Form 4 · Chapter 6

Applications of Linear Inequalities

Turn word problems into linear inequalities in two variables, shade the feasible region and read off integer solutions.

From words to inequalities

Many real-life limits are written as linear inequalities in two variables. Read each condition and match a phrase to a symbol: at least / minimum gives $\geq$, at most / maximum gives $\leq$, more than gives $>$, and less than gives $<$. If a shop buys $x$ chairs and $y$ tables, the cost limit RM800 with chairs RM40 and tables RM100 becomes $40x+100y\leq 800$.

Key formula

General form: $ax+by\leq c$ (or with $\geq,<,>$). Non-negative real quantities also need $x\geq 0$ and $y\geq 0$.

Feasible region and solutions

Draw each boundary line $ax+by=c$. Use a solid line for $\leq$ or $\geq$ and a dashed line for $<$ or $>$. Shade the side that satisfies the inequality (test the point $(0,0)$ when it is not on the line). The overlap of all shaded regions is the feasible region; every point in it satisfies all conditions at once.

Worked example

A stall makes $x$ cups of tea and $y$ cups of coffee. It has at most 12 cups total, and coffee is at least twice the tea. Write the inequalities and find the maximum coffee if $x\geq 1$.
Conditions: $x+y\leq 12$, $y\geq 2x$, $x\geq 1$, $y\geq 0$. Testing integers, $x=1$ gives $y\leq 11$ and $y\geq 2$, so max $y=11$. Check: $1+11=12\leq 12$ and $11\geq 2$. Maximum coffee $=11$ cups.

Remember

  • Solid line includes the boundary; dashed line excludes it.
  • Test $(0,0)$ to decide which side to shade.
  • Integer solutions are lattice points inside or on the feasible region.

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