1. ## Optimization problems

Here's the problem:
What are the dimensions of the soup can of greatest volume that can be made with 50 square inches of tin? (The entire can, including the top and bottom, are made of tin.)
and what's its volume?
Now, I know the volume of the space enclosed by a cylinder is $\displaystyle V= \pi r^2h$
What I don't know is how to differentiate this in respect to it's height. Up to now, I've just had to differentiate a few x's. Now, I guess I would just imagine the h as an x, but how would I differentiate for r? I tried isolating the r, but then I end up with a V on the other side. Help me, brothers.

2. ## Re: Optimization problems

The surface area of a cylinder is $\displaystyle 2\pi r^2 + 2\pi r h$ Since you have 50 square inches... $\displaystyle 2\pi r^2 + 2\pi r h = 50$. Solve for r or h then sub it into $\displaystyle V=\pi r^2 h$. I trust you know what to do from there

3. ## Re: Optimization problems

In that case, let's see if I do know what to do, or if I do not.
$\displaystyle V=\pi r^2h$ and $\displaystyle 50=2\pi r^2+2\pi r^2h$
In this case $\displaystyle r=\frac{50}{2\pi (h+1)}$ Yes?
So, let's plug it in a derive.
$\displaystyle V'(h)=\pi ' \frac{50}{2\pi (h+1)}h + \pi \frac{50}{2\pi (h+1)}'h + \frac{50}{2\pi (h+1)}h'$
$\displaystyle \pi '= 0$

$\displaystyle \frac{50}{2\pi (h+1)}'= \frac{50'(2\pi (h+1)) - 50(2\pi (h+1))'}{(2\pi (h+1))^2}$

$\displaystyle [2\pi (h+1)]'= [2\pi]'(h+1) + 2\pi(h+1)' = 0 + 2\pi = 2\pi$

So back to the complex fraction we go...

$\displaystyle \frac{0-50(2\pi)}{(2\pi (h+1))^2}= \frac{100\pi}{(2\pi(h+1))^2}$

Now, back again:

$\displaystyle V'(h)= 0+\pi (\frac{100\pi}{2\pi (h+1)^2}) h + \frac{50}{2\pi(h+1)}= \frac{100\pi ^2h}{(2\pi(h+1))^2}+\frac{50}{2\pi(h+1)}$

Next, I'll set it equal to zero and solve.

$\displaystyle V'(h)=\frac{100\pi ^2h}{(2\pi(h+1))^2}+\frac{50}{2\pi(h+1)}=0$

$\displaystyle 100\pi ^2h+50=[2\pi(h+1)]^3$

Let's factor...

$\displaystyle 50(2\pi ^2h+1)=[2\pi ^2(h+1)]^3$

So tempting, it is, to confuse the similar looking terms...
I think I've made a mistake up to this point, but I have to go to work now. Typing out the problem like I have takes a lot of time, but it keeps me from going fast and making mistakes I would on paper. And it gives one of you the opportunity to tell me I've made a mistake, if one wished. Anyway, I'll be back to work on this when I'm done working. ~ttfn.

4. ## Re: Optimization problems

You have $\displaystyle 50= 2\pi r^2 + 2\pi r^2h$

then $\displaystyle r= \frac{50}{2\pi (h+1)}$ but shouldn't that be $\displaystyle r^2=\frac{50}{2\pi (h+1)}$

5. ## Re: Optimization problems

Yes, it should. I forgot to type it that way, but I worked it out that way.