Answer
The dimensions of a cylindrical can with a bottom but no top, of fixed volume ${V_0}$ with minimum surface area are $\left( {r,h} \right) = \left( {{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}},{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}}} \right)$,
where $r$ is the radius and $h$ is the height of the cylindrical can.
Work Step by Step
We have the surface area (with a bottom but no top):
$S\left( {r,h} \right) = 2\pi rh + \pi {r^2}$,
where $r$ is the radius and $h$ is the height of the cylindrical can.
Our task is to minimize $S$ subject to a fixed volume constraint $V\left( {r,h} \right) = \pi {r^2}h = {V_0}$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla S = \lambda \nabla V$ yields
$\left( {2\pi h + 2\pi r,2\pi r} \right) = \lambda \left( {2\pi rh,\pi {r^2}} \right)$
So, the Lagrange equations are
$2\pi h + 2\pi r = \lambda \left( {2\pi rh} \right)$, ${\ \ \ }$ $2\pi r = \lambda \left( {\pi {r^2}} \right)$
(1) ${\ \ \ \ }$ $h + r = 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
$\lambda = \frac{2}{r} = \frac{{h + r}}{{rh}}$
$2 = 1 + \frac{r}{h}$
$1 = \frac{r}{h}$
So, $h=r$.
Step 3. Solve for $r$ and $h$ using the constraint
Substituting $h=r$ in the constraint gives
$\pi {r^2}\left( r \right) = {V_0}$
${r^3} = \frac{{{V_0}}}{\pi }$
$r = {\left( {\frac{{{V_0}}}{\pi }} \right)^{1/3}}$
Since $h=r$, the Lagrange critical point is $\left( {r,h} \right) = \left( {{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}},{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}}} \right)$.
Step 4. Calculate the critical values
Using the critical point $\left( {r,h} \right) = \left( {{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}},{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}}} \right)$ we evaluate the extreme value of $S$ subject to the constraint ${V_0} = \pi {r^2}h$:
$S\left( {{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}},{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}}} \right) = 3{\pi ^{1/3}}{V_0}^{2/3}$
For a fixed constraint ${V_0}$ we may increase $r$ and at the same time decrease $h$ such that ${V_0}$ is constant. Since $S$ is increasing, we conclude that $3{\pi ^{1/3}}{V_0}^{2/3}$ is a minimum value of $S$ subject to the constraint.
Thus, the dimensions of a cylindrical can with a bottom but no top, of fixed volume ${V_0}$ with minimum surface area are $\left( {r,h} \right) = \left( {{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}},{{\left( {\frac{{{V_0}}}{\pi }} \right)}^{1/3}}} \right)$,
where $r$ is the radius and $h$ is the height of the cylindrical can.
