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The corner points of the bounded feasible region for an L.P.P are (0, 10), (5, 5), (15,15), and (0, 20). If the objective function is Z=px+qy,p,q>0Z=px+qy, p,q>0, then the condition on pp and qq so that the maximum of ZZ occurs at (15, 15) and (0, 20) is

Solution

✅ Correct Option: 2

Equal maximum at (15,15)(15,15) and (0,20)(0,20) gives 15p+15q=20q15p+15q = 20q.

So 15p=5q15p = 5q, i.e. q=3pq = 3p.

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