Q.Corner points of the feasible region determined by the system of linear constraints are $(0, 3)$, $(1, 1)$ and $(3, 0)$. Let $Z = px + qy$, where $p, q > 0$. Condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3, 0)$ and $(1, 1)$ is
(A) $p = 2q$
(B) $p = \dfrac{q}{2}$
(C) $p = 3q$
(D) $p = q$

Why the Corner Point Theorem matters here

In linear programming, corner points (also called extreme points) are the vertices of the feasible region — the points where two or more boundary lines meet. The Corner Point Theorem tells us that if an optimal solution exists, it occurs at one of these corner points. But even when we're just counting them, the key is to find all intersections of the constraint boundaries that satisfy all constraints.

The constraints here are:

• $x \geq 0$ (the $y$-axis and everything to its right)
• $y \geq 0$ (the $x$-axis and everything above it)
• $x + y \geq 4$ (the half-plane above the line $x + y = 4$)

The first two constraints restrict us to the first quadrant. The third constraint cuts off the region near the origin.

Step-by-step reasoning

1. Identify the boundary lines.

   Each inequality becomes an equality at its boundary:

   $x = 0$, $y = 0$, and $x + y = 4$.

2. Find all pairwise intersections of these lines.

   • Intersection of $x = 0$ and $y = 0$: $(0,0)$.
   • Intersection of $x = 0$ and $x + y = 4$: substitute $x = 0$ gives $0 + y = 4$, so $(0,4)$.
   • Intersection of $y = 0$ and $x + y = 4$: substitute $y = 0$ gives $x + 0 = 4$, so $(4,0)$.

   So we have three candidate points: $(0,0)$, $(0,4)$, and $(4,0)$.

3. Check which candidates satisfy all constraints.

   • At $(0,0)$: $x \geq 0$ ✓, $y \geq 0$ ✓, but $x + y \geq 4$ becomes $0 \geq 4$ ✗. So $(0,0)$ is not in the feasible region.
   • At $(0,4)$: $x \geq 0$ ✓, $y \geq 0$ ✓, $0 + 4 \geq 4$ ✓. So $(0,4)$ is a corner point.
   • At $(4,0)$: $x \geq 0$ ✓, $y \geq 0$ ✓, $4 + 0 \geq 4$ ✓. So $(4,0)$ is a corner point.

4. Are there any other corner points? …

The Corner Point Theorem (also called the Fundamental Theorem of Linear Programming) tells us that if a linear programming problem has an optimal solution, that solution must occur at a vertex (corner point) of the feasible region. This is because the objective function is linear — its level lines are straight lines, and as you slide them across the convex polygon of feasible points, the last point touched before leaving the region is always a corner.

So here, we don’t need to know the constraints. The three corner points given are the only candidates for both maximum and minimum of $Z$. We simply evaluate $Z$ at each point and compare.

1. At $(2, 72)$:

   $Z = 18(2) + 9(72) = 36 + 648 = 684$

2. At $(15, 20)$:

   $Z = 18(15) + 9(20) = 270 + 180 = 450$

3. At $(40, 15)$:

   $Z = 18(40) + 9(15) = 720 + 135 = 855$

Now arrange them in order:

• Minimum value: $450$ at $(15, 20)$
• Maximum value: $855$ at $(40, 15)$ …

The Corner Point Theorem (also called the Fundamental Theorem of Linear Programming) tells us that if a linear programming problem has an optimal solution (a maximum or minimum), that solution must occur at one of the corner points (vertices) of the feasible region. This is because the objective function is linear, and the feasible region is a convex polygon — the function’s value changes linearly as you move across the region, so the extreme values will always be at the boundaries, specifically at the corners.

So, to find where $z$ is maximum and where it is minimum, we don’t need to graph anything or solve inequalities. We simply plug each corner point into $z = 18x + 9y$ and compare the results.

1. Evaluate at $(2, 72)$

   $z = 18(2) + 9(72) = 36 + 648 = 684$

2. Evaluate at $(15, 20)$

   $z = 18(15) + 9(20) = 270 + 180 = 450$

3. Evaluate at $(40, 15)$

   $z = 18(40) + 9(15) = 720 + 135 = 855$

Now compare the three values:

• $684$ at $(2, 72)$
• $450$ at $(15, 20)$
• $855$ at $(40, 15)$

The largest is $855$ at $(40, 15)$ — that’s the maximum.

The smallest is $450$ at $(15, 20)$ — that’s the minimum. …

In Linear Programming Problems (LPPs), we aim to optimize (maximize or minimize) a linear objective function subject to a set of linear constraints. These constraints define a region in the coordinate plane called the feasible region. This region is always a convex polygon (or an unbounded convex region).

The core idea behind solving such problems is the Corner Point Theorem. This theorem provides a powerful shortcut:

Why does this work?

Imagine the objective function, say $P = 3x + 2y$, as a family of parallel lines $3x + 2y = k$, where $k$ is the value of the profit. As we change $k$, these lines shift parallel to each other. To maximize $P$, we want to find the line with the largest possible $k$ that still intersects the feasible region.

Since the feasible region is a convex polygon, the "last" point this moving line will touch before leaving the region entirely will always be one of its vertices (corner points). Similarly, for minimization, the "first" point touched will also be a vertex. This geometric intuition is why we only need to check the corner points.

Let's apply this to the given problem.

1. Identify the Objective Function and Corner Points:

   We are given the objective function $P = 3x + 2y$, which we need to maximize.

   The corner points of the feasible region are provided as:

   • $(0, 0)$
   • $(0, 8)$
   • $(2, 7)$
   • $(5, 4)$
   • $(6, 0)$

2. Evaluate the Objective Function at Each Corner Point:

   We substitute the coordinates $(x, y)$ of each corner point into the profit function $P = 3x + 2y$ to find the profit value at that point.

   • At $(0, 0)$:

     $P = 3(0) + 2(0) = 0 + 0 = 0$

   • At $(0, 8)$:

     $P = 3(0) + 2(8) = 0 + 16 = 16$

   • At $(2, 7)$: …

By the corner-point method, evaluate $z=4x+y$ at every corner of the feasible region:

$(0,0):\;0,\qquad (0,50):\;50,$ …