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Math problem
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| Cloudburst |
Here's a math problem I have to solve today. Any hints or solutions?
The lawn needs 750 kg nitrogen (N), 250 kg phosphor (P) and 600 kg potassium (K). You can buy 2 kinds of fertilizer G1 and G2 to a cost of 40 respectively 50 for a bag. One bag contains 25kg.
G1 has 20%N 10%P 10%K
G2 has 20%N 4%P 15%K
How many of each bag do you have to but to minimize the cost?
Call the number of bags you need to buy x1 respectively x2.
I hope you understand my translation. I am supposed to use Matlab later (I hate it)... |
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| Massive84 |
Am not sure about your question, didn't understand it well but il start with this.
calculate first how many bags it takes to get to the amount of KG needed.
Each bag contains 25 kg right?
G1 has 20% N = 25kg * 0.2 (1 is 100%, 0.2 is 20%) = 5 Kg
G1 has 10% P = 20% of 25kg is 5kg, so 10% is half so it's 2,5kg.
G1 has 10% K =also 2,5kg.
Now we need 750kg N
How many times does 5kg(thats how much Nitrogen G1 contain) fit in 750kg(what we need).
750 / 5 = 150, So if we need Nitrogen only we need 150 bags.
Now for Phosphor
G1 has 2,5kg P so, 250kg / 2,5kg = 100 bags.
Now for Potassium
G1 has 2,5kg K so, 600kg / 2,5kg = 240 bags.
If you buy 100 bags, you won't have enough for N and K, and if you buy 150, you won't have enough for K, so i think the answer for G1 is 240 bags.
gonne do G2 on a new post. |
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| Massive84 |
G2
G2 has 20% N = same as G1 5Kg.
G2 has 4% P = 25kg * 0,04 = 1kg
G2 has 15% K = 25kg * 0,15 = 3,75kg
We need 750KG N = Same as G1 150 abgs.
G2 has 1kg P = 250kg / 1kg = 250 bags.
G2 has 3,75kg = 600kg / 3,75kg = 160 bags.
So for G2 you need 250 bags at least.
G1 is cheaper with 10 bags. |
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| töbias |
Just off the top of my head without getting too hot and bothered with spreadsheets I'd estimate it to be:
51 bags of g1
126 bags of g2
giving 885kg N, 253.5 P and 600 K for a total cost of $8,340. |
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| Cloudburst |
| I think that's a good estimate. I'll work some on it... |
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| THE_Chris |
| Cant be bothered doing it right now (spent the whole day doing relativity :whip: ), but basically form an equation, get a linking equation, differentiate, let it = 0, and go from there. |
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| töbias |
| quote: | Originally posted by Cloudburst
I think that's a good estimate. I'll work some on it... |
What was the answer in the end? |
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