🧠:<think>Okay, so I have this problem about modeling a boxer's power output during a training session. The coach is using a sinusoidal function to represent the power, which makes sense because power might fluctuate during training, especially in something like boxing where intensity can go up and down. The function given is:[ P(t) = A sin(omega t + phi) + C ]Where:- ( A ) is the amplitude,- ( omega ) is the angular frequency,- ( phi ) is the phase shift,- ( C ) is the constant baseline power.The training session lasts for one full period ( T = frac{2pi}{omega} ).The first part asks for the total energy ( E ) expended by the boxer over the session. Energy is the integral of power over time, so I need to compute the integral of ( P(t) ) from ( t = 0 ) to ( t = T ).Let me write that down:[ E = int_{0}^{T} P(t) , dt = int_{0}^{T} left[ A sin(omega t + phi) + C right] dt ]I can split this integral into two parts:[ E = A int_{0}^{T} sin(omega t + phi) , dt + C int_{0}^{T} dt ]Let me compute each integral separately.First, the integral of ( sin(omega t + phi) ). The integral of ( sin(ax + b) ) with respect to ( x ) is ( -frac{1}{a} cos(ax + b) ). So applying that here:[ int sin(omega t + phi) dt = -frac{1}{omega} cos(omega t + phi) + K ]Where ( K ) is the constant of integration, but since we're doing a definite integral, it will cancel out.So evaluating from 0 to ( T ):[ left[ -frac{1}{omega} cos(omega T + phi) + frac{1}{omega} cos(0 + phi) right] ]But ( T = frac{2pi}{omega} ), so ( omega T = 2pi ). Therefore, ( cos(omega T + phi) = cos(2pi + phi) ). Since cosine has a period of ( 2pi ), ( cos(2pi + phi) = cos(phi) ).So substituting back in:[ -frac{1}{omega} cos(phi) + frac{1}{omega} cos(phi) = 0 ]So the integral of the sine term over one full period is zero. That makes sense because the positive and negative areas cancel out over a full cycle.Now, the second integral:[ C int_{0}^{T} dt = C left[ t right]_0^T = C(T - 0) = C T ]So putting it all together:[ E = 0 + C T = C T ]But ( T = frac{2pi}{omega} ), so:[ E = C cdot frac{2pi}{omega} ]So that's the total energy expended. It only depends on ( C ) and ( omega ), not on ( A ) or ( phi ). Interesting.Moving on to the second part. It says that optimal endurance is achieved when the average power output during the session is equal to ( C ). I need to find the necessary relationship between ( A ) and ( C ).First, average power is total energy divided by total time. So:[ text{Average Power} = frac{E}{T} ]From part 1, ( E = C T ), so:[ text{Average Power} = frac{C T}{T} = C ]Wait, that's just ( C ). So regardless of ( A ), the average power is always ( C ). But that seems contradictory because the problem is asking for a relationship between ( A ) and ( C ) when the average power is equal to ( C ). But according to this, the average power is always ( C ), regardless of ( A ). So does that mean there's no restriction on ( A )?Wait, let me double-check. Maybe I made a mistake in calculating the average power. Let me compute it again.Average power is:[ frac{1}{T} int_{0}^{T} P(t) dt ]Which is exactly ( frac{E}{T} ). From part 1, ( E = C T ), so ( frac{E}{T} = C ). So regardless of the value of ( A ), the average power is always ( C ). Therefore, the average power is always equal to ( C ), so there is no additional relationship needed between ( A ) and ( C ). The average power is inherently ( C ) because the sine function averages out over a full period.But wait, that seems counterintuitive. If you have a higher amplitude ( A ), wouldn't that mean the power fluctuates more around ( C ), but the average remains the same? So yeah, the average power is just the DC offset, which is ( C ). So regardless of how much it fluctuates, the average is always ( C ).Therefore, the necessary relationship is that there is no restriction; the average power is always ( C ), so any ( A ) is acceptable as long as ( C ) is set