Explanation Given: 2 x 1 + x 2 ≥ 20 2 x 1 + 5 x 2 ≤ 80 and z = 6 x 1 + 2 x 2 is maximized Explanation: Provided Maximize z = 6 x 1 + 2 x 2 Subject to: 2 x 1 + x 2 ≥ 20 (I) 2 x 1 + 5 x 2 ≤ 80 (II) Step 1: First, add slack variable and subtract surplus variable in order to get equality constraints. Subtract surplus variable (say s 1 ) from equation (I) to get 2 x 1 + x 2 − s 1 = 20 Add slack variable (say s 2 ) to equation (II) to get 2 x 1 + 5 x 2 + s 2 = 80 Step 2: Represent the system of linear equations obtained above in simplex tableau. This is the initial tableau: [ x 1 x 2 s 1 s 2 z s 1 2 1 − 1 0 0 20 s 2 2 5 0 1 0 80 z − 6 − 2 0 0 1 0 ] _ The solution from initial tableau is ( x 1 , x 2 , s 1 , s 2 ) = ( 0 , 0 , − 20 , 80 ) _ . This is not a feasible solution because the value of the surplus variable ( s 1 ) _ is negative. Step 3: If any basic variable has a negative value, identify the nonzero number in that variable’s column, and note that row. In the initial tableau, basic variable s 1 is negative and the nonzero number in this variable column is − 1 , which is in row 1, that is, in R 1 . Step 4: Now, in the row located in step 3, find the most positive entry in the left most column. The most positive entry in column 1 is 2 . Step 5: In the column found in step 4, pick a pivot by test ratio (or by quotients). Column 1 is a pivot column. Apply test ratio: 20 2 = 10 , 80 2 = 40 Therefore, 2 is the pivot because the smallest positive ratio is obtained by the element of R 1 , which follows that s 1 is the depart variable and x 1 is the entry variable. Therefore [ x 1 x 2 s 1 s 2 z x 1 2 1 − 1 0 0 20 s 2 2 5 0 1 0 80 z − 6 − 2 0 0 1 0 ] _ Since pivot is not 1 , make the pivot element 1 by R 1 → R 1 2

11th Edition

ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Finite Mathematics (11th Edition)

Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 2E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 4E Ch. 4.1 - For Exercises 5-8. (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - Introduce slack variables as necessary, then...Ch. 4.1 - Introduce slack variables as necessary, then write...

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x 1 ≥ 0 , x 2 ≥ 0 , and the value of z.

Solution Summary: The author explains how to maximize the value of underset_z=240.

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x1≥0, x2≥0, and the value of z.

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