Answer to Question #222592 in Operations Research for HANNA

Question

Problem 2: A small business enterprise makes dresses and trousers. To make a

dress requires 2 hours of cutting and 2 hours of sewing. To make a trousers requires

1 hour of cutting and 3 hours of sewing. The profit on a dress is 40 birr and on a

trouser 50 birr. The business has a maximum of 32 hours of cutting time and 48

hours of sewing time to operate per week.

Required:

A. Define the decision variables

B. Write down the constraints in terms of the variables.

C. Write down the Objective Function in terms of the variables.

D. Write down the standard form of the problem

E. Determine how many dresses and trousers should be made to maximize profit

and what the maximum profit is. Using simplex method

Answer 1

$\text{We define the decision variables by $x_1$ and $x_2$, where }\\\text{$x_1$ = number of dresses} \\\text{$x_2$ = number of trousers} \\2.\text{ The constraint on cutting is given by} \\ 2x_1+x_2 \leq32 \\\text{ The constraint on sewing is given by} \\ 2x_1+3x_2 \leq48 \\3.\text{ The objective function is given by;}\\z=40x_1+50x_2 \\4.\text{ The objective function in its standard form is;}\\z=40x_1+50x_2+0x_3+0x_4\\\text{Subject to:}\\2x_1+x_2+x_3=32\\2x_1+3x_2+x_4=48\\\text{with all variables non negative}\\5.\text{ Using Simplex method, we compute our first Tableau:} \\\begin{matrix}&& x_1&x_2&x_3&x_4\\&&40&50&0&0\\\hline x_3&0&2&1&1&0&32\\x_4 &0&2&3&0&1&48\\&&-40&-50&0&0&0 \end{matrix} \\\text{Using the simplex method, we locate the most negative number in the bottom row, and }\\\text{and label it our work column, after which we compute the smallest positive ratio by }\\\text{by dividing the numbers in the work column by the corresponding numbers in last column}\\\text{and we label it our work column. Next, we reduce the pivot element to 1 and other numbers }\\\text{to 0 in the work column to obtain the next tableau} \\\begin{matrix}&& x_1&x_2&x_3&x_4\\&&40&50&0&0\\\hline x_3&0&\frac{4}{3}&0&1&-\frac{1}{3}&16\\x_2 &50&\frac{2}{3}&1&0&\frac{1}{3}&16\\&&-\frac{20}{3}&0&0&\frac{50}{3}&800 \end{matrix} \\\text{Repeating the process above, we have the next tableau} \\\begin{matrix}&& x_1&x_2&x_3&x_4\\&&40&50&0&0\\\hline x_1&40&1&0&\frac{3}{4}&-\frac{1}{4}&12\\x_2 &50&0&1&-\frac{1}{2}&\frac{1}{6}&8\\&&0&0&5&\frac{45}{3}&880 \end{matrix} \\\text{Therefore $x_1=12$ and $x_2=8$ and z = 880 is the optimal solution}\\\text{12 dresses and 8 trousers should be made and the maximum value is 880 birr}$