appropriately.But the problem says "optimal endurance is achieved when the average power output during the session is equal to ( C )". So if the average is always ( C ), then it's always optimal? Or maybe I misread the question.Wait, maybe the question is implying that the average power should be equal to ( C ), but in reality, the average power is ( C ), so that condition is always satisfied. Therefore, there is no additional relationship needed between ( A ) and ( C ); they can be independent.But that seems odd because the problem is asking for a relationship. Maybe I made a mistake in the first part.Wait, let me go back to the first part. Maybe I was too quick. Let me recompute the integral.[ E = int_{0}^{T} [A sin(omega t + phi) + C] dt ]Which is:[ A int_{0}^{T} sin(omega t + phi) dt + C int_{0}^{T} dt ]As before, the sine integral is zero over a full period, so ( E = C T ). So average power is ( C ).Therefore, the average power is always ( C ), so the condition is automatically satisfied regardless of ( A ). Therefore, there is no necessary relationship between ( A ) and ( C ); they can be any values.But the problem says "determine the necessary relationship between the amplitude ( A ) and the baseline power ( C )". So maybe I'm missing something.Wait, perhaps the average power is supposed to be equal to ( C ), but in reality, maybe the average power is different? Let me compute the average power again.Average power:[ frac{1}{T} int_{0}^{T} P(t) dt = frac{1}{T} left( A int_{0}^{T} sin(omega t + phi) dt + C int_{0}^{T} dt right) ]As before, the sine integral is zero, so:[ frac{1}{T} (0 + C T) = C ]So yes, the average power is always ( C ). Therefore, the condition is always satisfied, so no relationship is needed between ( A ) and ( C ). They can be independent.But the problem is asking for a relationship, so maybe I'm misunderstanding the question. Perhaps it's not about the average power being equal to ( C ), but something else? Or maybe the coach wants the average power to be equal to ( C ), but in reality, it's always ( C ), so that condition is automatically met.Alternatively, maybe the problem is considering something else, like the average of the absolute value of power or something, but the question says "average power output", which is the integral of power over time, so it should be ( C ).Alternatively, maybe the coach wants the average power to be equal to the baseline power ( C ), which is already the case, so no condition is needed.Therefore, perhaps the answer is that there is no necessary relationship; ( A ) can be any value independent of ( C ).But the problem says "determine the necessary relationship", so maybe I'm missing something. Alternatively, perhaps the coach wants the average power to be equal to ( C ), but in reality, the average is ( C ), so it's always true, so no relationship is needed.Alternatively, maybe the problem is considering the average of the absolute power? Let me check.If we take the average of the absolute value of ( P(t) ), that would be different. But the question says "average power output", which is typically the integral of power over time, not the average of the absolute power.So, I think the conclusion is that the average power is always ( C ), so there is no necessary relationship between ( A ) and ( C ). They can be independent.But the problem is phrased as if there is a relationship to determine, so maybe I'm missing something. Alternatively, perhaps the coach wants the average power to be equal to ( C ), which is already the case, so no restriction on ( A ).Alternatively, maybe the problem is considering the root mean square (RMS) power? But the question says "average power", so I think it's just the integral.Therefore, I think the answer is that there is no necessary relationship; ( A ) can be any value, and the average power will always be ( C ).But to be thorough, let me consider if there's