19 A steel manufacturer produces four sizes of I beams small

[1.9] A steel manufacturer produces four sizes of I beams: small, medium, large, and extra large. These beams can be produced on any one of three machine types: A, B, and C. The lengths in feet of the I beams that can be produced on the machines per hour are summarized below MACHINE BEAM Small Medium Large Extra large 350 250 200 125 650 400 350 200 850 700 600 325 Assume that each machine can be used up to 50 hours per week and that the hourly operating costs of these machines are respectively $30.00, $50.00, and

Solution

Let, Machine A has been operated for x hours in a week, machine B operated for y hours in a week and machine C operated for z hours in a week.

Since a machine can be operated up to 50 hours per week thus intial condition is given as:

x,y,z <= 50

As given, small beams can be produced by machine A at a rate of 350 feet per hour, by machine B at a rate of 650 feet per hour and by machine C at a rate of 850 feet per hour. So for x, y & h hours of operations of machine A, machine B & Machine C respectively, total Small beam produced is given by:

350x+650y+850z

Also the requirement of small size beam per week is 12000 feet, so the equation for small beam is given by:

350x+650y+850z >=12000

Similarly for medium beam, equation is given by

250x+200y+700z>=6000

for large beam, equation is given by

200x+350y+600z>=5000

and for extra large beam, the equation is given by

125x+200y+325z>=7000

For every operation, to maximize the profit the operating cost is minimized. In the above problem, hourly operating cost for machine A is $30, for machine B is $50 and for machine C is $80. Thus the total operating cost for the week is given as,

P = 30x+50y+80z

So the LPP is as below:

Minimize P = 30x+50y+80z

Constraints:

350x+650y+850z >=12000

250x+200y+700z>=6000

200x+350y+600z>=5000

125x+200y+325z>=7000

and x,y,z <= 50