Q.Find the value of the following: Maximise $Z = 3x + 2y$ subject to $x + 2y \le 10$, $3x + y \le 15$, $x, y \ge 0$.

Concept: Linear Programming — finding the maximum of $Z$ under given constraints.

Step 1 — Convert constraints to equations and find corner points.

$x + 2y = 10$ and $3x + y = 15$. Solve:

From $3x + y = 15$, $y = 15 - 3x$. Substitute into $x + 2(15 - 3x) = 10$ gives $x + 30 - 6x = 10 \Rightarrow -5x = -20 \Rightarrow x = 4$, then $y = 3$.

Other corner points: $(0,0)$, $(5,0)$ (from $3x + y = 15$ with $y=0$), $(0,5)$ (from $x + 2y = 10$ with $x=0$).

Step 2 — Evaluate $Z = 3x + 2y$ at each corner. …

This is a linear programming problem — you’re asked to maximise a linear function (the objective) over a region defined by linear inequalities. The key idea is that the maximum (or minimum) of a linear function over a convex polygon always occurs at one of the vertices (corner points) of the feasible region. So we don’t need to check every point inside; we just find where the constraints intersect and evaluate $Z$ at those corners.

Let’s work through it.

1. Plot the constraints and find the feasible region.

   The constraints are:

   $x + 2y \le 10$

   $3x + y \le 15$

   $x \ge 0$, $y \ge 0$

   The non-negativity conditions mean we’re only in the first quadrant. Each inequality is a half-plane; the feasible region is their intersection — a polygon.

2. Find the corner points.

   These come from:

   • Intersection of each constraint line with the axes.

   • Intersection of the two constraint lines with each other.

   • For $x + 2y = 10$:

     If $x = 0$, then $y = 5$ → point $(0, 5)$.

     If $y = 0$, then $x = 10$ → point $(10, 0)$.

   • For $3x + y = 15$:

     If $x = 0$, then $y = 15$ → point $(0, 15)$.

     If $y = 0$, then $x = 5$ → point $(5, 0)$.

   • Intersection of the two lines:

     Solve $x + 2y = 10$ and $3x + y = 15$.

     From the second, $y = 15 - 3x$. Substitute into the first:

     $x + 2(15 - 3x) = 10$

     $x + 30 - 6x = 10$

     $-5x = -20$

     $x = 4$

     Then $y = 15 - 3(4) = 3$.

     So the intersection is $(4, 3)$.

   Also, the origin $(0, 0)$ is always a corner when $x, y \ge 0$.

3. Which of these are actually in the feasible region?

   Not every intersection with axes is feasible — it must satisfy all constraints.

   • $(0, 0)$: satisfies both inequalities. Feasible.
   • $(0, 5)$: check $3(0) + 5 = 5 \le 15$ → OK. Feasible.
   • $(0, 15)$: check $0 + 2(15) = 30 \le 10$? No. So $(0, 15)$ is not feasible.
   • $(5, 0)$: check $5 + 2(0) = 5 \le 10$ → OK. Feasible.
   • $(10, 0)$: check $3(10) + 0 = 30 \le 15$? No. So $(10, 0)$ is not feasible. …

Method: Solving a two-variable LPP graphically (pruning infeasible intercepts)

The full graphical method, focused on the step students most often botch: rejecting axis intercepts that fail the other constraints.

Steps

Step 1: Draw the constraints and shade.

Plot each line by intercepts; test the origin to choose the half-plane; restrict to the first quadrant with $x,y\ge0$.

Step 2: List candidate corners, then filter by ALL constraints. …

Common Mistakes

Mistake 1: Treating every axis intercept as a corner.

Why it's wrong: $(0,15)$ fails $x+2y\le10$ and $(10,0)$ fails $3x+y\le15$, so neither is feasible. Correct approach: test each intercept against all constraints and discard the ones that violate any.

Mistake 2: Overlooking the intersection vertex $(4,3)$. …