Solved Examples

How to Use This Section

Sixteen fully worked problems in four graded batches: straightforward bounded problems, then floor-constraint and unbounded problems, then double-objective and pathological cases (no optimum, no region), then a mixed batch with fractional corners and ties. Attempt each on paper — draw the region every single time — before reading the solution.

Every solution below follows the same skeleton: corners → table → conclusion, with the half-plane test inserted whenever the region is unbounded. That skeleton is the exam answer format.

Key Point: the most common LP errors are not arithmetic. They are: mis-shading a half-plane (test with the origin!), missing a corner, evaluating $Z$ at a point outside the region, and trusting the table over an unbounded region without the test. Each batch below deliberately walks through these traps.

Batch 1 — Bounded Regions (Easy)

Example 1: One constraint plus the axes

Maximise $Z = 3x + 4y$ subject to $x + y \leq 4$, $x, y \geq 0$.

Solution:

1. Region: the triangle with corners $(0,0)$, $(4,0)$, $(0,4)$.
2. Table: $Z = 0, 12, 16$.

Answer: maximum $Z = 16$ at $(0, 4)$ — the larger coefficient ($4$, on $y$) pulls the optimum to the $y$-intercept.

Example 2: A negative coefficient

Minimise $Z = -3x + 4y$ subject to $x + 2y \leq 8$, $3x + 2y \leq 12$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $(4,0)$, $(2,3)$ (from $x + 2y = 8$, $3x + 2y = 12$), $(0,4)$.
2. Table: $0, -12, 6, 16$.

Answer: minimum $Z = -12$ at $(4, 0)$ — negative objective values are perfectly legal; the table sorts them like any others.

Example 3: A fractional corner wins

Maximise $Z = 5x + 3y$ subject to $3x + 5y \leq 15$, $5x + 2y \leq 10$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $(2,0)$, $\left(\dfrac{20}{19}, \dfrac{45}{19}\right)$ (solving the two lines), $(0,3)$.
2. Table: $0, 10, \dfrac{235}{19}, 9$.

Answer: maximum $Z = \dfrac{235}{19}$ at $\left(\dfrac{20}{19}, \dfrac{45}{19}\right)$ — corners need not be whole numbers, and $\dfrac{235}{19} \approx 12.4 > 10$; never round a corner away.

Example 4: A clean bounded maximisation

Maximise $Z = 3x + 2y$ subject to $x + 2y \leq 10$, $3x + y \leq 15$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $(5,0)$, $(4,3)$, $(0,5)$.
2. Table: $0, 15, 18, 10$.

Answer: maximum $Z = 18$ at $(4, 3)$.

Batch 2 — Floors and Unbounded Regions (Medium)

Example 5: An unbounded minimum that DOES exist

Minimise $Z = 3x + 5y$ subject to $x + 3y \geq 3$, $x + y \geq 2$, $x, y \geq 0$.

Solution:

1. Corners of the unbounded region: $(3,0)$, $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$ (solving the two lines), $(0,2)$.
2. Table: $9, 7, 10$ — smallest $m = 7$.
3. Half-plane test (region unbounded): does $3x + 5y < 7$ meet the region? The open half-plane lies below the line $3x + 5y = 7$, and the whole feasible region sits on or above both floor lines — no common point.

Answer: minimum $Z = 7$ at $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$ — the test passed, so the smallest table entry is genuinely the minimum.

Example 6: Minimum along a whole edge

Minimise $Z = x + 2y$ subject to $2x + y \geq 3$, $x + 2y \geq 6$, $x, y \geq 0$, and show the minimum occurs at more than two points.

Solution:

1. Corners: $(6, 0)$ and $(0, 3)$ (the line $x + 2y = 6$ meets both axes inside the region; $(0,3)$ satisfies $2x + y = 3$ exactly).
2. Table: $Z(6,0) = 6$ and $Z(0,3) = 6$ — a tie.
3. The tie's meaning: the objective $x + 2y$ is parallel to the binding constraint $x + 2y \geq 6$, so every point of the segment joining $(6,0)$ and $(0,3)$ gives $Z = 6$; the unbounded-region test on $x + 2y < 6$ finds no common point.

Answer: minimum $Z = 6$ at every point of the segment from $(6, 0)$ to $(0, 3)$ — infinitely many optimal solutions.

Example 7: A shifted lower bound

Maximise $Z = 2x + 3y$ subject to $x + y \leq 4$, $x \geq 1$, $y \geq 0$.

Solution:

1. Corners: $(1, 0)$, $(4, 0)$, $(1, 3)$ — the constraint $x \geq 1$ slices off the strip left of $x = 1$.
2. Table: $2, 8, 11$.

Answer: maximum $Z = 11$ at $(1, 3)$ — constraints like $x \geq 1$ are vertical boundary lines, treated exactly like any other.

Example 8: Unbounded, yet the minimum survives — on an edge

Minimise $Z = x + y$ subject to $x + y \geq 1$, $x, y \geq 0$.

