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    Exercise 6C - Question 8
    Solve the linear programming problem posed in Exercise 6A, Question 5

    maximise P=6x+7y
    subject to;
    x+2y<=500
    2x+y<=500
    x+y<=300
    y<=3x
    x>=50
    y>=0

    I got the same vertex, the intersection of x+y=300 and 2x+y=500
    which gives you the values x=200 and y=100 (correct according to the answers in the textbook)

    However in the textbook they have the value of P as P=2000 and I got P=1900 using the same values of x and y

    maximise P =6x+7y

    P=6(200)+7(100)
    P=1200+700
    P=1900

    Can someone please help me find the value of P, I can't see how its 2000
 
 
 
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Updated: October 20, 2015
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