Question

Minimize Z = 3x + 2y by graphical method under the following constraints: \(x + 2y \leq 10\), \(3x + y \leq 15\), \(x \geq 0\), \(y \geq 0\).

Hint:

For LPP graphical solutions, always find the intersection points of the constraint lines accurately. Create a table to list the corner points and the corresponding values of the objective function Z. This organized approach minimizes calculation errors and clearly shows the optimal solution.

Answer 1

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The graphical method involves plotting the constraints to identify the feasible region, finding the coordinates of the corner points of this region, and then evaluating the objective function Z at these corner points. The minimum (or maximum) value of Z will occur at one of these points.
Step 2: Detailed Explanation:
1. Graph the Constraints:
We treat the inequalities as equations to draw the boundary lines.
Line 1: \(x + 2y = 10\). It passes through (10, 0) and (0, 5).
Line 2: \(3x + y = 15\). It passes through (5, 0) and (0, 15).
Line 3: \(x = 0\) (the y-axis).
Line 4: \(y = 0\) (the x-axis).
The inequalities \(x \geq 0\) and \(y \geq 0\) restrict the feasible region to the first quadrant. For the other two inequalities, shading towards the origin (0,0) satisfies them.
2. Identify the Feasible Region and Corner Points:
The feasible region is the polygon formed by the intersection of these half-planes. The corner points (vertices) are:
O (0, 0): The origin.
A (5, 0): The x-intercept of the line \(3x + y = 15\).
C (0, 5): The y-intercept of the line \(x + 2y = 10\).
B: The intersection of lines \(x + 2y = 10\) and \(3x + y = 15\).
To find point B, we solve the system of equations: \[ x + 2y = 10 \quad \text{---(i)} \] \[ 3x + y = 15 \quad \text{---(ii)} \] From (ii), \(y = 15 - 3x\). Substitute this into (i): \[ x + 2(15 - 3x) = 10 \] \[ x + 30 - 6x = 10 \] \[ -5x = -20 \implies x = 4 \] Substitute \(x=4\) back into \(y = 15 - 3x\): \[ y = 15 - 3(4) = 15 - 12 = 3 \] So, point B is (4, 3).
3. Evaluate Z at each Corner Point:
The objective function is Z = 3x + 2y. At O(0, 0): \(Z = 3(0) + 2(0) = 0\) At A(5, 0): \(Z = 3(5) + 2(0) = 15\) At B(4, 3): \(Z = 3(4) + 2(3) = 12 + 6 = 18\) At C(0, 5): \(Z = 3(0) + 2(5) = 10\) Step 3: Final Answer:
Comparing the values of Z, the minimum value is 0, which occurs at the corner point O(0, 0).