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If the corner points of the bounded feasible region for a linear programming problem (LPP) are (0, 2), (3, 0), (6, 0), (6, 8) and (0,5), then which of the following are correct for the objective function Z=4x+6yZ = 4x + 6y?

(A) The minimum value of the objective function occurs at (0, 2) and (3, 0) only.

(B) The minimum value of the objective function occurs at the mid-point of the line segment joining the points (0, 2) and (3, 0) only.

(C) The minimum value of the objective function occurs at every point of the line segment joining the points (0, 2) and (3, 0).

(D) The difference between the maximum value and minimum value of the objective function is 60.

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

The objective function is Z=4x+6yZ = 4x + 6y.

Evaluating ZZ at each corner point:

Z(0,2)=4(0)+6(2)=12Z(0, 2) = 4(0) + 6(2) = 12

Z(3,0)=4(3)+6(0)=12Z(3, 0) = 4(3) + 6(0) = 12

Z(6,0)=4(6)+6(0)=24Z(6, 0) = 4(6) + 6(0) = 24

Z(6,8)=4(6)+6(8)=72Z(6, 8) = 4(6) + 6(8) = 72

Z(0,5)=4(0)+6(5)=30Z(0, 5) = 4(0) + 6(5) = 30

Minimum Z=12Z = 12 (at both (0,2)(0, 2) and (3,0)(3, 0))

Maximum Z=72Z = 72 (at (6,8)(6, 8))


Statement (A): Since both corner points give the same minimum value, the minimum occurs at every point on the line segment joining them — not just the endpoints. This is incorrect.


Statement (B): The midpoint does give Z=12Z = 12, but so does every other point on that segment — not the midpoint alone. This is incorrect.


Statement (C): Any point on the segment joining (0,2)(0, 2) and (3,0)(3, 0) can be written as (3t,  2−2t)(3t,\; 2 - 2t) where t∈[0,1]t \in [0, 1].

Z=4(3t)+6(2−2t)Z = 4(3t) + 6(2 - 2t)

=12t+12−12t= 12t + 12 - 12t

=12= 12 for every tt

Since Z=12Z = 12 (the minimum) for every point on this segment, Statement (C) is correct.


Statement (D):

Zmax⁡−Zmin⁡=72−12=60Z_{\max} - Z_{\min} = 72 - 12 = 60

Statement (D) is correct.


Therefore, the correct answer is (C) and (D) only.

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