8.2 Linear Inequalities & Linear Programming

Linear Inequalities & Linear Programming

Linear programming is the part of analytic geometry that turns straight lines into decision-making. A manufacturer with limited labour and material, a farmer with limited land and water, or a diet planner with a cost limit all face the same question: which combination of two quantities gives the best result while respecting every constraint? The tools are the straight-line skills from Section 8.1 plus a few new rules about inequalities.

---

1. Linear Inequalities in Two Variables

A linear inequality in $x$ and $y$ has one of the forms

$ax + by < c, \quad ax + by \le c, \quad ax + by > c, \quad ax + by \ge c$

Its solution set is a half-plane: all points on one side of the boundary line $ax + by = c$.

Graphing Procedure

1. Draw the boundary line $ax + by = c$ using its intercepts. Use a solid line for $\le$ or $\ge$ (boundary points are included) and a dashed line for $<$ or $>$ (boundary excluded).
2. Choose a test point not on the line, usually the origin $(0, 0)$.
3. Substitute it into the inequality. If the statement is true, shade the side containing the test point; otherwise shade the other side.

Example: Graph $2x + 3y \le 12$. The boundary $2x + 3y = 12$ passes through $(6, 0)$ and $(0, 4)$ and is drawn solid. Testing $(0, 0)$ gives $0 \le 12$, which is true, so the half-plane containing the origin (below and to the left of the line) is the solution region.

If the line passes through the origin, use another test point such as $(1, 0)$ or $(0, 1)$.

Shortcut for Standard Forms

• $x \ge 0$ is the right half-plane including the $y$-axis; $y \ge 0$ is the upper half-plane including the $x$-axis. Together they give the first quadrant, the usual non-negativity constraints.
• For $y \le mx + c$, shade below the line; for $y \ge mx + c$, shade above it (when the inequality is solved for $y$).

---

2. Systems of Inequalities and the Feasible Region

The solution of a system of linear inequalities is the intersection of the individual half-planes. In linear programming this common region is called the feasible region, and each point in it is a feasible solution.

• The feasible region is always convex: the segment joining any two of its points lies entirely inside it.
• When it is bounded, it is a convex polygon. Its corners are called vertices or corner points.
• Each vertex is the intersection of two boundary lines, found by solving the two equations simultaneously.
• A system may also have an empty feasible region (inconsistent constraints) or an unbounded one.

Example: Find the vertices of the region defined by $x \ge 0$, $y \ge 0$, $x + y \le 6$, $x + 2y \le 8$.

• $x + y = 6$ meets the $x$-axis at $(6, 0)$.
• $x + 2y = 8$ meets the $y$-axis at $(0, 4)$.
• $x + y = 6$ and $x + 2y = 8$ intersect where $y = 2$, $x = 4$, so at $(4, 2)$.

The feasible region is the quadrilateral with vertices $(0, 0)$, $(6, 0)$, $(4, 2)$ and $(0, 4)$. Check each vertex against every constraint: for example, $(6, 0)$ gives $6 + 0 = 6 \le 8$ in the second constraint, so it is feasible; the point $(8, 0)$, where $x + 2y = 8$ meets the $x$-axis, violates $x + y \le 6$ and is not a vertex.

---

3. Linear Programming: Objective Function and Corner-Point Theorem

A linear programming problem asks for the maximum or minimum of a linear objective function

$z = px + qy$

subject to linear constraints. The key result is:

Corner-Point Theorem: If the feasible region is bounded, the objective function attains both its maximum and its minimum at vertices of the region.

The reason is geometric: the lines $px + qy = k$ for different $k$ are parallel. Sliding such a line across the region, the last point it touches is always a corner (or a whole edge).

Corner-Point Method

1. Graph the constraints and identify the feasible region.
2. Find the coordinates of every vertex.
3. Evaluate $z$ at each vertex.
4. The largest value is the maximum; the smallest is the minimum.

Continuing the example, maximize $z = 3x + 4y$ over the region above:

The maximum is $z = 20$ at $(4, 2)$; the minimum is $z = 0$ at $(0, 0)$.

Special Cases

• Multiple optima: if the objective line is parallel to an edge, both endpoints of that edge give the same optimal value, and so does every point between them. For $z = x + y$ in the example, $(6, 0)$ and $(4, 2)$ both give $z = 6$.
• Unbounded region: a minimum may still exist (common in cost problems with $\ge$ constraints), but a maximum may not. Check whether $z$ can increase without limit along an unbounded edge.

---

4. Formulating a Linear Programming Problem

Word problems are converted into mathematics in four steps:

1. Decision variables: let $x$ and $y$ be the quantities to be decided.
2. Constraints: translate each resource limit into an inequality.
3. Non-negativity: add $x \ge 0$ and $y \ge 0$.
4. Objective: write the quantity to maximize (profit, output) or minimize (cost, time).

Worked example: A workshop makes chairs and tables. A chair needs 2 hours of carpentry and 1 hour of polishing; a table needs 3 hours of carpentry and 3 hours of polishing. Each week 36 carpentry hours and 24 polishing hours are available. Profit is Rs. 500 per chair and Rs. 900 per table. How many of each should be made?

• Variables: $x$ chairs, $y$ tables.
• Constraints: $2x + 3y \le 36$ (carpentry), $x + 3y \le 24$ (polishing), $x \ge 0$, $y \ge 0$.
• Objective: maximize $P = 500x + 900y$.

Vertices: $(0, 0)$; $(18, 0)$ from $2x = 36$; $(0, 8)$ from $3y = 24$; and the intersection of $2x + 3y = 36$ with $x + 3y = 24$, which gives $x = 12$, $y = 4$.

The best plan is 12 chairs and 4 tables per week, for a profit of Rs. 9,600.

---

5. Quick Checks for MCQs

• A vertex claimed in an answer option must satisfy every constraint; substituting it takes seconds and eliminates wrong options.
• The optimum is never at an interior point for a linear objective.
• Changing the objective function changes which vertex is optimal, but never the feasible region itself.
• A strict inequality ($<$) excludes the boundary, so a "maximum" may not be attained; Intermediate problems normally use $\le$ and $\ge$ to avoid this.