Partial Solutions to Brainfood #7

These are partial answers so that you can check to see if you are on target. We expect your actual answers to be "fleshed out" a bit more.
  1. Let r be the radius of the cylinder's base and h be the cylinders length. Then the surface area of the tank is
    S(r,h) = 4 Pi r^2 + 2 Pi r h
    and the volume of the tank is
    V(r,h) = 4 Pi/3 r^3 + Pi r^2 h.
    Then if lambda is a Lagrange multiplier, the solution will occur for values of (r,h,lambda) where
    grad(S) = lambda grad(V)
    V(r,h) = 500.

    Solving these equations simultaneously, we find that the minimum surface area occurs when (r,h)=( (375/Pi)^(1/3), 0).

  2. The figure below shows the geometry. Sure enough, the constraint curve (red) is tangent to the level sets for surface area (white) when h=0. So the geometry and algebra match!

  3. Dear Boss,

    You asked me to investigate the problem of building a cylindrical storage tank that holds 500 cubic meters of liquid. In an attempt to minimize our production costs, I attempted to design a tank that requires the least possible surface area.

    The problem of minimizing some function (like our surface area) while satisfying some constraint (such as a constant volume) is a standard problem in multivariable calculus. The usual approach is the so-called "Method of Lagrange Multipliers." I can supply the details if need be, but perhaps it is sufficient to note that this method allows us to look for the radius and length of a tank that has the least surface area. The general idea is to stay on a curve corresponding to the volume constraint, and look for maximum and minimum values of the cost function along this curve.

    Upon completing my analysis, I was surprised to find that the tank that minimizes surface area would be a spherical tank of radius 4.92 meters. Unfortunately, our customer explicitly specified a cylindrical tank. I recommend contacting the customer to


    P. Lagrange

Last modified: Fri Oct 25 10:48:53 1996