0
$\begingroup$

Given is the function $f(x,y) = 3x^2y+4y^3-3x^2-12y^2+1$

I'm looking for the extrema points. Therefore I calculated $f_x(x,y)= 6xy-6x$ and $f_y(x,y)=3x^2+12y^2-24y$ and set them to zero to find the possible points. I got the points: $P_1=(2,1)$, $P_2=(-2,1)$, $P_3=(0,0)$, $P_4=(0,2)$

Now I want to check if they are extrema points or not. Therefore I used the hessian matrix $H_f = \left[\begin{array}{c} f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\\ \end{array}\right] = \left[\begin{array}{c} 6y-6 & 6x\\ 6x & 24y-24 \\ \end{array}\right]$

When I calculate the determinante of the matrix for every point, I get:

$P_1=(2,1): det(H_f)= (6*1-6) *(24*1-24) -(6*2*6*2)=0-12*12$ which is $< 0$ and should therefore be a Maximum.

$P_2=(-2,1): det(H_f)= 0 -(-12*(-12))$ which is $< 0$ and should therefore be a Maximum.

$P_3=(0,0): det(H_f)= (-6) *(-24)$ which is $> 0$ and should therefore be a Minimum.

$P_4=(0,2): det(H_f)= (6*24)$ which is $> 0$ and should therefore be a Minimum.

But when I type my function in wolfram alpha to check my result, it only mentions the points $(0,0)$ as a Maximum and $(0,2)$ as a Minimum. In my calculation $(0,0)$ is a Minimum and I got two more extrema points. So what have I done wrong?

$\endgroup$
  • 1
    $\begingroup$ I would say, $f_{yy}=24y-24$ $\endgroup$ – georg Sep 20 '18 at 13:12
  • $\begingroup$ Thanks, I've corrected it, as it was a typo. $\endgroup$ – mrs fourier Sep 21 '18 at 10:09
2
$\begingroup$

When the determiant of the hessian is negative, it means the eigenvalues are of opposite sign and hence it is indefinite. It is a saddle point.

At point $(0,0)$ the $(1,1)$-entry of the Hessian is negative. Since the determinant is positive, both of the eigenvalues are negative. The hessian is negative defintie and hence it is a maximum point.

At point $(0,0)$ the $(1,1)$-entry of the Hessian is positive. Since the determinant is positive, both of the eigenvalues are positive. The hessian is positive defintie and hence it is a minimum point.

$\endgroup$
1
$\begingroup$

Given: $f(x,y)=3x^2y+4y^3-3x^2-12y^2+1$

FOC: $$\begin{cases}f_x=6xy-6x=0\\ f_y=3x^2+12y^2-24y=0\end{cases} \Rightarrow \\ 1) \ \begin{cases}x=0 \\ 12y^2-24y=0\end{cases} \ \ \text{or} \ \ 2) \begin{cases}y=1\\ 3x^2-12=0\end{cases} \Rightarrow \\ 1) \ (x_1,y_1)=(0,0); (x_2,y_2)=(0,2);\\ 2) \ (x_3,y_3)=(-2,1); (x_4,y_4)=(2,1).$$ SOC: $$H=\begin{vmatrix}6y-6&6x \\ 6x&24y-24\end{vmatrix}$$ At $(0,0)$: $H_1=-6<0; H_2=144>0 \Rightarrow f(0,0)=1$ is maximum;

At $(0,2)$: $H_1=6>0; H_2=144>0 \Rightarrow f(0,2)=-15$ is minimum;

At $(-2,1)$: $H_1=0; H_2=-144<0 \Rightarrow f(-2,1)=-7$ is a saddle point;

At $(2,1)$: $H_1=0; H_2=-144<0 \Rightarrow f(2,1)=-7$ is a saddle point.

Reference: Second derivative test.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.