Find the dimensions of a right circular cylindrical can with both a top and a bottom that holds 8 cubic cm and is constructed with the least amount of material possible.

Radius of can= cm Height of can= cm Area of Material = cm^2

This is what I have tried doing: \begin{align*} 8&= \pi r^2h \\ h&= \dfrac{8}{\pi r^2} \\ m&=2\pi r^2 +2\pi rh \\ m&= 2\pi r^2 + 2 \pi r\dfrac{8}{\pi r^2} \\ \frac{dy}{dx}&= 4\pi r - \dfrac{16\pi}{ r^2} =0. \end{align*}

I'm not sure what to do next..

  • $\begingroup$ Why dont you try the usual way of finding maxima and minima of a function by the 2nd derivative test and so on??Maybe that will tell you something. Also a transformation into cylindrical coordinates might help $\endgroup$ – Vishesh Oct 17 '13 at 6:54
  • 2
    $\begingroup$ Welcome to math.SE, and thanks for your thoughtful questions. It is time for you to start using TEX (MathJaX) notation for your posts. It is easier than you think, and a good skill to have in the future. Read notation, and there are two links there, quick reference is a good one. $\endgroup$ – J. W. Perry Oct 17 '13 at 6:56
  • $\begingroup$ I tried the second derivative test and I got a max of -2 and no minimum .. What do I do with that? $\endgroup$ – Sherry Eskandarian Oct 17 '13 at 6:59
  • $\begingroup$ I suppose that dy/dx stands for dm/dx. I suggest you first check your derivative. Then, since you look for an extremum, search for the value of "r" which makes the derivative equal to zero. Just for safety, compute the second derivative and proove that you obtained a minimum. $\endgroup$ – Claude Leibovici Oct 17 '13 at 9:08

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