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Hi there
Question on J Farms Ltd from Kaplan in relation to minimising costs.
Details – Jfarms ltd can buy two types of fertilisers which contain following % of chemicals,
Type X – Nitrates 18%, Phosphates 5%, Potash 2%
Type Y – Nitrates 3%, Phosphates 2%, Potash 5%
For a certain crop the following minimum quantities (Kg) are required:
Nitrates 100Kg, Phosphates 50Kg, Potash 40Kg
Type X costs $10 per kg and type y costs $5 per kg. J farms currently buys 1000Kg of each type and wishes to minimise its costs on fertilisers.
I was fine with tasks A and partly B but got stuck on finding the optimal solution.
I didn’t understand how to get to the following figures;
Point B – Solving 0.18x + 0.03y = 100Kg & 0.05x + 0.02y = 50
This gives x = 238.10 and y = 1904.80 (How were these figures calculated??)Objective function is Z = 10x + 5y
X is Type X
Y is Type YHave you watched my free lectures, because I explain in the lectures how to solve two simultaneous equations together 🙂
I have yes thanks, but your lecture didn’t cover example of when the objective is to minimise costs.
Can you help with this question?
Kind Regards
I understood from your first post that you were happy at arriving at the equations at point B and that your problem was in solving the two equations together. For that it makes no difference whether it is maximising or minimising.
I cannot check the equations themselves because I do not have the Kaplan Kit (only the BPP Revision Kit).You can solve the equations together in various ways (if you were taught a different way at school and remember it, then do it that way – all methods give the same final answer), What I do is as follows:
Multiply the second equation by 1.5 (so as to get the same number of y’s in both equations.
This gives: 0.075x + 0.03y = 75
If you subtract each term in this equation from each term in the other equation, then you get:
0.105x + 0 = 25
So x = 25/0.105 = 238.10
Put this in either equation and if you put it in the first equation you get:
42.858 + 0.03y = 100
So y = 1904.8
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