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Explanation:

Since P has to be produced in more number than Q and since time taken to produce P is least on M2, to maximize the output utilize the entire time available on M2 for producing P. Number of units of P produced in this time =(8×60)8 = 60 units. Now since the number of units of Q should be one-third that of P, we should manufacture 20 units of Q. To manufacture this on M1, it would take (20 × 6) = 120 min. So there are still (480 – 120) = 360 min of M1 to be utilized. Now for every 3 units of P that is manufactured, we have to manufacture 1 unit of Q. To run one such cycle on M1, it would take (3 × 10 + 1 × 6) = 36 min. Hence in 360 min, we have 10 such cycles and utilize all the idle time of M1. Hence, to maximize the output under the given condition it is possible to have no idle time on any of the machines.

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