Example 2. If $2 x^{2}+3 y^{2}=1$, find the extremum of $x+y$.



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Example 2. If $2 x^{2}+3 y^{2}=1$, find the extremum of $x+y$.
Solve $2 x^{2}+3 y^{2}=1$; the graph is an ellipse (see Figure 2). For the family of lines $x+y=c$ that have common points with the ellipse $2 x^{2}+3 y^{2}=1$, the lines for which $c$ takes extreme values must be tangent to the ellipse.
From $x+y=c$, we get $y=c-x$.
Substituting $y=c-x$ into $2 x^{2}+3 y^{2}=1$, we get
$$
5 x^{2}-6 c x+\left(3 c^{2}-1\right)=0 \text {. }
$$

If $x+y$ takes an extreme value, then
$$
\begin{aligned}
\Delta & =(-6 c)^{2}-4 \times 5 \times\left(3 c^{2}-1\right)=0, \\
\text { thus } c & = \pm \frac{5 \sqrt{6}}{6} .
\end{aligned}
$$

Therefore, the maximum value of $x+y$ is $c=\frac{5 \sqrt{6}}{6}$, and the minimum value of $x+y$ is $c=-\frac{5 \sqrt{6}}{6}$.