Q.Determine the maximum value of $Z = 11x + 7y$ subject to the constraints: $2x + y \le 6$, $x \le 2$, $x \ge 0$, $y \ge 0$.

Maximise $Z=11x+7y$ over $2x+y\le 6$, $x\le 2$, $x\ge 0$, $y\ge 0$ by testing the corner points.

Corners of the feasible region:

• $(0,0)$
• $(2,0)$ — from $x=2$, $y=0$
• $(2,2)$ — from $x=2$ and $2x+y=6$
• $(0,6)$ — from $x=0$ and $2x+y=6$

Values of $Z=11x+7y$:

• $(0,0):\;0$
• $(2,0):\;22$
• $(2,2):\;22+14=36$
• $(0,6):\;0+42=42$

The largest is $42$.

Set-up

We maximise $Z=11x+7y$ subject to

By the corner-point theorem the maximum of a linear objective over a bounded region occurs at a vertex, so we only need the vertices.

Step 1 — Find the corner points

The boundary lines are $x=0$, $y=0$, $x=2$ and $2x+y=6$ (intercepts $(3,0),(0,6)$).

• $x=0,\,y=0\Rightarrow (0,0)$.
• $x=2,\,y=0\Rightarrow (2,0)$.
• $x=2$ in $2x+y=6\Rightarrow 4+y=6\Rightarrow y=2\Rightarrow (2,2)$.
• $x=0$ in $2x+y=6\Rightarrow y=6\Rightarrow (0,6)$.

So the feasible region is the quadrilateral $(0,0),(2,0),(2,2),(0,6)$.

Step 2 — Evaluate $Z=11x+7y$

Step 3 — Pick the best

The values are $0,22,36,42$; the maximum is $42$, at $(0,6)$. Check $(0,6)$: $2(0)+6=6\le 6$ and $0\le 2$ — feasible.

Method: Corner-Point Method when One Constraint Cuts Off a Vertex

Use this for a bounded maximisation where a simple bound (like $x\le k$) trims the region, so some "obvious" intersection points are actually infeasible.

Steps

Step 1: Plot all boundaries, including the cutting bound.

Draw each constraint line and the non-negativity axes. A vertical/horizontal bound such as $x\le k$ or $y\le k$ slices across a sloping line.

Step 2: Find candidate intersections — then keep only feasible ones.

Solve each relevant pair of lines. Crucially, an axis-intercept of a sloping line (e.g. where $2x+y=6$ meets the $x$-axis) may lie outside the region because it violates the cutting bound. Discard any intersection that breaks any constraint.

Step 3: Evaluate $Z=ax+by$ at the surviving corners.

Tabulate $Z$ at each genuine vertex of the trimmed polygon and select the optimum.

Common Mistakes

Mistake 1: Using $(3,0)$ as a corner.

Why it's wrong: the line $2x+y=6$ meets the $x$-axis at $(3,0)$, but $x=3$ violates $x\le2$, so $(3,0)$ is outside the feasible region. Correct approach: the bound $x\le2$ cuts the region — the true corner on that edge is $(2,2)$, from $x=2$ in $2x+y=6$.

Mistake 2: Not testing every candidate against all constraints.

Why it's wrong: an intersection of two lines can still break a third constraint and so not be a real vertex. Correct approach: check each candidate corner against $2x+y\le6$, $x\le2$, $x\ge0$, $y\ge0$ before evaluating $Z=11x+7y$.