Let's go through each part step-by-step: **Given:** - The rate at which unprocessed gravel arrives: \( C(t) = 90 + 45 \cos \left( \frac{\pi t}{8} \right) \), where \( t \) is in hours, \( 0 \leq t \leq 8 \). - The plant processes gravel at a constant rate of 100 tons/hour. - At \( t = 0 \), the plant has 500 tons of unprocessed gravel. --- ### a) Find \( C(5) \) using correct units. Interpret your answer in the context of the problem. \[ C(5) = 90 + 45 \cos \left( \frac{\pi \times 5}{8} \right) \] Calculate \( \frac{\pi \times 5}{8} \): \[ \frac{5\pi}{8} \approx 1.9635 \text{ radians} \] \[ \cos(1.9635) \approx -0.3827 \] \[ C(5) = 90 + 45 \times (-0.3827) = 90 - 17.2215 = 72.7785 \] **Units:** tons/hour **Interpretation:** At \( t = 5 \) hours, the rate at which unprocessed gravel arrives at the plant is approximately **72.8 tons per hour**. --- ### b) Find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday. We want the integral of \( C(t) \) from \( t = 0 \) to \( t = 8 \): \[ \text{Total gravel} = \int_{0}^{8} C(t)\,dt = \int_{0}^{8} \left[ 90 + 45 \cos \left( \frac{\pi t}{8} \right) \right] dt \] Break into two parts: \[ = \int_{0}^{8} 90\,dt + \int_{0}^{8} 45 \cos \left( \frac{\pi t}{8} \right) dt \] First term: \[ \int_{0}^{8} 90\,dt = 90 \times 8 = 720 \] Second term: Let \( u = \frac{\pi t}{8} \implies dt = \frac{8}{\pi} du \), as \( t = 0 \rightarrow u = 0 \), \( t = 8 \rightarrow u = \pi \): \[ \int_{0}^{8} 45 \cos \left( \frac{\pi t}{8} \right) dt = 45 \int_{t=0}^{t=8} \cos \left( \frac{\pi t}{8} \right) dt \] \[ = 45 \int_{u=0}^{u=\pi} \cos(u) \cdot \frac{8}{\pi} du = 45 \cdot \frac{8}{\pi} \int_{u=0}^{u=\pi} \cos(u) du \] \[ = \frac{360}{\pi} [\sin(u)]_{0}^{\pi} \] \[ = \frac{360}{\pi} (\sin(\pi) - \sin(0)) = \frac{360}{\pi} (0 - 0) = 0 \] So, total gravel: \[ \boxed{720 \text{ tons}} \] **Interpretation:** **720 tons** of unprocessed gravel arrive at the plant during the 8-hour workday. --- ### c) Is the amount of unprocessed gravel at the plant increasing or decreasing at time \( t = 5 \) hours? The rate of change of unprocessed gravel is: \[ \text{Rate of change} = C(5) - \text{processing rate} = 72.78 - 100 = -27.22 \] **Interpretation:** Since this is negative, the amount of unprocessed gravel **is decreasing** at \( t = 5 \) hours. --- ### d) What is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? Justify your answer. Let \( G(t) \) be the amount of unprocessed gravel at time \( t \): \[ G(t) = \text{Initial} + \int_{0}^{t} \left[ C(x) - 100 \right] dx \] \[ = 500 + \int_{0}^{t} [90 + 45 \cos(\frac{\pi x}{8}) - 100 ] dx \] \[ = 500 + \int_{0}^{t} [-10 + 45 \cos(\frac{\pi x}{8}) ] dx \] \[ = 500 - 10t + 45 \int_{0}^{t} \cos(\frac{\pi x}{8}) dx \] Let’s compute the integral: Let \( u = \frac{\pi x}{8} \implies dx = \frac{8}{\pi} du \) \[ \int \cos(\frac{\pi x}{8}) dx = \frac{8}{\pi} \sin(\frac{\pi x}{8}) \] \[ 45 \int_{0}^{t} \cos(\frac{\pi x}{8}) dx = 45 \cdot \frac{8}{\pi} [\sin(\frac{\pi x}{8})]_{x=0}^{x=t} = \frac{360}{\pi} (\sin(\frac{\pi t}{8}) - 0) \] So, \[ G(t) = 500 - 10t + \frac{360}{\pi} \sin(\frac{\pi t}{8}) \] To find the maximum, set the derivative \( G'(t) \) to zero: \[ G'(t) = -10 + \frac{360}{\pi} \cos(\frac{\pi t}{8}) \cdot \frac{\pi}{8} = -10 + 45 \cos(\frac{\pi t}{8}) \] Set \( G'(t) = 0 \): \[ -10 + 45 \cos(\frac{\pi t}{8}) = 0 \] \[ \cos(\frac{\pi t}{8}) = \frac{2}{9} \] \[ \frac{\pi t}{8} = \arccos(\frac{2}{9}) \implies t = \frac{8}{\pi} \arccos(\frac{2}{9}) \] Compute \( t \): \[ \arccos(\frac{2}{9}) \approx 1.348 \text{ radians} \] \[ t = \frac{8}{\pi} \times 1.348 \approx 3.435 \text{ hours} \] Now, plug \( t \) back into \( G(t) \): \[ G(3.435) = 500 - 10(3.435) + \frac{360}{\pi} \sin\left(\frac{\pi \times 3.435}{8}\right) \] Calculate \( \frac{\pi \times 3.435}{8} \approx 1.348 \) radians \[ \sin(1.348) \approx 0.975 \] \[ \frac{360}{\pi} \times 0.975 \approx 111.7 \] \[ G(3.435) = 500 - 34.35 + 111.7 = 577.35 \] **Final Answer:** > The maximum amount of unprocessed gravel at the plant during the workday is approximately **577.4 tons**, which occurs at about **3.44 hours** after the workday begins. --- Let me know if you need any clarifications or further breakdowns!

1. At \( t = 5 \) hours, the rate of gravel arriving is approximately **72.8 tons/hour**. 2. The total amount of unprocessed gravel that arrives during the 8-hour day is **720 tons**. 3. At \( t = 5 \) hours, the amount of unprocessed gravel is **decreasing**. 4. The maximum amount of unprocessed gravel at the plant during the day is approximately **577.4 tons**, occurring around **3.44 hours** after the start.