Systems of linear inequalities in two variables

A school club wants to sell as much as possible, but it has limited time, limited ingredients and limited stalls. Every limit is one inequality, and all the limits must hold at the same time. Together they form a system. In this lesson we draw what the system allows and then pick the best choice.

Systems and their solution region

Each inequality gives a half-plane (see Linear inequalities in two variables). A pair solves the system only if it lies in all the half-planes, so:

the solution region of a system is the intersection of the solution regions of its inequalities.

The region is often a polygon. Its corners are called vertices; each vertex is where two boundary lines meet.

Optimising over a polygon region

Suppose we want the largest (or smallest) value of $F=ax+by$ over a polygon region. For each number $k$, the points with $F=k$ form a line, and all these lines are parallel. Sliding the line across the polygon, the last point where it touches the polygon is a corner.

So we only need to: (1) find the vertices, (2) compute $F$ at each, (3) compare.

Key terms

Summary

• A solution of a system satisfies all the inequalities; the solution region is the intersection of the half-planes.
• To test a pair, substitute it into every inequality; one failure is enough to reject it.
• The vertices of the region are the intersections of pairs of boundary lines that satisfy all the other inequalities.
• Over a bounded polygon, $ax+by$ attains its maximum and minimum at vertices: compute it at each vertex and compare.
• In real problems add the natural conditions ($x\ge 0$, $y\ge 0$) and read the answer in context.

Practice

Show solutions

1. $(2, 1)$: $3\ge 2$ and $1\le 3$, both true: a solution. $(0, 0)$: $0\ge 2$ is false: not a solution. $(5, 1)$: $6\ge 2$ is true but $4\le 3$ is false: not a solution.
2. $(1, 1)$: $3<4$ and $0>-1$ are true: a solution. $(2, 0)$: $4<4$ is false. $(0, 2)$: $2<4$ is true but $-2>-1$ is false. Only $(1, 1)$.
3. A triangle with vertices $(0, 0)$, $(3, 0)$, $(0, 3)$. Values of $2x+y$: $0$, $6$, $3$. The largest value is $6$, at $(3, 0)$.
4. The vertices are $(1, 0)$, $(4, 0)$, $(4, 3)$, $(1, 3)$. Values of $2x+y$: $2$, $8$, $11$, $5$. The largest is $11$ at $(4, 3)$ and the smallest is $2$ at $(1, 0)$.
5. Vertices: $(0, 0)$; $(6, 0)$; $(0, 4)$ (from $x+3y=12$); and the meeting of $x+y=6$ and $x+3y=12$: subtracting, $2y=6$, $y=3$, $x=3$, so $(3, 3)$. Values of $5x+4y$: $0$, $30$, $16$, $27$. The largest is $30$ at $(6, 0)$.
6. No. A pair would need $x+y\le 1$ and $x+y\ge 3$, so $3\le x+y\le 1$, which is impossible. The two half-planes do not overlap; the region is empty.
7. Count by the value of $x$: $x=0$ gives $y=0,\dots,4$ ($5$ pairs); $x=1$: $4$; $x=2$: $3$; $x=3$: $2$; $x=4$: $1$. Total $5+4+3+2+1=15$.
8. The system is $x\ge 0,\ y\ge 0,\ x+y\le 30,\ x\le 20,\ y\le 20$. Vertices: $(0, 0)$, $(20, 0)$, $(20, 10)$ ($x=20$ in $x+y=30$), $(10, 20)$ ($y=20$ in $x+y=30$), $(0, 20)$. Profit $F=4x+3y$: $0$, $80$, $110$, $100$, $60$. The largest profit is $110$ thousand with $20$ sandwiches and $10$ drinks.
9. The triangle lies in the first quadrant: $x\ge 0$, $y\ge 0$. The side from $(4, 0)$ to $(0, 2)$ lies on $x+2y=4$, and the origin satisfies $x+2y\le 4$. So the system is $x\ge 0,\ y\ge 0,\ x+2y\le 4$. (Check: $(4, 0)$ and $(0, 2)$ both give $4\le 4$.)
10. The vertices are $(0, 0)$, $(6, 0)$, $(3, 3)$, $(0, 4)$ and $F=x-y$ takes the values $0$, $6$, $0$, $-4$. The largest value is $6$ at $(6, 0)$ and the smallest is $-4$ at $(0, 4)$.