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0580/41 May/Jun 2023, Question 8: Linear Inequalities

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 12 marks

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The question, in short

From a tailor's weekly dress and shirt production constraints, form inequalities, graph the feasible region, and use it to find minimum output and maximum profit.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 3 marks

Write the three remaining inequalities in x and y describing shirts, total items, and shirts relative to dresses.

  1. 1No more than 7 shirts: y ≤ 7
  2. 2Less than 14 altogether: x + y < 14
  3. 3Shirts more than 2/3 of dresses: y > (2/3)x

Answery ≤ 7, x + y < 14, y > (2/3)x

Part (b) · 6 marks

On the grid, draw four boundary lines, shade unwanted regions, and label the feasible region R.

  1. 1Draw x = 4 and y = 7 as solid lines (≥ and ≤ include the line)
  2. 2Draw x + y = 14 as a dashed line through (0, 14) and (14, 0)
  3. 3Draw y = (2/3)x as a dashed line through (0, 0) and (12, 8)
  4. 4Shade left of x = 4, above y = 7, above x + y = 14 and below y = (2/3)x; label the unshaded region R

AnswerLines x = 4 (solid), y = 7 (solid), x + y = 14 (dashed), y = (2/3)x (dashed); correct shading everywhere but region R, with R labelled

Part (c) · 1 mark

Using the diagram, state the smallest number of dresses and shirts as an ordered pair.

  1. 1Smallest x in R is 4; then y > (2/3) × 4 = 2.67, so smallest y = 3
  2. 24 dresses and 3 shirts

Answer(4, 3)

Part (d) · 2 marks

With profit $10 per dress and $6 per shirt, use the diagram to find the greatest possible weekly profit.

  1. 1Profit = 10x + 6y; test whole-number points in R near x + y = 13
  2. 2(7, 6): 70 + 36 = 106; (6, 7): 60 + 42 = 102; (8, 5) not in R since 5 < 2/3 × 8
  3. 3Largest profit = $106

Answer106

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Also searched as: 0580/41/M/J/23 · 0580_s23_qp_41 · 0580 s23 41 · question 8.
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