Question
Begin the discussion by responding to the following prompt:
A fishing boat with a single passenger becomes stranded on a deserted island. Luckily, on the boat is a 5-gallon jug full of fresh water. The man will need to think carefully about how to best ration the water as there are no other sources of fresh water on the island.
The man thinks of three options and creates datasets and equation models to represent the first three days of each option.
For each Dataset, use the equation models provided to find
the amount of water remaining on day 6, and
the day on which the water jug will run empty.
For each Dataset, explain mathematically why the option is a good/bad choice for the man. Use words such as exponential, quadratic, linear, increasing, decreasing.
Practically speaking, what do you think is the best option? Why?
Dataset 1: Drink a little water to start, and then drink more and more each day. Equation model: y equals negative one-tenth x squared plus five.
Time (days) 0 1 2 3
Amount of water (gal.) 5 4.9 4.6 4.1
Dataset 2: Drink exactly half of the total amount of water available each day. Equation model: y equals five left parenthesis one-half right parenthesis start superscript x end superscript.
Time (days) 0 1 2 3
Amount of water (gal.) 5 2.5 1.25 0.625
Dataset 3: Drink exactly 4 cups of water each day. Equation model: y equals five left parenthesis one-half right parenthesis start superscript x end superscript.
Time (days) 0 1 2 3
Amount of water (gal.) 5 4.75 4.5 4.25
Post a detailed response to the Data Analysis Discussion prompt. Then respond to two classmates' posts.
A fishing boat with a single passenger becomes stranded on a deserted island. Luckily, on the boat is a 5-gallon jug full of fresh water. The man will need to think carefully about how to best ration the water as there are no other sources of fresh water on the island.
The man thinks of three options and creates datasets and equation models to represent the first three days of each option.
For each Dataset, use the equation models provided to find
the amount of water remaining on day 6, and
the day on which the water jug will run empty.
For each Dataset, explain mathematically why the option is a good/bad choice for the man. Use words such as exponential, quadratic, linear, increasing, decreasing.
Practically speaking, what do you think is the best option? Why?
Dataset 1: Drink a little water to start, and then drink more and more each day. Equation model: y equals negative one-tenth x squared plus five.
Time (days) 0 1 2 3
Amount of water (gal.) 5 4.9 4.6 4.1
Dataset 2: Drink exactly half of the total amount of water available each day. Equation model: y equals five left parenthesis one-half right parenthesis start superscript x end superscript.
Time (days) 0 1 2 3
Amount of water (gal.) 5 2.5 1.25 0.625
Dataset 3: Drink exactly 4 cups of water each day. Equation model: y equals five left parenthesis one-half right parenthesis start superscript x end superscript.
Time (days) 0 1 2 3
Amount of water (gal.) 5 4.75 4.5 4.25
Post a detailed response to the Data Analysis Discussion prompt. Then respond to two classmates' posts.
Answers
GPT-4o mini
**Data Analysis Discussion: Rationing Water on a Deserted Island**
**Dataset 1: Drink Increasing Amounts of Water (Quadratic Model)**
The equation model for this dataset is \( y = -\frac{1}{10}x^2 + 5 \), which is a quadratic function representing a parabolic decline in water consumption. By substituting values for \( x \) (days), we can find the amount of water remaining on day 6:
\[
y(6) = -\frac{1}{10}(6^2) + 5
\]
\[
= -\frac{1}{10}(36) + 5
\]
\[
= -3.6 + 5 = 1.4 \text{ gallons}
\]
To find the day when the water jug runs empty:
\[
0 = -\frac{1}{10}x^2 + 5
\]
\[
\frac{1}{10}x^2 = 5
\]
\[
x^2 = 50 \implies x = \sqrt{50} \approx 7.07 \text{ days}
\]
This option results in a decreasing amount of water consumed but increases the quantity consumed each day initially. This is a bad choice because while the man starts with a reasonable amount of water, the rapid increase in daily consumption means the water will run out after about 7 days, which is not sustainable, especially in the early days.
**Dataset 2: Drink Half the Amount Each Day (Exponential Model)**
The equation model for this dataset is \( y = 5 \left( \frac{1}{2} \right)^x \). This is an exponential decay model. To find the amount of water remaining on day 6, we can calculate:
\[
y(6) = 5 \left( \frac{1}{2} \right)^6
\]
\[
= 5 \times \frac{1}{64} = 0.078125 \text{ gallons}
\]
To find the day when the jug runs empty:
\[
0 = 5 \left( \frac{1}{2} \right)^x
\]
In practical terms, this function never truly reaches zero, but we can find how many days it would take for the water to be negligible (less than 0.1 gallon):
\[
5 \left( \frac{1}{2} \right)^x < 0.1
\]
\[
\left( \frac{1}{2} \right)^x < 0.02
\]
\[
x > \log_{1/2}(0.02) \approx 5.64 \text{ days}
\]
This option is bad because while the water lasts longer in terms of days, the quantity becomes almost unusable around day 6. There is a drastic reduction daily, making it impractical when needing sufficient hydration.
