Let's solve the problem **step by step**: --- ### **Given:** The cumulative number of components assembled by the student is: \[ Q(t) = 1 + 10t - 0.5t^2 \] where \( Q(t) \) = number of components, \( t \) = hours. --- ### **(a) When does the number of components assembled by the student increase most rapidly?** This is asking for the **time when the rate of assembly is at its maximum**, i.e., when the derivative \( Q'(t) \) is maximized. The rate is the **first derivative**: \[ Q'(t) = \frac{dQ}{dt} = 10 - t \] To find the maximum, we check if this rate has a maximum. Since \( Q'(t) \) is a straight-line (decreasing linearly), it is **maximum at the smallest value of \( t \)** in the relevant interval (i.e., at the start). But let's check the **second derivative** (for concavity check): \[ Q''(t) = \frac{d^2Q}{dt^2} = -1 \] This is **always negative**, so \( Q'(t) \) is **decreasing**. So, the rate is highest at \( t = 0 \) (the start), but the **question wants the time when the number of components assembled increases most rapidly**. This is generally interpreted as the **maximum rate**, which happens at \( t = 0 \). **But** let's check the context: If we want the time when the assembly rate is maximum (i.e., the rate starts to decrease after that), it's at \( t = 0 \). #### **Answer (a):** \[ \boxed{0.0\ \text{hours}} \] --- ### **(b) How many components were assembled at that time?** Plug \( t = 0 \) into \( Q(t) \): \[ Q(0) = 1 + 10 \times 0 - 0.5 \times (0)^2 = 1 \] #### **Answer (b):** \[ \boxed{1.0\ \text{component}} \] --- ### **(c) What was the student’s rate of assembling at that time?** This is \( Q'(0) \): \[ Q'(t) = 10 - t \implies Q'(0) = 10 - 0 = 10 \] #### **Answer (c):** \[ \boxed{10.0\ \text{components/hour}} \] --- ### **Summary Table:** | Part | Answer | |------|----------------------------------| | (a) | \( 0.0 \) hours | | (b) | \( 1.0 \) component | | (c) | \( 10.0 \) components per hour | --- **If you need the rest of the question (about employer’s intervention), let me know!**

Let's analyze the problem step-by-step with detailed calculations and explanations. --- ### **Given:** The cumulative number of components assembled by the student is modeled by the function: \[ Q(t) = 1 + 10t - 0.5t^2 \] where: - \( Q(t) \) = number of components assembled at time \( t \) (hours), - \( t \) = time in hours. --- ### **Part (a): When does the number of components assembled increase most rapidly?** **Objective:** Find the time when the **rate of assembly** is at its **maximum**. --- ### **Step 1: Find the rate of change of components** The rate of assembly at time \( t \) is given by the first derivative: \[ Q'(t) = \frac{dQ}{dt} = \frac{d}{dt} (1 + 10t - 0.5t^2) = 0 + 10 - t = 10 - t \] This tells us that the **rate decreases linearly** with \( t \). --- ### **Step 2: Find when the rate is maximum** Since \( Q'(t) = 10 - t \) is a **linear decreasing function**, its maximum occurs at the **smallest** value of \( t \). Assuming \( t \geq 0 \), the maximum rate occurs at: \[ t = 0 \] --- ### **Step 3: Confirm that the rate decreases after \( t = 0 \)** The second derivative: \[ Q''(t) = \frac{d^2Q}{dt^2} = -1 \] which is **negative**, indicating the function \( Q'(t) \) is **concave downward** — the rate of increase **starts at a maximum at \( t=0 \)** and then decreases. --- ### **Final answer for (a):** \[ \boxed{\text{The number of components increases most rapidly at } t=0 \text{ hours}} \] --- ### **Part (b): How many components were assembled at that time?** Calculate \( Q(t) \) at \( t=0 \): \[ Q(0) = 1 + 10 \times 0 - 0.5 \times 0^2 = 1 + 0 - 0 = 1 \] --- ### **Final answer for (b):** \[ \boxed{\text{Number of components assembled at } t=0: \quad 1} \] --- ### **Part (c): What was the student’s rate of assembling at that time?** Calculate \( Q'(0) \): \[ Q'(0) = 10 - 0 = 10 \] which means: \[ \boxed{\text{Rate at } t=0: \quad 10 \text{ components/hour}} \] --- ### **Summary of Results:** | Part | Explanation | Final Answer | |---------|------------------------|------------------------------| | (a) | Time when assembly rate is maximum | **0 hours** | | (b) | Components assembled at that time | **1 component** | | (c) | Rate of assembly at that time | **10 components/hour** | --- **If you have further questions or need additional details, feel free to ask!**

Based on the detailed analysis of the problem, here are the multiple-choice options for each part along with the correct answers: --- ### **Part (a): When does the number of components assembled increase most rapidly?** **Options:** A) 0 hours B) 5 hours C) 10 hours D) 20 hours **Correct Answer:** **A) 0 hours** --- ### **Part (b): How many components were assembled at that time?** **Options:** A) 0 components B) 1 component C) 10 components D) 20 components **Correct Answer:** **B) 1 component** --- ### **Part (c): What was the student’s rate of assembling at that time?** **Options:** A) 0 components/hour B) 5 components/hour C) 10 components/hour D) 20 components/hour **Correct Answer:** **C) 10 components/hour** --- **Summary:** | Part | Correct Choice | Explanation | |---------|------------------------|------------------------------| | (a) | **A)** 0 hours | Max rate at start, \( t=0 \) | | (b) | **B)** 1 component | \( Q(0) = 1 \) | | (c) | **C)** 10 components/hour | \( Q'(0) = 10 \) | --- Let me know if you'd like me to generate options for additional parts or clarify anything further!

Certainly! Here are the **final answers** to each part of the problem, along with detailed reasoning: --- ### **Part (a): When does the number of components assembled increase most rapidly?** **Analysis:** - The rate of assembly is given by \( Q'(t) = 10 - t \). - Since this is a decreasing linear function, its maximum occurs at the smallest value of \( t \), which is at **\( t=0 \)**. - The second derivative \( Q''(t) = -1 < 0 \) confirms the rate decreases after \( t=0 \). **Final Answer:** **At \( t = 0 \) hours.** --- ### **Part (b): How many components were assembled at that time?** **Calculation:** \[ Q(0) = 1 + 10 \times 0 - 0.5 \times 0^2 = 1 \] **Final Answer:** **1 component.** --- ### **Part (c): What was the student’s rate of assembling at that time?** **Calculation:** \[ Q'(0) = 10 - 0 = 10 \] **Final Answer:** **10 components/hour.** --- ### **Summary of Final Answers:** | Part | Final Answer | |---------|------------------------------| | (a) | **0 hours** | | (b) | **1 component** | | (c) | **10 components/hour** | --- If you need these in multiple-choice format or further clarification, let me know!

Certainly! Here are the final answers: **(a)** 0 hours **(b)** 1 component **(c)** 10 components per hour