Explanation Given: Find x 1 ≥ 0 and x 2 ≥ 0 such that x 1 + x 2 ≤ 100 2 x 1 + 3 x 2 ≤ 75 x 1 + 4 x 2 ≥ 50 and z = 5 x 1 − 3 x 2 is maximized. Explanation: The following information is provided: Maximize z = 5 x 1 − 3 x 2 Subject to: x 1 + x 2 ≤ 100 (I) 2 x 1 + 3 x 2 ≤ 75 (II) x 1 + 4 x 2 ≥ 50 (III) Step 1: First, add slack variable and subtract surplus variable in order to get equality constraints. Add slack variable (say s 1 , s 2 ) to equations (I) and (II) to get x 1 + x 2 + s 1 = 100 2 x 1 + 3 x 2 + s 2 = 75 Subtract surplus variable (say s 2 ) from equation (III) to get x 1 + 4 x 2 − s 2 = 50 Step 2: Represent the system of linear equations obtained above in the simplex tableau. This is the initial tableau: [ x 1 x 2 s 1 s 2 s 3 z s 1 1 1 1 0 0 0 100 s 2 2 3 0 1 0 0 75 s 3 1 4 0 0 − 1 0 50 z − 5 3 0 0 0 1 0 ] _ The solution from the initial tableau is ( x 1 , x 2 , x 3 , s 1 , s 2 ) = ( 0 , 0 , 0 , 100 , 75 , - 50 ) 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 3 is negative and the nonzero number in this variable column is − 1 , which is in row 3, that is, in R 3 . Step 4: Now, in the row located in step 3, find the most positive entry in the leftmost column. The most positive entry in column 1 is 2 . Step 5: From the column found in step 4, pick a pivot by the test ratio (or by quotients). Column 1 is the pivot column. Apply test ratio: 100 1 = 100 , 75 2 = 37.5 , 50 1 = 50 Therefore, 2 is the pivot because the smallest positive ratio is obtained by this element of R 2 , which follows that s 2 is the depart variable and x 1 is the enter variable. Therefore, [ x 1 x 2 s 1 s 2 s 3 z s 1 1 1 1 0 0 0 100 x 1 2 3 0 1 0 0 75 s 3 1 4 0 0 − 1 0 50 z − 5 3 0 0 0 1 0 ] _ Since pivot is not 1 , make the pivot element 1 by R 2 → R 2 2 . This leads to [ x 1 x 2 s 1 s 2 s 3 z s 1 1 1 1 0 0 0 100 x 1 1 3 2 0 1 2 0 0 75 2 s 3 1 4 0 0 − 1 0 50 z − 5 3 0 0 0 1 0 ] _ Step 6: Now make other entry in the pivot column zerobyperforming the following operations: R 1 → R 1 − R 2 R 3 → R 3 − R 2 R 4 → R 4 + 5 R 2 This leads to [ x 1 x 2 s 1 s 2 s 3 z s 1 0 − 1 2 1 − 1 2 0 0 125 2 x 1 1 3 2 0 1 2 0 0 75 2 s 3 0 5 2 0 − 1 2 − 1 0 25 2 z 0 21 2 0 5 2 0 1 375 2 ] _ Step 7: Repeat steps 3 to 6 if there is another negative basic variable present in the tableau, otherwise solve the problem by the simplex method for maximization

11th Edition

ISBN: 9780321979438

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

Publisher: PEARSON

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Chapter 4.4, Problem 13E

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x1≥0, x2≥0, and the value of z by using the simplex method.

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Chapter 4.4, Problem 12E

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Chapter 4 Solutions

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 by using the simplex method.

Solution Summary: The author explains the simplex method to determine the value of x_1ge 0, and z by adding slack and subtracting surplus variables to get equality constraints.

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x1≥0, x2≥0, and the value of z by using the simplex method.

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