# Linear Programming Question

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#1
Hi there, I have the following question:

For a certain manufacturing process a manufacturer requires 1000kg of nickel and 1600kg of tin. He obtains this by buying scrap metal and scrap cars. One tonne of scrap metal yields 100kg of nickel and 400kg of tin, while one tonne of scrap cars yields 100kg of nickel and 100kg of tin. If the scrap metal costs £15 per tonne and scrap cars £5 per tonne, how many tonnes of each should he buy and what will be his capital outlay?

I have only been given one example of linear programming in class, but its for working out maximum profit. This question is a bit different, I can't work out how to arrange it to solve it. Im guessing here I am required to work out the minimum cost that achieves at least the required amount of nickel and tin.

So would I be right in saying the following;
if Nickel = xA and Tin = xB

Minimise 1000xA + 1600xB

Subject to: 100xA + 400xB = 15
and 100xA + 100xB = 5

So xB = 1/30 or 0.0333
and xA = 1/60 or 0.016666

Please could someone confirm that I am on the right track

Many thanks
Jackie
0
#2
I think this goes wrong when i started to solve the 2 equations by using the simultaneous approach - thats incorrect right, i was going in to early for that
0
7 years ago
#3
(Original post by jackie11)
Hi there, I have the following question:

For a certain manufacturing process a manufacturer requires 1000kg of nickel and 1600kg of tin. He obtains this by buying scrap metal and scrap cars. One tonne of scrap metal yields 100kg of nickel and 400kg of tin, while one tonne of scrap cars yields 100kg of nickel and 100kg of tin. If the scrap metal costs £15 per tonne and scrap cars £5 per tonne, how many tonnes of each should he buy and what will be his capital outlay?

I have only been given one example of linear programming in class, but its for working out maximum profit. This question is a bit different, I can't work out how to arrange it to solve it. Im guessing here I am required to work out the minimum cost that achieves at least the required amount of nickel and tin.

So would I be right in saying the following;
if Nickel = xA and Tin = xB

Minimise 1000xA + 1600xB

Subject to: 100xA + 400xB = 15
and 100xA + 100xB = 5

So xB = 1/30 or 0.0333
and xA = 1/60 or 0.016666

Please could someone confirm that I am on the right track

Many thanks
Jackie
No

Your variables are scrap metal and scrap cars and you are trying to minimise the cost of buying them
1
7 years ago
#4
Hi Jackie
You're not quite on a right track. The solution should be as follows:
scrap metal = xA and scrap car = xB
min 15xA + 5xB
s.t. 100xA + 100xB = 1000
400xA + 100xB = 1600
1
#5
ok thank you both so much for your help, i had it completely wrong, thanks again
0
#6
Ok ive drawn my graph but i dont know which bit is my feasible region? Or is it where the two lines intersect?
0
7 years ago
#7
(Original post by jackie11)
ok thank you both so much for your help, i had it completely wrong, thanks again
No problem

Try to think these things through without worrying about algebra first.

They are real, practical problems ... If you put yourself in the position of the manufacturer then it will be easier to formulate the problem

They will usually relate to maximising profit or production OR minimising expenditure
0
7 years ago
#8
(Original post by jackie11)
Ok ive drawn my graph but i dont know which bit is my feasible region? Or is it where the two lines intersect?
When you drew your lines did you shade out the unacceptable regions?

And you have 4 lines
0
7 years ago
#9
Change = signs in equations into >=. Should help
0
#10
Ok ive drawn my graph but i dont know which bit is my feasible region? Or is it where the two lines intersect?
(Original post by TenOfThem)
No problem

Try to think these things through without worrying about algebra first.

They are real, practical problems ... If you put yourself in the position of the manufacturer then it will be easier to formulate the problem

They will usually relate to maximising profit or production OR minimising expenditure
Ok thank you for the tips

(Original post by TenOfThem)
When you drew your lines did you shade out the unacceptable regions?

And you have 4 lines
I have four lines including the xA and xB axis
I did a plot check and I want the area to the up and right of this two lines, so the area is not contained completely??

(Original post by rollerboller)
Change = signs in equations into >=. Should help
Thank you
0
#11
Ok I just need to solve the simultaneous equations where these lines intersect, which is at (2, 8) so he will need to purchase 2 tonnes of scrap metal and 8 tonnes of scrap cars, which will cost £70, is this correct?
0
7 years ago
#12
(Original post by jackie11)
Ok I just need to solve the simultaneous equations where these lines intersect, which is at (2, 8) so he will need to purchase 2 tonnes of scrap metal and 8 tonnes of scrap cars, which will cost £70, is this correct?
Since the requirements were exact this is correct

You only have a feasible point not a feasible region
0
#13
ok thank you very much
0
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