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Using simplex method solve the following L.P.P Max. $Z = 4x_1 +3x_2 + 6x_3$,

Using simplex method solve the following L.P.P

Max. $Z = 4x_1 +3x_2 + 6x_3$, subject to $2x_1 + 3x_2 +2x_3 ≤ 440 ; 4x_1 +3x_3≤470 ; 2x_1 + 5x_2 ≤ 430$

$x_1 , x_2 , x_3 ≥ 0$

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We express the given problem in standard form ,taking slack variables

$s_1, s_2, s_3$

$z – 4x_1 – 3x_2 – 6x_3 + 0s_1 +0s_2+ 0s_3 = 0\\ 2x_1 + 3x_2 + 2x_3 +s_1 +0s_2+0s_3 = 440 \\ 4x_1 +0x_2 +3x_3 +0s_1 +s_2+0s_3 = 470 \\ 2x_1 + 5x_2+0x_3+0s_1 +0s_2+s_3 = 430$

Now we form the simplex table as

Iteration no. Basic variables $\frac{coefficients of}{x_1,x_2,x_3,s_1,s_2,s_3}$ R.H.S solution Ratio
0 z -4, -3, -6, 0, 0, 0 0
$s_2$ leaves $s_1$ 2, 3, 2, 1, 0, 0 440 440/2=220
$x_3$ enters $s_2$ 4, 0, 3*, 0, 1, 0 470 470/3=156.67
$s_3$ 2, 5, 0, 0, 0, 1 430 430/0 = -
1 z 4, -3, 0, 0, 2, 0 940
$s_1$ leaves $s_1$ -2/3, 3*, 0, 1, - 2/3, 0 380/3 380/9=42.2
$x_2$ enters $x_3$ 4/3, 0, 1, 0, 1/3, 0 470/3
$s_3$ 2, 5, 0, 0, 0, 1 430 430/5=86
2 z 10/3, 0, 0, 1, 4/3, 0 3200/3
$x_2$ -2/9, 1, 0, 1/3, -2/9, 0 380/9
$x_3$ 4/3, 0, 1, 0, 1/3, 0 470/3
$s_3$ 28/9, 0, 0, -5/3, 10/9, 1 1970/9

Since all the elements in the row of z becomes positive ,we get solution from R.H.S.

$x_2=380/9 , x_3 = 470/3 , x_1=0 , Z_{Max} = 3200/3$