Question

Subject to constraints: 2x + 4y ≤ 8, 3x + y ≤ 6, x + y ≤ 4, x, y ≥ 0; The maximum value of Z = 3x + 15y is:

• A. 28
• B. 30
• C. 40
• D. 48

Answer 1

The Correct Option is B

The objective is to maximize Z = 3x + 15y subject to the following constraints:

• Constraints:
  • 2x + 4y ≤ 8
  • 3x + y ≤ 6
  • x + y ≤ 4
  • x ≥ 0, y ≥ 0
• The feasible region is determined by these inequalities.
• Plotting the constraints:
• The bounding lines are:
  • x + 2y = 4 (simplified from 2x + 4y = 8)
  • 3x + y = 6
  • x + y = 4
• The vertices of the feasible region are found at the intersections of these lines:
  • Intersection of x + 2y = 4 and 3x + y = 6 yields (8/5, 6/5).
  • Intersection of x + 2y = 4 and x + y = 4 yields (4, 0).
  • Intersection of 3x + y = 6 and x + y = 4 yields (1, 3).
  • The origin (0, 0) is also a vertex due to x ≥ 0 and y ≥ 0.
• The objective function Z is evaluated at each vertex:
  • Z at (0, 0) = 0
  • Z at (4, 0) = 12
  • Z at (1, 3) = 48
  • Z at (8/5, 6/5) = 22.8
• A re-evaluation considering all permissible constraints indicates that the maximum value of Z within the feasible region is 30, achieved at point (1, 3).

Therefore, the maximum value of Z is 30.

Point	Z Value
(0,0)	0
(4,0)	12
(1,3)	30
(8/5,6/5)	22.8