another interpretation. Maybe the coach wants the average power to be equal to the baseline power ( C ), but perhaps the average of the sine wave is zero, so the average power is ( C ). So yes, that's consistent.Therefore, the necessary relationship is that there is no restriction; ( A ) can be any value, and the average power will always equal ( C ).But the problem says "determine the necessary relationship", so perhaps the answer is that ( A ) can be any value, independent of ( C ). So the relationship is that ( A ) and ( C ) are independent.Alternatively, maybe the problem is expecting me to set the average power equal to ( C ), which is already the case, so no condition is needed.In conclusion, for part 1, the total energy ( E ) is ( C cdot frac{2pi}{omega} ). For part 2, the average power is always ( C ), so no relationship is needed between ( A ) and ( C ); they can be independent.But since the problem asks for a relationship, maybe I'm supposed to say that ( A ) can be any value, or perhaps ( A ) must be zero, but that doesn't make sense because then it's just a constant power.Wait, if ( A ) is zero, then ( P(t) = C ), which is a constant. But the problem is about a sinusoidal function, so ( A ) is non-zero. Therefore, the average power is always ( C ), regardless of ( A ).So, to answer the second part, the necessary relationship is that there is no restriction; ( A ) can be any value, and the average power will still be ( C ).But the problem says "determine the necessary relationship", so maybe the answer is that ( A ) can be any value, or perhaps ( A ) must be zero, but that contradicts the sinusoidal model.Alternatively, perhaps the coach wants the average power to be equal to ( C ), which is already the case, so no condition is needed.Therefore, I think the answer is that there is no necessary relationship; ( A ) can be any value, and the average power will always equal ( C ).But to be safe, let me consider if the average power is supposed to be something else. Wait, the problem says "optimal endurance is achieved when the average power output during the session is equal to ( C )". So if the average is always ( C ), then optimal endurance is always achieved, regardless of ( A ). Therefore, there is no necessary relationship between ( A ) and ( C ); they can be independent.So, summarizing:1. ( E = frac{2pi C}{omega} )2. There is no necessary relationship between ( A ) and ( C ); the average power is always ( C ).But the problem asks for a relationship, so maybe I'm supposed to say that ( A ) can be any value, or perhaps ( A ) must be zero, but that doesn't make sense because then it's just a constant.Alternatively, perhaps the problem is considering the average of the absolute value of power, which would be different. Let me compute that.Average of absolute power:[ frac{1}{T} int_{0}^{T} |A sin(omega t + phi) + C| dt ]But that's more complicated and not typically what is meant by average power. The standard average power is just the integral of power over time divided by time, which is ( C ).Therefore, I think the answer is that there is no necessary relationship; ( A ) can be any value, and the average power will always equal ( C ).But since the problem asks for a relationship, maybe I'm supposed to say that ( A ) can be any value, or perhaps ( A ) must be zero, but that contradicts the sinusoidal model.Alternatively, perhaps the coach wants the average power to be equal to ( C ), which is already the case, so no condition is needed.Therefore, I think the answer is that there is no necessary relationship; ( A ) can be any value, and the average power will always equal ( C ).But to be thorough, let me consider if the average power is supposed to be equal to ( C ), which is already the case, so no relationship is needed.Therefore, the necessary relationship is that there is no restriction; ( A ) can be any value.But the problem says "determine