Inequalities and Linear Programming

Solve linear and quadratic inequalities and combine them, shade a region from inequalities, find its corners, and find the greatest or least value of an expression in it.

In the exam for Edexcel Further Pure Maths 4PM1 · 14 real past-paper questions below

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Plans with limits

A bakery has only so much flour and so many hours. Each limit is an inequality. Together they mark out every plan the bakery can actually make. Which plan earns the most? That question is called linear programming.

Here, “programming” means planning, not computer code. The plan you want is always at a corner of the allowed region, and this lesson shows you why.

First: solve inequalities in one letter, x. These are the 4PM1 questions that end with “both”.
Then: inequalities in x and y. You draw lines, shade a region R, find its corners and test an expression P.

Key idea

Linear programming grew up in the 1930s and 1940s to plan factories and wartime supplies. Leonid Kantorovich, one of its founders, shared the 1975 Nobel Prize in Economics for this work.

Inequalities in x

A linear inequality is solved like an equation. Do the same thing to both sides, with one extra rule.

Rule

If you multiply or divide both sides by a negative number, flip the inequality sign. For example, $- 2 x < 6$ gives $x > - 3$ .

Worked example

1
Find the set of values of x for which $3 (x - 2) < x + 4$ .
2
$3 x - 6 < x + 4$ — Expand the bracket.
3
$2 x < 10$ — Take x from both sides and add 6.
4
$x < 5$ — Divide by 2. It is positive, so the sign stays.

For a quadratic inequality, collect everything on one side first, so the other side is 0. The $x^{2}$ terms may be on both sides.

Worked example

1
Find the set of values of x for which $4 x^{2} - 5 x + 3 < x^{2} + 2 x + 1$ .
2
$3 x^{2} - 7 x + 2 < 0$ — Move every term to the left.
3
$(3 x - 1) (x - 2) = 0 \Rightarrow x = \frac{1}{3}, 2$ — The critical values.
4
$\frac{1}{3} < x < 2$ — The parabola $3 x^{2} - 7 x + 2$ is below 0 between its roots.

Rule

When the x² number is positive: less than 0 gives one piece, between the critical values; greater than 0 gives two pieces, outside them, like $x < \frac{1}{3}$ or $x > 2$ .

Worked example

1
Find the values of x for which both $2 (x + 1) > 5 - x$ and $3 x^{2} - 7 x + 2 < 0$ .
2
$3 x > 3 \Rightarrow x > 1$ — Solve the linear one: $2 x + 2 > 5 - x$ .
3
$\frac{1}{3} < x < 2$ — The quadratic one, from before.
4
$1 < x < 2$ — “Both” means the overlap. Draw both on a number line and keep the part covered twice.

Common mistake

Dividing by a negative and keeping the sign. $- 4 x > 8$ gives $x < - 2$ .

Key idea

For “both”, draw the two answers on one number line. The answer is only where they overlap.

Draw and shade R

Now the inequalities have x and y. Each one is a line on the graph, with the allowed side shaded. The region where all of them are true is called R.

Steps

1
1. Draw each line. Two quick points: put $x = 0$ , then put $y = 0$ .
2. Use a solid line for $\leq$ or $\geq$ , and a dashed line for $<$ or $>$ .
3. Pick a test point not on the line, usually (0, 0). Put it into the inequality.
4. If it is true, R is on the side of the test point. If it is false, R is on the other side.
5. Shade, and label the region R.

Worked example

1
Show the region R where $y \leq x + 2$ , $x + 2 y \leq 10$ and $y \geq 1$ .
2
$y = x + 2 : (0, 2), (- 2, 0)$ — Two points for the first line.
3
$x + 2 y = 10 : (0, 5), (10, 0)$ — Put x = 0: 2y = 10. Put y = 0: x = 10.
4
$y = 1$ — A horizontal line through (0, 1).
5
$0 \leq 2, 0 \leq 10$ — Put (0, 0) into the first two. Both are true, so R is on the origin side of these two lines.
6
$0 \geq 1 is false$ — (0, 0) fails here, so R is above the line y = 1.
7
$R: a triangle$ — The three lines close off a triangle. Label it R.

Rule

4PM1 accepts shading the region you want or the parts you do not want. Either way, R must be one closed region and clearly labelled.

Common mistake

Plotting $x + 2 y = 10$ through (0, 10). With $x = 0$ , you get $2 y = 10$ , so $y = 5$ .

Key idea

Every line splits the page in two. One test point tells you which half you keep.

Corners of R

The corners of R are called vertices. Each one is where two boundary lines meet. You can read them from an accurate graph, but solving the equations is safer.

Worked example

1
Find the vertices of the triangle R from before.
2
$y = x + 2, y = 1 \Rightarrow x = - 1$ — Put y = 1 into the first line. Vertex (−1, 1).
3
$x + 2 (x + 2) = 10$ — Substitute $y = x + 2$ into $x + 2 y = 10$ .
4
$3 x + 4 = 10 \Rightarrow x = 2, y = 4$ — Vertex (2, 4).
5
$x + 2 (1) = 10 \Rightarrow x = 8$ — Put y = 1 into $x + 2 y = 10$ . Vertex (8, 1).

Common mistake

Using a crossing point that is not a corner of R. Only points where two lines meet on the edge of R count.

Key idea

Vertices come from simultaneous equations: two lines, one point. Write each one as coordinates.

Greatest and least values

The question gives an expression such as $P = x + 3 y$ and asks for its greatest or least value for points in R. Here is why the answer is at a vertex.

Each value of P is a straight line, like $x + 3 y = 6$ or $x + 3 y = 9$ . All these lines are parallel. As P grows, the line slides across the page. The last point of R it touches is always a corner.

Steps

1
1. Find every vertex of R.
2. Work out P at each vertex.
3. The biggest result is the greatest value. The smallest is the least value.

Worked example

1
For points in R, find the greatest value of $P = x + 3 y$ .
2
$(- 1, 1) : P = - 1 + 3 = 2$ — Put the first vertex into P.
3
$(2, 4) : P = 2 + 12 = 14$ — The second vertex.
4
$(8, 1) : P = 8 + 3 = 11$ — The third vertex.
5
$P_{m a x} = 14$ — The greatest value is 14, at (2, 4).

Worked example

1
For points in R, find the least value of $Q = 3 x - y$ .
2
$Q = - 4, 2, 23$ — At (−1, 1), (2, 4) and (8, 1).
3
$Q_{m i n} = - 4$ — The least value is −4, at (−1, 1).

Rule

A second method is the objective line: draw a line $x + 3 y = k$ and slide it parallel across R. Testing every vertex is quicker and safer, and the mark scheme accepts both.

Common mistake

Testing only one or two vertices, or giving the point instead of the value. Test them all, then write the value of P.

Key idea

Corners only. Put each vertex into P, then pick the largest or smallest.

Which step? Read the question

These questions follow the same pattern every year. Read each part and use the right tool.

You see	You do
A linear inequality in x	Solve like an equation; flip the sign if you divide by a negative
A quadratic inequality	Collect to one side, find the critical values, sketch the parabola
“Both … and …”	Draw both on a number line; keep the overlap
“Draw the line with equation …”	Two points: put x = 0, then y = 0
“Show, by shading, the region R”	Test (0, 0) for each line, shade, label R
“Find the greatest (or least) value of P”	Find every vertex, put each into P, compare

Common mistake

Answering “least” with the greatest value. Read which one the question asks for.

Key idea

Inequalities give the region. The region gives the corners. The corners give the answer.

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