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D1 OCR linear programming problem

Hi I have a question on this, how did the mark scheme get f>/20? I get why d>/10, when you sub g=40 into g</4d you get d>/10, and then I thought it would just be f>10 as it has been greater than d.
d1.png

its part iv) of question 2

thanks!
(edited 7 years ago)
Original post by Christina Tiana
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If f was 10, then d is at most 10 and g is at most 40. Total is at most 60, BUT you need 120.

Min feasble f occurs when g=4d and d=f, AND the sum of the three = 120.
(edited 7 years ago)
Original post by ghostwalker
If f was 10, then d is at most 10 and g is at most 40. Total is at most 60, BUT you need 120.

Min feasble f occurs when g=4d and d=f, AND the sum of the three = 120.


Thank you for the explanation, I have some confusion: if f was 20, it doesn't add up to 120? 40+20+20=80
Original post by ghostwalker
If f was 10, then d is at most 10 and g is at most 40. Total is at most 60, BUT you need 120.

Min feasble f occurs when g=4d and d=f, AND the sum of the three = 120.


Oh is it because it adds up to 60 initially, but then it has to times by 2, where g=80, d=20 and f=20?
Original post by Christina Tiana
Thank you for the explanation, I have some confusion: if f was 20, it doesn't add up to 120? 40+20+20=80


If that's g,d,f in that order, then g can be 80 giving 80+20+20=120

Our two inequalites g<=4d, and d <= f, can be put together as:

g/4 <= d <= f


As you can see f is constraining all the others.

For f to be a minimum, the others need to be maximized, in order to get our 120.

so, g/4 = d = f for minimum f.

Then g+d+f becomes 4f+f+f = 6f which equals 120, and f=20 is the minimum feasible value.

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