6・74 Let $R$ be the triangular region in the plane with vertices $A(4,1), B(-1,-6), C(-3,2)$ (including the interior and boundary of the triangle, as shown in the figure). Try to find the maximum and minimum values of the function $4 x-3 y$ as $(x, y)$ varies over $R$ (you need to prove your conclusion).
[Solution] Let $\lambda=4 x-3 y$. It is evident that when $\lambda$ is fixed and $(x, y)$ varies, we obtain a straight line on the plane. By letting $\lambda$ vary, we get a series of parallel lines on the plane. On each of these lines, the value of $4 x-3 y$ is the same. When the line passes through point $A$, $\lambda=13$, at which time the line passes through $\left(\frac{13}{4}, 0\right)$. When the line passes through point $B$, $\lambda=14$, at which time the line passes through $\left(\frac{7}{2}, 0\right)$. When the line passes through point $C$, $\lambda=-18$, at which time the line passes through $\left(-\frac{9}{2}, 0\right)$. Generally, the line $\lambda=4 x-3 y$ intersects the $x$-axis at $\left(x^{\prime}, 0\right)=\left(\frac{\lambda}{4}, 0\right)$, and $\lambda=4 x^{\prime}$ is proportional to $x^{\prime}$. Since $-\frac{9}{2} \leqslant x^{\prime} \leqslant \frac{7}{2}$, it follows that $-18 \leqslant \lambda \leqslant 14$. The maximum and minimum values sought are 14 and -18, respectively.