1. Meaning of the Corner Point Method
The corner point method is a simple way to find the optimal (maximum or minimum) value of the objective function in a linear programming problem. The idea is that when the feasible region is a polygon, the best solution always lies at one of its corner points.
These corner points are also called vertices of the feasible region.
2. Why the Optimal Value Occurs at a Corner
A linear objective function forms parallel straight lines when plotted on a graph. As these parallel lines move across the feasible region, the last point they touch before leaving the region is always a corner point.
This is why it is enough to check only the corner points instead of checking every point inside the region.
3. Steps in the Corner Point Method
The method follows a clear and simple process:
3.1. Step 1: Draw the Feasible Region
Plot all constraints on the graph and shade the region that satisfies all of them. The shape formed is the feasible region.
3.2. Step 2: Identify the Corner Points
Locate the points where the boundary lines intersect each other or the axes. These points form the vertices of the feasible region.
You can find corner points by:
- reading directly from the graph, or
- solving pairs of equations algebraically.
3.3. Step 3: Evaluate the Objective Function at Each Corner Point
Substitute the coordinates of each corner point into the objective function \( Z = ax + by \).
This gives the value of \( Z \) at that corner.
3.4. Step 4: Choose the Best Value
If the goal is to maximize \( Z \), choose the largest value obtained. If the goal is to minimize \( Z \), choose the smallest value.
The point where this best value occurs is the optimal solution.
4. Example of the Corner Point Method
Consider the feasible region defined by:
\( x + y \le 6 \)
\( x \ge 0 \)
\( y \ge 0 \)
Objective function:
\( Z = 5x + 3y \)
4.1. Finding the Corner Points
From the graph, the feasible region has the following corners:
- \( (0, 0) \)
- \( (6, 0) \)
- \( (0, 6) \)
4.2. Evaluating the Objective Function
- At \( (0, 0) \):
\( Z = 5(0) + 3(0) = 0 \)
- At \( (6, 0) \):
\( Z = 5(6) + 3(0) = 30 \)
- At \( (0, 6) \):
\( Z = 5(0) + 3(6) = 18 \)
4.3. Choosing the Optimal Value
The maximum value of \( Z \) is 30, which occurs at the point \( (6, 0) \). So the optimal solution is:
- \( x = 6 \)
- \( y = 0 \)
- Maximum \( Z = 30 \)