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 $q$ so that the minimum value of $Z$ occurs at $(3,0)$ and $(1,1)$ is . . . . . . .
- A$p = 2q$
- B$p = \frac{q}{2}$
- C$p = 3q$
- D$p = q$
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Determine graphically the minimum value of the objective function
$Z = -50x + 20y$ .....$(1)$
subject to the constraints:
${2x - y \geqslant -5}$ .....$(2)$
${3x + y \geqslant 3}$ .....$(3)$
${2x - 3y \leqslant 12}$ .....$(4)$
${x \geqslant 0, y \geqslant 0}$ .....$(5)$
$Z = -50x + 20y$ .....$(1)$
subject to the constraints:
${2x - y \geqslant -5}$ .....$(2)$
${3x + y \geqslant 3}$ .....$(3)$
${2x - 3y \leqslant 12}$ .....$(4)$
${x \geqslant 0, y \geqslant 0}$ .....$(5)$
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View SolutionThe corner points of the feasible region determined by the system of linear constraints are $(2, 72)$,$(15, 20)$,and $(40, 15)$. Let $Z = 6x + 3y$ be the objective function. The minimum value of $Z$ occurs at:
The objective function of a Linear Programming Problem $(LPP)$ is
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