For $F_1(x,y)=\frac{1}{4}{x}^4+x^{2}y+y^2$ and $F_2(x,y)=x^3+xy-x,$ find the second derivative matrices $A_1$ and $A_2$ :

$A=\left[\begin{array}{cc}de{l}^2\frac{F}{d}el{x}^2& de{l}^2\frac{F}{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}\\ de{l}^2\frac{F}{d}\text{e}\text{l}\text{y}\text{d}\text{e}\text{l}\text{x}& de{l}^2\frac{F}{d}el{y}^2\end{array}\right]$

$A_1$ is positive definite, so $F_1$ is concave up $(=$ convex$).$ Find the minimum point of $F_1$ and the saddle point of $F_2($look where first derivatives are zero$).$