The pressure of the oil in a reservoir tends to drop with time. By taking sample pressure readings for a particular oil reservoir, petroleum engineers have found that the change in pressure is given by the function \( P(t)=t^{3}-8 t^{2}+16 t \), where \( t \) is time in years from the date of the first reading. (a) Find the following: \( P(0), P(2), P(3), P(5) \) (b) Graph \( P(t) \) (c) Over what periods is the graph increasing and decreasing. A. Increases from 0 to \( 4 / 3 \) years and again after the 4 th year. Decreases from \( 4 / 3 \) to 4 years. B. Decreases from 0 to \( 4 / 3 \) years and again after the 5 th year. Increases from \( 4 / 3 \) to 5 years. C. Increases from 0 to \( 4 / 3 \) years and again after the 5 th year. Decreases from \( 4 / 3 \) to 5 years. D. Decreases from 0 to \( 4 / 3 \) years and again after the 4 th year. Increases from \( 4 / 3 \) to 4 years.

To determine over which periods the pressure \( P(t) = t^{3} - 8t^{2} + 16t \) is increasing or decreasing, we'll analyze the first derivative of \( P(t) \), denoted as \( P'(t) \). This derivative will help us identify the critical points and the behavior of the function. ### Step 1: Find the First Derivative \( P'(t) \) \[ P(t) = t^{3} - 8t^{2} + 16t \] Differentiate \( P(t) \) with respect to \( t \): \[ P'(t) = \frac{d}{dt}(t^{3}) - \frac{d}{dt}(8t^{2}) + \frac{d}{dt}(16t) = 3t^{2} - 16t + 16 \] ### Step 2: Find Critical Points by Solving \( P'(t) = 0 \) Set the first derivative equal to zero and solve for \( t \): \[ 3t^{2} - 16t + 16 = 0 \] Use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Where \( a = 3 \), \( b = -16 \), and \( c = 16 \): \[ t = \frac{16 \pm \sqrt{(-16)^{2} - 4 \times 3 \times 16}}{2 \times 3} = \frac{16 \pm \sqrt{256 - 192}}{6} = \frac{16 \pm \sqrt{64}}{6} \] \[ t = \frac{16 \pm 8}{6} \] So, the critical points are: \[ t = \frac{24}{6} = 4 \quad \text{and} \quad t = \frac{8}{6} = \frac{4}{3} \] ### Step 3: Determine Intervals of Increase and Decrease We'll test the sign of \( P'(t) \) in the intervals defined by the critical points: 1. **Interval \( (0, \frac{4}{3}) \):** Choose \( t = 1 \): \[ P'(1) = 3(1)^2 - 16(1) + 16 = 3 - 16 + 16 = 3 > 0 \] **Conclusion:** \( P(t) \) is **increasing** in this interval. 2. **Interval \( \left(\frac{4}{3}, 4\right) \):** Choose \( t = 2 \): \[ P'(2) = 3(2)^2 - 16(2) + 16 = 12 - 32 + 16 = -4 < 0 \] **Conclusion:** \( P(t) \) is **decreasing** in this interval. 3. **Interval \( (4, \infty) \):** Choose \( t = 5 \): \[ P'(5) = 3(5)^2 - 16(5) + 16 = 75 - 80 + 16 = 11 > 0 \] **Conclusion:** \( P(t) \) is **increasing** in this interval. ### Step 4: Summarize the Results - **Increasing:** From \( 0 \) to \( \frac{4}{3} \) years and again after the **4th** year. - **Decreasing:** From \( \frac{4}{3} \) to **4 years**. ### **Final Answer:** **Option A** is correct. > **A.** Increases from 0 to \( \frac{4}{3} \) years and again after the 4th year. Decreases from \( \frac{4}{3} \) to 4 years.