Linear Inequalities in Two Variables
How to Find the best value in a feasible region
Use this to find the maximum or minimum of a quantity subject to linear inequality constraints.
Before you start
- Shading a region from several inequalities
- Finding where two lines intersect
- Substituting coordinates into an expression
When to use it
Use this to find the maximum or minimum of a quantity subject to linear inequality constraints.
The steps
- Graph all the inequalities and shade the feasible region.
- Identify the corner points (vertices) of the region.
- Write the quantity to be optimised as an expression.
- Substitute each corner point into the expression.
- Choose the corner that gives the largest or smallest value, as asked.
Worked example
For x ≥ 0, y ≥ 0, x + y ≤ 6 and y ≤ 4, find the maximum value of P = 3x + 2y.
- Graph the four inequalities and shade the feasible region they enclose.
- Identify the corner points of the region: (0, 0), (6, 0), (2, 4) and (0, 4).
- Write the quantity to be optimised: P = 3x + 2y.
- Substitute each corner: P(0,0) = 0; P(6,0) = 18; P(2,4) = 6 + 8 = 14; P(0,4) = 8.
- The largest value is 18, at the corner (6, 0).
A second example, with a twist
This is a minimisation set in a real cost context, with an 'at least' constraint, so you pick the smallest value at a corner. A caterer prepares x rice trays and y noodle trays, needing at least 6 trays in total, at least 1 rice tray and at least 2 noodle trays.
Each rice tray costs RM5 and each noodle tray RM4. Minimise the total cost C = 5x + 4y.
- Write the constraints x ≥ 1, y ≥ 2 and x + y ≥ 6, then shade the feasible region.
- Identify the corner points of that region: (1, 5) and (4, 2).
- Write the quantity to be optimised: C = 5x + 4y.
- Substitute each corner: C(1,5) = 5 + 20 = 25; C(4,2) = 20 + 8 = 28.
- The question asks for the minimum, so choose the smaller value, 25, at (1, 5).
Formula pages
Practise this in a KBAT problem
Frequently asked questions
Why only check the corners, not points inside?
For a linear expression, the largest and smallest values over a polygon always occur at a corner, never strictly inside. So testing the corner points is enough.
This is why you find the vertices first: they are the only places where the maximum or minimum can sit.
How do I find the exact corner coordinates?
Each corner is where two boundary lines cross. Solve those two equations together, usually by substitution or elimination, to get the exact x and y.
Do not just read them off the graph, because a sketch can be slightly off and small errors change the optimum you choose.
How do I know whether to pick the biggest or smallest?
Read the question carefully. Words like 'maximum', 'greatest' or 'most profit' mean pick the largest value; words like 'minimum', 'least' or 'lowest cost' mean pick the smallest.
Work out the expression at every corner first, then choose the value that matches what the question is asking for.