Q.For the linear programming problem (LPP), the objective function is $Z=4x+3y$ and the feasible region determined by a set of constraints is shown in the graph: (Note: The figure is not to scale.) Which of the following statements is true?
(A) Maximum value of $Z$ is at $R(40,0)$.
(B) Maximum value of $Z$ is at $Q(30,20)$.
(C) Value of $Z$ at $R(40,0)$ is less than the value at $P(0,40)$.
(D) The value of $Z$ at $Q(30,20)$ is less than the value at $R(40,0)$.

Concept: Corner Point Theorem – For a linear programming problem with a bounded feasible region, the optimal value of the objective function occurs at one of the corner points of the region.

Step 1: Identify the corner points from the graph.

The feasible region has vertices $P(0,40)$, $Q(30,20)$, and $R(40,0)$.

Step 2: Evaluate $Z = 4x + 3y$ at each corner point.

• At $P(0,40)$: $Z = 4(0) + 3(40) = 120$
• At $Q(30,20)$: $Z = 4(30) + 3(20) = 120 + 60 = 180$
• At $R(40,0)$: $Z = 4(40) + 3(0) = 160$

Step 3: Compare the values.

$Z$ is maximum at $Q(30,20)$ with value 180. …

The Corner Point Theorem (also called the Fundamental Theorem of Linear Programming) is the key idea here. It states that if a linear programming problem has an optimal solution and the feasible region is a bounded convex polygon, then the optimum occurs at one of the vertices (corner points) of that region. This saves us from checking every single point inside the region — we only need to evaluate the objective function at the corners.

In the given graph, the feasible region is a triangle with vertices $P(0,40)$, $Q(30,20)$, and $R(40,0)$. Let’s evaluate $Z=4x+3y$ at each.

1. At $P(0,40)$:

   $Z = 4(0) + 3(40) = 0 + 120 = 120$

2. At $Q(30,20)$:

   $Z = 4(30) + 3(20) = 120 + 60 = 180$

3. At $R(40,0)$:

   $Z = 4(40) + 3(0) = 160 + 0 = 160$

Now compare the values: $120$, $180$, $160$. The largest is $180$ at $Q(30,20)$. So the maximum value of $Z$ is at $Q$. …

Method: Corner-Point Test for the Optimal Vertex

Use this when a bounded feasible region's vertices are known (or readable from a graph) and you must decide which corner optimises $Z=ax+by$, or judge which comparison statement about the corners is true.

Steps

Step 1: List every corner point.

By the Corner-Point Theorem, the optimum of a linear objective on a bounded region always sits at a vertex — so the interior can be ignored. Collect all vertices of the region.

Step 2: Evaluate $Z=ax+by$ at each corner.

Substitute each vertex's coordinates into $Z$ and tabulate the values. Do this for every corner — never assume the answer from a coordinate.

Step 3: Compare, then test the claim. …

Common Mistakes

Mistake 1: Assuming the corner with the largest $x$ gives the maximum.

Why it's wrong: $R(40,0)$ has the biggest $x$, but $Z(R)=160$ while $Z(Q)=4(30)+3(20)=180$ — the interior-edge corner $Q$ wins because $y$ carries weight $3$. Correct approach: evaluate $Z=4x+3y$ at every corner and compare, don't judge by a single coordinate.

Mistake 2: Mis-comparing two corner values in the statements. …