Word Problem.

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A Mining Company was founded in 1894 and the mine’s initial production was the extraction of 100kg/year of silver. Each following year saw a steady increase of 60kg/year until the silver production peaked at 700 kg/year. Production remained at this level until 1914, when an event caused the mine to close indefinitely. The company then stopped its mining operations on the claim during the First World War (1914–1918). In 1920, the company opened up a new second mine which, in the first year, saw a silver yield of 700kg/year. However, with every subsequent year, production decreased by 8%. For economic reasons, the new mine had to close when the production fell below 300kg/year.

1. When did the first mine reach the peak production of 700 kg/year?

2. How long was the second mine operating for?

3. Compare the production of both mines over the first five years. Find the total production figures of the two mines until that ceased operation. which of the two mines was more successful? Provide evidence and explain the reasons behind your decision.

4. Calculate the percentage of gold remaining in the ground after all mining operations ceased.

first steps?
 

Dr.Peterson

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First step: read through the problem (twice, first for an overview, then writing down the data).

Second step: I'd probably sketch a graph, as a way to see the whole story at once.

Third step: Answer each question, by reading through the problem again with that question in mind, while looking at the graph.

You probably have some more specific questions if you've already done any of this! Feel free to show work and ask them.
 

tkhunny

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Model the production.
What sort of shape might the models have?
1) +60 / year
2) -8% / year

Go!

I'm particularly excited to see what you get for #4.
 

Jomo

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1894--100kg for the year

Then increased 60kg each year until it reached 700kg/yr

So how much did it increase if it went from 100kg to 700kg? How many 60kg's is this increase??? If this is too hard then,

100kg + 60kg = 160kg = 1895
160kg + 60kg = 220kg = 1896
220kg + 60kg = 280kg = 1897
...
640kg + 60kg = 700kg =????

You need to figure out some way of doing this problem on your own. Good luck.
 
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@Jomo 1904 I wrote them all out! Note this is sequences and series so I should probably use a formula?
 
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Jomo

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@Jomo 1904 I wrote them all out! Note this is sequences and series so I should probably use a formula?
OK, good. As you become better in Algebra you will see better methods. It takes time to see the best way to do a problem but you need to think of someway of doing this. Good job. Now can you do the rest of the parts?
 

Jomo

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Here is a better way of doing this. You went from 100kg to 700kg which is a 600kg increase. Since you only increased 60kg per year, simply compute 600/60 to get 10 years. Now 1894 + 10 = 1904. Is this clear?
 
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Before I move on. What formula should I have used If I know a=100 d=60 n=?
 
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Question two. First-term is 700kg, I will deduce 8% until I get near/at 300kg
years123456789101112
amount (kgs)700644592.48545.082501.475461.357424.4484390.4925359.253330.513304.072279.746

Production fell below 300kg/year after 12 years? is there a better way to do this with a formula?
 

Jomo

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Year 1: 700
Year 2 : 700(.8)
Year 3 [700(.8)](.8)=700(.8)^2
year 3: [700(.8)^2](.8) = 700(.8)^3
...
year n: 700(.8)^(n-1)

Now solve 700(.8)^(n-1) < 300
So (.8)^(n-1) < 300/700 = 3/7~.4286

If you know about logs you can use that or you can simply multiply .8 by itself until you first go below .4286.

Is this clear. If yes, then how do you get n?
 
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wait so

it takes 5 years for the
years12345
amount(kg)700560448358.4286.72
 
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Question one. 1904

Question two. 5 years?
 
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Let An be the production rate in year (AD 1893 + n).
A1 = 100 kg/yr
An=An−1+An=An−1+60 kg/yr for n∈2,3,4,…,Nn∈2,3,4,…,N, where AN=AN= 700 kg/yr.
If you plot A vs n, you should be able to convince yourself that
An=(100 kg/yr)+(60 kg/yr)⋅(n−1)An=(100 kg/yr)+(60 kg/yr)⋅(n−1) within this range.
AN=700 kg/yr→N=11AN=700 kg/yr→N=11.

2. Let BnBn be the production rate of the second mine in year (AD 1919 + n).
B1 = 700 kg/yr
An 8 % decrease each year corresponds to
Bn=0.92⋅Bn−1Bn=0.92⋅Bn−1 for n∈2,3,4,…,Mn∈2,3,4,…,M, where BM<BM< 300 kg/yr.
This is the same as
Bn=(700 kg/yr)⋅(0.92)n−1Bn=(700 kg/yr)⋅(0.92)n−1
Finding M:
300≤700(0.92M−1)→0.92M−1≥3/7→M−1≤log0.92(3/7)=ln(3/7)ln(0.92)=10.2→M=12300≤700(0.92M−1)→0.92M−1≥3/7→M−1≤log0.92(3/7)=ln(3/7)ln(0.92)=10.2→M=12.

How do I do question 3 and 4 @tkhunny @Jomo
 

JeffM

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Production of silver (not gold) declines by 8% (not 80%) by year. Therefore production each year is 92% of the previous year's.
 
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Production of silver (not gold) declines by 8% (not 80%) by year. Therefore production each year is 92% of the previous year's.
which part is this for?
 

JeffM

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which part is this for?
Question 3. Production in year 1 of the second mine was 700. The next year it was reduced by 8% or 56, meaning production was 644, not 560. But you can simplify things by multiplying 700 by 0.92, which equals 644. You had the right idea in post # 11, but were using 80% rather than 8%.
 
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@JeffM Wait so that means post #9 is correct.

Second Mine
years12345
amount (kgs)700644592.48545.082501.475
2,983kgs over 5 years

Table for the First mine.
Year12345
Amount (kg)100160220280340
1100kg total over 5 years

I guess based off my tables the second mine was more successful.
 

JeffM

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@JeffM Wait so that means post #9 is correct.

Second Mine
years12345
amount (kgs)700644592.48545.082501.475
2,983kgs over 5 years

Table for the First mine.
Year12345
Amount (kg)100160220280340
1100kg total over 5 years

I guess based off my tables the second mine was more successful.
Nothing in the problem talks about profitability.

As I think jomo said, you can solve the problem easily using logs.

\(\displaystyle \text {Production in year } y = 700 * 0.92^{(x-1)}.\)

\(\displaystyle 300 = 700 * 0.92^{(x -1)} \implies log(300) = log(700 * 0.92^{(x-1)}) = log(700) + (x - 1) * log(0.92) \implies\)

\(\displaystyle x = \dfrac{log(300) - log(700) + log(0.92)}{log(0.92)} \approx 11.1.\)

The same answer you got but quicker.

Let's check.

\(\displaystyle 700 * 0.92^{(11-1)} \approx 304 \text { and } 700 * 0.92^{(12-1)} \approx 280.\)

Knowing how to use logs is handy for exponential equations.
 
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4. Calculate the percentage of silver remaining in the ground after all mining operations ceased. How do I go about this? @JeffM
 

JeffM

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4. Calculate the percentage of silver remaining in the ground after all mining operations ceased. How do I go about this? @JeffM
This is a terrible question because the problem (at least as stated) gives no clue about how much was there to begin with.

So, as tkhunny hinted in post #2, the question (at least as stated) is technically unanswerable.

Thus, the first thing to do is to go back to the statement of the problem and see if you missed something.

If not, you will have to guess what the question means. My guess is that it is asking you to assume that, approximately, everything that is there is in the process of being extracted. So the total there originally is equal to what was extracted before mining ceased and what could be extracted if mining continued until the amount left was approximately zero. (You cannot get to zero by 8% reductions, but a balance below 10 would give a total accurate to considerably better than 1% given that total extraction is already in the thousands.)
 
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