Intuition behind Lagrangian multiplier in this problem What shape should a rectangular box with a specific volume be in order to minimize its surface area?
Let the lengths of the box's edges be x, y, and z. Then the constraint of constant volume is simply g(x,y,z) = xyz - V = 0, and the function to minimize is f(x,y,z) = 2(xy+xz+yz).
Now applying lagrangian method (making differntial of the lagrangian equal to zero), (using L as the lagrangian multiplier)
we get these 4 equations,
2y + 2z - Lyz = 0
2x + 2z - Lxz = 0
2y + 2x - Lxy = 0
xyz = V
Solving these, I get 
x = y = z = 4/L
I understand that x = y = z is the minimum of our surface area minimization problem. But what does the L mean here, in terms of volume and surface area?  
 A: It means nothing. In order to see why, suppose that you decided to work, not with your function $g$, but with $g^\ast(x,y,z)=-g(x,y,z)=V-xyz$. What difference would that make? You will get the same answer, of course, but the new $L$ will have the opposite sign.
What what if you work with $g_k(x,y,z)=kxyz-kv$, with $k>0$? Again, you will get the same answer, but not the same $L$.
A: If you consider the problem of maximizing or minimizing $f$ subject to a constraint of the form $g=c$, but vary $c$, then you get a function $f^*(c)$.  You can ask yourself: what is the derivative of this function?  How much extra $f$ do we get by increasing the level of $g$?  In your case, how does the minimum surface area of the box change if we change the volume?
Suppose the constrained critical point for the value $c$ occurs at $(x(c),y(c),z(c))$.  So $f^*(c) = f(x(c),y(c),z(c))$.  By the chain rule:
$$
    \frac{df^*}{dc} = \frac{\partial f}{\partial x}\frac{dx}{dc} + \frac{\partial f}{\partial x}\frac{dy}{dc} + \frac{\partial f}{\partial z}\frac{dz}{dc} 
$$
By the Lagrange multiplier theorem, and again the chain rule:
\begin{align*}
    \frac{df^*}{dc} &= \lambda\frac{\partial g}{\partial x}\frac{dx}{dc} + \lambda\frac{\partial g}{\partial x}\frac{dy}{dc} + \lambda\frac{\partial g}{\partial z}\frac{dz}{dc}  \\
    &= \lambda \frac{d}{dc}g(x(c),y(c),z(c))
\end{align*}
But now $g(x(c),y(c),z(c)) = c$, by design.  So the derivative on the right-hand side is $1$.  In summary, the derivative of the extreme value, as a function of the constraint level, is the multiplier.
$$
    \frac{df^*}{dc} = \lambda
$$
It makes a lot of sense in economics problems where the decision variables are a mix of goods, the objective function is utility, and the constraint function is total costs.  Assuming we continually adjust consumption to maximize utility subject to a fixed budget, the multiplier is the marginal utility of money.
A: The optimization problem is: 
$\hspace{3cm}$  Minimize $f(x,y,z)=2(xy+yz+zx)$ subjec to $xyz=V$.
The Lagrange function is:
$\hspace{4cm}$ $L(x,y,z,\lambda)=2(xy+yz+zx)+\lambda(V-xyz)$.
$\lambda$ is a dummy variable that can be used to answer additional question:
$\hspace{3cm}$ How much will $f(x,y,z)$ change if $V$ changes by $k$ units?
Answer: It will change approximately by $\lambda k$. Indeed, consider $L(x,y,z,\lambda,V)$:
$$\Delta L=L_{V}\Delta V=\lambda k.$$
A: The idea behind the Lagrange Multipliers Method is that at the extreme values the level surface of the function to be optimized and the graph of the constraint are tangent. That makes the two normal vectors  to have the same direction. The constant $L$ is the proportion of the two normal vectors.   
A: The correct problem handling is as follows
First the Lagrangian
$$
L(x,y,\lambda) = 2(x y+x z+y z)+\lambda(V-x y z)
$$
Second the stationary points determination by solving
$$
\nabla L = \{L_x,L_y,L_z,L_{\lambda}\} = 0
$$
or
$$
-\lambda y z + 2 (y + z) = 0\\
-\lambda x z + 2 (x + z) = 0\\
-\lambda x y + 2 (x + y) = 0\\
 V - x y z = 0
$$
Four equations with four unknowns. After solving we have
$$
x = V^{1/3}, y = V^{1/3}, z = V^{1/3}, \lambda = 4/V^{1/3}
$$
without ambiguities. 
