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Solve the following linear programming problem by simplex method: Maximize

    $z= 3x_1+5x_2+4x_3$

subject to

    $2x_1+3x_2\leq8 \\ 2x_2+5x_3\leq10 \\ 3x_1 + 2x_2 + 4x_3 \leq 15 \\ x_1,x_2,x_3 \geq 0.$
asked Dec 21, 2017 by randomisation (1,920 points)  

1 Answer

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Adding slack variables, $x_4, x_5, x_6$, we get the simplex table as follows:

Base$C_b$$b$$x_1$$x_2$$x_3$$x_4$$x_5$$x_6$$b/a$
$x_4$08230100$\frac{8}{3}$
$x_5$0100250105
$x_6$015324001$\frac{15}{2}$
Z0-3-5-4000


Taking pivot variable as indicated, 

Base$C_b$$b$$x_1$$x_2$$x_3$$x_4$$x_5$$x_6$$b/a$
$x_2$5$\frac{8}{3}$$\frac{2}{3}$10$\frac{1}{3}$00-
$x_5$0$\frac{14}{3}$$\frac{-4}{3}$05$\frac{-2}{3}$10$\frac{14}{15}$
$x_6$0$\frac{29}{3}$$\frac{5}{3}$04$\frac{-2}{3}$01$\frac{29}{12}$
Z$\frac{40}{3}$$\frac{1}{3}$0-4$\frac{5}{3}$00


Again taking pivot variable as indicated,

Base$C_b$$b$$x_1$$x_2$$x_3$$x_4$$x_5$$x_6$$b/a$
$x_2$5$\frac{8}{3}$$\frac{2}{3}$10$\frac{1}{3}$004
$x_3$4$\frac{14}{15}$$\frac{-4}{15}$01$\frac{-2}{15}$$\frac{1}{5}$0$\frac{-7}{2}$
$x_6$0$\frac{89}{15}$$\frac{41}{15}$00$\frac{-2}{15}$$\frac{-4}{15}$1$\frac{89}{41}$
Z$\frac{256}{15}$$\frac{-11}{15}$00$\frac{17}{15}$$\frac{4}{5}$0
 


Again taking the indicated element as pivot variable, we get the simplex table as:

Base$C_b$$b$$x_1$$x_2$$x_3$$x_4$$x_5$$x_6$$b/a$
$x_2$5$\frac{50}{41}$010$\frac{15}{41}$$\frac{8}{41}$$\frac{-10}{41}$-
$x_3$4$\frac{62}{41}$001$\frac{-6}{41}$$\frac{5}{41}$$\frac{4}{41}$-
$x_1$3$\frac{89}{41}$100$\frac{-2}{41}$$\frac{-12}{41}$$\frac{15}{41}$-
Z$\frac{765}{15}$000$\frac{45}{41}$$\frac{24}{41}$$\frac{11}{41}$
 


This gives us the optimal solution as 

    $z = \frac{765}{41}$

at

    $x_1 = \frac{89}{41} \\ x_2 = \frac{50}{41} and \\ x_3 = \frac{62}{41}$

answered Dec 21, 2017 by randomisation (1,920 points)  
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