Question

Find the minimum value of the objective function \( Z = -50x + 20y \) by graphical method under the following constraints :
\( 2x - y \geq -5 \)
\( 3x + y \geq 3 \)
\( 2x - 3y \leq 12 \)
\( x \geq 0, y \geq 0 \)

Hint:

For an unbounded feasible region, always test whether a better value for the objective function can be found by moving further into the unbounded area. If so, no optimal solution exists. If not, the optimal solution occurs at a corner point.

Answer 1

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). We need to find the minimum value of a linear objective function within a feasible region defined by a set of linear inequalities. The minimum value, if it exists, will occur at one of the corner points (vertices) of the feasible region.
Step 2: Identifying the Feasible Region and Corner Points:
First, we graph the lines corresponding to each constraint:
\( L_1: 2x - y = -5 \implies y = 2x+5 \)
\( L_2: 3x + y = 3 \implies y = 3-3x \)
\( L_3: 2x - 3y = 12 \implies y = \frac{2}{3}x - 4 \)
The feasible region is the area in the first quadrant (\(x \geq 0, y \geq 0\)) that satisfies all inequalities:
\( y \leq 2x+5 \) (Below \(L_1\))
\( y \geq 3-3x \) (Above \(L_2\))
\( y \geq \frac{2}{3}x - 4 \) (Above \(L_3\))
The vertices of the feasible region are the points of intersection of these boundary lines. Point A: Intersection of \( L_2: 3x+y=3 \) and \( y \)-axis (\( x=0 \)).
\( 3(0) + y = 3 \implies y=3 \). So, A = (0, 3). Point B: Intersection of \( L_2: 3x+y=3 \) and \( x \)-axis (\( y=0 \)).
\( 3x + 0 = 3 \implies x=1 \). So, B = (1, 0). Point C: Intersection of \( L_3: 2x-3y=12 \) and \( x \)-axis (\( y=0 \)).
\( 2x - 3(0) = 12 \implies x=6 \). So, C = (6, 0).
The feasible region is unbounded in the first quadrant.
Step 3: Evaluating the Objective Function at Corner Points:
We evaluate the objective function \( Z = -50x + 20y \) at each corner point.
At A(0, 3): \( Z = -50(0) + 20(3) = 60 \)
At B(1, 0): \( Z = -50(1) + 20(0) = -50 \)
At C(6, 0): \( Z = -50(6) + 20(0) = -300 \)
Step 4: Final Answer:
The minimum value of the objective function at the corner points of the feasible region is -300, which occurs at the point (6, 0).