Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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} $

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.