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A window has the shape of a rectangle of height $h$ surmounted by a semi-circle of radius $r$. The area of the window is given by $A = 2rh + \frac{1}{2}\pi r^{2}$

NOT drawn to scale.

If the perimeter of the window is constant, prove that the area is a maximum when $h$ and $r$ are equal.

My attempt:

Let $P$ be the perimeter of the window. i.e: $P = 2h +2r+ \pi r$ $$h = \frac{(P-2r-\pi r)}{2}$$ I substituted my $h$ value into the area equation given.

$$A = 2r(\frac{P-2r-\pi r}{2}) + \frac{1}{2} \pi r^{2}$$ $$ A = Pr -2r^{2} - \frac{1}{2} \pi r^{2}$$

The maximum value of $A$ given is when $A' = 0$

i.e: $P-4r-\pi r = 0$ $$ r = \frac{P}{\pi +4} $$ $$ h = \frac{P}{\pi + 4} $$ By the second derivative test, $A'' = -4 - \pi < o$, So I have proved that the area is maximum when $r$ and $h$ are equal.

Is this solution correct? I have a feeling that I did not finish it off correctly. I Would appreciate it if someone can elaborate on my solution.

Aspiring Mathlete
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  • Your solution is correct. Which are your doubts? – Anatoly Nov 16 '14 at 16:33
  • I was just confused on how I actually proved that this is the maximum area. I had a feeling that it was pointless to work out the value of $r$ because the second derivative consists of constants only. – Aspiring Mathlete Nov 16 '14 at 16:35
  • You are right: You do not need the exact value of $r$ in order to show that there is a critical point which is furthermore a maximum. The reason is that $A$ is a quadratic function of $r$. You do however need the exact value of $r$ in order to show that this maximum is attained when $r=h$. – Martin Nov 16 '14 at 16:45

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The fact that the second derivative consists of negative constants is not so unusual. It tells you that the first derivative is always decreasing, and in particular that is positive for $r<\frac{P}{\pi+4}$ and negative for $r>\frac{P}{\pi+4}$. As a result, we conclude that $r=\frac{P}{\pi+4}$ is a maximum of the original function.

Anatoly
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