Graphing variables

Question

The cost to run Machine 1 for an hour is $2. During that hour, Machine 1 produced 240 bolts and 100 nuts. The cost to run Machine 2 for an hour is $2.40. During that hour Machine 2 produces 160 bolts and 160 nuts. How long should each machine run to produce an order of at least 2080 bolts and 1520 nuts to achieve minimum operating cost.

 

Declaration or variables?

 

Constraints?

 

Objective function?

Answer 1

 

The variables are:

    x = number of hours to run machine 1

    y = number of hours to run machine 2

 

Constraints:

    For one, you need to run them long enough to produce the required number of bolts:

                     240x + 160y ≥ 2080  (1)  ("machine 1 produces 240 bolts/hr", "machine 2 produces 160 bolts/hr")

 

    A second constraint is that the machines must run long enough to produce the required number of nuts:

                     100x  + 160y ≥ 1520   (2)    (similar reasoning as first constraint, using nuts/hr values)

 

 

The objective function is to minimize cost while meeting production goal:  

              Cost = 2x + 2.4y  

 

 

Temporarily set inequality (1) to  240x + 160y = 2080, and then re-write as

 

                               y = 13 - 1.5x  (1')

 

Same for (2):

                                y = 9.5 - 0.625x (2')

 

 

Graph (1') (starts at (0, 13) and slopes down to (8.6667, 0)).   Shade the region ABOVE this line.  The shaded region represents  the inequality (1).

 

Graph (2')  (starts at (0, 9.5) and slopes down to (15.2, 0)).   Shade the region ABOVE this line.   The shaded region represents the inequality (2).

 

The region where the two shaded regions overlap is the feasible region.   In this case it is unbounded.  The solution is always at one of the corners, and in this case that corner is at (4,7) 

 

At (4,7)  you will find the following:

                 Number of bolts produced =  4*240 + 7*160  = 2080

                 Number of nuts produced = 4*100 + 7*160 = 1520

                 Cost = 2*4 + 2.4*7 = 24.8

 

In this case the production output requirement is exactly met for both bolts and nuts, but it didn't have to be that way.  As long as more than enough were produced, it would solve the stated problem.