# Homework Help: Optimization Problem

1. Nov 29, 2009

### niravana21

1. The problem statement, all variables and given/known data
Here is the exact problem, in order to avoid confusion

3. The attempt at a solution
I know that I have to find an expression for "h" and substitute it back into the Volume, but the way I do it, it just becomes way too messy to find the derivative.

Any ideas?

2. Nov 29, 2009

### xeno_gear

What way do you do it?

3. Nov 29, 2009

### niravana21

well I tried to set the Area function equal to "H". Then I took that expression and substituted it back into the Volume function for "H".

Now the next step is to find the derivative, but I can't do that, since the new expression is messy. That's basically where I need the help.

4. Nov 29, 2009

### xeno_gear

Don't set A equal to h, set A equal to 1, then solve for h. You want to set A = 1 since you have constant surface area 1. Then you plug that expression for h into V. What you end up with there shouldn't be too bad.

5. Nov 29, 2009

### niravana21

sorry that's what I meant. But the expression for H in terms of R becomes complicated with the square root and quotient.

here's what I get for h in terms of r:

h=sqrt[(1/pi^2r^2) - (r^2)]

now I get stuck at trying to find the derivative of that.

6. Nov 29, 2009

### xeno_gear

right, but it might help to rewrite it:

\begin{align*} h &= \sqrt{\frac{1}{\pi^2 r^2} - r^2}\\ &= \sqrt{\frac{1}{\pi^2 r^2} - \frac{\pi^2 r^4}{\pi^2 r^2}}\\ &= \sqrt{\frac{1 - \pi^2r^4}{\pi^2r^2}}\\ &= \frac{\sqrt{1 - \pi^2r^4}}{\pi r} \end{align*}

So then
\begin{align*} V &= \frac{1}{3}\pi r^2 h\\ &= \frac{1}{3}\pi r^2 \frac{\sqrt{1 - \pi^2r^4}}{\pi r}\\ &= \frac{1}{3}r \sqrt{1 - \pi^2r^4} \end{align*}

Still a bit of a pain, but not as bad as before.

7. Nov 29, 2009

### niravana21

wow that does simply it! Now I find the derivative. But then would if I set the derivative equal to zero and find a critical point?

Also I'm having some difficulty of still finding the derivative :(
After simplification I get:
(-6πr^4+1-π^2 r^4)/(3√(1-π^2 r^4 ))

which I think is wrong since this online derivative calculator gave me this:
http://www.numberempire.com/cgi-bin/render2.cgi?\nocache%20\LARGE%20\frac{\partial%20f}{\partial%20x}%20%3D%20-{{3\%2C\pi^2\%2Cx^4-1}\over{3\%2C\sqrt{1-\pi\%2Cx^2}\%2C\sqrt{\pi\%2Cx^2%2B1}}} [Broken]

Last edited by a moderator: May 4, 2017
8. Nov 29, 2009

### xeno_gear

Hmm.. Here's what I've got:

\begin{align*} V &= \frac{1}{3}r\sqrt{1 - \pi^2r^4}\\ V' &= \frac{1}{3}\sqrt{1 - \pi^2r^4} + \frac{1}{3}r\frac{1}{2}(1 - \pi^2r^4)^{-1/2}(-4\pi^2r^4)\\ &= \frac{1}{3}\sqrt{1 - \pi^2r^4} - \frac{2\pi^2r^4}{3\sqrt{1 - \pi^2r^4}}\\ &= \frac{1 - \pi^2r^4 - 2\pi^2r^4}{3\sqrt{1 - \pi^2r^4}}\\ &= \frac{1 - 3\pi^2r^4}{3\sqrt{1 - \pi^2r^4}}\\ \end{align*}

This is the same thing the derivative calculator got, but it factored the denominator. It looks like you only made a minor mistake. Anyway, to find critical points, you want to know where the derivative is 0 or undefined. The derivative will be undefined when its denominator is 0, but in problems like these, the critical point you're looking for usually comes from setting the derivative equal to 0 and solving.