Answer to Question #178318 in Linear Algebra for Maria Cristina Lacsa

Question #178318

The Solfel Dynamic Frames creates two kinds of frame, wooden frame and metal frame. Each frame is processed under two machines, M1 and M2. Machine M1 has a maximum of 250 hours to be used per production and machine M2 has a maximum of 200 hours. Each wooden frame requires 2 hours to be processed in M1 and 2.5 hours in M2. Each metal frame requires 1.5 hours in M1 and 1 hour in M2. Profit is P 500 for a wooden frame and P 400 for a metal frame. Determine the number of wooden frame and metal frame to be produced to obtain a maximum profit if wooden frame must not exceed 20 pcs for each production. 

Expert's answer

If the company produces wooden frames, each P1000 of profit requires 4 hours of work in M1 and 5 hours in M2. If the company produces metal frames, each P1000 of profit requires 3.75 hours of work in M1 and 2.5 hours in M2.

As production of metal frames is much more economical, the company should only produce metal frames. With the given restrictions, the company can produce not more than min{250/1.5, 200/1}=166.67 metal frames. As the number 166.67 is not integer, one should take 166. But then the company has some excessive resource of the both machines:

M1: 250-166x1.5=1

M2: 200-166x1.0=34

This resource should be used for the production of wooden frames. Moreover, the production of metal frames can be somewhat reduced, in order to use the maximum resourses of both the machines.

Calculate. (MF denotes the number of Metal Frames, WF the number of Wooden Frames)

If MF=166, then WF=0, Profit=166x400=66400. Excessive resource of M1 is 1.

If MF=165, then WF=1, Profit=165x400 +1x500=66500. Excessive resource of M1 is 0.5.

If MF=164, then WF=2, Profit=164x400 +2x500=66600. Excessive resource of M1 is 0.

Therefore, the profit would be at a maximum of P66600 if the company produced 164 Metal frames and 2 wooden frames per production.

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