In solving the LP problem: "Minimize z = 6x + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To identify the redundant constraints,we analyze the feasible region defined by the given constraints:$1$. $x \geq 6$$2$. $y \geq 2$$3$. $2x + y \

Vedclass · Accepted Answer

$2x + y \geq 10, x \geq 0, y \geq 0$

Vedclass · Answer

(B) To identify the redundant constraints,we analyze the feasible region defined by the given constraints:$1$. $x \geq 6$$2$. $y \geq 2$$3$. $2x + y \geq 10$$4$. $x \geq 0, y \geq 0$If $x \geq 6$ and $y \geq 2$,then $2x + y \geq 2(6) + 2 = 12 + 2 = 14$. Since $14 > 10$,the constraint $2x + y \geq 10$ is automatically satisfied whenever $x \geq 6$ and $y \geq 2$ are satisfied.Similarly,if $x \geq 6$ and $y \geq 2$,then $x \geq 0$ and $y \geq 0$ are automatically satisfied.Thus,the constraints $2x + y \geq 10, x \geq 0, y \geq 0$ are redundant because they do not restrict the feasible region further than $x \geq 6$ and $y \geq 2$ already do.

In solving the $LP$ problem: "Minimize $z = 6x + 10y$ subject to $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0$." The redundant constraints are $....$

Similar Questions

Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = 5x + 10y$
subject to $x + 2y \leq 120, x + y \geq 60, x - 2y \geq 0, x, y \geq 0$.

For the $LP$ problem,"Maximize $z = x + 4y$ subject to $3x + 6y \leq 6$,$4x + 8y \geq 16$ and $x \geq 0, y \geq 0$."

If an $LPP$ admits an optimal solution at two consecutive vertices of a feasible region,then:

The corner points of the bounded feasible region are $(60,0), (120,0), (60,40), (40,20)$ and $(20,30)$. For the objective function $z=5x+10y$:
$(i)$ Maximum value of $z$.
$(ii)$ Minimum value of $z$.
$(iii)$ Maximum value of $z$ occurs at.
$(iv)$ Minimum value of $z$ occurs at.

Vedclass Test Series

Exam Paper Generator

Online Exam Module