# 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 - Mathematics

Sum

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.

Concept: Graphical Method of Solving Linear Programming Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.2 | Q 28 | Page 33