Let's break down and solve both questions step-by-step: --- **11) Isoquant for Three Practice Questions** **Restating the Problem:** - Practice questions for Econ 100 can be created using either: - \( a \) hours of TA time *or* - \( b \) hours of instructor time, - or any convex combination (linear combination) of the two. **Let:** - Let \( Q \) = number of practice questions (here, \( Q = 3 \)) - Let \( T \) = TA hours - Let \( I \) = Instructor hours For **one** question: - \( a \) TA hours *or* \( b \) instructor hours For **three** questions: - \( 3a \) TA hours *or* \( 3b \) instructor hours The **isoquant** is the locus of all combinations of TA and instructor hours that yield 3 questions: \[ \frac{T}{3a} + \frac{I}{3b} = 1 \] Or, rearranged: \[ T/a + I/b = 3 \] Or, \[ T = 3a - (a/b)I \] **Graph:** - X-axis: Instructor hours (\( I \)) - Y-axis: TA hours (\( T \)) - The isoquant is a straight line between (0, 3a) and (3b, 0). **Technical Rate of Substitution (TRS):** - TRS = slope of the isoquant = Marginal rate at which TA hours can be substituted for instructor hours while keeping output constant - From the equation above, slope = \(-a/b\) - So, TRS = \( a/b \) (in absolute value). --- **12) New TA with \( a' < a \)** **Restating the Problem:** - New TA requires \( a' < a \) hours per question (more efficient). - Instructor requirement unchanged. - Isoquant for 3 questions is now: \[ T/a' + I/b = 3 \] Or, \[ T = 3a' - (a'/b)I \] - Endpoints: (0, 3a') and (3b, 0) **Technical Rate of Substitution (TRS):** - New TRS = \( a'/b \) (in absolute value) - Since \( a' < a \), the TRS has **decreased**. **Relative Value Change:** - TA hours are now more "efficient" relative to instructor hours (since \( a' < a \)), so the value (cost in terms of instructor hours) of 1 TA hour has **decreased**. --- ### **Final Answers** #### **11)** - **Isoquant equation:** \( T/a + I/b = 3 \) - **Graph:** A straight line from (0, 3a) to (3b, 0). Title: "Graph D2" - **TRS:** \( a/b \) (in absolute value) #### **12)** - **New isoquant equation:** \( T/a' + I/b = 3 \) - **Graph:** A straight line from (0, 3a') to (3b, 0) on "Graph D2" - **New TRS:** \( a'/b \) (in absolute value) - **Change:** The relative value of TA hours compared to instructor hours has **decreased** (TA is now more efficient). --- **Let me know if you'd like to see a sketch of the graphs!**