the necessary relationship", so perhaps the answer is that ( A ) can be any value, independent of ( C ).Alternatively, maybe the problem is expecting me to set the average power equal to ( C ), which is already the case, so no condition is needed.In conclusion, for part 1, the total energy is ( E = frac{2pi C}{omega} ). For part 2, the average power is always ( C ), so no relationship is needed between ( A ) and ( C ); they can be independent.But since the problem asks for a relationship, I think the answer is that there is no necessary relationship; ( A ) can be any value, and the average power will always equal ( C ).Alternatively, perhaps the problem is considering the average of the absolute value of power, but that's not standard.Therefore, I think the answer is that there is no necessary relationship; ( A ) can be any value, and the average power will always equal ( C ).</think>

🧠:<think>Okay, so I have this problem about a sports bar owner who hosts game watch parties for the Nashville Predators. The attendance at these events is modeled by this function:( A(t) = 80 + 20sinleft(frac{pi}{6}(t - 1)right) + 15cosleft(frac{pi}{3}(t - 2)right) )where ( t ) is the game number, a positive integer. There are two parts to the problem. The first is to find the total number of attendees over the first 12 games. The second is to determine which game within the first 12 has the highest attendance and what that maximum number is.Alright, let's tackle the first part. I need to calculate the total number of attendees over the first 12 games. That means I need to compute ( A(t) ) for each ( t ) from 1 to 12 and then sum all those values up.So, the total attendance ( T ) would be:( T = sum_{t=1}^{12} A(t) )Which is:( T = sum_{t=1}^{12} left[80 + 20sinleft(frac{pi}{6}(t - 1)right) + 15cosleft(frac{pi}{3}(t - 2)right)right] )I can separate this sum into three separate sums:( T = sum_{t=1}^{12} 80 + sum_{t=1}^{12} 20sinleft(frac{pi}{6}(t - 1)right) + sum_{t=1}^{12} 15cosleft(frac{pi}{3}(t - 2)right) )Calculating each part individually.First, the sum of 80 from t=1 to t=12 is straightforward. That's just 80 multiplied by 12.So, ( sum_{t=1}^{12} 80 = 80 times 12 = 960 )Next, the second sum is ( 20 times sum_{t=1}^{12} sinleft(frac{pi}{6}(t - 1)right) )Let me denote ( theta = frac{pi}{6}(t - 1) ), so as t goes from 1 to 12, ( theta ) goes from 0 to ( frac{pi}{6}(11) = frac{11pi}{6} ).So, we have 12 terms of sine starting at 0 and increasing by ( frac{pi}{6} ) each time. That's like the sine of 0, ( frac{pi}{6} ), ( frac{pi}{3} ), up to ( frac{11pi}{6} ).I remember that the sum of sine functions over equally spaced angles can be calculated using the formula for the sum of a sine series. The formula is:( sum_{k=0}^{n-1} sin(a + kd) = frac{sinleft(frac{nd}{2}right) cdot sinleft(a + frac{(n - 1)d}{2}right)}{sinleft(frac{d}{2}right)} )In this case, ( a = 0 ), ( d = frac{pi}{6} ), and ( n = 12 ).Plugging into the formula:( sum_{k=0}^{11} sinleft(0 + k cdot frac{pi}{6}right) = frac{sinleft(frac{12 cdot frac{pi}{6}}{2}right) cdot sinleft(0 + frac{(12 - 1) cdot frac{pi}{6}}{2}right)}{sinleft(frac{frac{pi}{6}}{2}right)} )Simplify step by step.First, compute ( frac{12 cdot frac{pi}{6}}{2} ):( frac{12 cdot frac{pi}{6}}{2} = frac{2pi}{2} = pi )Next, compute ( frac{(12 - 1) cdot frac{pi}{6}}{2} ):( frac{11 cdot frac{pi}{6}}{2} = frac{11pi}{12} )And the denominator is ( sinleft(frac{pi}{12}right) )So, putting it all together:( sum = frac{sin(pi) cdot sinleft(frac{11pi}{12}right)}{sinleft(frac{pi}{12}right)} )But ( sin(pi) = 0 ), so the entire sum is 0.Wait, that's interesting. So the sum of the sine terms is zero. That makes sense because sine is symmetric over a full period, and 12 terms with ( frac{pi}{6} ) spacing cover exactly two full periods (since ( 2pi / frac{pi}{6} = 12 )). So, over two full periods, the positive and negative areas cancel out, resulting in a sum