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MHT CET202016 Oct 2020Evening ShiftMathematicsLinear ProgrammingActual

The maximum value of Z =3 x+5 y , subject to 3 x+2 y 18, x 4, y 6x, y 0 is

Options

  1. A30
  2. B27
  3. C36
  4. D32

Correct answer

C. 36

Step-by-step solution

Point of intersection of x=4 and 3 x+2 y=18 is Q (4,3) Point of intersection of y =6 and 3 x +2 y =18 is P (2,6) Point D (4,0) and C (0,6) are as shown. The feasible region of th given L.P.P. is shaded portion CPQ D O. We have to maximize Z=3 x+5 y Now, Z at C (0,6)=3(0)+5(6)=30 Z at P (2,6) =3(2)+5(6)=36 Z at Q (4,3)=3(4)+5(3)=27Z at D(4,0)=3(4)+5(0)=12 Z at O (0,0)=3(0)+5(0)=0 Clearly the maximum value of Z is 36 at P (2,6)

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