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.
Calculate the partial derivative, local minima/maxima, and saddle points.
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0Note 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. – 2012-11-24
2 Answers
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:
- $f_x(x_0,y_0)>0,f_y(x_0,y_0)>0$: $(x_0,y_0)$ is local minimum
- $f_x(x_0,y_0)<0,f_y(x_0,y_0)<0$: $(x_0,y_0)$ is local maximum
- $f_x(x_0,y_0)>0,f_y(x_0,y_0)<0$ 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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3-1 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 $f_{yy}$. – 2012-11-24
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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0aside: the reasons $f$or 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. – 2012-11-24