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How do I find the maxima, minima and saddle point for this equation?

$f(x, y) = (3x^2) +(x^2)y - x(y^2) + 1$

I have already found that

$f'(x) = 6x+2xy-(y^2)$

$f'(y) = (x^2)-2xy$

and also found the gradient to be

$= [6x+2xy - (y^2), (x^2) - 2xy]$

This is the part I am stuck at, how do I find these stationary points I mentioned above? I do know that I have to find point from the equations in the gradient, and then insert these points in another equation in order to find wether delta is smaller or bigger than zero for maxima and minima, and double derivative of equation to be bigger than 0 if it has saddle point, but how? I struggle here.

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  • $\begingroup$ Hello @peterovic, welcome to MSE. It is highly recommended to use MathJax to typeset mathematical formulas in post. A tutorial can be found here. $\endgroup$
    – Ernie060
    Apr 30, 2019 at 21:01
  • $\begingroup$ See math.stackexchange.com/questions/1360060/… and other related problems in the handy list at right for the methods to use. $\endgroup$
    – amd
    Apr 30, 2019 at 21:13

2 Answers 2

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Now,You have to check the Hessian,$H$.

Here,$$H=\frac{1}{4}f_{xx}(x_0,y_0).f_{yy}(x_0,y_0)-f_{xy}(x_0,y_0)^2[\text{found from Hessian matrics.}]$$

$i)$ if $H\lt 0$,then $(x_0,y_0)$is a saddle point.

$ii)$ if $H\gt 0$,then $(x_0,y_0)$is a stable/critical point(maxima,minima).

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So, in order to find the critical points, you must solve the system$$\left\{\begin{array}{l}6x+2xy-y^2=0\\x^2-2xy=0.\end{array}\right.$$The second equation is equivalent to $x(x-2y)=0$. There are therefore two possibilities: $x=0$ or $x=2y$. If $x=0$, the first equation becomes $-y^2=0$. So, the only equation in this case is $(0,0)$. If $x=2y$, then the first equation becomes $12y+3y^2=0$, which has two solutions: $y=0$, and $y=-4$. Conclusion: there are two critical points: $(0,0)$ and $(-8,-4)$.

The Hessian of $f$ at a point $(x,y)$ is $\left[\begin{smallmatrix}2 y+6 & 2 x-2 y \\ 2 x-2 y & -2 x\end{smallmatrix}\right]$. In particular, the Hessian at $(-8,-4)$ is $\left[\begin{smallmatrix}-2 & -8\\ -8 &16\end{smallmatrix}\right]$. Since its determinant is negative, it has a positive eigenvalue and a negative one. Therefore, $(-8,-4)$ is a saddle point.

At $(0,0)$ it's more complex. One of the eigenvalues is $0$ and the other one is $6$. In fact, it is a saddle point: $f(x,x)>f(0,0)$ if $x\neq0$ and $f\left(\frac{y^2}{6+2y},y\right)<f(0,0)$ for each $y\neq0$.

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