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Linear Inequalities in Two Variables · Form 4

Linear Inequalities in Two Variables: Worked Examples (KBAT)

This set tackles real-world optimisation: reading a worded situation, setting up the full system of inequalities, then testing each corner point to find a maximum or minimum. It suits students aiming for full marks on application questions that hide the technique inside a story.

Written by the spmmath.com.my editorial team.

Worked example 1

A bakery makes chocolate cakes (x) and vanilla cakes (y) in a day. Each chocolate cake uses 200 g of flour and each vanilla cake uses 100 g, with 1200 g of flour available.

Each cake takes 1 hour of baking time, and 8 hours are available. Profit is RM4 per chocolate cake and RM3 per vanilla cake.

How many of each should the bakery make to maximise profit, and what is the maximum profit?

  1. Let x = chocolate cakes and y = vanilla cakes.
  2. Flour: 200x + 100y ≤ 1200, which simplifies to 2x + y ≤ 12.
  3. Time: x + y ≤ 8; also x ≥ 0 and y ≥ 0.
  4. Find the corner points of the feasible region: (0, 0), (6, 0), (0, 8) and the intersection of the two lines.
  5. Solve 2x + y = 12 and x + y = 8: subtracting gives x = 4, then y = 4, so (4, 4).
  6. Profit P = 4x + 3y. Evaluate: (0,0)→0, (6,0)→24, (0,8)→24, (4,4)→16 + 12 = 28.
  7. The largest value is 28, at (4, 4).

Worked example 2

A school hires large buses (x), each carrying 40 students, and small buses (y), each carrying 20 students, for a trip. At least 240 students must be carried, no more than 10 buses in total may be hired, and at most 5 large buses are available.

A large bus costs RM300 and a small bus RM200. Find how many of each bus minimises the cost, and state the minimum cost.

  1. Let x = large buses and y = small buses.
  2. Capacity: 40x + 20y ≥ 240, which simplifies to 2x + y ≥ 12.
  3. Total buses: x + y ≤ 10; large buses: x ≤ 5; also x ≥ 0 and y ≥ 0.
  4. Corner points come from the boundary lines: 2x + y = 12, x + y = 10 and x = 5.
  5. 2x + y = 12 ∩ x + y = 10 gives x = 2, y = 8 → (2, 8).
  6. x = 5 ∩ x + y = 10 gives y = 5 → (5, 5); x = 5 ∩ 2x + y = 12 gives y = 2 → (5, 2).
  7. Cost C = 300x + 200y. Evaluate: (2,8)→2200, (5,5)→2500, (5,2)→1900.
  8. The smallest value is 1900, at (5, 2); check capacity: 40(5) + 20(2) = 240 ≥ 240.

Worked example 3

A tailor makes shirts (x) and dresses (y). A shirt uses 1 m of cloth and a dress uses 2 m, with 40 m of cloth available.

A shirt takes 2 hours and a dress takes 1 hour, with 50 hours available. To fill an order, at least 6 dresses must be made.

Profit is RM15 per shirt and RM25 per dress. How many of each should the tailor make to maximise profit, and what is the maximum profit?

  1. Let x = shirts and y = dresses.
  2. Cloth: x + 2y ≤ 40; time: 2x + y ≤ 50.
  3. Order: y ≥ 6; also x ≥ 0.
  4. Corner points: (0, 6) and (0, 20) on the y-axis, plus two intersections.
  5. x + 2y = 40 ∩ 2x + y = 50: from the first, x = 40 − 2y; substitute to get 80 − 3y = 50, so y = 10, x = 20 → (20, 10).
  6. 2x + y = 50 ∩ y = 6 gives 2x = 44, x = 22 → (22, 6).
  7. Profit P = 15x + 25y. Evaluate: (0,6)→150, (0,20)→500, (22,6)→480, (20,10)→300 + 250 = 550.
  8. The largest value is 550, at (20, 10).

Worked example 4

A furniture workshop makes x bookshelves and y cabinets in a week. Each bookshelf uses 3 units of wood and each cabinet uses 2 units; at most 36 units of wood are available.

Each bookshelf needs 2 hours of labour and each cabinet needs 4 hours; at most 40 hours of labour are available. The profit is RM50 per bookshelf and RM60 per cabinet.

Find how many of each should be made to give the maximum profit, and state that profit.

  1. Wood: 3x + 2y ≤ 36. Labour: 2x + 4y ≤ 40, which simplifies to x + 2y ≤ 20. Also x ≥ 0, y ≥ 0.
  2. Find the corner points of the region. Solve 3x + 2y = 36 and x + 2y = 20 together: subtract to get 2x = 16, so x = 8, then y = 6 → (8, 6).
  3. Other corners: (0, 10) from x + 2y = 20, (12, 0) from 3x + 2y = 36, and (0, 0).
  4. Test profit P = 50x + 60y at each corner: (0,0)→0; (12,0)→600; (0,10)→600; (8,6)→400 + 360 = 760.
  5. The largest profit is RM760 at (8, 6).

Worked example 5

A school must transport 180 students using x vans and y minibuses. Each van seats 15 and each minibus seats 30, and they must seat all 180 students.

