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Linear Programming Test - 10

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Linear Programming Test - 10
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  • Question 1
    1 / -0

    The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y

    Compare the quantity in Column A and Column B

    Column A                Column B

    Maximum of Z          325

    Solution
    Corner Points Corresponding value of Z = 4x + 3y

    (0,0)

    (0.40)

    (20,40)

    (60,20)

    (60,0)

    0

    120

    200

    300 (maximum)

    240

    Hence, maximum value of Z = 300 < 325.

    So, the quantity in column B is greater.

     

  • Question 2
    1 / -0

    Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. The Minimum value of F occurs at

    Solution
    Corner Points Corresponding value of F = 4x + 6y

    (0,2)

    (3,0)

    (6,0)

    (6,8)

    (0,5)

    12 (minimum)

    12 (minimum)

    24

    72 (maximum)

    30

    Hence, minimum value of F occurs at any points on the line segment joining the points (0, 2) and (3, 0).

     

  • Question 3
    1 / -0

    Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. Maximum of F - Minimum of F is equal to

    Solution
    Corner Points Corresponding value of F = 4x + 6y

    (0,2)

    (3,0)

    (6,0)

    (6,8)

    (0,5)

    12 (minimum)

    12 (minimum)

    24

    72 (maximum)

    30

    maximum of F - minimum of F

    = 72 - 12 = 60

     

  • Question 4
    1 / -0

    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. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is

    Solution
    Corner points Corresponding value of Z = px + qy; q > 0

    (0,3)

    (1,1)

    (3,0)

    3q

    p + q

    3p

    So, condition of p and q, so that the minimum of Z occurs at (3,0) and (1,1) is p + q = 3p

    ⇒ 2p = q

     

    ∴ p = q/2

     

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