Q.Maximise $Z = 3x + 4y$, subject to the constraints: $x + y \le 1$, $x \ge 0$, $y \ge 0$.

Concept: Linear Programming Graphical Method — the optimum lies at a corner of the feasible region.

Steps:

1. Plot the constraint $x + y \le 1$ as a line through $(1,0)$ and $(0,1)$. The feasible region is the triangle with vertices $(0,0)$, $(1,0)$, and $(0,1)$ (since $x, y \ge 0$).

2. Evaluate $Z = 3x + 4y$ at each corner:

   • At $(0,0)$: $Z = 0$
   • At $(1,0)$: $Z = 3$
   • At $(0,1)$: $Z = 4$

3. The maximum value is $4$, occurring at $(0,1)$.

The graphical method for linear programming works because the optimal value of a linear objective, subject to linear constraints, always occurs at a corner (vertex) of the feasible region — provided the region is bounded. Here, the constraints are few and simple, so we can visualise everything on the $xy$-plane.

The constraint $x + y \le 1$ is a half-plane below the line $x + y = 1$. Together with $x \ge 0$ and $y \ge 0$, this forms a triangle with vertices at the origin and the two intercepts of the line. The objective $Z = 3x + 4y$ is a family of parallel lines; increasing $Z$ shifts the line outward. The last corner touched as we push outward gives the maximum.

Let’s go step by step.

1. Plot the constraints.

   The line $x + y = 1$ meets the $x$-axis at $(1,0)$ and the $y$-axis at $(0,1)$. The inequality $x + y \le 1$ means we take the region below this line (including the line itself). The non-negativity constraints $x \ge 0$, $y \ge 0$ restrict us to the first quadrant.

   The feasible region is the triangle with vertices:

   • $A(0,0)$
   • $B(1,0)$
   • $C(0,1)$

2. Evaluate the objective at each vertex.

   At $A(0,0)$: $Z = 3(0) + 4(0) = 0$

   At $B(1,0)$: $Z = 3(1) + 4(0) = 3$

   At $C(0,1)$: $Z = 3(0) + 4(1) = 4$

3. Compare values.

   The largest value among $\{0, 3, 4\}$ is $4$, occurring at $(0,1)$.

Since the feasible region is bounded and the objective is linear, the maximum is indeed at a vertex. No need to check interior points or edges — the corner point theorem guarantees it.

Method: Maximising a Linear Objective over a Triangular Region

Use this for a compact two-variable maximisation whose feasible region is a small triangle formed by one sloping constraint and the two axes.

Steps

Step 1: Draw the constraint and close the region with the axes.

Plot the sloping line (via its intercepts) and combine it with $x\ge0,\,y\ge0$. With a single upper-bound constraint like $x+y\le c$, the region is the triangle whose vertices are the origin and the line's two axis-intercepts.

Step 2: List all three corners — do not skip the origin.

The vertices are $(0,0)$ and the two intercepts. The origin is a genuine corner and must be included.

Step 3: Evaluate $Z=ax+by$ at each corner and choose the largest.

By the Corner-Point Theorem the maximum is at a vertex. A quick sense-check: with a trade-off constraint $x+y\le c$, the objective leans toward whichever variable has the larger coefficient, so the optimum tends to sit on that variable's axis — but confirm by computing all three values.

Common Mistakes

Mistake 1: Checking only the intercepts and skipping the origin.

Why it's wrong: the corners are $(0,0)$, $(1,0)$ and $(0,1)$; ignoring $(0,0)$ drops a genuine vertex. Correct approach: with $x+y\le1$ and $x,y\ge0$, list all three triangle vertices before evaluating.

Mistake 2: Assuming the larger $x$-term dominates.

Why it's wrong: the coefficient of $x$ is $3$ and of $y$ is $4$; under the trade-off $x+y\le1$ the objective prefers $y$, so the maximum is at $(0,1)$ with $Z=4$, not $(1,0)$ with $Z=3$. Correct approach: compare $Z=3x+4y$ at every corner rather than guessing from the $x$-coefficient.