At most 10 vehicles are available. Each van trip costs RM60 and each minibus trip costs RM150.

Find the number of vans and minibuses that gives the minimum total cost, and state that cost.

  1. Seats: 15x + 30y ≥ 180, which simplifies to x + 2y ≥ 12. Vehicles: x + y ≤ 10. Also x ≥ 0, y ≥ 0.
  2. Find the corner points. Solve x + 2y = 12 and x + y = 10 together: subtract to get y = 2, then x = 8 → (8, 2).
  3. Other corners: (0, 6) from x + 2y = 12 with x = 0, and (0, 10) from x + y = 10 with x = 0.
  4. Test cost C = 60x + 150y: (0,6)→900; (0,10)→1500; (8,2)→480 + 300 = 780.
  5. The lowest cost is RM780 at (8, 2).

Worked example 6

A food stall prepares x plates of nasi lemak and y plates of mee goreng each morning. The rice-and-ingredient limit gives 2x + y ≤ 60, and the cooking-time limit gives x + 2y ≤ 60.

To meet regular demand the stall must prepare at least 10 plates of nasi lemak. The profit is RM3 per plate of nasi lemak and RM4 per plate of mee goreng.

Find the number of each to prepare for maximum profit, and state that profit.

  1. The constraints are 2x + y ≤ 60, x + 2y ≤ 60, x ≥ 10 and y ≥ 0.
  2. Solve 2x + y = 60 and x + 2y = 60 together: from the first y = 60 − 2x, substitute to get x + 120 − 4x = 60, so 3x = 60, x = 20, y = 20 → (20, 20).
  3. Other corners: (30, 0) from 2x + y = 60 with y = 0; (10, 25) from x + 2y = 60 with x = 10; and (10, 0).
  4. Test profit P = 3x + 4y: (10,0)→30; (30,0)→90; (10,25)→130; (20,20)→60 + 80 = 140.
  5. The largest profit is RM140 at (20, 20).

Worked example 7

A company advertises using online ads (x) and newspaper ads (y). Each online ad reaches 300 people and costs RM40; each newspaper ad reaches 100 people and costs RM90.

The campaign must reach at least 1800 people. The total number of ads used must not exceed 12, and at most 7 online ads may be run.

Find the number of each type of ad that minimises the total cost, and state the minimum cost.

  1. Set up the inequalities: 300x + 100y ≥ 1800 → 3x + y ≥ 18; x + y ≤ 12; x ≤ 7; x ≥ 0, y ≥ 0
  2. Find the corner points of the feasible region: (3, 9), (7, 5), (7, 0), (6, 0)
  3. Objective: minimise C = 40x + 90y
  4. At (3, 9): C = 40(3) + 90(9) = 120 + 810 = 930
  5. At (7, 5): C = 40(7) + 90(5) = 280 + 450 = 730
  6. At (7, 0): C = 40(7) + 90(0) = 280
  7. At (6, 0): C = 40(6) + 90(0) = 240
  8. The minimum cost is RM240, at x = 6 online ads and y = 0 newspaper ads

Worked example 8

A jeweller makes bracelets (x) and necklaces (y) each day. Each bracelet uses 2 bead-sets and takes 1 hour to make; each necklace uses 5 bead-sets and takes 2 hours to make.

At most 40 bead-sets and at most 18 hours are available each day. To fulfil standing orders, at least 3 necklaces must be made each day.

The profit is RM12 per bracelet and RM25 per necklace. Find the number of bracelets and necklaces that maximises the daily profit, and state the maximum profit.

  1. Set up the inequalities: 2x + 5y ≤ 40; x + 2y ≤ 18; y ≥ 3; x ≥ 0
  2. Find the corner points of the feasible region: (0, 3), (0, 8), (10, 4), (12, 3)
  3. Objective: maximise P = 12x + 25y
  4. At (0, 3): P = 12(0) + 25(3) = 75
  5. At (0, 8): P = 12(0) + 25(8) = 200
  6. At (10, 4): P = 12(10) + 25(4) = 120 + 100 = 220
  7. At (12, 3): P = 12(12) + 25(3) = 144 + 75 = 219
  8. The maximum profit is RM220, at x = 10 bracelets and y = 4 necklaces

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What do KBAT questions on linear inequalities in two variables usually ask for?

KBAT questions embed the inequalities in a real constraint problem, for example, a budget or a mix of two products, and ask you to set up the inequalities yourself from a written scenario before graphing them. The final step often asks for the best combination within the feasible region, so read the situation carefully to decide whether you need a maximum or minimum value.

How do I turn a word problem into the correct inequality without a diagram to copy?

Underline every number and constraint word ("at most", "at least", "not more than") in the question, assign variables to the two unknowns, and turn each constraint into one inequality, one sentence at a time. Writing constraints separately before combining them on one graph reduces the chance of missing one entirely.

How do I find the maximum or minimum value once I have the feasible region?

The optimum value of a linear expression over a feasible region always occurs at one of its corner points, so list the coordinates of every corner, substitute each into the expression you're optimising, and compare the results. Never assume the answer lies inside the region, always check the vertices.

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