Finding the Absolute Maximum for a Multivariable Function Without an Interval

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Homework Statement


I found the local maximum for the equation g(x,y)=3xe^y - x^3 - e^(3y), which is (1, 0)...now I am supposed to show that this is not the absolute maximum for g(x,y). I don't know how to find an absolute maximum without an interval! Could anyonw shed some light on this?


Homework Equations


I have solved for the partial derivatives with respect to x, y, xx, yy, and xy (all needed to find the local max).


The Attempt at a Solution


I really don't know where to go from here, all I think we have learned is the extreme value theorem where you compare the critical points to the boundary values, but I am not given any boundary values, just that x and y are continuous for all values.
 
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If there is no boundary, then there may not be a max or min. But if they are, then they occur where the gradient is 0.

Find all points at which [/itex]\nabla g= \vec{0}[/itex] and evaluate them. The largest of those values is the "global maximum" which, I think, is what you are calling the "absolute maximum". In this case the second derivatives are not needed since it does not matter whether they are local maxima or not.
 
But the question was showing the point was not an absolute max. All that is necessary is to show for some (x,y) the function gets greater. Given that the function is

g(x,y)=3xey - x3 - e(3y)

one can readily see that limit as x --> -oo of g(x,0) = oo so there is no global max.
 
thanks LCKurtz, that's what i ended up doing..i graphed it on my computer and then figured that out :))
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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