Pearson Edexcel International Advanced Level Mathematics: Linear programming: formulation, graphical solution and integer solutions (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover linear programming for Pearson Edexcel International Advanced Level Mathematics (YMA01), Specification Issue 3, April 2019: unit D1, section 5, topics 5.1 to 5.3. The specification states that D1 is optional for IAS and IAL Mathematics and Further Mathematics. Learn the methods from the study guide and the revision notes. See also the course hub, the checklist and all free 10-minute diagnostics.

In every question x and y are non-negative.

Questions

1. A shop sells x hardback and y paperback books. Write each condition as an inequality in x and y.

(a) At least 15 hardbacks are in stock. [1] (b) The shop holds at most 120 books in total. [1] (c) Hardbacks are no more than 25% of all the books. [1] (d) Hardbacks need 3 cm of shelf, paperbacks 2 cm, and 240 cm is available. [1]

2. A tutoring centre runs x group sessions and y one-to-one sessions each week. A group session needs 2 tutor-hours and a one-to-one session needs 1 tutor-hour. There are 36 tutor-hours. At most 30 sessions can be held in total. At least 5 group sessions must run. There must be no more than twice as many one-to-one sessions as group sessions. Income is £60 per group session and £35 per one-to-one session. Formulate a linear program to maximise income. [7]

3. Maximise P = 5x + 3y subject to 2x + y ≤ 16, x + 2y ≤ 14 and x + y ≥ 9. [5]

4. The feasible region for a problem has vertices (0, 0), (10, 0), (9, 3), (3, 9) and (0, 10). It is bounded by x + y ≤ 12, 3x + y ≤ 30 and x + 3y ≤ 30. The objective is to maximise P = 7x + 4y.

(a) Find the gradient of the line 7x + 4y = 28. [1] (b) Use the ruler method to explain which vertex is optimal. [2] (c) Find the maximum value of P. [1]

5. A café blends x kg of bean A and y kg of bean B. Minimise the cost C = 8x + 6y pounds, subject to 3x + y ≥ 18, x + 2y ≥ 16 and x + y ≤ 20. [6]

6. A club hires x large boats and y small boats. The cost in pounds is C = 40x + 30y, to be minimised, subject to 3x + y ≥ 30, x + 2y ≥ 24 and x + y ≤ 20. The numbers of boats are integers.

(a) Ignoring the integer condition, find the minimum cost and the point where it occurs. [3] (b) Find the minimum cost when x and y must be integers, justifying your answer. [3]

7. A caterer bakes x small, y medium and z large cakes. Oven time is 4, 6 and 10 minutes per cake respectively, with 480 minutes available. Mixing time is 5, 8 and 12 minutes per cake, with 600 minutes available. At least half of all the cakes must be small. Profit is 60p, 100p and 170p per cake respectively. Formulate a linear program to maximise profit. [6]

8. Maximise P = 6x + 5y subject to 3x + 2y ≤ 48 (machine hours), x + y ≤ 20 (labour hours), x ≥ 4 (contract) and x + 3y ≤ 50 (material). The optimal solution is x = 8, y = 12.

(a) Show that (8, 12) satisfies all four constraints and state which are binding. [2] (b) State how much material is unused and by how much x exceeds its minimum. [2]

9. A bakery makes x batches of bread and y batches of cakes each day. Bread needs 3 oven-hours and 1 preparation-hour per batch. Cakes need 2 oven-hours and 3 preparation-hours per batch. There are 60 oven-hours and 45 preparation-hours. At least 12 batches must be made. Profit is £8 per batch of bread and £9 per batch of cakes.

