The corner points of the feasible region determined by the following system of linear inequalities:
`2x+yge10, x+3yle15,x,yge0` are `(0,0),(5,0),(3,4)` and `(0,5)`. Let `Z=px+qy`, where `p,qge0`. Condition on `p` and `q` so that the maximum of `Z` occurs at both `(3,4)` and `(0,5)` is:
(a) `p=q` (b) `p=2q` (c) `p=eq` (d) `q=3p`

Explanation

Since, maximum of Z=px+qy occurs at both points (3,4) and (0,5)


Hence value of Z=px+qy must be equal at both the points.


∴p×3+q×4=p×0+q×5


⇒3p+4q=5q


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