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Question 94 of 100

Q.Solve the following linear programming problem: Maximise: z=150x+250yz = 150x + 250y; Subject to: 4x+y≤404x+y \le 40, 3x+2y≤603x+2y \le 60, x≥0x \ge 0, y≥0y \ge 0

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 4mImportance★★★★★
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Find the feasible region's corner points and evaluate zz at each.

Constraints: 4x+y≤404x+y\le40, 3x+2y≤603x+2y\le60, x≥0x\ge0, y≥0y\ge0.

Corner points of the feasible region:

  • (0,0)(0,0)
  • (10,0)(10,0): where 4x+y=404x+y=40 meets the xx-axis (check: 3(10)+2(0)=30≤603(10)+2(0)=30\le60 ✓)
  • (0,30)(0,30): where 3x+2y=603x+2y=60 meets the yy-axis (check: 4(0)+30=30≤404(0)+30=30\le40 ✓)
  • Intersection of 4x+y=404x+y=40 and 3x+2y=603x+2y=60: from the first, y=40−4xy=40-4x; substituting, 3x+2(40−4x)=60⇒3x+80−8x=60⇒−5x=−20⇒x=4, y=243x+2(40-4x)=60 \Rightarrow 3x+80-8x=60 \Rightarrow -5x=-20 \Rightarrow x=4,\ y=24, giving (4,24)(4,24). …

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