[Math] How to find the dimensions of a max area

geometrytrigonometry

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?

Best 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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