A factory manufactures three products, A, B, and C.

Question

A factory manufactures three products, A, B, and C. Each product requires the use of two machines, Machine I and Machine II. The total hours available, respectively, on Machine I and Machine II per month are 7,080 and 8,120. The time requirements and profit per unit for each product are listed below.

	A	B	C
Machine I	3	10	12
Machine II	7	7	15
Profit	$12	$15	$20

How many units of each product should be manufactured to maximize profit, and what is the maximum profit?

Start by setting up the linear programming problem, with A, B, and C representing the number of units of each product that are produced.

 

Maximize P=

subject to:

   ≤ 7,080

   ≤ 8,120

Enter the solution below. If needed round numbers of items to 1 decimal place and profit to 2 decimal places.

The maximum profit is $ when the company produces:

 units of product A

 units of product B

 units of product C

Answer 1

This is a linear programming problem to maximize the target function P = 12A + 15B + 20C with the constraints 3A + 10B + 15C ≤ 7080, and 7A + 7B + 15C ≤ 8120, A, B, C ≥ 0. Let s and t be slack variables so that we can standardize the problem as

3A + 10B + 15C + s = 7080,

7A + 7B + 15C + t = 8120, A, B, C, s, t ≥ 0.

To determine the basic solutions, we choose 3 of the 5 variables and set them equal to zero. There are 10 diffrent ways to have such a choice, but the one that A = B = C = 0 is obviously inapproapriate. In each such choice, we have a system of linear equations with 2 unknowns, which can be solved readily. For instance, if we set C = s = t = 0, then we can solve for A and B, with A = 4520/7, B = 3600/7, and P = 15462+6/7. Among the 9 basic solutions, there are 5 not feasible as one of the variable is negative. Among the 4 remaining feasible solutions, the one with C = 0 = s = t gives the optimal solution. Since the number of units has to be an integer, we round them up as

A = 645, B = 514, C = 0, and P = 15450.