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?
Best Answer
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.