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linear programming question

for part a, I got 6-x-y

I am confused on part b)i), I simplified 5x+3y+2(6-x-y) but that gives me 3x+y+12. If I substitute -x-y I get 3x+y but I am not sure why I should substitute -x-y and not 6-x-y.

Any help will be much appreciated
(edited 2 years ago)
Reply 1
Original post by machau
for part a, I got 6-x-y

I am confused on part b)i), I simplified 5x+3y+2(6-x-y) but that gives me 3x+y+12. If I substitute -x-y I get 3x+y but I am not sure why I should substitute -x-y and not 6-x-y.

Any help will be much appreciated


You want to maximize it. What difference does the +12 make?
(edited 2 years ago)
Reply 2
Original post by mqb2766
You want to maximize it. What difference does the +12 make?

Sorry I'm not sure I understand, the +12 will increase P by 12?
Original post by machau
Sorry I'm not sure I understand, the +12 will increase P by 12?

Suppose f is a function of x and y. If you find x and y that maximize f, they're also going to maximize f+12...
Reply 4
Original post by DFranklin
Suppose f is a function of x and y. If you find x and y that maximize f, they're also going to maximize f+12...

oh right, yes I understand now. thank you :smile:
Reply 5
Original post by machau
oh right, yes I understand now. thank you :smile:

The important thing to realize is that you're searching for the values of x and y that maximize the function. You're not directly interested in finding the value of the function itself. You could multiply the function by a positive constant, and the location of the maximum would be unchanged. Multiplying the function by a negative constant turns a maximization problem into a minimization problem etc.
(edited 2 years ago)
Reply 6
Original post by mqb2766
The important thing to realize is that you're searching for the values of x and y that maximize the function. You're not directly interested in finding the value of the function itself. You could multiply the function by a positive constant, and the location of the maximum would be unchanged. Multiplying the function by a negative constant turns a maximization problem into a minimization problem etc.

thank you!

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