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Simultaneous equations and inequalities
Lesson 29 of 75
Challenge

Integer solutions in regions

Objective. Find and optimise integer solutions in feasible regions.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

Integer solutions in regions extend feasible-region work by requiring whole-number coordinate points, often to optimise a quantity like cost or profit subject to constraints. After shading the feasible region defined by the inequalities, list the lattice points (integer x and y pairs) that lie within or on its boundary, then evaluate the objective expression at each to find the optimum. For instance, within x + y ≤ 5, x ≥ 1, y ≥ 1 the integer point maximising x + 2y is found by checking candidates like (1,4).

Worked example

Find the integer solution maximising P = x + 3y subject to x ≥ 0, y ≥ 0 and x + y ≤ 4.

3 … 0
Illustrated card showing 3 … 0
  1. 1List integer points with x + y ≤ 4, x ≥ 0, y ≥ 0
  2. 2To maximise P = x + 3y, favour larger y since it has the bigger coefficient
  3. 3Largest y with x ≥ 0 is y = 4 when x = 0, giving P = 0 + 12 = 12
  4. 4Check nearby points like (1,3): P = 1 + 9 = 10, which is smaller, confirming (0,4) is best

Answer: x = 0, y = 4, giving maximum P = 12

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Common mistake. Students sometimes find the optimum of the continuous region using a vertex and assume it is already an integer solution without checking it lies on whole-number coordinates. Always test nearby lattice points inside the region if the vertex is non-integer.

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