So we have function $f(x,y) = e^{xy}$ this has the constraint $x^3+y^3=16$ By use of Lagrange Multipliers or the Lagranian we find there is only one critical point at (2,2).
What confuses me alot is that i can't quite grasp what have we exactly found at point (2,2) ?
The constrainted maximum that is closest to the unconstrainted global maximum? This makes no sense since this function has no global maximum nor global minimum.
On
It's a point of local extrema for the function taking into account the constraint.
To verify whether it is a maximum or a minimum you have to calculate the value of "f" at that point and compare with all the others possible max/min values.
From a geometrical point of view it's the point where the gradient of $f(x,y)$ ant the gradient of the constrain function $\phi(x,y)=x^3+y^3-16$ are parallel.
Here is a nice graphical explanation:
Courtesy of Wolfram, here's a contour plot of $z=e^{xy}$ (in the white region, $z$ is large; in the red region, it's close to $0$) along with $x^3+y^3=16$ (the blue curve):
If you are standing at the solution $(2,2)$, you can see that by walking along the constraint, you will be going downhill very quickly. You have found the maximum of $e^{xy}$ subject to the given constraint.
Also note that the point $(2,2)$ is precisely where the tangent of $x^3+y^3=16$ coincides with the tangent of the level curves of $e^{xy}$.