Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — a maximisation problem
Maximise P=4x+3y subject to x+y≤10, 2x+y≤16, x≥0, y≥0.
1Draw both boundary lines and shade the feasible region.The origin satisfies both, so keep the side containing it.
2Vertex (0,0): P=0
3Vertex (0,10), where x+y=10 meets the y-axis. Check 2(0)+10≤16 ✓ Then P=30.Always check the other constraints.
4Vertex (8,0), where 2x+y=16 meets the x-axis. Check 8≤10 ✓ Then P=32.
5Intersection: subtracting the equations gives x=6, so y=4. Then P=24+12=36.
6Maximum P=36 at (6,4).The interior vertex wins here — which is why all four must be tested.
Example 2 — a minimisation problem
Minimise C=3x+2y subject to x+y≥8, 3x+y≥12, x,y≥0.
1With ≥ constraints the feasible region lies away from the origin and is unbounded above.A minimum still exists; a maximum would not.
2Vertex (0,12): check 0+12≥8 ✓ Then C=24.
3Vertex (8,0): check 3(8)≥12 ✓ Then C=24.
4Intersection: x+y=8 and 3x+y=12 give 2x=4, so x=2, y=6.
5C=6+12=18
6Minimum C=18 at (2,6).Lower than both axis vertices.
Example 3 — formulating from words
A workshop makes tables and chairs. Each table needs 4 hours and each chair 1 hour, with 40 hours available. At most 12 tables can be made. Write the constraints.
1Define the variables: let x be the number of tables and y the number of chairs per week.In words, with units — worth marks on its own.
2Labour: 4x+y≤40'Available' means an upper limit.
3Tables: x≤12'At most' is again ≤.
4Non-negativity: x≥0, y≥0Easy to omit, and always required.
5Both are counts, so an integer solution will be needed.State this — the graphical optimum may be fractional.