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Let $f(x,y)=x^4-8x^2+y^4-18y^2$

Find the set of global minimizers of f?
Does f have a global maximizer?Justify?

I first calculated the gradient of f and then let partial derivative of x and y to be equal to 0.
Thereby the critical points I found are (0,0)(0,3)(0,-3)(6,0)(6,3)(6,-3).

I think a global maximizer doesn't exist as when limit of function goes to infinity from both x and y ,function goes to infinity.
But how to find if global minimizers are there?

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The global minimizer has to be one of these (0,0)(0,3)(0,-3)(6,0)(6,3)(6,-3). Check each to see which one makes $f$ the least value. Because of symmetry, you can save some work.

But....

Are you sure of your answers?

$$ 4 x^3 - 16x = 0 \Rightarrow x=0, \pm 2 \\ 4 y^3 - 36 y=0 \Rightarrow y=0, \pm 3 $$ So you have 9 critica points: (0,0), (0,3), (0,-3), (2,0), (2,3), (2,-3), (-2,0), (-2,3), (-2,-3)

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  • $\begingroup$ I have made a mistake.After computing the values of f at these critical points, do I have to check if these critical points are minimizers.What happens if the function value is lowest at a point which is actually a saddle point.Then I can't take that as a global minimizer right? $\endgroup$ – clarkson Jan 8 '14 at 1:41
  • $\begingroup$ The minimum can never be at a saddle point. You can eliminate saddle points by looking at the second derivatives. I any case you will find, due to symmetry, multiple global minima (in fact any critical point with $x\ne 0$ and $y\ne0$. $\endgroup$ – user44197 Jan 8 '14 at 2:04
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I think you can find the global minimizers without the use of the derivative: just note that $$ f(x,y)= x^4-8x^2+y^4-18y^2= (x^4-8x^2+16)+(y^4-18y^2+81)-97= (x^2-4)^2+(y^2-9)^2-97 $$ Then, for every $x,y \in\mathbb{R}$, we have that $f(x,y)\geq -97$ and at $x=\pm 2, y=\pm 3$, we have $f(x,y)=-97$

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  • $\begingroup$ I didn't see that $\endgroup$ – clarkson Jan 8 '14 at 1:42

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