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Chapter 4: Problem 14

Solve each linear programming problem by the simplex method. $$ \begin{array}{ll} \text { Maximize } & P=5 x+4 y \\ \text { subject to } & 3 x+5 y \leq 78 \\ & 4 x+y \leq 36 \\ & x \geq 0, y \geq 0 \end{array} $$

Short Answer

Expert verified
The solution to the linear programming problem is \(x = 9\), \(y = 0\), with the maximum value of the objective function \(P = 45\).

Step by step solution

01

Convert the inequalities to equations by introducing slack variables

To convert the inequalities to equations, we will introduce slack variables s1 and s2: Equation 1: \(3x + 5y + s_1 = 78\) Equation 2: \(4x + y + s_2 = 36\)
02

Set up the initial simplex tableau

Based on the new equations and the objective function (P), we can set up the initial simplex tableau: $$ \begin{array}{c|cccc|c} & x & y & s_1 & s_2 & b \\ \hline s_1 & 3 & 5 & 1 & 0 & 78 \\ s_2 & 4 & 1 & 0 & 1 & 36 \\ \hline -P & -5 & -4 & 0 & 0 & 0 \\ \end{array} $$
03

Find the pivot column and pivot row

The pivot column is the one with the most negative value in the objective function row, thus column x (second column). The pivot row is determined by dividing the b-column values by their corresponding pivot column values, and selecting the row with the smallest nonnegative result: Row 1: 78 / 3 = 26 Row 2: 36 / 4 = 9 The pivot row is Row 2, and the pivot element is 4.
04

Perform the pivot operation

To perform the pivot operation, make the pivot element 1 by dividing the entire pivot row by the pivot element: New Row 2 = Row 2 / 4 $$ \begin{array}{c|cccc|c} & x & y & s_1 & s_2 & b \\ \hline s_1 & 3 & 5 & 1 & 0 & 78 \\ s_2 & 1 & 1/4 & 0 & 1/4 & 9 \\ \hline -P & -5 & -4 & 0 & 0 & 0 \\ \end{array} $$ Now, eliminate the pivot column in other rows by replacing the row with the sum of that row times a scalar, added to the pivot row times a scalar: New Row 1 = Row 1 - (3 * New Row 2) New Objective Function Row = Objective Function Row - (-5 * New Row 2) $$ \begin{array}{c|cccc|c} & x & y & s_1 & s_2 & b \\ \hline s_1 & 0 & 13/4 & 1 & -3/4 & 51 \\ s_2 & 1 & 1/4 & 0 & 1/4 & 9 \\ \hline -P & 0 & 1/4 & 0 & 5/4 & 45 \\ \end{array} $$
05

Check the optimality

Since all the coefficients in the objective function row are nonnegative, this solution is optimal.
06

Read the solution from the tableau

The final tableau tells us the values of the variables: x = 9 y = 0 P = 45 So the solution of the linear programming problem is \(x = 9\), \(y = 0\), with the maximum value of the objective function \(P = 45\).

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Most popular questions from this chapter

Use the technique developed in this section to solve the minimization problem. $$ \begin{aligned} \text { Minimize } & C=2 x-3 y-4 z \\ \text { subject to } &-x+2 y-z \leq 8 \\ & x-2 y+2 z \leq 10 \\ & 2 x+4 y-3 z \leq 12 \\ & x \geq 0, y \geq 0, z \geq 0 \end{aligned} $$

Solve each linear programming problem by the simplex method. $$ \begin{array}{ll} \text { Maximize } & P=3 x+4 y \\ \text { subject to } & x+y \leq 4 \\ & 2 x+y \leq 5 \\ & x \geq 0, y \geq 0 \end{array} $$

Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{aligned} \text { Minimize } & C=30 x+12 y+20 z \\ \text { subject to } & 2 x+4 y+3 z \geq 6 \\ & 6 x+z \geq 2 \\ 6 y+2 z & \geq 4 \\ x & \geq 0, y \geq 0, z \geq 0 \end{aligned} $$

Consider the linear programming problem $$ \begin{array}{lr} \text { Maximize } & P=3 x+2 y \\ \text { subject to } & x-y \leq 3 \\ x & \leq 2 \\ & x \geq 0, y \geq 0 \end{array} $$ a. Sketch the feasible set for the linear programming problem. b. Show that the linear programming problem is unbounded. c. Solve the linear programming problem using the simplex method. How does the method break down?

Determine whether the given simplex tableau is in final form. If so, find the solution to the associated regular linear programming problem. If not, find the pivot element to be used in the next iteration of the simplex method. $$ \begin{array}{rrrrrrrr|r} x & y & z & s & t & u & v & P & \text { Constant } \\ \hline \frac{5}{2} & 3 & 0 & 1 & 0 & 0 & -4 & 0 & 46 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 9 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 12 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 6 \\ \hline-180 & -200 & 0 & 0 & 0 & 0 & 300 & 1 & 1800 \end{array} $$
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