum must be at the maximum allowed by the constraint; that is, the value of $h$ that maximizes $A$ is $h=\frac{160}{\pi+6}$, and since we found this value by setting $w=h$, the value of $w$ is the same.

This answer makes sense intuitively, as the maximum area of a figure with fixed perimeter is often found by making it as regular as possible.

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