Question

Minimize Z = 3x + 2y by graphical method under the following constraints:
x + y \( \ge \) 8,
3x + 5y \( \le \) 15,
x \( \ge \) 0, y \( \ge \) 0

Hint:

When solving an LPP graphically, always shade the regions carefully. If the shaded areas for all constraints do not overlap, the feasible region is empty, and the problem has no solution. It's a good first step to quickly check if the regions could possibly overlap.

Answer 1

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The goal is to find the minimum value of an objective function Z, subject to a set of linear constraints. The graphical method involves plotting the constraints to find the feasible region (the set of all points satisfying all constraints) and then evaluating the objective function at the corner points of this region.
Step 2: Key Formula or Approach:
1. Treat each inequality as an equation to plot the boundary lines.
2. Determine the region represented by each inequality (by testing a point like (0,0)).
3. Identify the common region that satisfies all constraints simultaneously (the feasible region).
4. If a feasible region exists, find its vertices (corner points). The optimal solution (minimum or maximum) will occur at one of these vertices.
Step 3: Detailed Explanation or Calculation:
Let's analyze the constraints:
Constraint 1: \( x + y \ge 8 \). The boundary line is \( x+y=8 \). This line passes through (8,0) and (0,8). Since \( 0+0 \ge 8 \) is false, the region is the half-plane that does not contain the origin.
Constraint 2: \( 3x + 5y \le 15 \). The boundary line is \( 3x+5y=15 \). This line passes through (5,0) and (0,3). Since \( 3(0)+5(0) \le 15 \) is true, the region is the half-plane that contains the origin.
Constraint 3 & 4: \( x \ge 0, y \ge 0 \). This restricts the solution to the first quadrant.
Now, let's visualize the feasible region.
- The region for \( 3x + 5y \le 15 \) (along with \( x \ge 0, y \ge 0 \)) is a triangle with vertices at (0,0), (5,0), and (0,3). All points in this region are close to the origin.
- The region for \( x + y \ge 8 \) (along with \( x \ge 0, y \ge 0 \)) is the area above the line connecting (8,0) and (0,8). All points in this region are far from the origin.
There is no point (x, y) that can simultaneously be close to the origin (satisfying \( 3x + 5y \le 15 \)) and far from the origin (satisfying \( x + y \ge 8 \)). The two regions defined by the constraints are disjoint.
Therefore, there is no common region that satisfies all the given constraints. The feasible region is an empty set.
Step 4: Final Answer:
Since there is no feasible region (the intersection of the constraints is an empty set), there is no point that satisfies all the given conditions. Therefore, there is no solution to this minimization problem.