**Dataset 3: Drink 4 Cups Each Day (Linear Model)**
In this dataset, the man drinks a constant amount of water, which is equivalent to 1 gallon per day (4 cups). Thus, the dataset can be expressed as a linear function with a slope of -1. To find how much water remains on day 6:
\[
y(6) = 5 - 1 \cdot 6 = -1 \text{ gallons}
\]
The jug will run empty on day 5:
\[
5 - 1 \cdot 5 = 0 \text{ gallons}
\]
This option allows for a steady decline in water usage. The practicality here is that while it does run out in a matter of days, it ensures the man has a constant supply without the risk of running low too quickly. However, since he runs out on day 5, it is not a viable long-term solution.
**Conclusion: Best Option**
Upon analysis, while each option has strengths and weaknesses, none of them provides a sustainable long-term solution. However, drinking the 4 cups linearly seems to be the best choice in a practical sense, because it ensures a predictable and manageable water rationing plan, at least for the first five days, which might give the man time to devise another strategy or signal for help.
In a survival scenario, even an uncertain amount of stable hydration might be preferable over unsustainable exponential or quadratic increases which lead to quicker depletion. Therefore, the linear option allows the man to maintain hydration with enough predictability to make a calculated decision as he assesses his situation.
**Dataset 1: Drink Increasing Amounts of Water (Quadratic Model)**
The equation model for this dataset is \( y = -\frac{1}{10}x^2 + 5 \), which is a quadratic function representing a parabolic decline in water consumption. By substituting values for \( x \) (days), we can find the amount of water remaining on day 6:
\[
y(6) = -\frac{1}{10}(6^2) + 5
\]
\[
= -\frac{1}{10}(36) + 5
\]
\[
= -3.6 + 5 = 1.4 \text{ gallons}
\]
To find the day when the water jug runs empty:
\[
0 = -\frac{1}{10}x^2 + 5
\]
\[
\frac{1}{10}x^2 = 5
\]
\[
x^2 = 50 \implies x = \sqrt{50} \approx 7.07 \text{ days}
\]
This option results in a decreasing amount of water consumed but increases the quantity consumed each day initially. This is a bad choice because while the man starts with a reasonable amount of water, the rapid increase in daily consumption means the water will run out after about 7 days, which is not sustainable, especially in the early days.
**Dataset 2: Drink Half the Amount Each Day (Exponential Model)**
The equation model for this dataset is \( y = 5 \left( \frac{1}{2} \right)^x \). This is an exponential decay model. To find the amount of water remaining on day 6, we can calculate:
\[
y(6) = 5 \left( \frac{1}{2} \right)^6
\]
\[
= 5 \times \frac{1}{64} = 0.078125 \text{ gallons}
\]
To find the day when the jug runs empty:
\[
0 = 5 \left( \frac{1}{2} \right)^x
\]
In practical terms, this function never truly reaches zero, but we can find how many days it would take for the water to be negligible (less than 0.1 gallon):
\[
5 \left( \frac{1}{2} \right)^x < 0.1
\]
\[
\left( \frac{1}{2} \right)^x < 0.02
\]
\[
x > \log_{1/2}(0.02) \approx 5.64 \text{ days}
\]
This option is bad because while the water lasts longer in terms of days, the quantity becomes almost unusable around day 6. There is a drastic reduction daily, making it impractical when needing sufficient hydration.
**Dataset 3: Drink 4 Cups Each Day (Linear Model)**
In this dataset, the man drinks a constant amount of water, which is equivalent to 1 gallon per day (4 cups). Thus, the dataset can be expressed as a linear function with a slope of -1. To find how much water remains on day 6:
\[
y(6) = 5 - 1 \cdot 6 = -1 \text{ gallons}
\]
The jug will run empty on day 5:
\[
5 - 1 \cdot 5 = 0 \text{ gallons}
\]
This option allows for a steady decline in water usage. The practicality here is that while it does run out in a matter of days, it ensures the man has a constant supply without the risk of running low too quickly. However, since he runs out on day 5, it is not a viable long-term solution.
**Conclusion: Best Option**
Upon analysis, while each option has strengths and weaknesses, none of them provides a sustainable long-term solution. However, drinking the 4 cups linearly seems to be the best choice in a practical sense, because it ensures a predictable and manageable water rationing plan, at least for the first five days, which might give the man time to devise another strategy or signal for help.
In a survival scenario, even an uncertain amount of stable hydration might be preferable over unsustainable exponential or quadratic increases which lead to quicker depletion. Therefore, the linear option allows the man to maintain hydration with enough predictability to make a calculated decision as he assesses his situation.