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4PM1/02 Nov 2023, Question 4: Inequalities and Linear Programming

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 8 marks

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The question, in short

Draw three straight lines on given axes, shade the feasible region for three inequalities, then find extreme values of P = 2y − 3x over that region.

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

Part (a) · 3 marks

Plot the lines y = −x − 1, y = 3x − 8 and y = (x + 8) ÷ 2 on the grid.

  1. 1y = −x − 1 passes through (−1, 0) and (0, −1)
  2. 2y = 3x − 8 passes through (2, −2) and (3, 1)
  3. 32y = x + 8 passes through (−2, 3) and (0, 4)

AnswerThree straight lines: y = −x − 1 through (−1, 0) and (0, −1); y − 3x + 8 = 0 through (3, 1) and (2, −2); 2y = x + 8 through (−2, 3) and (0, 4)

Part (b) · 1 mark

Shade the region satisfying y ≥ −x − 1, y ≥ 3x − 8 and 2y ≤ x + 8.

  1. 1R lies above y = −x − 1, above y = 3x − 8 and below 2y = x + 8
  2. 2It is the closed triangle bounded by the three lines

AnswerR is the triangle with vertices (−10/3, 7/3), (7/4, −11/4) and (24/5, 32/5): above y = −x − 1, to the left of/above y = 3x − 8 and below 2y = x + 8

Part (c) · 4 marks

Find the maximum and minimum values of P = 2y − 3x at the vertices of region R.

  1. 1Vertices of R: (−10/3, 7/3), (7/4, −11/4), (24/5, 32/5)
  2. 2P = 2y − 3x at each vertex: 44/3, −43/4, −8/5
  3. 3Greatest P = 44/3 at (−10/3, 7/3)
  4. 4Least P = −43/4 = −10.75 at (7/4, −11/4)

Answer14.4 to 15, −11 to −10.4

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