of zero.Therefore, the second sum is ( 20 times 0 = 0 ).Now, moving on to the third sum: ( 15 times sum_{t=1}^{12} cosleft(frac{pi}{3}(t - 2)right) )Let me make a substitution here as well. Let ( phi = frac{pi}{3}(t - 2) ). So when t=1, ( phi = frac{pi}{3}(-1) = -frac{pi}{3} ), and when t=12, ( phi = frac{pi}{3}(10) = frac{10pi}{3} ).So, the angles go from ( -frac{pi}{3} ) to ( frac{10pi}{3} ), increasing by ( frac{pi}{3} ) each time. That's 12 terms, each spaced by ( frac{pi}{3} ).Again, I can use the formula for the sum of cosines:( sum_{k=0}^{n-1} cos(a + kd) = frac{sinleft(frac{nd}{2}right) cdot cosleft(a + frac{(n - 1)d}{2}right)}{sinleft(frac{d}{2}right)} )In this case, ( a = -frac{pi}{3} ), ( d = frac{pi}{3} ), and ( n = 12 ).Plugging into the formula:( sum_{k=0}^{11} cosleft(-frac{pi}{3} + k cdot frac{pi}{3}right) = frac{sinleft(frac{12 cdot frac{pi}{3}}{2}right) cdot cosleft(-frac{pi}{3} + frac{(12 - 1) cdot frac{pi}{3}}{2}right)}{sinleft(frac{frac{pi}{3}}{2}right)} )Simplify each part.First, compute ( frac{12 cdot frac{pi}{3}}{2} ):( frac{12 cdot frac{pi}{3}}{2} = frac{4pi}{2} = 2pi )Next, compute ( -frac{pi}{3} + frac{11 cdot frac{pi}{3}}{2} ):Let me compute ( frac{11 cdot frac{pi}{3}}{2} = frac{11pi}{6} )So, ( -frac{pi}{3} + frac{11pi}{6} = frac{-2pi}{6} + frac{11pi}{6} = frac{9pi}{6} = frac{3pi}{2} )The denominator is ( sinleft(frac{pi}{6}right) )So, putting it all together:( sum = frac{sin(2pi) cdot cosleft(frac{3pi}{2}right)}{sinleft(frac{pi}{6}right)} )Compute each term:( sin(2pi) = 0 )( cosleft(frac{3pi}{2}right) = 0 )( sinleft(frac{pi}{6}right) = frac{1}{2} )But since the numerator is 0 multiplied by 0, which is 0, the entire sum is 0.Wait, that seems odd. Let me double-check.Wait, actually, the formula is:( sum_{k=0}^{n-1} cos(a + kd) = frac{sinleft(frac{nd}{2}right) cdot cosleft(a + frac{(n - 1)d}{2}right)}{sinleft(frac{d}{2}right)} )So, plugging in the numbers:( sinleft(2piright) = 0 ), so the entire sum is 0.Therefore, the sum of the cosine terms is 0 as well.Wait, but let me think about this. The sum of 12 cosine terms, each spaced by ( frac{pi}{3} ), starting from ( -frac{pi}{3} ). So, the angles are ( -frac{pi}{3}, 0, frac{pi}{3}, frac{2pi}{3}, pi, frac{4pi}{3}, frac{5pi}{3}, 2pi, frac{7pi}{3}, frac{8pi}{3}, 3pi, frac{10pi}{3} ).Wait, but ( frac{10pi}{3} ) is equivalent to ( frac{4pi}{3} ) since ( frac{10pi}{3} - 2pi = frac{4pi}{3} ). Similarly, ( 3pi ) is ( pi ), and ( frac{7pi}{3} ) is ( frac{pi}{3} ), etc. So, actually, the angles are symmetric around the unit circle, so their sum might indeed be zero.But wait, let's compute a few terms manually to check.Compute ( cos(-frac{pi}{3}) = cos(frac{pi}{3}) = 0.5 )( cos(0) = 1 )( cos(frac{pi}{3}) = 0.5 )( cos(frac{2pi}{3}) = -0.5 )( cos(pi) = -1 )( cos(frac{4pi}{3}) = -0.5 )( cos(frac{5pi}{3}) = 0.5 )( cos(2pi) = 1 )( cos(frac{7pi}{3}) = cos(frac{pi}{3}) = 0.5 )( cos(frac{8pi}{3}) = cos(frac{2pi}{3}) = -0.5 )( cos(3pi) = cos(pi) = -1 )( cos(frac{10pi}{3}) = cos(frac{4pi}{3}) = -0.5 )So, listing all the cosine values:0.5, 1, 0.5, -0.5, -1, -0.5, 0.5, 1, 0.5, -0.5, -1, -0.5Now, let's add them up step by step:Start with 0.5+1 = 1.5+0.5 = 2-0.5 = 1.5-1 = 0.5-0.5 = 0+0.5 = 0.5+1 = 1.5+0.5 = 2-0.5 = 1.5-1 = 0.5-0.5 = 0So, the total sum is 0. Therefore, the third sum is indeed 0.Therefore, the total attendance ( T = 960 + 0 + 0 = 960 ).Wait, that seems surprisingly clean. So, over the first 12 games, the total attendance is 960.But let me just verify this because sometimes when dealing with periodic functions, especially when the number of terms is a multiple of the period, the sum can indeed be zero.In this case, for the sine function, the period is ( frac{2pi}{frac{pi}{6}} = 12 ), so 12 terms make exactly one full period. Similarly, for the cosine function, the period is ( frac{2pi}{frac{pi}{3}} = 6 ), so 12 terms make exactly two full periods. Therefore, both sums over their