Question

a company manufactures and sells two models of lamps, a1 and a2. the labor time involved in manufacturing model a1 is 20 minutes and 30 minutes for model a2. the mechanical (machine) work involved for manufacturing model a1 is 20 minutes and 10 minutes for model a2. the labor time available per month is 110 hours (6,600 minutes), and the machine is available for 70 hours (4,200 minutes) per month. due to a defect, 30 a1 lamps and 20 a2 lamps are not ready to be sold. if the profit per model is $15 and $10 for a1 and a2 respectively, determine the quantities of each lamp that should be manufactured to obtain the maximum profit. a. 150 units of model a1 and 120 units of model a2 b. 210 units of model a1 and 60 units of model a2 c. 30 units of model a1 and 20 units of model a2 d. 200 units of model a1 and 20 units of model a2

Explanation:

Step 1: Define Variables

Let \( x \) be the number of A1 lamps and \( y \) be the number of A2 lamps.

Step 2: Set Up Constraints

Labor Time Constraint:

Labor time for A1: 20 minutes, A2: 30 minutes. Total labor time available: 6600 minutes.
So, \( 20x + 30y \leq 6600 \) (simplify: \( 2x + 3y \leq 660 \))

Machine Time Constraint:

Machine time for A1: 20 minutes, A2: 10 minutes. Total machine time available: 4200 minutes.
So, \( 20x + 10y \leq 4200 \) (simplify: \( 2x + y \leq 420 \))

Defect Constraint:

Due to defect, \( x \geq 30 \) (since 30 A1 are defective, need to make at least 30 to have sellable ones) and \( y \geq 20 \) (20 A2 are defective, need to make at least 20 to have sellable ones).

Step 3: Objective Function (Profit)

Profit \( P = 15x + 10y \) (we need to maximize this).

Step 4: Check Each Option

Option A: \( x = 150, y = 120 \)

• Labor: \( 20(150) + 30(120) = 3000 + 3600 = 6600 \) (within 6600)
• Machine: \( 20(150) + 10(120) = 3000 + 1200 = 4200 \) (within 4200)
• Profit: \( 15(150) + 10(120) = 2250 + 1200 = 3450 \)

Option B: \( x = 210, y = 60 \)

• Labor: \( 20(210) + 30(60) = 4200 + 1800 = 6000 \) (within 6600)
• Machine: \( 20(210) + 10(60) = 4200 + 600 = 4800 \) (exceeds 4200, invalid)

Option C: \( x = 30, y = 20 \)

• Labor: \( 20(30) + 30(20) = 600 + 600 = 1200 \) (within 6600)
• Machine: \( 20(30) + 10(20) = 600 + 200 = 800 \) (within 4200)
• Profit: \( 15(30) + 10(20) = 450 + 200 = 650 \) (low profit)

Option D: \( x = 200, y = 20 \)

• Labor: \( 20(200) + 30(20) = 4000 + 600 = 4600 \) (within 6600)
• Machine: \( 20(200) + 10(20) = 4000 + 200 = 4200 \) (within 4200)
• Profit: \( 15(200) + 10(20) = 3000 + 200 = 3200 \) (less than Option A)

Answer:

A. 150 units of model A1 and 120 units of model A2