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Solve the Following Linear Programming Problem Graphically: Minimize Z = 6 X + 3 Y Subject to the Constraints: 4 X + Y ≥ 80 X + 5 Y ≥ 115 3 X + 2 Y ≤ 150 X ≥ 0 , Y ≥ 0 - CBSE (Arts) Class 12 - Mathematics

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Question

 Solve the following linear programming problem graphically:
Minimize  z = 6 x + 3 y
Subject to the constraints:

4 x + \[y \geq\] 80
x + 5 \[y \geq\] 115 

3 x + 2 \[y \leq\] 150
\[x \geq\] 0  , \[y \geq\] 0

Solution

The given constraints are 

4x + y  \[\geq\] 80

x + 5y  \[\geq\]  115

3x + 2y  \[\geq\]  150

x,y  \[\geq\]  0 

Converting the given inequations into equations, we get
4x + y = 80, x + 5y = 115, 3x + 2= 150, x = 0 and = 0
These lines are drawn on the graph and the shaded region ABC represents the feasible region of the given LPP.

It can be observed that the feasible region is bounded. The coordinates of the corner points of the feasible region are A(2, 72), B(15, 20) and C(40, 15).
The values of the objective function, Z at these corner points are given in the following table:
 

Corner Point Value of the Objective Function Z = 6x + 3y
A(2, 72) Z = 6 × 2 + 3 × 72 = 228
B(15, 20) Z = 6 × 15 + 3 × 20 = 150
C(40, 15) Z = 6 × 40 + 3 × 15 = 285

From the table, is minimum at = 15 and y = 20 and the minimum value of is 150.
Thus, the minimum value of Z is 150.

  Is there an error in this question or solution?

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Solution Solve the Following Linear Programming Problem Graphically: Minimize Z = 6 X + 3 Y Subject to the Constraints: 4 X + Y ≥ 80 X + 5 Y ≥ 115 3 X + 2 Y ≤ 150 X ≥ 0 , Y ≥ 0 Concept: Graphical Method of Solving Linear Programming Problems.
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