NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Simultaneous equations and inequalities

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A local business offers two training packages for employees. Package A costs $120 per person and includes 4 hours of training.

Package B costs $180 per person and includes 6 hours of training. The company has a budget of $2100 and needs at least 30 hours of training.

Let x be the number of employees in Package A and y be the number in Package B. The company wants to find the maximum number of employees that can be trained while staying within budget and meeting the minimum training hours requirement.

Solve the system of inequalities: 120x + 180y ≤ 2100 and 4x + 6y ≥ 30, and determine the maximum possible value of x + y. Provide the exact value as a simplified fraction or integer.

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Worked solution

  1. The company has a budget of $2100 and needs at least 30 hours of training. Let x be the number of employees in Package A ($120 each, 4 hours), and y be the number in Package B ($180 each, 6 hours). The constraints are: 120x + 180y ≤ 2100 (budget) and 4x + 6y ≥ 30 (minimum training hours). The goal is to maximise x + y, the total number of employees trained.
  2. Simplify the budget inequality: divide all terms by 60 to reduce coefficients. 120x/60 + 180y/60 ≤ 2100/60 → 2x + 3y ≤ 35.
  3. Simplify the training hours inequality: divide all terms by 2. 4x/2 + 6y/2 ≥ 30/2 → 2x + 3y ≥ 15.
  4. Now the system is: 2x + 3y ≤ 35 and 2x + 3y ≥ 15. Both inequalities share the same left-hand side. This means the feasible region lies between 15 and 35 for 2x + 3y.
  5. Let S = 2x + 3y. Then 15 ≤ S ≤ 35. We want to maximise x + y. Express x + y in terms of S and y: from S = 2x + 3y, solve for x: 2x = S - 3y → x = (S - 3y)/2. Then x + y = (S - 3y)/2 + y = (S - 3y + 2y)/2 = (S - y)/2.
  6. To maximise x + y = (S - y)/2, we need to minimise y for a given S. Since S is bounded between 15 and 35, and x and y must be non-negative integers, we test integer values of S from 35 downwards to find the maximum x + y.
  7. Start with S = 35 (maximum allowed). Then x + y = (35 - y)/2. To maximise this, minimise y. Try y = 0: x = 35/2 = 17.5 → not integer. y = 1: x = (35 - 3)/2 = 32/2 = 16 → integer. So (x, y) = (16, 1). Check if 4x + 6y ≥ 30: 4(16) + 6(1) = 64 + 6 = 70 ≥ 30 → valid. x + y = 17.
  8. Try S = 34. Then x = (34 - 3y)/2. Try y = 0: x = 17 → integer. Check 4(17) + 6(0) = 68 ≥ 30 → valid. x + y = 17. Same as before. Try y = 2: x = (34 - 6)/2 = 28/2 = 14 → x + y = 16 < 17. So maximum remains 17.
  9. Try S = 33. y = 1: x = (33 - 3)/2 = 30/2 = 15 → x + y = 16 < 17. y = 3: x = (33 - 9)/2 = 24/2 = 12 → x + y = 15. No improvement.
  10. Try S = 35 with y = 3: x = (35 - 9)/2 = 26/2 = 13 → x + y = 16 < 17. y = 5: x = (35 - 15)/2 = 20/2 = 10 → x + y = 15. All less than 17.
  11. Try S = 35, y = 1 → x = 16 → x + y = 17. Try y = 3: x = 13 → x + y = 16. Try y = 5: x = 10 → x + y = 15. Try y = 7: x = (35 - 21)/2 = 14/2 = 7 → x + y = 14. All less than 17. So 17 is maximum.
  12. Verify (x, y) = (16, 1): cost = 120(16) + 180(1) = 1920 + 180 = 2100 ≤ 2100 → within budget. Training = 4(16) + 6(1) = 64 + 6 = 70 ≥ 30 → meets requirement. Total employees = 17. No higher value found. Maximum x + y is 17.

Reference answer: 17

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