(a) Formulate a linear program to maximise profit P. [4] (b) Find the coordinates of the vertices of the feasible region. [3] (c) Ignoring the integer condition, find the maximum profit. [2] (d) The batches are whole numbers. Find the best production plan and its profit, justifying your answer. [3]

10. A gardener buys x bags of compost A and y bags of compost B. Compost A has 5 units of nitrogen and 2 units of potash per bag. Compost B has 2 units of nitrogen and 3 units of potash per bag. At least 60 units of nitrogen and 48 units of potash are needed. Storage holds at most 30 bags. Bag A costs £7 and bag B costs £5.

(a) Formulate a linear program to minimise the cost C. [4] (b) Find the minimum cost when fractions of a bag are allowed. [3] (c) The bags are sold whole. Find the cheapest whole-number purchase, justifying your answer. [3]

Answers

1. (a) x ≥ 15 [1] [1] (b) x + y ≤ 120 [1] [1] (c) x ≤ 0.25(x + y), so 4x ≤ x + y, giving 3x ≤ y [1] [1] (d) 3x + 2y ≤ 240 [1] [1] Examiner insight: For a percentage condition, write the total x + y before simplifying.

2. Let x = group sessions and y = one-to-one sessions per week [1]. Maximise I = 60x + 35y [1]. Tutor time: 2x + y ≤ 36 [1]. Total: x + y ≤ 30 [1]. Minimum: x ≥ 5 [1]. Ratio: y ≤ 2x [1]. x and y are non-negative integers [1]. [7] Examiner insight: Each sentence usually gives one constraint, so tick them off. The integer condition is easy to forget.

3. (4, 5) from x + 2y = 14 and x + y = 9 [1]; (6, 4) from 2x + y = 16 and x + 2y = 14 [1]; (7, 2) from 2x + y = 16 and x + y = 9 [1]. P: (4, 5) gives 35; (6, 4) gives 42; (7, 2) gives 41 [1]. The maximum is P = 42 at x = 6, y = 4 [1]. [5] Examiner insight: Show the pair of lines used for each vertex, so the coordinates can be followed.

4. (a) Gradient = −7/4 [1]. [1] (b) The line 7x + 4y = 28 meets the axes at (4, 0) and (0, 7); slide a ruler parallel to it away from the origin [1]. Its gradient −7/4 lies between −3 (3x + y = 30) and −1 (x + y = 12), so the last vertex touched is (9, 3) [1]. [2] (c) P = 7(9) + 4(3) = 75 [1]. [1] Examiner insight: Say which way you move the ruler and why the last point touched is the answer.

5. From the intercepts, the vertices of R are (0, 18), (0, 20), (4, 6), (16, 0) and (20, 0) [1]. (4, 6) from x = 16 − 2y: 48 − 6y + y = 18, so y = 6 and x = 4 [1] [1]. C: (0, 18) gives 108; (0, 20) gives 120; (16, 0) gives 128; (20, 0) gives 160; (4, 6) gives 68 [1]. The minimum is C = 68 [1], by blending 4 kg of A and 6 kg of B at a cost of £68 [1]. [6] Examiner insight: With “at least” constraints, check that your vertex list includes those on the axes.

6. (a) Solve 3x + y = 30 and x + 2y = 24: y = 30 − 3x, so x + 60 − 6x = 24, giving x = 7.2 [1], y = 8.4 [1]. Cost = 40(7.2) + 30(8.4) = £540 [1]. [3] (b) (7, 8) gives 3x + y = 29 < 30, not feasible. (8, 8) is feasible with cost 560, and (7, 9) is feasible with cost 550 [1]. Every other feasible point costs more than 540, and costs are multiples of 10, so no integer plan can cost less than 550 [1]. The minimum is £550, with 7 large and 9 small boats [1]. [3] Examiner insight: Rounding to (7, 8) is infeasible and (8, 8) is dearer. State why 550 cannot be beaten.

7. Let x, y, z be the numbers of small, medium and large cakes [1]. Maximise P = 60x + 100y + 170z [1]. Oven: 4x + 6y + 10z ≤ 480 [1]. Mixing: 5x + 8y + 12z ≤ 600 [1]. “At least half small”: x ≥ ½(x + y + z), so x ≥ y + z [1]. x, y, z non-negative integers [1]. [6] Examiner insight: If you simplify the oven inequality, divide every term, giving 2x + 3y + 5z ≤ 240.

