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

The key idea is that the maximum of a linear objective under linear constraints occurs at a corner point of the feasible region.

Step 1 – Feasible region

The constraints $x + y \le 4$, $x \ge 0$, $y \ge 0$ define a right triangle with vertices at $(0,0)$, $(4,0)$, and $(0,4)$.

Step 2 – Evaluate $Z$ at each corner

• At $(0,0)$: $Z = 3(0) + 4(0) = 0$
• At $(4,0)$: $Z = 3(4) + 4(0) = 12$
• At $(0,4)$: $Z = 3(0) + 4(4) = 16$

Step 3 – Compare

The largest value is $16$ at $(0,4)$.

The core idea here is that in linear programming, the maximum (or minimum) of a linear objective function under linear constraints always occurs at a corner point of the feasible region — provided the region is bounded. This is the corner point theorem. So instead of checking every possible point (which is infinite), we only need to examine the vertices of the region formed by the constraints.

Let’s build the feasible region step by step.

1. Plot the constraints.

   The inequality $x + y \le 4$ describes all points on or below the line $x + y = 4$.

   The conditions $x \ge 0$ and $y \ge 0$ restrict us to the first quadrant.

   So the feasible region is the triangle with vertices at $(0,0)$, $(4,0)$, and $(0,4)$.

2. Identify the corner points.

   These are the intersections of the boundary lines:

   • Intersection of $x = 0$ and $y = 0$: $(0,0)$
   • Intersection of $y = 0$ and $x + y = 4$: $(4,0)$
   • Intersection of $x = 0$ and $x + y = 4$: $(0,4)$

   There is no fourth corner because the line $x + y = 4$ meets the axes exactly at these two points.

3. Evaluate the objective function at each corner.

   $Z = 3x + 4y$:

   • At $(0,0)$: $Z = 3(0) + 4(0) = 0$
   • At $(4,0)$: $Z = 3(4) + 4(0) = 12$
   • At $(0,4)$: $Z = 3(0) + 4(4) = 16$

4. Compare the values.

   The largest is $16$ at $(0,4)$.

Method: Graphical Corner-Point Method (Bounded Maximisation)

The standard route for maximising a linear objective $Z=ax+by$ subject to two-variable linear constraints.

Steps

Step 1: Plot each constraint line.

Replace every inequality by an equation and draw the line, usually via its axis intercepts. Include the non-negativity lines $x=0$ and $y=0$.

Step 2: Shade the correct half-plane and find the feasible region.

Test the origin in each inequality: if it holds, keep the origin's side; otherwise the other. The feasible region is the overlap of all the kept half-planes — for constraints like $x+y\le c$ with $x,y\ge0$ it is a closed polygon in the first quadrant.

Step 3: Read off the corner points.

The vertices are where boundary lines cross — solve the relevant pairs of equations (or read them from the graph).

Step 4: Evaluate $Z$ at every corner and pick the best.

By the Corner-Point Theorem the maximum sits at a vertex, so tabulate $Z=ax+by$ at each and take the largest.

Common Mistakes

Mistake 1: Assuming the maximum is where $x$ is largest.

Why it's wrong: $(4,0)$ gives $Z=12$, but $(0,4)$ gives $Z=3(0)+4(4)=16$ — larger, because $y$'s coefficient ($4$) beats $x$'s ($3$). Correct approach: evaluate $Z=3x+4y$ at all three corners $(0,0),(4,0),(0,4)$ and take the biggest.

Mistake 2: Forgetting the origin as a corner.

Why it's wrong: skipping $(0,0)$ can lose the minimum (and, in other problems, the maximum). Correct approach: list every vertex of the triangle, including the origin, before evaluating.