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's define the variables: - Let \( x \) be the number of Boomers produced. - Let \( y \) be the number of XL2000s produced. ### Income Function The income generated from selling the MP3 players can be expressed as: \[ \text{Income} = 14x + 1y \] ### 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**: \[ 32x + 16y \leq 4448 \implies 2x + y \leq 278 \quad \text{(dividing by 16)} \] 2. **Electronics constraint**: \[ 2x + 6y \leq 486 \implies x + 3y \leq 243 \quad \text{(dividing by 2)} \] 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. **Intersection of \( 2x + y = 278 \) and \( x + 3y = 243 \)**: - From \( 2x + y = 278 \), we can express \( y \): \[ y = 278 - 2x \] - Substitute into \( x + 3y = 243 \): \[ x + 3(278 - 2x) = 243 \implies x + 834 - 6x = 243 \implies -5x = -591 \implies x = 118.2 \] - Substitute \( x \) back to find \( y \): \[ y = 278 - 2(118.2) = 41.6 \] 2. **Intersection of \( 2x + y = 278 \) and \( 47x + 117y = 9525 \)**: - From \( 2x + y = 278 \): \[ y = 278 - 2x \] - Substitute into \( 47x + 117y = 9525 \): \[ 47x + 117(278 - 2x) = 9525 \implies 47x + 32526 - 234x = 9525 \implies -187x = -23001 \implies x = 122.9 \] - Substitute \( x \) back to find \( y \): \[ y = 278 - 2(122.9) = 32.2 \] 3. **Intersection of \( x + 3y = 243 \) and \( 47x + 117y = 9525 \)**: - From \( x + 3y = 243 \): \[ x = 243 - 3y \] - Substitute into \( 47x + 117y = 9525 \): \[ 47(243 - 3y) + 117y = 9525 \implies 11421 - 141y + 117y = 9525 \implies -24y = -1896 \implies y = 79 \] - Substitute \( y \) back to find \( x \): \[ x = 243 - 3(79) = 6 \] ### Evaluating the Vertices Now we have the vertices of the feasible region: 1. \( (118.2, 41.6) \) 2. \( (122.9, 32.2) \) 3. \( (6, 79) \) ### Calculating Income at Each Vertex 1. For \( (118.2, 41.6) \): \[ \text{Income} = 14(118.2) + 1(41.6) = 1654.8 + 41.6 = 1696.4 \] 2. For \( (122.9, 32.2) \): \[ \text{Income} = 14(122.9) + 1(32.2) = 1719.6 + 32.2 = 1751.8 \] 3. For \( (6, 79) \): \[ \text{Income} = 14(6) + 1(79) = 84 + 79 = 163 \] ### Conclusion The maximum income occurs at \( (122.9, 32.2) \). Thus, Liana should sell approximately: - **Boomers**: 123 (rounding to the nearest whole number) - **XL2000s**: 32 (rounding to the nearest whole number) **Best income**: \( \$ 1751.8 \)