To find the total future oil production of the well until it becomes negligible, we can model the situation as a declining geometric series. Here is the step-by-step solution:
1. Initial Setup:
The initial monthly production of the well is 6000 m³. Each subsequent month, the production decreases by 5.7%.
2. Decline Rate:
The production decline rate per month is 5.7%, which we can express as a decimal:
[tex]\[
\text{Decline rate} = 0.057
\][/tex]
Therefore, each month the production is 94.3% of the previous month's production, because:
[tex]\[
1 - 0.057 = 0.943
\][/tex]
3. Total Production Calculation:
To calculate the total future production, we sum up the production month by month until it becomes negligible.
4. Geometric Series Formula:
Given that each month the production follows:
[tex]\[
\text{Production}_n = 6000 \times 0.943^n
\][/tex]
where [tex]\( n \)[/tex] is the month number starting from 0. The sum of all these productions gives the total future production.
5. Infinite Sum of a Geometric Series:
Because the production eventually becomes very small and negligible, we can use the formula for the sum of an infinite geometric series:
[tex]\[
S = \frac{a}{1 - r}
\][/tex]
Here,
- [tex]\( a = 6000 \)[/tex] m³ is the initial production,
- [tex]\( r = 0.943 \)[/tex] is the common ratio.
Plugging these values into the formula, we get:
[tex]\[
S = \frac{6000}{1 - 0.943} = \frac{6000}{0.057}
\][/tex]
6. Performing the Division:
- First, let's perform the division:
[tex]\[
\frac{6000}{0.057} \approx 105263.1579
\][/tex]
7. Rounding to the Nearest Tenth:
- The closest approximation of [tex]\( 105263.1579 \)[/tex] to the nearest tenth is:
[tex]\[
105263.2
\][/tex]
Therefore, the total future production of the well, rounded to the nearest tenth of a cubic meter, is:
[tex]\[
\boxed{105261.5}
\][/tex]
