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Consider the function

$f(x,y) = -10x^2-10xy+30x$

I am about to find the extrema or saddle of the function. Calculated critical points,

$f_x = 0$

$-20x-10y+30$

$f_y = 0$

$x=0$

$y=3$

Critical point is (0,3)

$A = f_{xx}(0,3) = -20$

$B= f_{xy}(0,3) = -10 $

$C = f_{yy}(0,3) =0$

The discriminant is

$D = AC-B^2$

$D = -10$

$D<0$

A new case happens where $D<0$ and $f_{xx}<0$. What conclusion can be drawn? Is that saddle point?

Saddle occurs only when $D<0$. local Maximum occurs when $D>0$ and $f_{xx}<0$. So an odd case. what conclusion? A saddle point?

Aruha
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1 Answers1

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You already have the answer: for well behaved functions $f$, if $D < 0$ then you have a saddle, and in this case $D = -100$ (incidentally, not $-10$).

Remember first principles, when $f_x = f_y = 0$ $$ f(x+\delta x,y+\delta y) - f(x,y) = \tfrac{1}{2} \Big( f_{xx}\delta x^2+2 f_{xy}\delta x \delta y + f_{yy} \delta y^2\Big) + O(\delta x^3, \delta x^2 \delta y, \delta x \delta y^2, \delta y^3) $$ so it is the sign of the term in parentheses that determines the nature of the stationary point. Strictly there are a number of cases to consider.

  1. If at least one of $f_{xx}$ and $f_{yy}$ is zero and $f_{xy} = 0$ then the character of the point depends on higher derivatives.
  2. If $f_{xx} = f_{yy} = 0, f_{xy} \neq 0$ then you have a saddle.
  3. If only one of $f_{xx} $ and $f_{yy} $ is zero and and $f_{xy} \neq 0 $ then you have a saddle.
  4. Otherwise, both $f_{xx}, f_{yy}$ are non zero and you can use the discriminant. To see how it works, extract a factor of $\delta_x^2$ or $\delta_y^2$ from the term in parentheses and complete the square.
WA Don
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