Solve the following linear programming problem graphically.Maxim

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 Multiple Choice QuestionsShort Answer Type

1. Solve the following Linear Programming Problems graphically:
Maximise    Z = 3x + 4y
subject to the constraints:    x + y ≤ 4,  x ≥ 0, y ≥ 0
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 Multiple Choice QuestionsLong Answer Type

2. Solve the following linear programming problem graphically:
Maximise    Z = 4x + y
subject to the constraints: x + y ≤ 50,  3x + y ≤ 90,  x ≥ 0, y ≥ 0
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3. Find the maximum value of f = x + 2 y subject to the constraints:
2x + 3 y ≤ 6
x + 4 y ≤ 4
x, y ≥ 0
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4. Maximize z = 9 x + 3 y subject to the constraints
2x + 3y ≤ 13
2x + y ≤ 5
x, y ≥ 0
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5. Solve the following Linear Programming Problems graphically:
Minimise    Z = - 3x + 4 y
subject to the constraints: x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
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6. Solve the following Linear Programming Problems graphically:
Minimise    Z = 5x + 3y
subject to the constraints: 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
116 Views

7. Solve the following Linear Programming Problems graphically:
Maximise    Z = 3x + 2y
subject to the constraints: x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0, 
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8. Maximize z = 4x + 1y such that x + 2y ≤ 20, x + y ≤ 15, x ≥ 0, y ≥ 0.
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9. Solve the following linear programming problem graphically.
Maximize z = 11x + 5y
subject to the constraints
3x + 2y ≤ 25,   x + y ≤ 10,  x, y ≥ 0


We are to maximize
z = 11x + 5y
subject to the constraints
3x + 2y  ≤ 25
x + y ≤ 10
x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 3x + 2y = 25
For x = 0,   2y = 25   or   straight y space equals 25 over 2
For y = 0,  3x = 25   or   straight x equals 25 over 3
therefore space space space space space space line space meets space OX space in
straight A open parentheses 25 over 3 comma space 0 close parentheses space and space OY space is space straight L open parentheses 0 comma space 25 over 2 close parentheses.
Again we draw the graph of x + y = 10
For x = 0, y = 10
For y = 0, x = 10
∴  line meets OX in B (10, 0) and OY in M (0, 10).

Since feasible region is the region which satisfies all the constraints
∴   OACM is the feasible region and corner points are
straight O left parenthesis 0 comma space 0 right parenthesis comma space space space straight A open parentheses 25 over 3 comma space 0 close parentheses comma space space straight C space left parenthesis 5 comma space 5 right parenthesis comma space space space straight M left parenthesis 0 comma space 10 right parenthesis.
At  straight O left parenthesis 0 comma space 0 right parenthesis comma space space space space space space space space space straight z space equals space 11 space left parenthesis 0 right parenthesis space plus space 5 space left parenthesis 0 right parenthesis space equals space 0 plus 0 space equals 0

At  straight A open parentheses 25 over 3 comma space 0 close parentheses comma space space straight z space equals space 11 space open parentheses 25 over 3 close parentheses space plus space 5 space left parenthesis 0 right parenthesis space equals space 275 over 3 plus 0 space equals space 275 over 3 space equals space 91 2 over 3

At space straight C left parenthesis 5 comma space 5 right parenthesis comma space space space straight z space equals 11 space left parenthesis 5 right parenthesis space plus space 5 space left parenthesis 5 right parenthesis space equals space 55 space plus space 25 space equals space 80
At space straight M left parenthesis 0 comma space 10 right parenthesis comma space space straight z space equals space 11 space left parenthesis 0 right parenthesis space plus space 5 space left parenthesis 10 right parenthesis space equals space 0 space plus space 5 space 0 space equals space 50
therefore space space space maximum space value space space equals space 91 2 over 3 at space open parentheses 25 over 3 comma space 0 close parentheses.

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10. Maximize z = 30x + 19y such that x + y ≤ 24, x + 1 halfy ≤ 16, x ≥ 0, y ≥ 0.
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