Q.Minimise $Z = 3x + 2y$ subject to the constraints: $x + y \ge 8$, $3x + 5y \le 15$, $x \ge 0,\ y \ge 0$.

Concept: Linear Programming Constraints – The feasible region is the intersection of all half-planes defined by the constraints. If no point satisfies all constraints simultaneously, the problem is infeasible.

Step 1 – Write constraints clearly

$x + y \ge 8$ (region above the line $x+y=8$)

$3x + 5y \le 15$ (region below the line $3x+5y=15$)

$x \ge 0,\ y \ge 0$ (first quadrant)

Step 2 – Check for common region

The line $x+y=8$ has intercepts $(8,0)$ and $(0,8)$.

The line $3x+5y=15$ has intercepts $(5,0)$ and $(0,3)$. …

Why this happens — the core idea

In Linear Programming, every constraint cuts the plane into two halves: one that satisfies it, one that doesn't. The feasible region is the intersection of all these half-planes. If that intersection is empty, no point satisfies all constraints simultaneously — the problem has no solution.

Here, the first constraint demands that $x+y$ be at least 8. The second demands that $3x+5y$ be at most 15. Since both $x$ and $y$ are non-negative, these two conditions pull in opposite directions. Let's see exactly why they cannot both hold.

Step-by-step reasoning

1. Write down the constraints clearly

2. Interpret constraint (1)

   The line $x+y=8$ has intercepts $(8,0)$ and $(0,8)$. The inequality $x+y\ge 8$ means we want points on or above this line. Since $x,y\ge 0$, the smallest possible $x+y$ in the first quadrant is $0$ (at the origin), but here we need it to be at least 8. So the feasible region for (1) is the half-plane that starts at the line and goes outward, away from the origin.

3. Interpret constraint (2)

   The line $3x+5y=15$ has intercepts $(5,0)$ and $(0,3)$. The inequality $3x+5y\le 15$ means we want points on or below this line. This half-plane includes the origin, because $3(0)+5(0)=0 \le 15$.

4. Check if the two half-planes overlap

   The first half-plane lies above the line $x+y=8$. The second lies below the line $3x+5y=15$. For an overlap to exist, there must be some point $(x,y)$ with $x,y\ge 0$ that is simultaneously above the first line and below the second.

   Let’s test the extreme point of the second constraint: the point $(5,0)$ lies on $3x+5y=15$. At this point, $x+y=5$, which is less than 8 — so it fails (1).

   The point $(0,3)$ also lies on the second line. Here $x+y=3$, again less than 8.

   In fact, every point that satisfies (2) has $x+y$ at most? Let's find the maximum possible $x+y$ under (2) and non-negativity.

5. Maximise $x+y$ subject to $3x+5y\le 15$, $x,y\ge 0$

   This is a small LP itself. The feasible region for (2) alone is a triangle with vertices $(0,0)$, $(5,0)$, $(0,3)$. The value of $x+y$ at these vertices:

   • $(0,0)$: $0$
   • $(5,0)$: $5$ …

Method: Graphical Method — Detecting an Empty Feasible Region

Not every LP has a solution. This method shows how the graphical technique reveals an infeasible problem, where two constraints pull in opposite directions and leave no common region.

Steps

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

Use intercepts for each line and the origin test for the direction, keeping $x\ge 0,\ y\ge 0$. Watch the relative positions: a $\ge$ constraint pushes the region away from the origin, a $\le$ constraint keeps it near the origin.

Step 2: Check whether the shaded half-planes actually overlap.

The feasible region is the common area of all half-planes. If the "away from origin" region of one constraint lies entirely beyond the "near origin" region of another, there is no overlap.

Step 3: Confirm algebraically with a bounding argument. …

Common Mistakes

Mistake 1: Solving $x+y=8$ and $3x+5y=15$ and quoting that intersection.

Why it's wrong: it gives $x=12.5,\ y=-4.5$, which violates $y\ge 0$ — and the inequality directions oppose each other anyway, so the point is not feasible. Correct approach: a line intersection means nothing unless it satisfies every inequality and the non-negativity conditions.

Mistake 2: Not realising the two half-planes never overlap.

Why it's wrong: $x+y\ge 8$ needs points far from the origin, but every point with $3x+5y\le 15$ and $x,y\ge 0$ has $x+y\le 5$. Since $5<8$, the region is empty. Correct approach: compare the constraints' reach — if one caps $x+y$ below what the other demands, the problem is infeasible.

Mistake 3: Forcing a numeric minimum for an empty region. …