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# D1 Linear Programming Question watch

1. 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

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Updated: October 20, 2015
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