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MHT CET202014 Oct 2020Morning ShiftMathematicsLinear ProgrammingActual

The maximum value of Z =3 x+5 y , subject to x+4 y 24, y 4, x 0, y 0 is

Options

  1. A20
  2. B120
  3. C72
  4. D44

Correct answer

C. 72

Step-by-step solution

array |l|l|l| +4 y=24 & A(24,0) & B(0,6) =4 & - & C(0,4) array Feasible region is OADC Objective function is Z=3 x+5 yZ(A)=3(24)+0 =72Z(D)=3 8+5 4=24+20=44Z(C)=3 0+5 4=20

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