Calculation of extrema points

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?

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

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.

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.