Answer to Question #306820 in Operations Research for MESERER

Question #306820

A firm manufactures two products; the net profit on product 1 is Birr 3 per unit and Birr 5

per unit on product 2. The manufacturing process is such that each product has to be

processed in two departments D1 and D2. Each unit of product1 requires processing for 1

minute at D1 and 3 minutes at D2; each unit of product 2 requires processing for 2 minutes

at D1 and 2 minutes at D2. Machine time available per day is 860 minutes at D1 and 1200

minutes at D2. How much of product 1 and 2 should be produced every day so that total

profit is maximum. (solve with graphical method


1
Expert's answer
2022-03-07T21:29:01-0500

The given data is summarized in the following table


"\\begin{array}{|c|c|c|c|}\n\\hline\n& D_1 & D_2 & \\text{Profit}\\\\\n\\hline\n\\text{Product 1} & 1 & 3 & 3\\\\\n\\text{Product 2} & 2 & 2 & 5\\\\\n\\hline\n\\text{Available Time} & 860 & 1200 &\\\\\n\\hline\n\\end{array}"


Let "x_1, x_2" be the number of units of Product 1 and Product 2 to be produced respectively to get maximum profit. Then "x_1, x_2 \\ge 0".

The objective function is "z = 3x_1+ 5x_2" and the contraints are

"\\begin{aligned}\nx_1+ 2x_2 &\\le 860\\\\\n3x_1 + 2x_2 & \\le 1200\n\\end{aligned}" .

Hence the linear programming model for the given problem is

"\\text{Maximize } z = 3x_1+ 5x_2\\\\\n\\text{subject to}\\\\\n\\begin{aligned}\nx_1+ 2x_2 &\\le 860\\\\\n3x_1 + 2x_2 & \\le 1200\\\\\nx_1, x_2 &\\geq 0\n\\end{aligned}"


To solve the problem graphically, we consider the constraints as equations and draw straight lines. The graph of the equations is given below.




The feasible region is bounded by the extreme points OABC. Point B is the point of intersection obtained by solving the two equations. The values at these extreme points are tabulated below.


"\\begin{array}{|c|c|}\n\\hline\n\\text{Extreme point} & \\text{Value of }z = 3x_1 + 5x_2\\\\\n\\hline\nO(0,0) & 0\\\\\nA(400,0) & 1200\\\\\nB(170,345) & 2235\\\\\nC(0,430) & 2150\\\\\n\\hline\n\\end{array}"


From the table, the maximum value occurs at B(170, 345), and the maximum value is z = 2235.


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