Skip to main contentSkip to solution

The optimal value of linear programming problem maximum Z=3x+4yZ = 3x + 4y, subject to,

x+3y≤12x + 3y \leq 12

x+y≥8x + y \geq 8

x,y≥0x, y \geq 0 is

Solution

✅ Correct Option: 3

Corner points: (8,0)(8, 0), (12,0)(12, 0), (6,2)(6, 2).

Z(8,0)=24Z(8,0) = 24; Z(12,0)=36Z(12,0) = 36; Z(6,2)=18+8=26Z(6,2) = 18 + 8 = 26.

Maximum is 3636 at (12,0)(12, 0).

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question