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I am having trouble finding the partial derivative. And clues or hints regarding said problem and how to find saddle points/local maxima is appreciated.

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Note that if you set the derivatives to zero and solve, you will get the only stationary point $x=y=-\sqrt[3]{2}$. If you substitute this into the second derivative, you will get the matrix $\pmatrix{2 & 1 \\ 1 & 2}$, which is positive definite, hence this is the only local minimum. – copper.hat Nov 24 '12 at 20:50

Partial derivatives are calculated by regarding the function as a function in only one argument and considering the other variables as constants.

So you obtain $f_x(x,y)=\frac{d}{dx}(y x -2 x^{-1}-2/y)=y+2x^{-2}$. Saddle points and local maxima/minima are always at places where both derivatives vanish simultaneously. To decide which is which, you have to look at the second derivatives:

  1. $f_x(x_0,y_0)>0,f_y(x_0,y_0)>0$: $(x_0,y_0)$ is local minimum
  2. $f_x(x_0,y_0)<0,f_y(x_0,y_0)<0$: $(x_0,y_0)$ is local maximum
  3. $f_x(x_0,y_0)>0,f_y(x_0,y_0)<0$ or$f_x(x_0,y_0)<0,f_y(x_0,y_0>0$: $(x_0,y_0)$ is a saddle point.
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how can i find the x and y coordinates of the maxima/minima? – gt2 Nov 24 '12 at 20:03
-1 You need to look at the eigenvalues of the second derivative at the stationary points. generally, it is not sufficient to just look at $f_{xx}$ and $f_{yy}$. – copper.hat Nov 24 '12 at 20:37

I often find that the best way to go about optimization problems is to go about mapping the derivatives out, and then plugging in points:

  • fx=y+2/(x^2)
  • fxx=-4/(x^3)
  • fy=x+2/(y^2)
  • fyy=-4/(y^3)
  • fxy=0

Now you plugin the desired points into your new equations and evaluate. Values wherein fx and fy are equal to 0 are said to be critical points. These are the extrema you are looking for and can be classified using the following equation:

  • D=[(fxx)(fyy)]-[(fxy)^2]

If D is positive and both fxx and fyy are negative at the chosen values, then the critical point is a maximum. If D is positive and both fxx and fyy are positive at the chosen values, then the critical point is a minimum. If D is negative at the chosen values, then the critical point is a saddle. If D is zero then the nature of the critical point is indeterminable, it could be any kind of point, geometric analysis would be necessary.

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aside: the reasons for the meaning of these values for D stem from the second derivative. if both fxx and fyy are positive, then the function is increasing at that point, and since these points were determined to be critical points, then the point must be a minimum, and vice-versa for the negation. The only time D is negative is when fxx and fyy have opposite signs, so in one direction the function increases, but the function decreases in the other, hence a saddle. – Wes Nov 24 '12 at 20:24

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