P(x)=-0.004x^3+.025x^2+0.55x+8 1 . Use the model to predict the hourly number of items produced in a 4 hour shift 2. According to the model, how many hours makes the most productive shift m? That is, at how many hours is the hotly production at a maximum? 3. At the shift length found in part B, how many items are produced per hour?

1. Predict the hourly number of items produced in a 4-hour shift:
P(4) = -0.004(4)^3 + 0.025(4)^2 + 0.55(4) + 8
(4)^3 = 64
(4)^2 = 16
P(4) = -0.004(64) + 0.025(16) + 0.55(4) + 8
P(4) = -0.256 + 0.4 + 2.2 + 8
P(4) = 10.344
The hourly number of items produced in a 4-hour shift is approximately 10.344.

2. Determine the shift length that results in the maximum hourly production:
P'(x) = \frac{d}{dx} \left( -0.004x^3 + 0.025x^2 + 0.55x + 8 \right) = -0.012x^2 + 0.05x + 0.55
Solve the quadratic equation:
-0.012x^2 + 0.05x + 0.55 = 0
x = \frac{-0.05 \pm \sqrt{(0.05)^2 - 4(-0.012)(0.55)}}{2(-0.012)}
x = \frac{-0.05 \pm \sqrt{0.0025 + 0.0264}}{-0.024}
x = \frac{-0.05 \pm \sqrt{0.0289}}{-0.024}
x = \frac{-0.05 \pm 0.17}{-0.024}
x_1 = \frac{-0.05 + 0.17}{-0.024} = \frac{0.12}{-0.024} = -5
x_2 = \frac{-0.05 - 0.17}{-0.024} = \frac{-0.22}{-0.024} = 9.17
Since a shift cannot be negative hours, we disregard \( x_1 = -5 \). The maximum productivity occurs at \( x \approx 9.17 \) hours.

3. Calculate the hourly production at the most productive shift length:
P(9.17) = -0.004(9.17)^3 + 0.025(9.17)^2 + 0.55(9.17) + 8
(9.17)^3 \approx 770.9
(9.17)^2 \approx 84.1
P(9.17) = -0.004(770.9) + 0.025(84.1) + 0.55(9.17) + 8
P(9.17) = -3.084 + 2.1025 + 5.0435 + 8
P(9.17) \approx 12.061

Therefore, the hourly production at the most productive shift length (approximately 9.17 hours) is about 12.061 items per hour.