The solution set for minimizing the function | vedclass

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

Question

(C) To minimize $z = x + y$ subject to the constraints:$1$) $x + y \geqslant 2$$2$) $x + 2y \leqslant 8$$3$) $y \leqslant 3$$4$) $x, y \geqslant 0$Fir

Vedclass · Accepted Answer

infinitely many points

Vedclass · Answer

(C) To minimize $z = x + y$ subject to the constraints:$1$) $x + y \geqslant 2$$2$) $x + 2y \leqslant 8$$3$) $y \leqslant 3$$4$) $x, y \geqslant 0$First,we identify the feasible region by plotting the lines:- $x + y = 2$ passes through $(2, 0)$ and $(0, 2)$.- $x + 2y = 8$ passes through $(8, 0)$ and $(0, 4)$.- $y = 3$ is a horizontal line.The vertices of the feasible region are found by the intersection of these lines:- Intersection of $x + y = 2$ and $x = 0$ is $(0, 2)$.- Intersection of $x + y = 2$ and $y = 0$ is $(2, 0)$.- Intersection of $x + 2y = 8$ and $y = 0$ is $(8, 0)$ (but this is outside the region $y \leqslant 3$ and $x+y \geqslant 2$ constraints).- Intersection of $x + 2y = 8$ and $y = 3$ is $x + 6 = 8 \implies x = 2$,so $(2, 3)$.- Intersection of $x = 0$ and $y = 3$ is $(0, 3)$.The corner points of the feasible region are $(0, 2), (2, 0), (2, 3), (0, 3)$.Evaluating $z = x + y$ at these points:- At $(0, 2)$,$z = 0 + 2 = 2$.- At $(2, 0)$,$z = 2 + 0 = 2$.- At $(2, 3)$,$z = 2 + 3 = 5$.- At $(0, 3)$,$z = 0 + 3 = 3$.The minimum value is $2$,which occurs at all points on the line segment connecting $(0, 2)$ and $(2, 0)$. Since a line segment contains infinitely many points,the solution set contains infinitely many points.

The solution set for minimizing the function $z = x + y$ with constraints $x + y \geqslant 2$,$x + 2y \leqslant 8$,$y \leqslant 3$,$x, y \geqslant 0$ contains

Similar Questions

The difference between the maximum and minimum values of the objective function $Z = 3x + 5y$, subject to the constraints $x + 3y \leq 60$, $x + y \geq 10$, $x - y \leq 0$, and $x, y \geq 0$ is

The maximum value of $Z=10 x+25 y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$ is

The objective function $z = 4x + 5y$ subject to the constraints $2x + y \geq 7$,$2x + 3y \leq 15$,$y \leq 3$,$x \geq 0$,and $y \geq 0$ has a minimum value at which point?

The shaded part of the given figure indicates the feasible region. Then the constraints are

The minimum value of $Z = 3x + y$, subject to the constraints $2x + 3y \leq 6$, $x + y \geq 1$, $x \geq 0$, $y \geq 0$ is....

Vedclass Test Series

Exam Paper Generator

Online Exam Module