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Find the maximum and minimum values of Z = 2x+y, subject to the constraints x+3y?6,x-3y?3,3x+4y?24,-3x+2y?6,5x+y?5,x ?0 and y?0 The answer given in the book is:Maximum Z=14 1/3 at(84/13,15/13) and minimum Z=3 1/14 at(9/14,25/14)
Asked by Manoj | 23 Jun, 2013, 03:21: PM
if you draw a graph for the given 4 constraint equation --  x+3y?6,x-3y?3,3x+4y?24,-3x+2y?6,5x+y?5,x ?0 and identify the polygon enclosed by the equations, you will find that its is a pentagon, with A(9/2,1/2); B(84/13,15/13); C(4/3, 5); D(4/13,45/13); E(9/14,25/14)

Now since its an enclosed area, the minimum/maximum value of Z should be at one of the vertices.

hence, at A(9/2,1/2); Z = 2(9/2) + (1/2) =19/2
at B(84/13,15/13); Z = 2(84/13) + (15/13)= 181/13
at C(4/3, 5); Z = 2(4/3) + (5)) = 23/3
at D(4/13,45/13); Z = 2(4/13) + (45/13) = 53/13
at E(9/14, 25/14); Z = 2(9/14) + 25/14 = 43/14

So, the minimum value of Z is at E(9/14, 25/14) and the value is 43/14 = 3 1/14
And the maximum value of Z is at B(84/13, 15/13) and the value is 181/13 = 13 12/13

It seems like the answer you wrote for the maximum value of Z is wrong, even though the coordinates are correct.
Answered by | 23 Jun, 2013, 06:59: PM

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