# Math Help - Help

1. ## Help

consider a symmetric cross inscribed in a circle of radius r

http://img593.imageshack.us/img593/5237/cross.png

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 θʹ and find the value of θʹ that maximizes the area.
c) Show that the critical numbers of parts (a) and (b) yield the same maximum area. What is the area?

2. ## Re: Help

Let y be half the thickness of one arm. Then $y^2 = r^2 - x^2$

So the thickness of one arm is $2\sqrt{(r^2-x^2)}$

The area of the cross is twice the area of each large rectangle less the double counted square in the middle:

$A=8x\sqrt{(r^2-x^2)}-4y^2$

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

$\frac{dA}{dx}=8\sqrt{(r^2-x^2)}+4x(r^2-x^2)^{-\frac 12}(-2x)+8x=0$

$\therefore 8(r^2-x^2)-8x^2+8x\sqrt{(r^2-x^2)}=0$

$\therefore r^2-2x^2+x\sqrt{(r^2-x^2)}=0$

$\therefore (\frac rx)^2-2+\sqrt{(\frac rx)^2 -1}=0$

$\therefore \sqrt{(\frac rx)^2 -1}=2-(\frac rx)^2$

$\therefore (\frac rx)^2 -1=4-4(\frac rx)^2+(\frac rx)^4$

$\therefore (\frac rx)^4-5(\frac rx)^2+5=0$

$(\frac rx)^2=\frac{5 \pm \sqrt{25-20}}{2}$

I hope this is right! haven't had much time to check it.

3. ## Re: Help

How would you suggest doing part b and c?

4. ## Re: Help

One way would be to just substitute $x=r cos(\theta)$ and $y=r sin(\theta)$ into

$A=8x\sqrt{(r^2-x^2)}-4y^2$

then simplify and follow the same procedure as for part a.

5. ## Re: Help

I disagree with Kiwi_Dave's solution for part (b). The problem gives you $\theta$, which will not be the same as if you set $x = r\cos \theta, y = r\sin \theta$. In this instance, $\theta$ is a constant.

Let's give labels to the intersections of the cross with the circle. Starting at the point $(1,0)$ and going counterclockwise, the intersections of the cross with the circle are A,B,C,D,E,F,G,H. Let's call the center of the circle O. By symmetry, the areas of triangles OBC, ODE, OFG, and OHA are all equal. To calculate the area of triangle OBC, it is the following area:

$\dfrac{r^2}{2}\sin \theta$

Next, to figure out the area we have not counted yet requires we figure out the angle AOB. By symmetry, this angle will be $\dfrac{2\pi - 4\theta}{4} = \dfrac{\pi}{2}-\theta$. So, the area we are missing in the first quadrant is the area of triangle AOB minus the area of a right triangle. That right triangle has sides of length $x - y$.

So, let's calculate the area of triangle AOB:
$\dfrac{r^2}{2}\sin\left(\dfrac{\pi}{2}-\theta\right) = \dfrac{r^2}{2}\cos \theta$

Next, $x = r\cos\left(\dfrac{\theta}{2} \right), y = r\sin\left(\dfrac{\theta}{2}\right)$, so the right triangle we wish to remove has area:

$\dfrac{r^2}{2}\left(\cos\left(\dfrac{\theta}{2} \right) - \sin\left(\dfrac{\theta}{2} \right)\right)^2 = \dfrac{r^2}{2}(1-\sin \theta)$

So, the area we are missing is the difference of these two:
$\dfrac{r^2}{2}\left(\cos \theta + \sin \theta -1 \right)$

So, the total area as a function of $\theta$ is

$A(\theta) = 4r^2\sin \theta + 2r^2\cos \theta - 2r^2$

Then, taking the derivative to find critical numbers:

$A'(\theta) = 4r^2\cos \theta -2r^2\sin \theta$

This has critical numbers when $\tan \theta = 2$. This gives $\sin \theta = \dfrac{2}{\sqrt{5}}, \cos \theta = \dfrac{1}{\sqrt{5}}$.

To show that this is the same critical value you found for $x$, you just need to convert from $\theta$ to $x$. You have $x = r\cos\left(\dfrac{\theta}{2}\right)$. The half-angle formula for cosine is $\cos\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1+\cos \theta}{2}}$. So,

$x = r\sqrt{\dfrac{1+\tfrac{\sqrt{5}}{5}}{2}} = r\sqrt{\dfrac{5+\sqrt{5}}{10}}$

Solving what Kiwi_Dave found a bit more, you get the same answer.