Example 3.44. Investigate the function for local extremum

$$
f\left(x_{1}, x_{2}\right)=3 x_{1}^{2} x_{2}-x_{1}^{3}-\frac{4}{3} x_{2}^{3}
$$
Solution. We find the derivatives $f_{x_{1}}^{\prime}$ and $f_{x_{2}}^{\prime}$; solving the obtained system

$$
\left\{\begin{array}{l}
6 x_{1} x_{2}-3 x_{1}^{2}=0 \\
3 x_{1}^{2}-4 x_{2}^{2}=0
\end{array}\right.
$$

we find two points $P_{0}(6,3)$ and $P_{1}(0,0)$.

To establish the nature of the found stationary points, we will use the sufficient condition for an extremum and investigate the sign definiteness of the quadratic form, which, as is easily noticed, represents the second differential:

$$
\begin{gathered}
d^{2} f=f_{x_{1} x_{1}}^{\prime \prime} d x_{1}^{2}+f_{x_{1} x_{2}}^{\prime \prime} d x_{1} d x_{2}+f_{x_{2} x_{1}}^{\prime \prime} d x_{2} d x_{1}+f_{x_{2} x_{2}}^{\prime \prime} d x_{2}^{2}= \\
=\left(6 x_{2}-6 x_{1}\right) d x_{1}^{2}+6 x_{1} d x_{1} d x_{2}+6 x_{1} d x_{2} d x_{1}-8 x_{2} d x_{2}^{2}
\end{gathered}
$$

We form the matrix $\left(\begin{array}{cc}6 x_{2}-6 x_{1} & 6 x_{1} \\ 6 x_{1} & -8 x_{2}\end{array}\right)$. We compute the values of the corresponding minors according to (3.36):

$$
\begin{aligned}
& \Delta_{1}=\left.\left(6 x_{2}-6 x_{1}\right)\right|_{\substack{x_{1}=6 \\
x_{2}=3}}=-18, \\
& \Delta_{2}=\left|\begin{array}{cc}
6 x_{2}-6 x_{1} & 6 x_{1} \\
6 x_{1} & -8 x_{2}
\end{array}\right|=\left.\left(6\left(x_{2}-x_{1}\right)\left(-8 x_{2}\right)-\left(6 x_{1}\right)^{2}\right)\right|_{\substack{x_{1}=6 \\
x_{2}=3}}=648
\end{aligned}
$$

According to (3.37), we obtain that at point $P_{0}$ the function has a maximum $f_{\text {max }}=72$.

Now let's investigate the point $P_{1}(0,0)$. At this point $\Delta_{1}(0,0)=0$, $\Delta_{2}(0,0)=0$ and additional investigation is required. Since at $x_{1}=0$ and $x_{2}=0, f(0,0)=0$; at $x_{1}=0$, and at $x_{1}>0$ and $x_{2}=0, f\left(x_{1}, 0\right)=-x_{1}^{3}<0$, there are both positive and negative values of $f\left(x_{1}, x_{2}\right)$ in the neighborhood of point $P_{1}(0,0)$, meaning $P_{1}(0,0)$ is a saddle point.