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). c. Sketch the graph labeling key points. 2. **Formula and Rules:** - To find maxima and minima, find critical points by setting the first derivative $$B'(t)$$ to zero. - To find where the battery discharges most rapidly, find where the rate of decrease $$B'(t)$$ is minimum (most negative), or equivalently where the second derivative $$B''(t)$$ is zero and changes sign. 3. **Step a: Find $$B'(t)$$** Given $$B(t) = 100\left(1 - \frac{t}{10}e^{-0.2t}\right) = 100 - 10te^{-0.2t}$$ Differentiate: $$B'(t) = -10 \frac{d}{dt}\left(te^{-0.2t}\right)$$ Use product rule: $$\frac{d}{dt}(te^{-0.2t}) = e^{-0.2t} + t \cdot (-0.2)e^{-0.2t} = e^{-0.2t}(1 - 0.2t)$$ So, $$B'(t) = -10 e^{-0.2t}(1 - 0.2t)$$ 4. **Find critical points by setting $$B'(t) = 0$$:** $$-10 e^{-0.2t}(1 - 0.2t) = 0$$ Since $$e^{-0.2t} \neq 0$$ for all $$t$$, solve: $$1 - 0.2t = 0 \implies t = 5$$ 5. **Evaluate $$B(t)$$ at critical point 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. **Determine max and min:** - $$B(0) = 100$$ (maximum) - $$B(5) \approx 81.605$$ (minimum) - $$B(10) \approx 86.47$$ 7. **Step b: Find when battery discharges most rapidly:** The rate of decrease is greatest when $$B'(t)$$ is minimum (most negative). Find $$B''(t)$$: $$B'(t) = -10 e^{-0.2t}(1 - 0.2t)$$ Differentiate using product rule: $$B''(t) = -10 \frac{d}{dt}\left(e^{-0.2t}(1 - 0.2t)\right)$$ $$= -10 \left( -0.2 e^{-0.2t}(1 - 0.2t) + e^{-0.2t}(-0.2) \right)$$ $$= -10 e^{-0.2t} \left(-0.2(1 - 0.2t) - 0.2 \right)$$ Simplify inside brackets: $$-0.2(1 - 0.2t) - 0.2 = -0.2 + 0.04t - 0.2 = 0.04t - 0.4$$ So, $$B''(t) = -10 e^{-0.2t} (0.04t - 0.4)$$ 8. **Set $$B''(t) = 0$$ to find inflection points:** $$-10 e^{-0.2t} (0.04t - 0.4) = 0$$ $$0.04t - 0.4 = 0 \implies t = 10$$ 9. **Analyze $$B'(t)$$ at $$t=10$$:** $$B'(10) = -10 e^{-2} (1 - 2) = -10 e^{-2} (-1) = 10 e^{-2} \approx 1.353 > 0$$ Since $$B'(t)$$ changes from negative to positive around $$t=10$$, the steepest descent (most negative $$B'(t)$$) occurs before $$t=10$$. Check $$B'(t)$$ at $$t=5$$: $$B'(5) = -10 e^{-1} (1 - 1) = 0$$ Check $$B'(t)$$ at $$t=0$$: $$B'(0) = -10 e^{0} (1 - 0) = -10 < 0$$ Since $$B'(t)$$ goes from -10 at 0 to 0 at 5 and then positive at 10, the most rapid decrease is at $$t=0$$. 10. **Step c: Sketch and label key points:** - Maximum battery level: $$B(0) = 100$$ at $$t=0$$ - Minimum battery level: $$B(5) \approx 81.605$$ at $$t=5$$ - Point of steepest descent: at $$t=0$$ where $$B'(0) = -10$$ (maximum negative slope) **Final answers:** - 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.