The Acme Class Ring Company designs and sells two types of rings: the VIP and the SST. They can produce up to 24 rings each day using up to 60 total man-hours of labor. It takes 3 man-hours to make one VIP ring, versus 2 man-hours to make one SST ring. How many of each type of ring should be made daily to maximize the company's profit, if the profit on a VIP ring is \( \$ 40 \) and on an SST ring is \( \$ 30 \) ? A. 14 VIP and 14 SST B. 12 VIP and 12 SST C. 10 VIP and 14 SST D. 14 VIP and 10 SST

To solve this problem, we need to set up a linear programming model to maximize the profit from producing VIP and SST rings, given the constraints on production and labor hours. ### Step 1: Define the Variables Let: - \( x \) = number of VIP rings produced - \( y \) = number of SST rings produced ### Step 2: Define the Objective Function The profit from producing the rings can be expressed as: \[ P = 40x + 30y \] We want to maximize \( P \). ### Step 3: Define the Constraints 1. **Production Constraint**: The total number of rings produced cannot exceed 24. \[ x + y \leq 24 \] 2. **Labor Constraint**: The total man-hours used cannot exceed 60. \[ 3x + 2y \leq 60 \] 3. **Non-negativity Constraints**: The number of rings produced cannot be negative. \[ x \geq 0, \quad y \geq 0 \] ### Step 4: Solve the System of Inequalities We will solve the inequalities to find the feasible region and then evaluate the objective function at the vertices of this region. 1. From the first constraint: \[ y \leq 24 - x \] 2. From the second constraint: \[ 2y \leq 60 - 3x \implies y \leq 30 - \frac{3}{2}x \] ### Step 5: Find the Intersection Points To find the vertices of the feasible region, we need to find the intersection of the lines defined by the constraints. 1. Set \( x + y = 24 \) and \( 3x + 2y = 60 \): - From \( y = 24 - x \), substitute into the second equation: \[ 3x + 2(24 - x) = 60 \] \[ 3x + 48 - 2x = 60 \implies x = 12 \] \[ y = 24 - 12 = 12 \] So one vertex is \( (12, 12) \). 2. Set \( x + y = 24 \) and \( y = 0 \): \[ x + 0 = 24 \implies x = 24 \] So another vertex is \( (24, 0) \). 3. Set \( 3x + 2y = 60 \) and \( y = 0 \): \[ 3x + 0 = 60 \implies x = 20 \] So another vertex is \( (20, 0) \). 4. Set \( 3x + 2y = 60 \) and \( x = 0 \): \[ 0 + 2y = 60 \implies y = 30 \] So another vertex is \( (0, 30) \). ### Step 6: Evaluate the Objective Function at Each Vertex 1. At \( (12, 12) \): \[ P = 40(12) + 30(12) = 480 + 360 = 840 \] 2. At \( (24, 0) \): \[ P = 40(24) + 30(0) = 960 + 0 = 960 \] 3. At \( (20, 0) \): \[ P = 40(20) + 30(0) = 800 + 0 = 800 \] 4. At \( (0, 30) \): \[ P = 40(0) + 30(30) = 0 + 900 = 900 \] ### Step 7: Determine the Maximum Profit The maximum profit occurs at \( (24, 0) \) with a profit of \( 960 \). However, this violates the production constraint of 24 rings. Now, we need to check the feasible combinations from the options provided: - A. \( (14, 14) \) - B. \( (12, 12) \) - C. \( (10, 14) \) - D. \( (14, 10) \) ### Step 8: Check Each Option 1. **Option A: \( (14, 14) \)** - Total rings: \( 14 + 14 = 28 \) (exceeds 24) 2. **Option B: \( (12, 12) \)** - Total rings: \( 12 + 12 = 24 \) - Labor: \( 3(12) + 2(12) = 36 + 24 = 60 \) (valid) - Profit: \( 40(12) + 30(12) = 840 \) 3. **Option C: \( (10, 14) \)** - Total rings: \( 10 + 14 = 24 \) - Labor: \( 3(10) + 2(14) = 30 + 28 = 58 \) (valid) - Profit: \( 40(10) + 30(14) = 400 + 420 = 820 \) 4. **Option D: \( (14, 10) \)** - Total rings: \( 14 + 10 = 24 \) - Labor: \( 3(14) + 2(10) = 42 + 20 = 62 \) (exceeds 60) ### Conclusion The best option that maximizes profit while satisfying all constraints is **Option B: 12 VIP and 12 SST**.