甽6. If $x+3y=10$, find the extremum of $x^{2}+3y^{2$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. 

6. If $x+3y=10$, find the extremum of $x^{2}+3y^{2}$.
Solve $x+3 y$
$=10$ as the line in Figure 6. Let $x^{2}+3 y^{2}=c$, then
$$
\frac{x^{2}}{(\sqrt{c})^{2}}+\frac{y^{2}}{\left(\sqrt{\frac{c}{3}}\right)^{2}}=1 \text {. }
$$

From Figure 6, we know that when the line $x+3 y=10$ is tangent to the ellipse $x^{2}+3 y^{2}=c$, the semi-axes of the ellipse are the shortest. We have
$$
\left\{\begin{array}{l}
x+3 y=10 \\
x^{2}+3 y^{2}=c
\end{array}\right.
$$

We get $c=25$, i.e., the minimum value of $x^{2}+3 y^{2}$ is 25. (Author's affiliation: No. 100 High School, Tianjin City)