Solution:

1. Corners: $(1, 0)$ and $(0, 1)$, both giving $Z = 1$.
2. Test: $x + y < 1$ is exactly the excluded side of the floor constraint — no common point with the region.

Answer: minimum $Z = 1$, attained at every point of the segment joining $(1,0)$ and $(0,1)$ — an unbounded region with a tie and a genuine minimum, all in the simplest possible problem.

Batch 3 — Double Objectives and Pathologies (Medium-Hard)

Example 9: Minimise and maximise together

Minimise and maximise $Z = 5x + 10y$ subject to $x + 2y \leq 120$, $x + y \geq 60$, $x - 2y \geq 0$, $x, y \geq 0$.

Solution:

1. Corners: $(60, 0)$, $(120, 0)$, $(60, 30)$, $(40, 20)$.
2. Table: $300, 600, 600, 400$.
3. Read both ends: minimum $300$; maximum $600$ tied at $(120, 0)$ and $(60, 30)$.

Answer: minimum $Z = 300$ at $(60, 0)$; maximum $Z = 600$ at every point of the segment joining $(120, 0)$ and $(60, 30)$.

Example 10: Another double, with the tie at the minimum

Minimise and maximise $Z = x + 2y$ subject to $x + 2y \geq 100$, $2x - y \leq 0$, $2x + y \leq 200$, $x, y \geq 0$.

Solution:

1. Corners: $(0, 50)$, $(20, 40)$, $(50, 100)$, $(0, 200)$.
2. Table: $100, 100, 250, 400$.

Answer: minimum $Z = 100$ at every point of the segment joining $(0, 50)$ and $(20, 40)$ (the objective is parallel to the floor constraint); maximum $Z = 400$ at $(0, 200)$.

Example 11: No maximum exists

Maximise $Z = -x + 2y$ subject to $x \geq 3$, $x + y \geq 5$, $x + 2y \geq 6$, $y \geq 0$.

Solution:

1. Corners of the unbounded region: $(6, 0)$, $(4, 1)$, $(3, 2)$; values $-6, -2, 1$ — largest $M = 1$.
2. Half-plane test: does $-x + 2y > 1$ meet the region? Yes — for large $y$ (with $x = 3$, any $y > 2$ works), the region climbs without limit and $Z$ grows with it.

Answer: $Z$ has no maximum — every floor constraint leaves the region open upward, and the objective rewards exactly that direction.

Example 12: No feasible region

Maximise $Z = x + y$ subject to $x - y \leq -1$, $-x + y \leq 0$, $x, y \geq 0$.

Solution:

1. Read the two constraints: $x - y \leq -1$ says $y \geq x + 1$; $-x + y \leq 0$ says $y \leq x$.
2. Contradiction: no point can satisfy $y \geq x + 1$ and $y \leq x$ simultaneously.

Answer: there is no feasible region and hence no maximum — recognising the contradiction algebraically saves the whole graph.

Batch 4 — Mixed Practice (Medium)

Example 13: Two floors, one cheap corner

Minimise $Z = 5x + 7y$ subject to $2x + y \geq 8$, $x + 2y \geq 10$, $x, y \geq 0$.

Solution:

1. Corners: $(0, 8)$, $(2, 4)$ (solving the two lines), $(10, 0)$.
2. Table: $56, 38, 50$; smallest $m = 38$.
3. Test (unbounded): $5x + 7y < 38$ shares no point with the region.

Answer: minimum $Z = 38$ at $(2, 4)$.

Example 14: A comfortable maximisation

Maximise $Z = 10x + 6y$ subject to $3x + y \leq 12$, $2x + 5y \leq 34$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $(4,0)$, $(2,6)$ (solving the lines), $\left(0, \dfrac{34}{5}\right)$.
2. Table: $0, 40, 56, \dfrac{204}{5} = 40.8$.

Answer: maximum $Z = 56$ at $(2, 6)$.

Example 15: A designed tie

Maximise $Z = 2x + 4y$ subject to $x + 2y \leq 8$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $(8,0)$, $(0,4)$.
2. Table: $0, 16, 16$ — tie at both nonzero corners.
3. Why: $Z = 2x + 4y = 2(x + 2y)$ is a multiple of the constraint's left side — the objective line is parallel to the boundary $x + 2y = 8$.

Answer: maximum $Z = 16$ at every point of the segment joining $(8, 0)$ and $(0, 4)$ — whenever the objective is proportional to a binding constraint, expect a whole optimal edge.

Example 16: Fractional corner again, minimum this time

Maximise $Z = 3x + 5y$ subject to $x + 2y \leq 6$, $3x + 2y \leq 10$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $\left(\dfrac{10}{3}, 0\right)$, $(2, 2)$ (solving the lines), $(0, 3)$.
2. Table: $0, 10, 16, 15$.

Answer: maximum $Z = 16$ at $(2, 2)$ — the interior intersection beats both intercept corners, as it usually does when the objective weights both variables comparably.