Question

Reexamine Figure 4.9.1 in Example 3. Then use a root-finding application to determine when tank B contains more salt than tank A.

Short answer

The solution is

After$t=13.73$ minutes the tank B contains more salt than tank A

Step-by-step solution

Given information

Find when tank B contains more salt than tank A

The amount of salt in brine mixtures in tanks A and B are$x_1\left(t\right)and{x}_2\left(t\right).$

From the graph shown in Figure 4.8.1, we see that$x_1$ decreases with time. Whereas$x_2$ first increases and after that it decreases.

We also notice that after the point $x_1\left(t\right)=x_2\left(t\right)$, we have$x_2>x_1$

The solution obtained in the Example 3 from the book is given by

$$\begin{gathered} x_1\left(t\right)=\frac{25}{2}{e}^{-3t/25}+\frac{25}{2}{e}^{-t/25} \\ {x}_2\left(t\right)=-25e^{-3t/25}+25e^{-t/25} \end{gathered}$$

Now, we put$x_1=x_2$ to find when tank B contains more salt than tank A

$$\begin{gathered} x_1\left(t\right)=x_2\left(t\right) \\ \frac{25}{2}{e}^{-3t/25}+\frac{25}{2}{e}^{-t/25}=-25e^{-3t/25}+25e^{-t/25} \\ t=13.7327 \\ t=13.73minutes \end{gathered}$$

After$t=13.73$ minutes the tank B contains more salt than tank A

Conclusion

The solution is

After$t=13.73$ minutes the tank B contains more salt than tank A