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Question 91 of 100

Q.Solve the following L.P.P. by graphical method: Maximise Z=6x+4yZ = 6x + 4y subject to x≤2x \le 2, x+y≤3x + y \le 3, −2x+y≤1-2x + y \le 1, x≥0x \ge 0, y≥0y \ge 0

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 4mImportance★★★★★
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Plot the feasible region, find its corner points, and evaluate ZZ at each.

Constraints: x≤2x\le2, x+y≤3x+y\le3, −2x+y≤1-2x+y\le1, x≥0x\ge0, y≥0y\ge0.

Finding the corner points of the feasible region:

  • O=(0,0)O = (0,0) (origin)
  • Intersection of x=2x=2 and y=0y=0: A=(2,0)A=(2,0)
  • Intersection of x=2x=2 and x+y=3x+y=3: y=1⇒B=(2,1)y=1 \Rightarrow B=(2,1) [check −2x+y=−4+1=−3≤1-2x+y=-4+1=-3\le1 ✓]
  • Intersection of x+y=3x+y=3 and −2x+y=1-2x+y=1: subtracting, 3x=2⇒x=23,y=73⇒C=(23,73)3x=2 \Rightarrow x=\tfrac23, y=\tfrac73 \Rightarrow C=\left(\tfrac23,\tfrac73\right)
  • Intersection of −2x+y=1-2x+y=1 and x=0x=0: y=1⇒D=(0,1)y=1 \Rightarrow D=(0,1)

So the feasible region is the polygon O(0,0)→A(2,0)→B(2,1)→C(23,73)→D(0,1)→OO(0,0) \to A(2,0) \to B(2,1) \to C\left(\tfrac23,\tfrac73\right) \to D(0,1) \to O.

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