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How can I find the global min/max in this problem?
Find the critical points of the function :
f(x,y)=$2x^3-3x^2y-12x^2-3y^2$
and determine their type.
Are there any global min/max?

The critical points were: (0,0), (2,-2),(-4,-8) By using the hessian matrix I found their types:
(0,0) is a local maximum
(-4,-8) is a saddle
(2,-2) is a saddle.

But I do not understand how to find the global minimum or maximum.
I know that if the Hessian matrix is negative semi definite then any local max is a global max and if Hessian matrix is a positive semi definite then any local min is a global min.

Please help me to find the global minimum /maximum?

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look on x-axis, as x is increasing f(x,y) is also increasing similarly on y-axis as y is increasing ; f(x,y) is decreasing, so there is no global max, min.

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  • $\begingroup$ checking what happens to the function as x increases and y increases is that a method to find global min /max $\endgroup$ – clarkson Dec 30 '13 at 7:07
  • $\begingroup$ absolutely it's not the method but by way of contradiction if max occurs at a point then there should be no point where greater than max value occur. but here always we can have such a point. $\endgroup$ – Sry Dec 30 '13 at 7:12
  • $\begingroup$ So I have to check the behavior of the function (y->infinity,x=0) and (x->infinity,y=0) $\endgroup$ – clarkson Dec 30 '13 at 7:17
  • $\begingroup$ it depends on problem Mr.clarkson $\endgroup$ – Sry Dec 30 '13 at 7:18
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There isn't a global maximum or minimum, at least not when considered on $\mathbb{R}^{2}$. Set $y=0$ and you can easily see that your function neither has an upper nor lower bound.

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Note that the highest power of $x$ is 3. So if you make $x$ large positive then $f$ is also large positive (just set $y=0$). Similarly if you make $x$ large negative, $f$ will be large negative.

So no global maximima/minima

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