Maximize Z = 3x+5y, subject to the constraints x+2y?2000,x+y?1500,y?600,x?0 and y?0 The answer given in the book is:Maximum Z = 5500 at x = 1000 and y = 500

Asked by Manoj | 23rd Jun, 2013, 03:10: PM

Expert Answer:

if you draw a graph for the given 4 constraint equation --  x+2y?2000,x+y?1500,y?600,x?0 and y?0 and identify the polygon enclosed by the equations, you will find that its is a quadrilateral, with A(0,0); B(600,0); C(1000,500); D(1500,0)
 
Now since its an enclosed area, the maximum value of Z should be at one of the vertices. 
 
hence, at A(0,0); Z = 3(0) + 5(0) = 0
at B(600,0); Z = 3(600) + 5(0) = 1800
at C(1000,500); Z = 3(1000) + 5(500) = 5500
at D(1500,0); Z = 3(1500) + 5(0) = 4500
 
So, the maximum value of Z is at C(1000,500) and the value is 5500

Answered by  | 23rd Jun, 2013, 07:03: PM

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