Question #86113

A company produces three products, 1 P2 P , and P , from three raw materials A, B 3
and C. One unit of product P1 requires one unit of A, 3 unit of B and 2 units of C. A
unit of product P2 requires two units each of A and B and 3 units of C, while one
unit of P needs 2 units of A, 6 units of B and 3 4 units of C. The company has a daily
availability of 8 units of A, 12 units of B and 12 units of C. It is further known that
the profit per unit is ` 3, 2 and 5 for 1 P2 P , and P , respectively. How many units of 3
product P1 , product P2 and product P should the company manufacture to 3
maximize the profit? Formulate this problem as a Linear programming problem.

Expert's answer

Answer on Question #86113 – Math – Other

Question

A company produces three products P1, P2 and P3, from three raw materials A, B and C. One unit of product P1 requires one unit of A, 3 units of B and 2 units of C. A unit of product P2 requires two units each of A and B and 3 units of C, while one unit of P3 needs 2 units of A, 6 units of B and 4 units of C. The company has a daily availability of 8 units of A, 12 units of B and 12 units of C.

It is further known that the profit per unit is Rs.3, Rs.2 and Rs.5 for P1, P2 and P3, respectively. How many units of product P1, product P2 and product P3 should the company manufacture to maximize the profit? Formulate this problem as a Linear programming problem.

Solution

Let X1,X2X_{1}, X_{2} and X3X_{3} be the quantities of P1, P2 and P3, respectively.

The LP

Maximize p=3X1+2X2+5X3p = 3X_{1} + 2X_{2} + 5X_{3}

Subject to

X1+2X2+2X38X_{1} + 2X_{2} + 2X_{3}\leq 8

3X1+2X2+6X3123X_{1} + 2X_{2} + 6X_{3}\leq 12

2X1+3X2+4X3122X_{1} + 3X_{2} + 4X_{3}\leq 12

X1,X2,X30X_{1},X_{2},X_{3}\geq 0

Maximize p=3X1+2X2+5X3p = 3X_{1} + 2X_{2} + 5X_{3}

Subject to

X1+2X2+2X3+s1=8X_{1} + 2X_{2} + 2X_{3} + s_{1} = 8

3X1+2X2+6X3+s2=123X_{1} + 2X_{2} + 6X_{3} + s_{2} = 12

2X1+3X2+4X3+s3=122X_{1} + 3X_{2} + 4X_{3} + s_{3} = 12

X1,X2,X3,s1,s2,s30X_{1},X_{2},X_{3},s_{1},s_{2},s_{3}\geq 0

3X12X25X3+p=0-3X_{1} - 2X_{2} - 5X_{3} + p = 0


Pivot is 6



The maximum profit is 12:


X1=4,X2=0,X3=0X _ {1} = 4, X _ {2} = 0, X _ {3} = 0


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