Geometric Series to Solve for Year a Resource Will Be Depleted Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Simple derivative task, no function, only values given, how the graph might look?Simplify using Geometric seriesGeometric series for this problem..?Optimizing number of production runs?Solve an geometric seriesGeometric series functionfind the annual per capita spending for personal consumption in doller?Basic Geometric SeriesStarting index for geometric series testSolve geometric series equation with large terms
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Geometric Series to Solve for Year a Resource Will Be Depleted
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Simple derivative task, no function, only values given, how the graph might look?Simplify using Geometric seriesGeometric series for this problem..?Optimizing number of production runs?Solve an geometric seriesGeometric series functionfind the annual per capita spending for personal consumption in doller?Basic Geometric SeriesStarting index for geometric series testSolve geometric series equation with large terms
$begingroup$
The original problem is as follows:
A community has 300 million tons of a non-renewable resource. Annual consumption is 25 million tons per year. Consumption is expected to decrease by 10% each year. Will the resource ever be depleted?
I set up the geometric series as
Total usage = 25[1 + (0.90) + (0.90)^2 + ...]
= 25[1/1-0.90]
= 250 million tons
No, the resource will not be depleted at this rate.
The second part of the problem asks:
What is the minimum percentage they can decrease consumption by to guarantee the resource does not run out?
Total usage = 25[1 + x + x^2 + ...]
= 25[1/1-x] = 300
x = 0.916
Minimum we can decrease consumption by is 8.4%
The final part of the question asks:
Suppose that it is not possible to decrease consumption of this resource by the previously given amount. The population is only able to decrease consumption by 5% each year. After how many years will the resource run out?
But I'm not exactly sure how to set this up. I'm assuming that solving for the year is just solving for n in the series where the sum up to (0.95)^n = 300, but I'm not sure how to go about setting up this equation.
calculus geometric-series
edited Apr 1 at 19:06
Andrei
13.9k21330
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
$endgroup$
add a comment |
$begingroup$
The original problem is as follows:
A community has 300 million tons of a non-renewable resource. Annual consumption is 25 million tons per year. Consumption is expected to decrease by 10% each year. Will the resource ever be depleted?
I set up the geometric series as
Total usage = 25[1 + (0.90) + (0.90)^2 + ...]
= 25[1/1-0.90]
= 250 million tons
No, the resource will not be depleted at this rate.
The second part of the problem asks:
What is the minimum percentage they can decrease consumption by to guarantee the resource does not run out?
Total usage = 25[1 + x + x^2 + ...]
= 25[1/1-x] = 300
x = 0.916
Minimum we can decrease consumption by is 8.4%
The final part of the question asks:
Suppose that it is not possible to decrease consumption of this resource by the previously given amount. The population is only able to decrease consumption by 5% each year. After how many years will the resource run out?
But I'm not exactly sure how to set this up. I'm assuming that solving for the year is just solving for n in the series where the sum up to (0.95)^n = 300, but I'm not sure how to go about setting up this equation.
calculus geometric-series
edited Apr 1 at 19:06
Andrei
13.9k21330
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
$endgroup$
add a comment |
$begingroup$
The original problem is as follows:
A community has 300 million tons of a non-renewable resource. Annual consumption is 25 million tons per year. Consumption is expected to decrease by 10% each year. Will the resource ever be depleted?
I set up the geometric series as
Total usage = 25[1 + (0.90) + (0.90)^2 + ...]
= 25[1/1-0.90]
= 250 million tons
No, the resource will not be depleted at this rate.
The second part of the problem asks:
What is the minimum percentage they can decrease consumption by to guarantee the resource does not run out?
Total usage = 25[1 + x + x^2 + ...]
= 25[1/1-x] = 300
x = 0.916
Minimum we can decrease consumption by is 8.4%
The final part of the question asks:
Suppose that it is not possible to decrease consumption of this resource by the previously given amount. The population is only able to decrease consumption by 5% each year. After how many years will the resource run out?
But I'm not exactly sure how to set this up. I'm assuming that solving for the year is just solving for n in the series where the sum up to (0.95)^n = 300, but I'm not sure how to go about setting up this equation.
calculus geometric-series
edited Apr 1 at 19:06
Andrei
13.9k21330
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
$endgroup$
The original problem is as follows:
A community has 300 million tons of a non-renewable resource. Annual consumption is 25 million tons per year. Consumption is expected to decrease by 10% each year. Will the resource ever be depleted?
I set up the geometric series as
Total usage = 25[1 + (0.90) + (0.90)^2 + ...]
