Solve the following linear programming problem using graphical method
Minimize: $Z = 60 x + 30 y$
Subject to:
$2 x + 3 y \geq 120$
$2 x + y \geq 80$
$x \geq 0, y \geq 0$

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Solution

corner points $A = (0, 80)$ and $C = (60, 0)$ are found by inspection
Point B:
System: $2 x + 3 y = 120..... (1)$
$2 x + y = 80......... (2)$
$(1) - (2) = 2 y = 40$
$y = 20$
Substitute for $y = 20$ in (2):
$2 x + 20 = 80$
$2 x = 60$
$x = 30$
Point $B : (30, 20)$
Extreme Values
Corner point $O b j e c t i v e f u n c t i o n$ Z=60x+30y
$(0, 80)$ $60 (0) + 30 (80) = 2400$
$(30, 20)$ $60 (30) + 30 (20) = 2400$
$(60, 0)$ $60 (60) + 30 (0) = 3600$
From the table above, there are two minimum values for the objective function: $A = (0, 80)$ and $B = (30, 20)$ . In this situation, the objective function will have the same minimum value $(2, 400)$ at all points the boundary line segment A and B

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