Q.Find the value of the following: Minimise $Z = -3x + 4y$ subject to $x + 2y \le 8$, $3x + 2y \le 12$, $x \ge 0$, $y \ge 0$.

Concept: Linear Programming — Minimisation with non-negative constraints.

We minimise $Z = -3x + 4y$ under the given constraints.

Step 1 – Plot the feasible region

Constraints:

$x + 2y \le 8$

$3x + 2y \le 12$

$x \ge 0, y \ge 0$

Step 2 – Find corner points

Intersection of $x+2y=8$ and $3x+2y=12$: subtract to get $2x=4 \Rightarrow x=2$, then $y=3$.

Other corners: $(0,0)$, $(4,0)$ (from $3x+2y=12$ with $y=0$), $(0,4)$ (from $x+2y=8$ with $x=0$).

Step 3 – Evaluate $Z$ at each corner

$(0,0): Z = 0$

$(4,0): Z = -12$

$(0,4): Z = 16$

$(2,3): Z = -6 + 12 = 6$

The minimum value is $-12$ at $(4,0)$.

Why This Approach Works

Linear programming problems with two variables are solved graphically. The constraints define a polygon (the feasible region) in the $xy$-plane. The objective function $Z = -3x + 4y$ is linear, so its extreme values (minimum and maximum) must occur at the vertices (corner points) of this polygon — not in the interior. This is the corner point theorem.

Here, the coefficient of $x$ is negative ($-3$), so making $x$ large reduces $Z$. The coefficient of $y$ is positive ($+4$), so increasing $y$ increases $Z$. To minimise $Z$, we want $x$ as large as possible and $y$ as small as possible, within the constraints.

Step-by-Step Solution

1. Write down the constraints clearly

We have:

• $x + 2y \le 8$
• $3x + 2y \le 12$
• $x \ge 0$, $y \ge 0$

These are all linear inequalities. The non-negativity constraints ($x \ge 0$, $y \ge 0$) restrict us to the first quadrant.

2. Find the boundary lines and their intersection points

Convert each inequality to an equation to find the lines:

• Line 1: $x + 2y = 8$
• Line 2: $3x + 2y = 12$

Find where each line meets the axes:

• For Line 1: when $x=0$, $2y=8 \Rightarrow y=4$; when $y=0$, $x=8$. So points: $(0,4)$ and $(8,0)$.
• For Line 2: when $x=0$, $2y=12 \Rightarrow y=6$; when $y=0$, $3x=12 \Rightarrow x=4$. So points: $(0,6)$ and $(4,0)$.

Now find the intersection of the two lines:

Subtract the first equation from the second:

Substitute $x=2$ into $x + 2y = 8$:

So the intersection point is $(2, 3)$.

3. Identify the feasible region

The feasible region is the set of points satisfying all constraints. Since both inequalities are "$\le$", the region lies below both lines (and in the first quadrant). The corner points of this polygon are:

• $(0,0)$ — origin
• $(4,0)$ — from Line 2 on the $x$-axis
• $(0,4)$ — from Line 1 on the $y$-axis
• $(2,3)$ — intersection of the two lines

4. Evaluate the objective function at each corner point

We compute $Z = -3x + 4y$ at each point:

5. Determine the minimum

The smallest value among these is $-12$ at $(4,0)$. Since the feasible region is bounded (a closed polygon), this is the global minimum.

Method: Solving a two-variable LPP graphically (minimisation)

This is the full graphical corner-point method — the technique behind every "linear programming problems class 12 with solutions" question. Here the objective mixes a negative and a positive coefficient.

Steps

Step 1: Draw each constraint line and shade its half-plane.

Replace every inequality by an equation, plot via axis intercepts, and test the origin to pick the correct side. The non-negativity conditions $x\ge0,\ y\ge0$ keep you in the first quadrant.

Step 2: Identify the feasible region and its true corners.

Corners are intersections of boundary lines that satisfy all constraints — discard any intersection that violates even one inequality.

Step 3: Evaluate and minimise.

Compute $Z$ at every valid corner and take the smallest. With a negative $x$-coefficient, expect the minimum where $x$ is largest and $y$ smallest — a useful check on the arithmetic. A bounded region guarantees this minimum is global.

Common Mistakes

Mistake 1: Accepting an axis intercept without checking the other constraint.

Why it's wrong: $(0,6)$ lies on $3x+2y=12$ but fails $x+2y\le8$, so it is not a corner. Correct approach: test every candidate against all constraints before treating it as a vertex.

Mistake 2: Mishandling the $-3x$ term.

Why it's wrong: forgetting the minus sign makes $(4,0)$ look like a large positive value instead of $-12$. Correct approach: a larger $x$ decreases $Z=-3x+4y$, so the minimum is at the largest-$x$, smallest-$y$ corner, $(4,0)$.