= 25[1/1-0.90]
= 250 million tons
No, the resource will not be depleted at this rate.
The second part of the problem asks:
What is the minimum percentage they can decrease consumption by to guarantee the resource does not run out?
Total usage = 25[1 + x + x^2 + ...]
= 25[1/1-x] = 300
x = 0.916
Minimum we can decrease consumption by is 8.4%
The final part of the question asks:
Suppose that it is not possible to decrease consumption of this resource by the previously given amount. The population is only able to decrease consumption by 5% each year. After how many years will the resource run out?
But I'm not exactly sure how to set this up. I'm assuming that solving for the year is just solving for n in the series where the sum up to (0.95)^n = 300, but I'm not sure how to go about setting up this equation.
calculus geometric-series
calculus geometric-series
edited Apr 1 at 19:06
Andrei
13.9k21330
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
edited Apr 1 at 19:06
Andrei
13.9k21330
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
edited Apr 1 at 19:06
Andrei
13.9k21330
edited Apr 1 at 19:06
Andrei
13.9k21330
edited Apr 1 at 19:06
Andrei
13.9k21330
13.9k21330
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
asked Apr 1 at 18:51
John ProctorJohn Proctor
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Welcome on math.stackexchange!
For the geometric series there is not only a formula for $sum_n=0^infty q^n$ but also for the finite sums, namely
$$
sum_n=0^k q^n=frac1-q^k+11-q
$$
and this holds for $q not =1$.
With this formula you can then calculate the year.
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
$endgroup$
$begingroup$
Thanks, that's exactly the formula I was looking for.
$endgroup$
– John Proctor
Apr 1 at 19:06
$begingroup$
You're welcome.
$endgroup$
– Jonas Lenz
Apr 1 at 19:07
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Welcome on math.stackexchange!
For the geometric series there is not only a formula for $sum_n=0^infty q^n$ but also for the finite sums, namely
$$
sum_n=0^k q^n=frac1-q^k+11-q
$$
and this holds for $q not =1$.
With this formula you can then calculate the year.
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
$endgroup$
$begingroup$
Thanks, that's exactly the formula I was looking for.
$endgroup$
– John Proctor
Apr 1 at 19:06
$begingroup$
You're welcome.
$endgroup$
– Jonas Lenz
Apr 1 at 19:07
add a comment |
$begingroup$
Welcome on math.stackexchange!
For the geometric series there is not only a formula for $sum_n=0^infty q^n$ but also for the finite sums, namely
$$
sum_n=0^k q^n=frac1-q^k+11-q
$$
and this holds for $q not =1$.
With this formula you can then calculate the year.
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
$endgroup$
$begingroup$
Thanks, that's exactly the formula I was looking for.
$endgroup$
– John Proctor
Apr 1 at 19:06
$begingroup$
You're welcome.
$endgroup$
– Jonas Lenz
Apr 1 at 19:07
add a comment |
$begingroup$
Welcome on math.stackexchange!
For the geometric series there is not only a formula for $sum_n=0^infty q^n$ but also for the finite sums, namely
$$
sum_n=0^k q^n=frac1-q^k+11-q
$$
and this holds for $q not =1$.
With this formula you can then calculate the year.
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
$endgroup$
Welcome on math.stackexchange!
For the geometric series there is not only a formula for $sum_n=0^infty q^n$ but also for the finite sums, namely
$$
sum_n=0^k q^n=frac1-q^k+11-q
$$
and this holds for $q not =1$.
With this formula you can then calculate the year.
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
answered Apr 1 at 18:58
Jonas LenzJonas Lenz
694215
694215
$begingroup$
Thanks, that's exactly the formula I was looking for.
$endgroup$
– John Proctor
Apr 1 at 19:06
$begingroup$
You're welcome.
$endgroup$
– Jonas Lenz
Apr 1 at 19:07
add a comment |
$begingroup$
Thanks, that's exactly the formula I was looking for.
$endgroup$
– John Proctor
Apr 1 at 19:06
$begingroup$
You're welcome.
$endgroup$
– Jonas Lenz
Apr 1 at 19:07
$begingroup$
Thanks, that's exactly the formula I was looking for.
$endgroup$
– John Proctor
Apr 1 at 19:06
$begingroup$
Thanks, that's exactly the formula I was looking for.
$endgroup$
– John Proctor
Apr 1 at 19:06
$begingroup$
You're welcome.
$endgroup$
– Jonas Lenz
Apr 1 at 19:07
$begingroup$
You're welcome.
$endgroup$
– Jonas Lenz
Apr 1 at 19:07
add a comment |
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