Lagrange multiplier problem - Why doesn't the method work? Maximize the distance to the origin of a function with four variables given the constraints: 
\begin{align}
xyza &= 1 \\
x + y + z + a &= 4
\end{align}
Here's my solution: We maximize $f(x,y,z,a) = x^2+y^2+z^2+a^2$ subject to the given constraint equations. The Lagrange multiplier method yields six equations with six variables.
\begin{align}
2x &= \lambda_1yza+\lambda_2\\
2y &= \lambda_1xza+\lambda_2\\
2z &= \lambda_1xya+\lambda_2\\
2a &= \lambda_1xyz+\lambda_2\\
xyza &= 1\\
x + y + z + a &= 4
\end{align}
Mathematica is unable to solve this system. Is there something wrong in the question (I made it up) itself? Thanks!
 A: The question is ill posed.
One of the first things you should do before applying Lagrange is to verify
that there is indeed a solution. There is no solution in this case.
Fix $z=-1$, then look for solutions of $x+y-1-{1 \over xy} = 4$. Multiplying across by $x$ gives
$x^2+x (y-5) -{1 \over y} = 0$, which has a solution
$x= {1 \over 2} (5-y -\sqrt{(y-5)^2+{4 \over y}})$.
Letting $y \to \infty$ there are solutions, hence the distance to the origin is unbounded.
A: If you forget the first constraint, and instead minimize the distance to the origin from the hyperplane $x + y + z + a = 4$, your equations will look like:
\begin{align*}
2x & = \lambda_1 \\
2y & = \lambda_1 \\
2z & = \lambda_1 \\
2a & = \lambda_1 \\
4 & = x + y + z + a
\end{align*}
This gives the unique solution $x = y = z = a = 1$.  (You might have guessed this, since $\langle 1, 1, 1, 1 \rangle$ is the normal vector and also represents a point on the plane.)
In any case, adding a second constraint cannot possibly decrease the minimum distance to the origin, and since $(1,1,1,1)$ also lies on the hypersurface represented by your first constraint, it must be the unique solution.
EDIT: Sorry, just realized you wanted to maximize.  Still, you can solve the system you wrote.  We know from equation 5 that $x,y,z,a$ are nonzero, so we can multiply the first equation by $x$, the second by $y$, etc.  Using the 5th equation to simplify gives:
\begin{align*}
2x^2 & = \lambda_2x \\
2y^2 & = \lambda_2y \\
2z^2 & = \lambda_2z \\
2a^2 & = \lambda_2a \\
\end{align*}
We can divide each by its respective variable (again since they are all nonzero) to get $x=y=z=a=\frac{\lambda_2}{2}$, and again we get the unique solution $(1,1,1,1)$.  This may be the only point at which your constraints intersect.
