Answer :
To find out when the reservoir has the most water for the given time range [tex]\(0 \leq t \leq 30\)[/tex] seconds, we need to follow these steps:
1. Define the water level function: Let's call the water level function [tex]\( W(t) \)[/tex]. This function gives the water level in the reservoir at time [tex]\( t \)[/tex]. The exact form of [tex]\( W(t) \)[/tex] should be provided within the problem statement, but for this solution, we'll symbolically assume it's given.
2. Find the first derivative [tex]\( W'(t) \)[/tex]: To determine the critical points where the water level might be at a maximum, we first need to differentiate the water level function with respect to time [tex]\( t \)[/tex]. This gives us the rate of change of water level at any given moment.
3. Set the derivative [tex]\( W'(t) \)[/tex] to zero and solve for [tex]\( t \)[/tex]: The critical points occur where the first derivative is zero. Solve the equation:
[tex]\[ W'(t) = 0 \][/tex]
These solutions are the critical points which potentially include local maxima, minima, or points of inflection.
4. Evaluate [tex]\( W(t) \)[/tex] at the critical points and endpoints: Substitute these critical points back into the original water level function [tex]\( W(t) \)[/tex] to find the water levels at these points. Also, evaluate the function at the boundaries of the interval (i.e., [tex]\( t = 0 \)[/tex] and [tex]\( t = 30 \)[/tex]).
5. Compare the values of [tex]\( W(t) \)[/tex]: The maximum water level will be the highest value obtained from the evaluations at the critical points and the endpoints.
Let's go through this process in more detail:
### Step-by-Step Solution
1. Assume the water level function [tex]\( W(t) \)[/tex] is given:
[tex]\[ W(t) = \text{{given function}} \][/tex]
For this solution, let's assume [tex]\( W(t) \)[/tex] is defined, and we proceed with finding its derivative.
2. Find the first derivative [tex]\( W'(t) \)[/tex]:
[tex]\[ W'(t) = \frac{d}{dt} W(t) \][/tex]
3. Set [tex]\( W'(t) = 0 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ W'(t) = 0 \][/tex]
Solving this equation will provide the critical points. Let's say we find the critical points to be [tex]\( t = t_1, t_2, \ldots \)[/tex].
4. Evaluate [tex]\( W(t) \)[/tex] at the critical points and boundaries [tex]\( t = 0 \)[/tex] and [tex]\( t = 30 \)[/tex]:
[tex]\[ \text{{Evaluate at }} t_1: W(t_1) \\ \text{{Evaluate at }} t_2: W(t_2) \\ \ldots \\ \text{{Evaluate at }} t = 0: W(0) \\ \text{{Evaluate at }} t = 30: W(30) \][/tex]
5. Compare these values to find the maximum:
[tex]\[ \text{{Max water level}} = \max \{ W(0), W(t_1), W(t_2), \ldots, W(30) \} \][/tex]
The time [tex]\( t \)[/tex] at which the maximum water level occurs corresponds to this maximum value.
So, the reservoir has the most water at the time [tex]\( t \)[/tex] which results in the highest value of [tex]\( W(t) \)[/tex]. We need to evaluate [tex]\( W(t) \)[/tex] at the critical points and endpoints and pick the largest one.
Without the explicit form of [tex]\( W(t) \)[/tex], we can't provide a specific numeric answer. However, this detailed process will lead you to the exact time [tex]\( t \)[/tex] when the reservoir holds the most water.
1. Define the water level function: Let's call the water level function [tex]\( W(t) \)[/tex]. This function gives the water level in the reservoir at time [tex]\( t \)[/tex]. The exact form of [tex]\( W(t) \)[/tex] should be provided within the problem statement, but for this solution, we'll symbolically assume it's given.
2. Find the first derivative [tex]\( W'(t) \)[/tex]: To determine the critical points where the water level might be at a maximum, we first need to differentiate the water level function with respect to time [tex]\( t \)[/tex]. This gives us the rate of change of water level at any given moment.
3. Set the derivative [tex]\( W'(t) \)[/tex] to zero and solve for [tex]\( t \)[/tex]: The critical points occur where the first derivative is zero. Solve the equation:
[tex]\[ W'(t) = 0 \][/tex]
These solutions are the critical points which potentially include local maxima, minima, or points of inflection.
4. Evaluate [tex]\( W(t) \)[/tex] at the critical points and endpoints: Substitute these critical points back into the original water level function [tex]\( W(t) \)[/tex] to find the water levels at these points. Also, evaluate the function at the boundaries of the interval (i.e., [tex]\( t = 0 \)[/tex] and [tex]\( t = 30 \)[/tex]).
5. Compare the values of [tex]\( W(t) \)[/tex]: The maximum water level will be the highest value obtained from the evaluations at the critical points and the endpoints.
Let's go through this process in more detail:
### Step-by-Step Solution
1. Assume the water level function [tex]\( W(t) \)[/tex] is given:
[tex]\[ W(t) = \text{{given function}} \][/tex]
For this solution, let's assume [tex]\( W(t) \)[/tex] is defined, and we proceed with finding its derivative.
2. Find the first derivative [tex]\( W'(t) \)[/tex]:
[tex]\[ W'(t) = \frac{d}{dt} W(t) \][/tex]
3. Set [tex]\( W'(t) = 0 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ W'(t) = 0 \][/tex]
Solving this equation will provide the critical points. Let's say we find the critical points to be [tex]\( t = t_1, t_2, \ldots \)[/tex].
4. Evaluate [tex]\( W(t) \)[/tex] at the critical points and boundaries [tex]\( t = 0 \)[/tex] and [tex]\( t = 30 \)[/tex]:
[tex]\[ \text{{Evaluate at }} t_1: W(t_1) \\ \text{{Evaluate at }} t_2: W(t_2) \\ \ldots \\ \text{{Evaluate at }} t = 0: W(0) \\ \text{{Evaluate at }} t = 30: W(30) \][/tex]
5. Compare these values to find the maximum:
[tex]\[ \text{{Max water level}} = \max \{ W(0), W(t_1), W(t_2), \ldots, W(30) \} \][/tex]
The time [tex]\( t \)[/tex] at which the maximum water level occurs corresponds to this maximum value.
So, the reservoir has the most water at the time [tex]\( t \)[/tex] which results in the highest value of [tex]\( W(t) \)[/tex]. We need to evaluate [tex]\( W(t) \)[/tex] at the critical points and endpoints and pick the largest one.
Without the explicit form of [tex]\( W(t) \)[/tex], we can't provide a specific numeric answer. However, this detailed process will lead you to the exact time [tex]\( t \)[/tex] when the reservoir holds the most water.
