Problem 12 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve each linear programming problem by the simplex method. $$ \begin{array}{lc} \text { Maximize } & P=5 x+3 y \\ \text { subject to } & x+y \leq 80 \\ & 3 x \leq 90 \\ & x \geq 0, y \geq 0 \end{array} $$

Step-by-Step Solution

Answer

The optimal solution for the given linear programming problem using the simplex method is achieved at \(x = 30\) units and \(y = 10\) units. The maximum profit is \(P = 10\) units.

Step 1: Write the problem in a standard form

Add slack variables to convert inequalities into equalities, and rewrite the problem in standard form: Maximize \( P = 5x + 3y \) subject to \( x + y + s_1 = 80 \) \( 3x + s_2 = 90 \) \( x, y, s_1, s_2 \geq 0 \) Now, we have 2 slack variables, making the total number of variables 4.

Step 2: Set up the simplex tableau

Construct a tableau for the problem. Arrange the coefficients of the variables and the objective function in the table, with slack variables added: | | x | y | \(s_1\) | \(s_2\) | RHS | |--|---|---|------|------|-----| | P | -5 | -3 | 0 | 0 | 0 | | \(s_1\) | 1 | 1 | 1 | 0 | 80 | | \(s_2\) | 3 | 0 | 0 | 1 | 90 |

Step 3: Find the pivot element

Determine the pivot element, starting from the most negative element in the first row. The smallest negative value is -5 (in the x-column). Find the minimum ratio for each row, only considering positive coefficients for the variables: Row 1: \( \frac{80}{1} = 80 \) Row 2: \( \frac{90}{3} = 30 \) The smallest ratio is 30 (in the second row of the x-column), so this is the pivot element.

Step 4: Perform a pivot operation

Row operations to eliminate the x variable in all rows except the pivot row will ensure that the pivot element is 1: New Rows: P: \( \frac{-5}{3} * row_2 + row_P = \) - 3y - \( \frac{1}{3}s_1 + s_2 = \) \(\frac{-420}{3} = - 140 \) \( s_1 \): \( -1 * row_2 + row_{s1} = \) - 2x + y + \( s_1 - s_2 = -10 \) \( s_2 \): \( \frac{1}{3} * row_2 \) New Tableau: | | x | y | \(s_1\) | \(s_2\) | RHS | |--|---|---|------|------|-----| | P | 0 | -3 | -1/3 | 1 | -140 | | \(s_1\) | 0 | 1 | 1 | -1 | 10 | | \(s_2\) | 1 | 0 | 0 | 1/3 | 30 |

Step 5: Repeat the pivot process

The y-column in the first row is still negative, hence we need another pivot operation. Row 1: \( \frac{10}{1} = 10 \) (only one eligible row) The pivot element is found in the y-column and the \( s_1 \) row. Perform a pivot operation: New Row: P: \( -3 * row_{s1} + row_P = \) \( 3s_1 - \frac{2}{3}s_2 = 10 \) New Tableau: | | x | y | \(s_1\) | \(s_2\) | RHS | |--|---|---|------|------|-----| | P | 0 | 0 | 3 | - 2/3 | 10 | | \(s_1\) | 0 | 1 | 1 | -1 | 10 | | \(s_2\) | 1 | 0 | 0 | 1/3 | 30 |

Step 6: Read the solution from the tableau

As there are no negative entries left in the first row, the simplex method is complete. The solution can be read from the tableau: \( x = 30, y = 10, P = 10 \) Thus, the optimal solution is achieved when x = 30 units, y = 10 units, with a maximum profit of P = 10 units.

Key Concepts

Linear Programming

Linear programming is a mathematical technique used to optimize a particular objective, like maximizing profit or minimizing cost, subject to certain constraints. In the original exercise, the goal is to maximize the linear objective function \( P = 5x + 3y \), which represents profit. The constraints are given by inequalities that define the feasible region where solutions exist.

The first step in solving a linear programming problem is to identify the objective function and the constraints. Then, translate these conditions into mathematical expressions. In our exercise, we have the constraints:

• \( x + y \leq 80 \)
• \( 3x \leq 90 \)
• \( x \geq 0 \), \( y \geq 0 \)

These inequalities define a space in which all potential solutions must lie. The essence of linear programming is finding the best solution within that defined space.

Optimization

Optimization in the context of linear programming involves choosing the best option from a set of possible solutions. The simplex method is a popular algorithmic approach for finding the optimal solution of linear programming problems. It works by moving from one feasible solution to another, at each step improving the value of the objective function.

In this example, the simplex method starts with a basic feasible solution and searches for optimality by performing pivot operations. These operations look to replace a 'non-basic' variable (like the coefficients \( x, y \)) by introducing a new variable that provides a better objective function value.

The goal is to transform the initial set of inequalities into equalities by introducing slack variables. This transformation allows the system to be represented in a matrix form, enabling the application of the simplex algorithm. The final tableau provides the best possible value of the objective function within the constraints.

Slack Variables

Slack variables are introduced in linear programming to transform inequality constraints into equality constraints. This transformation is crucial because the simplex method requires a standard form representation where all the constraints are equalities.

In our problem, slack variables \( s_1 \) and \( s_2 \) are added:

• \( x + y + s_1 = 80 \)
• \( 3x + s_2 = 90 \)

These slack variables represent the unused resources in the constraints. For instance, \( s_1 \) represents how much less than 80 the sum of \( x \) and \( y \) can be, and \( s_2 \) represents how much less than 90 the product of \( 3x \) can be.

The introduction of slack variables turns the inequalities into a linear system that is ready for the simplex method, thereby making the solution process smooth and algorithmically feasible.