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I have to compute optimal parametres of truncated cone so that its Volume is fixed (lets say it is 1) and its surface is minimal using Lagrange method

These are equations desribing my object: \begin{equation} V(R, r, h) = \frac{\pi}{3} h (R^2 + Rr + r^2) = 1 \end{equation}

My Cone doesnt have a top (its basically a cup) \begin{equation} S(R, r, h) = \pi(R + r) \sqrt{(R - r)^2 + h^2} + \pi r^2 \end{equation}

Lagrange equation: \begin{equation*} L(R, r, h, \lambda) = \pi(R + r) \sqrt{(R - r)^2 + h^2} + \pi r^2 - \lambda(\frac{\pi}{3} h (R^2 + Rr + r^2) - 1) \end{equation*}

I computed all the partial derivations, too (Im not sure if they are correct, so pardon any mistakes please): \begin{equation*} \frac{\partial}{\partial R} L(R, r, h, \lambda) = \frac{\pi(h^2 + 2R(R-r))}{\sqrt{h^2 + (R - r)^2}} - 2\lambda \frac{\pi}{3}hR - \lambda \frac{\pi}{3}hr= 0 \end{equation*}

\begin{equation*} \frac{\partial}{\partial r} L(R, r, h, \lambda) = \frac{\pi(h^2 + 2r(r-R))}{\sqrt{h^2 + (R - r)^2}} - \lambda \frac{\pi}{3}hR - 2\lambda \frac{\pi}{3}hr= 0 \end{equation*}

\begin{equation*} \frac{\partial}{\partial h} L(R, r, h, \lambda) = \frac{\pi h(r + R)}{\sqrt{h^2 + (R - r)^2}} - \lambda \frac{\pi}{3}R^2 - \lambda \frac{\pi}{3}Rr - \lambda \frac{\pi}{3}r^2= 0 \end{equation*}

\begin{equation*} \frac{\partial}{\partial \lambda} L(R, r, h, \lambda) = - \frac{\pi}{3} h (R^2 + Rr + r^2) + 1 = 0 \end{equation*}

The problem is, that this leads to a non-linear system of equations and I cant seem to be ale to solve them. Do you know how to find a correct solution (and how) for my parametres R, r, h?

Thank you for any help!

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up vote 0 down vote accepted

Ok, I computed these equations using Math software.. I guess its too difficult do do it by myself:

$r = 0$, $R = \frac{(3/\pi)^{1/3}}{\sqrt{2}}$, $h = 2(\frac{3}{\pi})^{1/3}$, $\lambda = 4(\frac{\pi}{3})^{1/3} $

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