Optimization using Lagrange multipliers: 55 gallon steel drum I am currently trying to figure out this particular problem with not a great understanding of Lagrange multipliers. I understand the process, but am not sure how the values are generated. Could anyone help me solve this with explanation behind the steps?

Using the dimensions in the figure below, use the method of Lagrange multipliers to optimize the steel drum(cylinder). Find the smallest possible surface area while maintaining (constraint) the desired volume of $57.20$ gallons or $13,213\;in^3$.

Minimize: $\;\; SA(r,h)=2\pi rh+2\pi r^2$
Constraint: $\;\; V(r,h)= \pi hr^2$
diameter $\;\;= 22\;\frac12$ inch,
height  $\;\; = 33\;\frac18$ inch.

Here is an outline of how to solve this problem.
Generally, Lagrange multipliers is not a recipe for solution and requires some extra thinking. This problem follows this pattern.
The problem is $\min \{ S(r,h) | V(r,h) = K \}$, where $K$ is some positive
constant, $S(r,h) = r^2+rh$ and $V(r,h) = h r^2$ (I stripped out irrelevant constants to simplify).
Step 1: The first, and typically overlooked, step is to convince yourself that there actually is a solution.
This problem can be solved directly without Lagrange multipliers. Note that in this case, $V(r,h) = K$ implies that $h>0$ and so we see that $h = {K \over r^2}$ and $S(r,{K \over r^2}) = r^2 + {K \over r}$. Plotting this shows that $S$ is unbounded below so technically it has no minimum.
Clearly a negative $r$ is aphysical, there is an implied constraint that $r \ge 0$.
We (well, me) still need to convince ourselves that there is a minimum. Note that if we pick any feasible $r,h$, for example, $(r,h) = (1,K)$, then any solution must have cost no larger than this so the problem is equivalent to $\min \{ S(r,h) | V(r,h) = K , S(r,h) \le S(1,K), r \ge 0\}$. It is straightforward to see that the feasible set is compact hence a solution exists. Hence we can drop the artificial $S(r,h) \le S(1,K)$ constraint.
To say a solution exists means that there is some point $(r^*,h^*)$ that is feasible and $\epsilon>0$ such that for any feasible $(r,h) \in B((r^*,h^*), \epsilon)$ that $S(r^*,h^*) \le S(r,h)$.
Furthermore, the $V$ constraint shows that at a solution we must have $r>0$
(and hence is not binding at the solution).
The whole point of this diatribe is that $(r^*,h^*)$ is a local solution to the original problem and so the Lagrange multiplier conditions hold at
this point.
Step 2: Actually use the multipliers.
We have the equations $h^*+2r^*+2 \lambda r^*h^* = 0$, $r^* + \lambda (r^*)^2 = 0$. We know that $r^* >0$ so the last equation gives $r^* = - {1 \over \lambda}$ and the
other equation gives $h^*=-{2 \over \lambda}$. Substituting into the
$V$ constraint gives $\lambda = -\sqrt[3]{2 \over K}$.
Hence $r^*= \sqrt[3]{K \over 2}$, $h^* = 2\sqrt[3]{K \over 2}$ with optimal surface area $S(r^*,h^*) = 3 \sqrt[3]{K^2 \over 4}$.
The numbers in the problem are a bit inconsistent, the volume constraint is $13,213$ cu.in. but the diameter and height given result in a volume of around $13,171$ cu.in.
