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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Similar Questions
For the linear programming constraints $x+2y \geq 10$,$3x+4y \leq 24$,$x \geq 0$,and $y \geq 0$,which of the following is not a corner point of the feasible region?
Medium
View SolutionCorner 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:
The coordinates of the corner points of the bounded feasible region are $(0,10), (5,5), (15,15)$,and $(0,20)$. The maximum value of the objective function $Z = 10x + 20y$ is:
The corner points of the feasible region determined by the system of linear constraints are $(0,0), (0,40), (20,40), (60,20), (60,0)$. The objective function is $z=4x+3y$. Compare the quantity in Column $A$ and Column $B$.
Column Value $A$. Maximum of $z$ $300$ $B$. Constant value $325$
| Column | Value |
| $A$. Maximum of $z$ | $300$ |
| $B$. Constant value | $325$ |
Medium
View SolutionDetermine the maximum value of $Z=3x+4y$ if the feasible region (shaded) for a $LPP$ is shown in the adjacent figure.
Medium
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