4-205 There is a mixture of copper and zinc weighing 400 grams, with a volume of 50 cubic centimeters. It is known that the specific gravity of copper is less than 9 grams/cubic centimeter and greater than 8.8 grams/cubic centimeter, and the specific gravity of zinc is less than 7.2 grams/cubic centimeter and greater than 7.1 grams/cubic centimeter. Try to find the range of the weight of copper and zinc in the mixture.
[Solution] Let the mixture contain $x$ grams of copper and $y$ grams of zinc, and denote the specific gravities of copper and zinc as $D_{1}$ and $D_{2}$, respectively. Then we have:
$$\left\{\begin{array}{l}x+y=400, \\ \frac{x}{D_{1}}+\frac{y}{D_{2}}=50 .\end{array}\right.$$
Solving this system of equations, we get:
$$\left\{\begin{array}{l}x=\frac{50 D_{1}\left(8-D_{2}\right)}{D_{1}-D_{2}}, \\ y=\frac{50 D_{2}\left(D_{1}-8\right)}{D_{1}-D_{2}} .\end{array}\right.$$
When $D_{2}$ is fixed, let $k_{1}=50\left(8-D_{2}\right)>0, k_{2}=50 D_{2}>0$.
$$\begin{aligned}
x-x^{\prime} & =k_{1}\left(\frac{D_{1}}{D_{1}-D_{2}}-\frac{D_{1}^{\prime}}{D_{1}^{\prime}-D_{2}}\right) \\
& =k_{1} \frac{D_{2}\left(D_{1}^{\prime}-D_{1}\right)}{\left(D_{1}-D_{2}\right)\left(D_{1}^{\prime}-D_{2}\right)},
\end{aligned}$$

From this, it is evident that the larger $D_{1}$ is, the smaller $x$ is, and the smaller $D_{1}$ is, the larger $x$ is.
$$\begin{aligned}
y-y^{\prime} & =k_{2}\left(\frac{D_{1}-8}{D_{1}-D_{2}}-\frac{D_{1}^{\prime}-8}{D_{1}^{\prime}-D_{2}}\right) \\
& =k_{2} \frac{\left(8-D_{2}\right)\left(D_{1}-D_{1}^{\prime}\right)}{\left(D_{1}-D_{2}\right)\left(D_{1}^{\prime}-D_{2}\right)}
\end{aligned}$$

From this, it is evident that the larger $D_{1}$ is, the larger $y$ is, and the smaller $D_{1}$ is, the smaller $y$ is.
When $D_{1}$ is fixed, similarly, we have:
$$\begin{aligned}
x-x^{\prime \prime} & =k_{3}\left(\frac{8-D_{2}}{D_{1}-D_{2}}-\frac{8-D_{2}^{\prime \prime}}{D_{1}-D_{2}^{\prime \prime}}\right) \\
& =k_{3} \frac{\left(D_{1}-8\right)\left(D_{2}^{\prime \prime}-D_{2}\right)}{\left(D_{1}-D_{2}\right)\left(D_{1}-D_{2}^{\prime \prime}\right)}, \text { where } k_{3}=50 D_{1}>0,
\end{aligned}$$
The larger $D_{2}$ is, the smaller $x$ is, and the smaller $D_{2}$ is, the larger $x$ is.
$$\begin{aligned}
y-y^{\prime \prime} & =k_{4}\left(\frac{D_{2}}{D_{1}-D_{2}}-\frac{D_{2}^{\prime \prime}}{D_{1}^{\prime \prime}-D_{2}}\right) \\
& =k_{4} \frac{D_{1}\left(D_{2}-D_{2}^{\prime \prime}\right)}{\left(D_{1}-D_{2}\right)\left(D_{1}-D_{2}^{\prime \prime}\right)}, \text { where } k_{4}=50\left(D_{1}-8\right) \\
& >0
\end{aligned}$$
The larger $D_{2}$ is, the larger $y$ is, and the smaller $D_{2}$ is, the smaller $y$ is.
In summary, when $D_{1}$ is the largest and $D_{2}$ is the largest, $x$ is the smallest and $y$ is the largest; when $D_{1}$ is the smallest and $D_{2}$ is the smallest, $x$ is the largest and $y$ is the smallest.
$$\text { Therefore, } \begin{aligned}
\frac{50 \times 9(8-7.2)}{9-7.2} & \leqslant x \leqslant \frac{50 \times 8.8(8-7.1)}{8.8-7.1}, \\
& \frac{50 \times 7.1(8.8-8)}{8.8-7.1} \leqslant y \leqslant \frac{50 \times 7.2(9-8)}{9-7.2} .
\end{aligned}$$

Calculating, we get 200 grams $\leqslant x \leqslant 233$ grams,
167 grams $\leqslant y \leqslant 200$ grams.