1. **State the problem:** We have two functions representing water flow rates into and out of a tank over time $t$ hours: $$H(t) = 12(0.93)^t$$ $$N(t) = 15(0.77)^t$$ We want to find the minimum amount of water in the tank from $t=0$ to $t=10$. 2. **Define the net water amount function:** The net water in the tank at time $t$ is the integral of the difference between inflow and outflow rates: $$W(t) = \int_0^t [H(x) - N(x)] \, dx$$ 3. **Find the net rate function:** $$R(t) = H(t) - N(t) = 12(0.93)^t - 15(0.77)^t$$ 4. **Find critical points:** The minimum water amount occurs where the derivative of $W(t)$, which is $R(t)$, changes sign. Set $R(t) = 0$: $$12(0.93)^t = 15(0.77)^t$$ Divide both sides by 15: $$\frac{12}{15}(0.93)^t = (0.77)^t$$ Simplify fraction: $$0.8(0.93)^t = (0.77)^t$$ Rewrite as: $$0.8 = \frac{(0.77)^t}{(0.93)^t} = \left(\frac{0.77}{0.93}\right)^t$$ Take natural logarithm: $$\ln(0.8) = t \ln\left(\frac{0.77}{0.93}\right)$$ Solve for $t$: $$t = \frac{\ln(0.8)}{\ln(0.77/0.93)} \approx \frac{-0.2231}{-0.1906} \approx 1.17$$ 5. **Evaluate $W(t)$ at critical point and endpoints:** Calculate $W(t) = \int_0^t R(x) dx = \int_0^t [12(0.93)^x - 15(0.77)^x] dx$ Use formula for integral of $a^x$: $$\int a^x dx = \frac{a^x}{\ln(a)} + C$$ So, $$W(t) = \left[ \frac{12(0.93)^x}{\ln(0.93)} - \frac{15(0.77)^x}{\ln(0.77)} \right]_0^t$$ Calculate each term: At $t=0$: $$W(0) = 0$$ At $t=1.17$: $$W(1.17) = \frac{12(0.93)^{1.17} - 12}{\ln(0.93)} - \frac{15(0.77)^{1.17} - 15}{\ln(0.77)}$$ Numerical values: $(0.93)^{1.17} \approx 0.917$ $(0.77)^{1.17} \approx 0.743$ Calculate: $$\frac{12(0.917) - 12}{\ln(0.93)} = \frac{11.004 - 12}{-0.07257} = \frac{-0.996}{-0.07257} \approx 13.72$$ $$\frac{15(0.743) - 15}{\ln(0.77)} = \frac{11.145 - 15}{-0.26136} = \frac{-3.855}{-0.26136} \approx 14.75$$ So, $$W(1.17) = 13.72 - 14.75 = -1.03$$ At $t=10$: $(0.93)^{10} \approx 0.484$ $(0.77)^{10} \approx 0.073$ Calculate: $$\frac{12(0.484) - 12}{\ln(0.93)} = \frac{5.81 - 12}{-0.07257} = \frac{-6.19}{-0.07257} \approx 85.33$$ $$\frac{15(0.073) - 15}{\ln(0.77)} = \frac{1.10 - 15}{-0.26136} = \frac{-13.90}{-0.26136} \approx 53.18$$ So, $$W(10) = 85.33 - 53.18 = 32.15$$ 6. **Conclusion:** The minimum water amount in the tank between $t=0$ and $t=10$ is approximately $-1.03$ units at $t \approx 1.17$ hours. This negative value indicates the tank is losing water overall at that time.