I have been asked to find the critical points of the function:
$$ f(x,y) = e^{xy} $$
with the constraint:
$$ x^3+y^3=16 $$
The first thing i do, is make my constraint into a function, by moving everything to the left side:
$$ g(x,y) = x^3+y^3-16 $$
I now take the partial derivative of both function, with respect to each variable: $$ \frac{\partial f}{\partial x} = e^{xy}y $$ $$ \frac{\partial f}{\partial y} = e^{xy}x $$ $$ \frac{\partial g}{\partial x} = 3x^2 $$ $$ \frac{\partial g}{\partial y} = 3y^2 $$
Now I set up my lagrange multiplier equations:
$$ e^{xy}y = 3x^2 \cdot \lambda $$ $$ e^{xy}x = 3y^2 \cdot \lambda $$
I solve for x in the first equations and for y in the second equations, and get:
$$ x = 0 $$ $$ y = 0 $$
If I put my values into the original constraint, I get:
$$ 0^3 + 0^3 = 16 $$
Which is false. Wolframs lagrange multiplier calculator tells me, that I should get a global maximum, but I haven't found any. Am I missing a step somewhere?
In problems like this it is customary to define a Lagrangian function
$$L\left(x\right)=f\left(x\right)-\lambda g\left(x\right)$$
where $f\left(x\right)=e^{xy}$ is your function and $g\left(x\right)=x^{3}+y^{3}-16$ is the constraint. Explicitly
$$L\left(x\right)=e^{xy}-\lambda \left(x^{3}+y^{3}-16\right)$$
Then the equations of exterma of $f$ with the constraint $g$ are
$$\begin{cases}0=\frac{\partial L}{\partial x}=ye^{xy}-3\lambda x^2\\0=\frac{\partial L}{\partial y}=xe^{xy}-3\lambda y^2\\0=\frac{\partial L}{\partial \lambda}=x^{3}+y^{3}-16\end{cases}$$
Now, there is no problem with your method. You just noticed that $x=y=0$ can't be a solution. Thus we can divide the first equation by the second to get
$$\frac{y}{x}=\frac{x^{2}}{y^{2}}$$
or
$$x^{3}=y^{3}$$
leading to $x=y$. Plugging this into the third equation gives you
$$x=y=2$$
as a solution.