1. **Problem statement:** We want to find the minimum number of pastries the café has at any time $t$ in the interval $0 \leq t \leq 6$ hours. Initially, there are 100 pastries already made. The rate of pastry production $R(t)$ is given by a piecewise linear graph: - From $t=0$ to $t=2$, $R(t)$ increases from 10 to 70 pastries/hour. - From $t=2$ to $t=6$, $R(t)$ decreases from 70 to 40 pastries/hour. 2. **Understanding the problem:** The number of pastries at time $t$, denoted $P(t)$, is the initial pastries plus the integral of the rate $R(t)$ from 0 to $t$: $$P(t) = 100 + \int_0^t R(x) \, dx$$ We want to find the minimum value of $P(t)$ for $0 \leq t \leq 6$. 3. **Step 1: Express $R(t)$ as piecewise linear functions.** From the graph: - For $0 \leq t \leq 2$, $R(t)$ goes from 10 to 70 linearly: $$R(t) = 10 + \frac{70 - 10}{2 - 0} (t - 0) = 10 + 30t$$ - For $2 < t \leq 6$, $R(t)$ goes from 70 to 40 linearly: $$R(t) = 70 + \frac{40 - 70}{6 - 2} (t - 2) = 70 - \frac{30}{4}(t - 2) = 70 - 7.5(t - 2)$$ 4. **Step 2: Find $P(t)$ by integrating $R(t)$ piecewise.** - For $0 \leq t \leq 2$: $$P(t) = 100 + \int_0^t (10 + 30x) dx = 100 + \left[10x + 15x^2\right]_0^t = 100 + 10t + 15t^2$$ - For $2 < t \leq 6$: First find $P(2)$: $$P(2) = 100 + 10(2) + 15(2)^2 = 100 + 20 + 60 = 180$$ Then integrate $R(t)$ from 2 to $t$: $$\int_2^t (70 - 7.5(x - 2)) dx = \left[70x - 7.5 \frac{(x - 2)^2}{2}\right]_2^t = 70(t - 2) - 3.75(t - 2)^2$$ So for $2 < t \leq 6$: $$P(t) = P(2) + 70(t - 2) - 3.75(t - 2)^2 = 180 + 70(t - 2) - 3.75(t - 2)^2$$ 5. **Step 3: Find the minimum of $P(t)$ on $[0,6]$.** Since $P(t)$ is continuous and differentiable piecewise, check critical points by setting derivative $P'(t) = R(t)$ to zero or check endpoints. - For $0 \leq t \leq 2$: $$P'(t) = R(t) = 10 + 30t$$ Set $P'(t) = 0$: $$10 + 30t = 0 \Rightarrow t = -\frac{1}{3}$$ Not in $[0,2]$, so no critical point here. - For $2 < t \leq 6$: $$P'(t) = R(t) = 70 - 7.5(t - 2)$$ Set $P'(t) = 0$: $$70 - 7.5(t - 2) = 0 \Rightarrow 7.5(t - 2) = 70 \Rightarrow t - 2 = \frac{70}{7.5} = \frac{280}{30} = \frac{28}{3} \approx 9.33$$ Not in $[2,6]$, so no critical point here. 6. **Step 4: Evaluate $P(t)$ at endpoints $t=0$ and $t=6$ to find minimum.** - At $t=0$: $$P(0) = 100 + 10(0) + 15(0)^2 = 100$$ - At $t=6$: $$P(6) = 180 + 70(6 - 2) - 3.75(6 - 2)^2 = 180 + 70(4) - 3.75(4)^2 = 180 + 280 - 3.75(16) = 180 + 280 - 60 = 400$$ 7. **Step 5: Check $P(t)$ at $t=2$ (junction point):** $$P(2) = 180$$ 8. **Step 6: Check $P(t)$ at $t=0$ is minimum 100 pastries.** Since $P(t)$ is increasing on $[0,2]$ and then continues increasing on $[2,6]$, the minimum number of pastries is at $t=0$ with 100 pastries. **Final answer:** The minimum number of pastries the café has at any time in $0 \leq t \leq 6$ is **100 pastries** at the start of the day.