Thread: Optimization and Differentials

Hi...I'm working on a problem for the minimization of materials for a can. This juice can should hold 163 mL, and I'm already stuck on the first step. When setting up the minimization equation I should be able to graph it and get an idea of about where the minimum should be. I've got 2*Pi*r^2 + 326/r so far, but the graph doesn't show anything that looks like a zero slope anywhere - it just looks like a standard sqrt function. So before I go further with this, what am I doing wrong?

Originally Posted by marqade

Hi...I'm working on a problem for the minimization of materials for a can. This juice can should hold 163 mL, and I'm already stuck on the first step. When setting up the minimization equation I should be able to graph it and get an idea of about where the minimum should be. I've got 2*Pi*r^2 + 326/r so far, but the graph doesn't show anything that looks like a zero slope anywhere - it just looks like a standard sqrt function. So before I go further with this, what am I doing wrong?

Did you start with $\displaystyle 163=\pi{r^2}h$

and $\displaystyle S=2\pi{r}(r+h)$?

I started with 163 = Pi*r^2*h;
A = 2*Pi*r^2 + 2*Pi*r*h;
h = 163/(Pi*r^2)

Originally Posted by marqade

I started with 163 = Pi*r^2*h;
A = 2*Pi*r^2 + 2*Pi*r*h;
h = 163/(Pi*r^2)

Thats correct...I just factored by suface area

Now imput that into yoru surface area to get

$\displaystyle S=2\pi{r}\bigg(r+\frac{163}{\pi{r^2}}\bigg)$

This means that $\displaystyle S'=2\pi{r}\bigg(1-\frac{326}{\pi{r^3}}\bigg)+2\pi\bigg(r+\frac{163}{ \pi{r^2}}\bigg)$

Now solve $\displaystyle S'=0$

If we cut this same can from hexagons instead, thereby saving even more material, could I write the area as a function of the radius as such: A(r) = 6*(1/2)(2/r)(r)? because 1/2 b*h = A; b = 2/r and the height = radius. Problem with this is that my variable cancel themselves out and I'm not sure what to do with that. Do I need to be using trig formulas?
x = (1/r*sin90)/(sin(45)) = sin(45)/r = (sqrt(2))/2r.