For a linear programming problem, the objective function is $z = px + qy$, where $p, q > 0$. If at the corner points $(0, 10)$ and $(5, 5)$ the values of $z$ are $90$ and $60$ respectively, then the relation between $p$ and $q$ is . . . . . . .
- A$q = 3p$
- B$p = 3q$
- C$q = 2p$
- D$p = 2q$
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Corner points of the feasible region determined by the system of linear constraints are $(0,3), (1,1)$ and $(3,0)$. Let $z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ such that the minimum of $z$ occurs at both $(3,0)$ and $(1,1)$ is:
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