1. **Problem statement:** We have a battery level function $$B(t) = 100\left(1 - \frac{t}{10} e^{-0.2t}\right)$$ for $$t \in [0,10]$$ hours. We need to find: a. The maximum and minimum battery levels during the first 10 hours. b. The time when the battery is discharging most rapidly (greatest rate of decrease). 2. **Formula and rules:** To find maxima and minima, we use the first derivative test: - Find $$B'(t)$$. - Solve $$B'(t) = 0$$ to find critical points. - Use the second derivative or test values to classify maxima or minima. For the rate of decrease, find when $$B'(t)$$ is minimum (most negative). 3. **Find the first derivative $$B'(t)$$:** $$B(t) = 100\left(1 - \frac{t}{10} e^{-0.2t}\right) = 100 - 10 t e^{-0.2t}$$ Differentiate: $$B'(t) = -10 \frac{d}{dt} \left(t e^{-0.2t}\right)$$ Use product rule: $$\frac{d}{dt} (t e^{-0.2t}) = e^{-0.2t} + t \cdot (-0.2) e^{-0.2t} = e^{-0.2t} (1 - 0.2 t)$$ So: $$B'(t) = -10 e^{-0.2t} (1 - 0.2 t)$$ 4. **Find critical points by solving $$B'(t) = 0$$:** $$-10 e^{-0.2t} (1 - 0.2 t) = 0$$ Since $$e^{-0.2t} \neq 0$$ for all $$t$$, solve: $$1 - 0.2 t = 0 \implies t = 5$$ 5. **Evaluate $$B(t)$$ at critical points and endpoints:** - At $$t=0$$: $$B(0) = 100\left(1 - 0\right) = 100$$ - At $$t=5$$: $$B(5) = 100\left(1 - \frac{5}{10} e^{-0.2 \times 5}\right) = 100\left(1 - 0.5 e^{-1}\right)$$ $$e^{-1} \approx 0.3679$$ $$B(5) \approx 100 (1 - 0.5 \times 0.3679) = 100 (1 - 0.18395) = 81.605$$ - At $$t=10$$: $$B(10) = 100\left(1 - 1 \times e^{-2}\right) = 100 (1 - e^{-2})$$ $$e^{-2} \approx 0.1353$$ $$B(10) \approx 100 (1 - 0.1353) = 86.47$$ 6. **Classify extrema:** - At $$t=0$$, $$B(0) = 100$$ (endpoint) - At $$t=5$$, $$B(5) \approx 81.605$$ (critical point) - At $$t=10$$, $$B(10) \approx 86.47$$ (endpoint) Since $$B(5) < B(10) < B(0)$$, the minimum battery level is at $$t=5$$, maximum at $$t=0$$. 7. **Find when battery discharges most rapidly:** The rate of decrease is greatest when $$B'(t)$$ is minimum (most negative). Recall: $$B'(t) = -10 e^{-0.2t} (1 - 0.2 t)$$ Find $$B''(t)$$ to locate extrema of $$B'(t)$$: $$B''(t) = -10 \frac{d}{dt} \left(e^{-0.2t} (1 - 0.2 t)\right)$$ Use product rule: $$\frac{d}{dt} \left(e^{-0.2t} (1 - 0.2 t)\right) = e^{-0.2t} \cdot (-0.2)(1 - 0.2 t) + e^{-0.2t} \cdot (-0.2) = e^{-0.2t} (-0.2 + 0.04 t - 0.2) = e^{-0.2t} (0.04 t - 0.4)$$ So: $$B''(t) = -10 e^{-0.2t} (0.04 t - 0.4)$$ Set $$B''(t) = 0$$: $$0.04 t - 0.4 = 0 \implies t = 10$$ 8. **Evaluate $$B'(t)$$ at $$t=10$$:** $$B'(10) = -10 e^{-2} (1 - 2) = -10 e^{-2} (-1) = 10 e^{-2} \approx 10 \times 0.1353 = 1.353 > 0$$ Since $$B'(t)$$ changes from negative to positive at $$t=5$$ and is positive at $$t=10$$, the minimum of $$B'(t)$$ (most negative) occurs at $$t=0$$ or near $$t=5$$. Check $$B'(0)$$: $$B'(0) = -10 e^{0} (1 - 0) = -10 < 0$$ Check $$B'(5)$$: $$B'(5) = -10 e^{-1} (1 - 1) = 0$$ Since $$B'(t)$$ goes from -10 at 0 to 0 at 5, the most rapid decrease is at $$t=0$$. 9. **Summary:** - Maximum battery level: $$100\%$$ at $$t=0$$ hours. - Minimum battery level: approximately $$81.6\%$$ at $$t=5$$ hours. - Battery discharges most rapidly at $$t=0$$ hours. 10. **Desmos function:** $$y = 100\left(1 - \frac{t}{10} e^{-0.2 t}\right)$$