3
algebra

A manufacturer makes two products, A and B. Product A requires 2 hours of labor, Product B requires 3 hours. Total labor available is 120 hours. They must make at least 10 of Product A. Which system models this?

A

2A+3B≤120,A≥10,A≥0,B≥02A + 3B \leq 120, A \geq 10, A \geq 0, B \geq 0

B

2A+3B≥120,A≥10,A≥0,B≥02A + 3B \geq 120, A \geq 10, A \geq 0, B \geq 0

C

3A+2B≤120,A≥10,A≥0,B≥03A + 2B \leq 120, A \geq 10, A \geq 0, B \geq 0

D

2A+3B≤120,B≥10,A≥0,B≥02A + 3B \leq 120, B \geq 10, A \geq 0, B \geq 0

Correct Answer: A

Choice A is the correct answer. Translate constraints.

  1. Labor: 2A+3B≤1202A + 3B \leq 120 (Total hours constraint)
  2. Quantity: A≥10A \geq 10 (At least 10 of A)
  3. Non-negativity: A≥0,B≥0A \geq 0, B \geq 0 (Cannot make negative products)

?�� Strategic Tip: Always include non-negativity constraints (x≥0,y≥0x \geq 0, y \geq 0) for real-world production problems.

Choice B is incorrect because≥120\geq 120 implies they must use more than available labor. Choice C is incorrect because it swaps the labor hours per product. Choice D is incorrect because it puts the minimum quantity on B instead of A.

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Linear Functions Set 76