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:

  • A
    $p = 2q$
  • B
    $p = \frac{q}{2}$
  • C
    $p = 3q$
  • D
    $p = q$

Explore More

Similar Questions

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 . . . . . . .

The solution for minimizing the function $z = x + y$ under an $L$.$P$.$P$. with constraints $x + y \geqslant 2$,$x + 2y \leqslant 8$,$y \leqslant 3$,$x, y \geqslant 0$ is

Maximize the function $Z = 11x + 7y$,subject to the constraints:
$x \leq 3, y \leq 2, x \geq 0, y \geq 0$

For the objective function $Z = 4x + y$ subject to the constraints $x + y \leq 50$,$3x + y \leq 90$,$x \geq 0$,$y \geq 0$,whose corner points of the feasible region are $(0,0)$,$(30,0)$,$(20,30)$,and $(0,50)$,the maximum value of $Z$ is . . . . . . .

The maximum value of $Z = 3x + 4y$ subject to the constraints $x + y \leq 4, x \geq 0, y \geq 0$ is . . . . . . .

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo