Liana manages a store that sells MP3 players. Each Boomer takes 32 hours to create the plastic pieces, 2 hours to create the electronics, and 47 hours of labor to assemble the final product. Each XL2000 requires 16 hours to create the plastic pieces, 6 hours to create the electronics, and 117 hours of labor to assemble the final product. The factory has 4448 hours to create the plastic pieces, 486 hours to create the electronics, and 9525 hours of labor to assemble the final product each week. If each Boomer generates \( \$ 14 \) in income, and each XL2000 generates \( \$ 1 \), how many of each of the MP3 players should Liana have the store sell each week to make the most income? Boomer: XL2000: Best income:

To determine how many of each MP3 player Liana should sell to maximize income, we can set up a linear programming problem. Let: - \( x \) = number of Boomers produced - \( y \) = number of XL2000s produced ### Objective Function The income generated from selling the MP3 players can be expressed as: \[ \text{Income} = 14x + 1y \] We want to maximize this income. ### Constraints We have the following constraints based on the hours available for each type of work: 1. **Plastic pieces constraint**: \[ 32x + 16y \leq 4448 \] 2. **Electronics constraint**: \[ 2x + 6y \leq 486 \] 3. **Labor constraint**: \[ 47x + 117y \leq 9525 \] 4. **Non-negativity constraints**: \[ x \geq 0, \quad y \geq 0 \] ### Solving the Constraints Now, we will solve the constraints to find the feasible region. 1. **Plastic pieces constraint**: \[ 32x + 16y \leq 4448 \implies 2x + y \leq 278 \] 2. **Electronics constraint**: \[ 2x + 6y \leq 486 \implies x + 3y \leq 243 \] 3. **Labor constraint**: \[ 47x + 117y \leq 9525 \] ### Finding Intersection Points To find the feasible region, we need to find the intersection points of the constraints. 1. **From \( 2x + y = 278 \) and \( x + 3y = 243 \)**: - Solve for \( y \) in terms of \( x \): \[ y = 278 - 2x \] Substitute into the second equation: \[ x + 3(278 - 2x) = 243 \implies x + 834 - 6x = 243 \implies -5x = -591 \implies x = 118.2 \] Substitute back to find \( y \): \[ y = 278 - 2(118.2) = 41.6 \] 2. **From \( 2x + y = 278 \) and \( 47x + 117y = 9525 \)**: - Substitute \( y = 278 - 2x \) into the labor constraint: \[ 47x + 117(278 - 2x) = 9525 \implies 47x + 32526 - 234x = 9525 \implies -187x = -23001 \implies x = 122.9 \] Substitute back to find \( y \): \[ y = 278 - 2(122.9) = 32.2 \] 3. **From \( x + 3y = 243 \) and \( 47x + 117y = 9525 \)**: - Solve for \( y \): \[ y = \frac{243 - x}{3} \] Substitute into the labor constraint: \[ 47x + 117\left(\frac{243 - x}{3}\right) = 9525 \implies 47x + 39(243 - x) = 9525 \implies 47x + 9477 - 39x = 9525 \implies 8x = 48 \implies x = 6 \] Substitute back to find \( y \): \[ y = \frac{243 - 6}{3} = 79 \] ### Evaluating the Vertices Now we evaluate the income at the vertices of the feasible region: 1. \( (118, 41) \) \[ \text{Income} = 14(118) + 1(41) = 1652 + 41 = 1693 \] 2. \( (122, 32) \) \[ \text{Income} = 14(122) + 1(32) = 1708 + 32 = 1740 \] 3. \( (6, 79) \) \[ \text{Income} = 14(6) + 1(79) = 84 + 79 = 163 \] ### Conclusion The maximum income occurs at \( (122, 32) \): - **Boomers**: 122 - **XL2000s**: 32 - **Best income**: \( \$ 1740 \) Thus, Liana should sell **122 Boomers** and **32 XL2000s** to maximize her income, which would be **\$1740**.