Example 1: If the coordinates of point $(x, y)$ satisfy $x^{2}+y^{2}+5 x \leqslant 0$, find the maximum and minimum values of $3 x+4 y$.


The constraint condition is $\left(x+\frac{5}{2}\right)^{2}+y^{2} \leqslant \frac{25}{4}$, and the points $(x, y)$ that satisfy this condition are on the circumference and inside the circle $\left(x+\frac{5}{2}\right)^{2}+y^{2} = \frac{25}{4}$.

Let the objective function be $3 \mathrm{x}+4 \mathrm{y}=\mathrm{m}$, then $y=-\frac{3}{4} x+\frac{m}{4}$, which represents a family of parallel lines with a slope of $-\frac{3}{4}$. As shown in the right figure: when the line is tangent to the circle, the intercept $\frac{\mathrm{m}}{4}$ 
achieves its extreme value, and accordingly, $\mathrm{m}$ also achieves its extreme value. By formula 1.1, we get $\left(-\frac{3}{4}\right)^{2} \times \frac{25}{4}$
$+\frac{25}{4}=\left[-\frac{3}{4} \cdot\left(-\frac{5}{2}+\frac{m}{4}\right)\right]^{2}$, which simplifies to $m^{2}+15 \mathrm{~m}$
$-100=0$. Solving this, we get $\mathrm{m}_{1}=5, \mathrm{~m}_{2}=-20$.
$\therefore 3 x+4 y$ has a maximum value of 5 and a minimum value of -20.