For the LPP- maximise z - x -4 y subject to the constraints x -2 y - 2- x -2 y - 8- x - y - 0A- Z max-16B- Has no feasible solutionC- z max-4D- z max-8

The correct option is B Has no feasible solution nbsp; From the graph, we can observe that there is no common region. Therefore, given LPP has no feasible so ...

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Standard XII

Mathematics

Law of Reciprocal

For the LPP; ...

Question

For the LPP; maximise $z = x + 4 y$ subject to the constraints $x + 2 y \leq 2, x + 2 y \geq 8, x, y \geq 0$

z_{max} = 4

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z_{max} = 8

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z_{max} = 16

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Has no feasible solution

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Solution

The correct option is D Has no feasible solution

From the graph, we can observe that there is no common region.
Therefore, given LPP has no feasible solution.

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Similar questions

Q. For the LPP; maximise

z = x + 4 y

subject to the constraints

x + 2 y \leq 2

x + 2 y \geq 8

x, y \geq 0

Q. For the LPP; maximize

z = 5 x - 2 y

subject to the constraints

3 x + 2 y \leq 6, 10 x + 9 y \geq 45, x, y \geq 0

Q. Maximise

Z = x + 2 y

Subject to the constraints

x + 2 y \geq 100

2 x - y \leq 0

2 x + y \leq 200

x, y \geq 0

.
Solve the given LPP graphically.

Q. If

(0, 0), (30, 0), (20, 30)

and

(0, 50)

are the vertices of the feasible region of the LPP
Maximize

Z = 60 x + 50 y

Subject to

x + y \geq 50, 3 x + y \leq 90

and

x, y \geq 0

, then

Z_{m a x} =

Q. Maximise

Z = - x + 2 y

, subject to the constraints:

x \geq 3, x + y \geq 5, x + 2 y \geq 6, y \geq 0

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For the LPP; maximise z=x+4y subject to the constraints x+2y≤2, x+2y≥8,x,y≥0

The correct option is D Has no feasible solution From the graph, we can observe that there is no common region. Therefore, given LPP has no feasible solution.

For the LPP; maximise $z = x + 4 y$ subject to the constraints $x + 2 y \leq 2, x + 2 y \geq 8, x, y \geq 0$

The correct option is D Has no feasible solution

From the graph, we can observe that there is no common region.
Therefore, given LPP has no feasible solution.