Answer
$\left( {r,h} \right) = \left( {3,6} \right)$
Work Step by Step
We have the surface area (including the top and bottom):
$S\left( {r,h} \right) = 2\pi rh + 2\pi {r^2}$
Our task is to minimize $S$ subject to the constraint $V = \pi {r^2}h = 54\pi $.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla S = \lambda \nabla V$ yields
$\left( {2\pi h + 4\pi r,2\pi r} \right) = \lambda \left( {2\pi rh,\pi {r^2}} \right)$
So, the Lagrange equations are
$2\pi h + 4\pi r = \lambda \left( {2\pi rh} \right)$, ${\ \ \ }$ $2\pi r = \lambda \left( {\pi {r^2}} \right)$
(1) ${\ \ \ \ }$ $h + 2r = rh\lambda $, ${\ \ \ }$ $2 = r\lambda $
Step 2. Solve for $\lambda$ in terms of $r$ and $h$
The second equation of (1) implies that $r \ne 0$ and $\lambda \ne 0$. As a result $h \ne 0$.
From equation (1) we obtain
$\lambda = \frac{2}{r} = \frac{{h + 2r}}{{rh}}$
$2 = 1 + 2\frac{r}{h}$
$1 = 2\frac{r}{h}$
So, $h = 2r$.
Step 3. Solve for $r$ and $h$ using the constraint
Substituting $h = 2r$ in the constraint gives
$\pi {r^2}\left( {2r} \right) = 54\pi $
${r^3} = 27$
$r = 3$
The critical point is $\left( {r,h} \right) = \left( {3,6} \right)$.
Step 4. Calculate the critical values
Using the critical point $\left( {r,h} \right) = \left( {3,6} \right)$ we evaluate the extreme value of $S$ subject to the constraint $V = \pi {r^2}h = 54\pi $:
$S\left( {3,6} \right) = 54\pi $
For a given constraint $V = 54\pi $ we may increase $r$ and at the same time decrease $h$ such that $V$ is constant. Since $S$ is increasing, we conclude that $S\left( {3,6} \right) = 54\pi $ is a minimum value subject to the constraint.
