Use LaGrange multipliers to find maximum and minimum values I am having trouble understanding how to solve the problem below.  Can anyone show me how to solve this?  Here is the problem definition:  
"Use LaGrange multipliers to find the maximum and minimum values of the function $f(x,y)=e^{xy}$ subject to the constraint $x^3+y^3=16$."
This is problem number 14.8.6 in the seventh edition of Stewart Calculus.  
My work so far:
Use $\nabla f=\lambda\nabla g$ and $g(x,y)=16$
$f_x=ye^{xy}=\lambda g_x=\lambda 3x^2$
$f_y=xe^{xy}=\lambda g_y=\lambda 3y^2$
Solving these equations gives me $x=1, y=1, \lambda=\frac{e}{3}$
However, I am confused because $f(1,1)=e$, but $g(1,1)=2\ne 16$
How do I finish this problem correctly?
 A: Start with the equations that you have derived:
\begin{eqnarray*}
ye^{xy}&=&3\lambda x^2,\\
xe^{xy}&=&3\lambda y^2,\\
x^3+y^3&=&16.
\end{eqnarray*}
As a first step, show that none of $\lambda, x$ or $y$ can be zero. (If one of them is zero, then the first two equations show that all three must be zero, contradicting the third equation.) This is useful as we now know that we can divide by these terms at will. 
Now comparing the first and second equations gives $x^3=y^3$, and since both are real, we get $x=y$. The third equation then gives $x=y=2$, and the first or second yields the value of $\lambda$. We find $f=e^4$ at this point, and it must be a maximum.
As regards the minimum, recall that the Lagrange multiplier method identifies the possible location of max/min points IF they exist. I don't think your example has a minimum: By taking $x^3 = N^3$ and $y^3=16-N^3$, where $N$ is a large positive number, the constraint $x^3+y^3=16$ is satisfied. But $xy\sim-N^2$ can be made arbitrarily large and negative, so that $e^{xy}$ can be made as close as we like to $0$, but of course would never equal zero. So we can say that the infimum
$$\inf\{e^{xy}:x^3+y^3=16\}=0,$$
but the minimum
$$\min\{e^{xy}:x^3+y^3=16\}$$
does not exist: for any candidate minimum at $(x_0,y_0)$, we can always find $(x_1,y_1)$ with $0<f(x_1,y_1)<f(x_0,y_0)$. 

Another approach is to eliminate $y$ and to treat this as a single variable max/min problem for $h(x)=\exp[x(16-x^3)^{1/3}]$. This gives another way of understanding why there is no minimum point of the function. 
