Exercise 1.7.24

• For $F_1$, we have $\frac{\partial F_1}{\mathit{\partial x}}=x^3+2\mathit{xy}$, $\frac{\partial F_1}{\mathit{\partial y}}=x^2+2y$, and $H_1=\left[\begin{array}{cc}\hfill 3x^2+2y\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2\hfill \end{array}\right]$

Since $F_1=\frac{1}{4}{x}^4+x^{2}y+y^2={(\frac{1}{2}{x}^2+y)}^2\ge 0$, let the derivatives equal to 0, we have $(x,y)\text{satisfy}{x}^2+2y=0$. it achieves minimum of $0$. At the minimum point we have $H_1=\left[\begin{array}{cc}\hfill 2x^2\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2\hfill \end{array}\right]$. One can notice that the determinants of $H_1$ are all larger than or equal to 0.

$H_1$ is semi-positive definite.

• For $F_2$, we have $\frac{\partial F_2}{\mathit{\partial x}}=3x^2+y-1$, $\frac{\partial F_2}{\mathit{\partial y}}=x$, and $H_2=\left[\begin{array}{cc}\hfill 6x\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$.

$H_2$ always has determinant less than 0 regardless of the $(x,y)$, so it’s not positive definite. let the first derivatives to zero, we see $(x,y)=(0,1)$ is the saddle point of $F_2$.