8. (a) 3(8) + 2(12) = 48; 8 + 12 = 20; 8 ≥ 4; 8 + 3(12) = 44 ≤ 50, so all are satisfied [1]. Machine hours and labour hours hold with equality, so they are binding [1]. [2] (b) Material unused: 50 − 44 = 6 [1]. x exceeds its minimum by 8 − 4 = 4 [1]. [2] Examiner insight: Interpret in context, using units from the question.

9. (a) Maximise P = 8x + 9y [1]. Oven: 3x + 2y ≤ 60 [1]. Preparation: x + 3y ≤ 45 [1]. Total: x + y ≥ 12, and x, y non-negative integers [1]. [4] (b) Axis intercepts: (12, 0) and (0, 12) from x + y = 12; (20, 0) from the oven line; (0, 15) from the preparation line [1]. Solve 3x + 2y = 60 and x + 3y = 45: x = 45 − 3y, so 135 − 9y + 2y = 60 [1], giving y = 75/7 and x = 90/7 [1]. [3] (c) P: (12, 0) gives 96; (0, 12) gives 108; (0, 15) gives 135; (20, 0) gives 160; (90/7, 75/7) gives 1395/7 ≈ 199.29 [1]. The maximum is £199.29 at that vertex [1]. [2] (d) The nearest integer point (13, 11) fails the oven limit: 3(13) + 2(11) = 61 > 60 [1]. (12, 11) is feasible: 58 ≤ 60, 45 ≤ 45, 23 ≥ 12, and P = 195 [1]. A better plan would have P from 196 to 199; every integer point on those lines breaks a limit, such as (11, 12) on 196, where 47 > 45. So the best plan is 12 batches of bread and 11 of cakes, profit £195 [1]. [3] Examiner insight: Show one infeasible rounded point, the best feasible point, and why nothing between it and the bound works.

10. (a) Minimise C = 7x + 5y [1]. Nitrogen: 5x + 2y ≥ 60 [1]. Potash: 2x + 3y ≥ 48 [1]. Storage: x + y ≤ 30, with x, y non-negative integers [1]. [4] (b) The vertices are (0, 30), (24, 0), (30, 0) and the meeting of the nitrogen and potash lines. Eliminating y gives 11x = 84 [1], giving x = 84/11 and y = 120/11 [1]. C: (0, 30) gives 150; (24, 0) gives 168; (30, 0) gives 210; the other vertex gives £108 [1]. [3] (c) Since the minimum 108 occurs only at a non-integer point, an integer plan costs at least £109. The integer points on 7x + 5y = 109 and 110 all fail the nitrogen or potash need, such as (7, 12) on 109, which gives 5x + 2y = 59 < 60 [1]. (8, 11) is feasible: 62 ≥ 60, 49 ≥ 48, 19 ≤ 30 [1]. Cost 7(8) + 5(11), so the cheapest is 8 bags of A and 11 of B, £111 [1]. [3] Examiner insight: Check every constraint for your chosen point, including storage, and state the purchase.

Where marks are usually lost

• Leaving out the definition of the variables in a formulation.
• Reversing a ratio, for example writing y ≤ 2x as 2y ≤ x.
• Missing the non-negativity and integer conditions.
• Missing a vertex on an axis when the constraints are “at least”.
• Moving the ruler towards the origin when maximising, or away from it when minimising.
• Rounding to an integer without testing every constraint.
• Stopping at the objective value without stating the plan.

Next steps

• Revisit the revision notes and the study guide.
• Tick off the topic on the checklist and see the course hub.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019 (first teaching September 2018), published by Pearson Education Limited. Unit D1, section 5: Linear programming, topics 5.1 to 5.3.