Explanation Given: Maximize 5 x 1 + 4 x 2 Subject to x 1 + 5 x 2 ≤ 70 , 3 x 1 + 2 x 2 ≤ 54 and x 1 ≥ 0 , x 2 ≥ 0 . Algorithm used: “Simplex algorithm: 1. Change the inequality into equalities by adding slack variables. Let M be a variable equal to the objective function, and below the constraint equations write an equation of the form − ( objective function ) + M = 0 . 2. Set up the initial simplex tableau. The slack variables (and M ) provide the initial basic feasible solution. 3. Check the bottom row of the tableau for optimality. If all the entries to the left of the vertical line are nonnegative, then the solution is optimal. If some are negative, then choose the variable x k for which the entry in the bottom row is as negative as possible. 4. Bring the variable x k into the solution. Do this by pivoting on the positive entry a p k for which nonnegative ratio b i a i k is the smallest. The new basic feasible solution includes an increased value for M . 5. Repeat the process, beginning at step 3, until all the entries in the bottom row are nonnegative.” Calculation: To compute the initial simplex tableau as follows. Add slack variable as follows. To change the objective function 5 x 1 + 4 x 2 into equation by introducing a new variable M where, M = 5 x 1 + 4 x 2 . Thus, the objective function can be written as, − 5 x 1 − 4 x 2 + M = 0 . The system of equations written as follows. x 1 + 5 x 2 + x 3 = 70 3 x 1 + 2 x 2 + x 4 = 54 − 5 x 1 − 4 x 2 − + M = 0 Use the above system of equations and obtain the initial simplex tableau as follows. [ x 1 x 2 x 3 x 4 M 1 5 1 0 0 70 3 2 0 1 0 54 − 5 − 4 0 0 1 0 ] Compute the basic feasible solution as follows. Look in row 1 and 2, the columns of the 2 × 2 identity matrix in these two rows, the basic variables are x 3 and x 4 . Thus, the basic solution is, x 1 = x 2 = 0 , x 3 = 70 , x 4 = 54 and M = 0 . This is not the optimal solution. Since only the slack variables are non zero. Thus, the bottom row is, M = 5 x 1 + 4 x 2 . The value of M will increase when any variable x 1 or x 2 increase. Since the coefficient of x 1 is largest of the two coefficients. So bring x 1 into the solution. Compute the pivot variable as follows. b i a i k = b 1 a 11 = 70 1 = 70 b i a i k = b 2 a 12 = 54 3 = 18 The smallest value is 18

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Chapter 9.3, Problem 10E

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To maximize: The equation 5x1+4x2 subject to the equations x1+5x2≤70 , 3x1+2x2≤54 and x1≥0,x2≥0 .

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Chapter 9.3, Problem 9E

Chapter 9.3, Problem 11E

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Answer the number questions with the following answers +/- 2 sqrt(2) +/- i sqrt(6) (-3 +/-3 i sqrt(3))/4 +/-1 +/- sqrt(6) +/- 2/3 sqrt(3) 4 -3 +/- 3 i sqrt(3)

1 Matching 10 points Factor and Solve 1)x3-216 0, x = {6,[B]} 2) 16x3 = 54 x-[3/2,[D]] 3)x4x2-42 0 x= [ +/-isqrt(7), [F] } 4)x+3-13-9x x=[+/-1.[H]] 5)x38x2+16x=0, x = {0,[K}} 6) 2x6-10x-48x2-0 x-[0, [M], +/-isqrt(3)) 7) 3x+2x²-8 x = {+/-i sqrt(2), {Q}} 8) 5x³-3x²+32x=2x+18 x = {3/5, [S]} [B] [D] [F] [H] [K] [M] [Q] +/-2 sqrt(2) +/- i sqrt(6) (-3+/-3 i sqrt(3))/4 +/- 1 +/-sqrt(6) +/- 2/3 sqrt(3) 4 -3 +/- 3 i sqrt(3) [S]

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To maximize: The equation 5 x 1 + 4 x 2 subject to the equations x 1 + 5 x 2 ≤ 70 , 3 x 1 + 2 x 2 ≤ 54 and x 1 ≥ 0 , x 2 ≥ 0 .

Solution Summary: The author explains how to maximize the inequality by adding slack variables.

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To maximize: The equation 5x1+4x2 subject to the equations x1+5x2≤70 , 3x1+2x2≤54 and x1≥0,x2≥0 .

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Chapter 9 Solutions