D1 - linear programming problem?

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iMacJack
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#1
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https://d0247e65b1a6ead8e6286cd43803...%20Edexcel.pdf

Question 8 - I don't even know where to begin

Help would be greatly appreciated, P.S. I have looked at the mark scheme and I still fail to understand!
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NotNotBatman
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What don't you understand?
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iMacJack
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(Original post by NotNotBatman)
What don't you understand?
How to formulate the problem
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NotNotBatman
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(Original post by iMacJack)
How to formulate the problem
When formulating a linear programming program, you need to first write the objective function, which is usually minimise (total) cost or maximise (total) profits, then list the constraints, by writing "subject to..."

All of the information is given in the question; work out the cost of Pack A and Pack B and multiply each by their quantities x and y; this gives the total cost.

For the constraints.

There must be at least 15000 flyers, the number of flyers in pack A is 20 and is 50 in pack B, so the number of flyers in total is 20x + 50y,(this is 20 times the number of Pack A's sold + 50 times the number of pack B's sold); this must be greater than 15000, so 20x +50y > 15000. Do the same with the other constraints.
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iMacJack
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(Original post by NotNotBatman)
When formulating a linear programming program, you need to first write the objective function, which is usually minimise (total) cost or maximise (total) profits, then list the constraints, by writing "subject to..."

All of the information is given in the question; work out the cost of Pack A and Pack B and multiply each by their quantities x and y; this gives the total cost.

For the constraints.

There must be at least 15000 flyers, the number of flyers in pack A is 20 and is 50 in pack B, so the number of flyers in total is 20x + 50y,(this is 20 times the number of Pack A's sold + 50 times the number of pack B's sold); this must be greater than 15000, so 20x +50y > 15000. Do the same with the other constraints.
I get the easy bits but I don't understand how they got the objective function as minimise c = 660x + 600y, and also I don't get how you do the between 40 and 60 percent bit?

Sorry to be a pain, you're being really helpful!
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NotNotBatman
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(Original post by iMacJack)
I get the easy bits but I don't understand how they got the objective function as minimise c = 660x + 600y, and also I don't get how you do the between 40 and 60 percent bit?

Sorry to be a pain, you're being really helpful!
15p per poster and 3p per flyer. If there is 40 posters and 20 flyers in Pack A, then the cost of producing 1 of pack A is  (15\times 40) +(3 \times 20) = 660, do the same with Pack B and you'll get 600.

The total number of packs is the number of A + the number of B, so (x+y).
You want x, the number of pack A's to be between 40% and 60% of (x+y), so form an inequality using fractions for the percentages.
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iMacJack
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(Original post by NotNotBatman)
15p per poster and 3p per flyer. If there is 40 posters and 20 flyers in Pack A, then the cost of producing 1 of pack A is  (15\times 40) +(3 \times 20) = 660, do the same with Pack B and you'll get 600.

The total number of packs is the number of A + the number of B, so (x+y).
You want x, the number of pack A's to be between 40% and 60% of (x+y), so form an inequality using fractions for the percentages.
Ahhhhhhhhhhhhhhh!!! You make it so much easier! Much appreciated!
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