# Thread: 3rd and final optimization problem

1. ## 3rd and final optimization problem

Trying to design an aluminum can containing a volume of 2,000 cubic centimeters.

Find the dimensions of the can that will minimize the amount of aluminum used.

The three formulas I'm working with will obviously
Volume: π r2 h=2000
Vertical surface area: 2
π r h
Lid and base surface area:
π r2 (x2)

I understand that basically what I'm trying to do is isolate one variable (for example make h=2000/
π r2) so that I can now substitute that equation in for the variable h in another formula.

That will result in a quadratic equation, from which I will find the derivative, find max and min, and figure out the volume at the minimum. Am I missing something? I keep attempting this, but I keep getting stuck after I
substitute 2000/πr^2 for h because I don't have a quadratic equation to work with.

2. ## Re: 3rd and final optimization problem

you don't always need a quadratic equation to solve a optimization problem,

when the volume is $\displaystyle V$,

\displaystyle \begin{align*}V&=\pi r^2h\\h&=\frac{V}{\pi r^2}\\\end{align*}

then,
\displaystyle \begin{align*}A&=2\pi rh +2\pi r^2\\&=2\pi r\left[\frac{V}{\pi r^2}\right]+2\pi r^2\\ &= 2 \left[ \frac{V}{r}+\pi r^2\right] \end{align*}

to find the maximum point you can use differentiation,

$\displaystyle \frac{dA}{dr}=2\left[2\pi r- \frac{V}{r^2}\right]$

equate this to zero to find an extrema.

3. ## Re: 3rd and final optimization problem

Once again you come to the rescue! Thank you so very much for explaining things in a way that I can easily understand them. You have already helped me more than my textbook has.

4. ## Re: 3rd and final optimization problem

I'm glad that I could be a help