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I've really been scratching my head over this optimization problem. "Consider a symmetric cross inscribed in a circle of radius $r$." The length from the center of the cross to the middle of one of its arms is $x$. Also, the angle between two line segments drawn from the cross's center to the vertices of one of its arms has a measure of $\theta$. Here's a diagram:

A diagram of the situation: a cross inscribed in a circle.

There are three parts to the problem: "(a) Write the area $A$ of the cross as a function of $x$ and find the value of $x$ that maximizes the area. (b) Write the area $A$ of the cross as a function of $\theta$ and find the value of $\theta$ that maximizes the area. (c) Show that the critical numbers of parts (a) and (b) yield the same maximum area. What is that area?"

So, let me show you what I've done so far. For part (a), I decided to break the cross into two middle rectangles and two side rectangles. I saw that a middle rectangle (from the center to the top) would have an area of

$$x \cdot 2 \sqrt{r^2 - x^2}$$

using the Pythagorean theorem. I worked out that a side rectangle (the remaining area on the right, adjacent to the middle rectangles) would have an area of

$$2 \sqrt{r^2-x^2} \cdot \left( x - \sqrt{r^2 - x^2} \right) .$$

So, the area of the cross is

$$A = 2 \bigg( x \cdot 2 \sqrt{r^2 - x^2} + 2 \sqrt{r^2 - x^2} \cdot \Big( x - \sqrt{r^2 - x^2} \Big) \bigg) = 8x \sqrt{r^2 - x^2} - 4r^2 + 4x^2 .$$

If my math is right there (fingers crossed), then I'll take the first derivative to locate a maximum.

$$A^\prime = 8 \sqrt{r^2 - x^2} + 8x \left( 1 \over 2 \right) \left( r^2 - x^2 \right)^{- {1 \over 2}} \left( -2x \right) + 8x.$$

I was a little unsure about what to do at this point. I plugged the $A^\prime$ equation into my graphing calculator, substituting $1^2$ for $r^2$ (for a radius of $1$). The graph crosses the $x$-axis at $x \approx 0.85$. Substituting $2^2$ for $r^2$ (for a radius of $2$) gives me $x \approx 1.70$. From this, I concluded that

$$A^\prime = 0 \; \mathbf{at} \; x \approx 0.85r.$$

Analysis of graphs of $A$ for various values of $r$ concludes that, indeed, maxima do appear at $x \approx 0.85r$. So, I have the function $A$ in terms of $x$, but I'm curious: What should my final answer be for the second part of (a)? All I have is $x \approx 0.85r$. Is that a sufficient answer?

As for part (b), I really have no idea how to write $A$ in terms of $\theta$. I know that $\text{area} = {1 \over 2} b \cdot c \cdot \sin A$ for triangles, but I really need help writing the area of this cross in terms of $\theta$.

Part (c) should be easy enough once I finish (b).

If you got to the end of this, I sincerely thank you for reading, and I would really appreciate an answer (and any corrections to my math). Thanks!

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You can simplify the formula for $A'$ if you set it to zero and multiply with $\sqrt{1-x^2}$ (with $r=1$). In this way you get $0=1-2x^2+8x\sqrt{1-x^2}$. This is still tricky, but you might want to try a tailor expansion until $x^2$ to simplify further. –  Konstantin Dec 18 '12 at 1:07
@Konstantin I computed $x\sqrt{1-x^2}=2x^2-1$. Square both sides and you have a quadratic in $x^2$. The solutions are $x^2={1\over2}\pm{\sqrt5\over10}$. –  David Mitra Dec 18 '12 at 1:22
@DavidMitra aww thanks –  Konstantin Dec 18 '12 at 1:28

2 Answers 2

up vote 5 down vote accepted

Without loss of generality we may assume that the radius is $1$: we can scale area by $r^2$ later.

By the formula you quoted, the area of the triangle enclosed by the two lines that form the angle $\theta$ is $\frac{1}{2}\sin\theta$. The area covered by one of the two full arms of the cross is therefore $4$ times this, which is $2\sin\theta$.

Double this to get the sum $4\sin\theta$ of the areas covered by the two full arms. Unfortunately, this sum counts the area of the middle square twice. So we will need to subtract the area of that square.

Note that by trigonometry, the horizontal segment at the top of the cross has length $2\sin(\theta/2)$. Thus the middle square has area $4\sin^2(\theta/2)$. Using the trigonometric identity $\cos 2\phi=1-2\sin^2\phi$, we find that the middle square has area $2-2\cos \theta$.

So the area of the cross is $4\sin\theta +2\cos\theta-2$. Maximizing should be straightforward.

Remarks: $1.$ We don't really need calculus. Look at the equivalent problem of maximizing $4\sin\2\theta+2\cos\theta$. Rewrite this as $$2\sqrt{5}\left(\frac{2}{\sqrt{5}}\sin \theta+\frac{1}{\sqrt{5}}\cos\theta\right),$$ and let $\psi$ be the angle whose cosine is $2/\sqrt{5}$ and whose sine is $1/\sqrt{5}$. Then our expression becomes $2\sqrt{5}\sin(\theta+\psi)$. The maximum possible value of the sine function is $1$. So the maximum area is $2\sqrt{5}-2$.

$2.$ We can use the above calculation to answer your question about $x$. Alternately, set the derivative equal to $0$, as you did. Manipulation will yield an explicit expression for the root.

$3.$ At a certain stage you were maximizing $8x\sqrt{r^2-x^2}-4r^2+4x^2$. Let $x=r\sin t$. We want to maximize $8r^2\sin t\cos t-4r^2+4r^2\sin^2 t$. Forget about the $r^2$ part, it is a constant multiplier. Now simplify to $8\sin t\cos t-4\cos^2 t$, differentiate. (Using double angle identities to simplify first is a good idea.) We can think of this as just a technical device to ensure we end up with a simple equation.

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Could you show me how $\sqrt{2 - 2\cos\theta} = 2\sin(\theta/2)$? –  Jackson Dec 18 '12 at 1:44
Recall the trig identity $\cos 2t=2\cos^2 t-1=1-2\sin^2 t$. Rewrite as $2\sin^2 t=1-\cos 2t$. Double. Get $4\sin^2 t=2-2\cos 2t$. Finally, let $t=\theta/2$. –  André Nicolas Dec 18 '12 at 1:47
Thank you so much! It took me a while, but I got everything figured out. I cannot thank you enough :) –  Jackson Dec 18 '12 at 2:07
I will add a last remark to my post in a few minutes. –  André Nicolas Dec 18 '12 at 2:15

I'm trying to solve the same problem, and got to the same concusions. The thing is, if you graph the function you get this :


as you can see, the graph takes negative values when x gets smaller, but if the function represents the area of the cross, how come that area is negative ??

I even made a little program to test this, here it is: http://www.khanacademy.org/cs/calculus-maxarea-circle-v00x/1512261081

does anybody have an idea of what's going on ?

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