Maximize Z= 3X1 + 4X2
subject to 5X, + 4X2 s 200
3X1 + 5X2 150
5X1 + 4X2 > 100
8X1 + 4X2 2 80
both
X1, X2 20
using graphical method of Linear
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) x1 = 2.5, x2 = 35 z = 3x1 + 4x2 Let us solve the equations 2x1 + x2 = 40 ... (1) 2x1 + 5x2 = 180 … (2) (1) – (2) ⇒ -4x2 = -140 x2 = 35 We have 2x1 + x2 = 40 2x1 + 35 = 40 2x1 = 5 x1 = 2.5
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