Lagrange Multiplier to find minimum value Find the minimum cost of a rectangular box of volume $180 \text{ cm}^3$ whose top and bottom cost $9$ cents per $\text{cm}^2$ and whose sides cost $5$ cents per $\text{cm}^2$.
My approach:
The given volume $V=180\text{ cm}^3$.
let the length $x$ and breadth $x$ height $h$.
$V=x^2y=180$ which implies $y=180/(x^2)$
Total cost $C=9x^2+9x^2+5\cdot (4xy)$ (since whose top and bottom cost $9$ cents per $\text{cm}^2$ and whose sides cost $5$ cents per $\text{cm}^2$)
$C=18x^2+20x\cdot \frac{180}{x^2}$ since $y=\frac{180}{x^2}$.
$C=18x^2+\frac{3600}{x}$
differentiating with respect to x
$C'=36x - \frac{3600}{x^2}$
For minimum cost we set $C'=0$
$36x - \frac{3600}{x^2}=0$ which implies $36x^3=3600$ which implies $x^3=\frac{3600}{36}$
$x=\sqrt[3]{\frac{3600}{36}}=4.641588834$
therefore the minimum cost of a rectangular box of volume is $x=4.641588834$ cents.
The answer is wrong but I have no ideas where is the problem. Any idea?
 A: Well, what you've done is not Lagrange Multipliers, but more of a substitution method for constrained optimization. See the difference between section 2.4 and 2.5 in the link I provide.

Let's continue from your methodology. You've found the correct value for $x$, However:
Note that $x$ is not the final answer since you are asked to find the minimum cost, not the corresponding length ($x$ is in $\text{cm}$, not cents).
Therefore, you must find evaluate $C$ by simply substituting $x$ into the function for the cost you derived:
$$C=18x^2+\frac{3600}{x}$$
That should give you the correct answer.

Alternatively, if you want to use Lagrange Multipliers, you have your function to minimize:
$$f(x,y)=18x^2+20xy$$
And your constraint:
$$g(x,y)=x^2y-180$$
Applying them gives the Lagrangian:
$$\mathcal{L}(x,y,\lambda)=f(x,y)-\lambda\cdot g(x,y)=18x^2+20xy-\lambda (x^2y-180)$$
Evaluating partial derivatives and setting each equal to zero gives the following system of equations:
$$\begin{cases} \frac{\partial \mathcal{L}}{\partial x}=36x+20y-2xy\lambda=0 \\ \frac{\partial \mathcal{L}}{\partial y}=20x-\lambda x^2=0 \\ \frac{\partial \mathcal{L}}{\partial \lambda}=-x^2y+180=0\end{cases}$$
Which you can solve, and then substitute the values of $x$ and $y$ to obtain the same answer for the cost $C$, which is equivalent to $f(x,y)$.