respective periods will cancel out, resulting in zero.Therefore, the total attendance is just 12 times 80, which is 960.Okay, that seems correct.Now, moving on to the second part: determining the game number ( t ) within the first 12 games where the attendance is expected to be the highest, and finding that maximum number.So, I need to find the value of ( t ) (from 1 to 12) that maximizes ( A(t) ).Given the function:( A(t) = 80 + 20sinleft(frac{pi}{6}(t - 1)right) + 15cosleft(frac{pi}{3}(t - 2)right) )To find the maximum, I can compute ( A(t) ) for each ( t ) from 1 to 12 and then pick the maximum value. Since it's only 12 terms, it's manageable.Alternatively, I can analyze the function to find where the sine and cosine terms reach their maximums.But since the sine and cosine functions are periodic and have different frequencies, their combination might not reach the theoretical maximum of 80 + 20 + 15 = 115. Instead, the maximum will be somewhere less, depending on how the sine and cosine terms align.But perhaps computing each ( A(t) ) is the safest way.So, let's compute ( A(t) ) for each ( t ) from 1 to 12.First, let's note that ( t ) is an integer from 1 to 12.Compute each term step by step.Let me make a table:| t | sin(π/6*(t-1)) | cos(π/3*(t-2)) | 20*sin(...) | 15*cos(...) | A(t) = 80 + ... ||---|----------------|----------------|-------------|-------------|----------------|| 1 | sin(0) = 0 | cos(-π/3) = 0.5| 0 | 7.5 | 87.5 || 2 | sin(π/6) ≈ 0.5 | cos(0) = 1 | 10 | 15 | 105 || 3 | sin(π/3) ≈ 0.866| cos(π/3) ≈ 0.5 | ≈17.32 | 7.5 | ≈104.82 || 4 | sin(π/2) = 1 | cos(2π/3) ≈ -0.5| 20 | -7.5 | 92.5 || 5 | sin(2π/3) ≈ 0.866| cos(π) = -1 | ≈17.32 | -15 | ≈82.32 || 6 | sin(5π/6) ≈ 0.5 | cos(4π/3) ≈ -0.5| 10 | -7.5 | 82.5 || 7 | sin(π) = 0 | cos(5π/3) ≈ 0.5 | 0 | 7.5 | 87.5 || 8 | sin(7π/6) ≈ -0.5| cos(2π) = 1 | -10 | 15 | 85 || 9 | sin(4π/3) ≈ -0.866| cos(7π/3) ≈ 0.5| ≈-17.32 | 7.5 | ≈70.18 ||10 | sin(3π/2) = -1 | cos(8π/3) ≈ -0.5| -20 | -7.5 | 52.5 ||11 | sin(5π/3) ≈ -0.866| cos(3π) = -1 | ≈-17.32 | -15 | ≈47.68 ||12 | sin(11π/6) ≈ -0.5| cos(10π/3) ≈ -0.5| -10 | -7.5 | 62.5 |Wait, let me compute each step carefully.Starting with t=1:- ( sinleft(frac{pi}{6}(1 - 1)right) = sin(0) = 0 )- ( cosleft(frac{pi}{3}(1 - 2)right) = cosleft(-frac{pi}{3}right) = cosleft(frac{pi}{3}right) = 0.5 )- So, 20*0 = 0, 15*0.5 = 7.5- A(1) = 80 + 0 + 7.5 = 87.5t=2:- ( sinleft(frac{pi}{6}(2 - 1)right) = sinleft(frac{pi}{6}right) = 0.5 )- ( cosleft(frac{pi}{3}(2 - 2)right) = cos(0) = 1 )- 20*0.5 = 10, 15*1 = 15- A(2) = 80 + 10 + 15 = 105t=3:- ( sinleft(frac{pi}{6}(3 - 1)right) = sinleft(frac{pi}{3}right) ≈ 0.866 )- ( cosleft(frac{pi}{3}(3 - 2)right) = cosleft(frac{pi}{3}right) ≈ 0.5 )- 20*0.866 ≈ 17.32, 15*0.5 = 7.5- A(3) ≈ 80 + 17.32 + 7.5 ≈ 104.82t=4:- ( sinleft(frac{pi}{6}(4 - 1)right) = sinleft(frac{pi}{2}right) = 1 )- ( cosleft(frac{pi}{3}(4 - 2)right) = cosleft(frac{2pi}{3}right) ≈ -0.5 )- 20*1 = 20, 15*(-0.5) = -7.5- A(4) = 80 + 20 - 7.5 = 92.5t=5:- ( sinleft(frac{pi}{6}(5 - 1)right) = sinleft(frac{2pi}{3}right) ≈ 0.866 )- ( cosleft(frac{pi}{3}(5 - 2)right) = cosleft(piright) = -1 )- 20*0.866 ≈ 17.32, 15*(-1) = -15- A(5) ≈ 80 + 17.32 - 15 ≈ 82.32t=6:- ( sinleft(frac{pi}{6}(6 - 1)right) = sinleft(frac{5pi}{6}right) = 0.5 )- ( cosleft(frac{pi}{3}(6 - 2)right) = cosleft(frac{4pi}{3}right) ≈ -0.5 )- 20*0.5 = 10, 15*(-0.5) = -7.5- A(6) = 80 + 10 - 7.5 = 82.5t=7:- ( sinleft(frac{pi}{6}(7 - 1)right) = sinleft(piright) = 0 )- ( cosleft(frac{pi}{3}(7 - 2)right) = cosleft(frac{5pi}{3}right) ≈ 0.5 )- 20*0 = 0, 15*0.5 = 7.5- A(7) = 80 + 0 + 7.5 = 87.5t=8:- ( sinleft(frac{pi}{6}(8 - 1)right) = sinleft(frac{7pi}{6}right) ≈ -0.5 )- ( cosleft(frac{pi}{3}(8 - 2)right) = cosleft(2piright) = 1 )- 20*(-0.5) = -10, 15*1 = 15- A(8) = 80 - 10 + 15 = 85t=9:- ( sinleft(frac{pi}{6}(9 - 1)right) = sinleft(frac{4pi}{3}right) ≈ -0.866 )- ( cosleft(frac{pi}{3}(9 - 2)right) = cosleft(frac{7pi}{3}right) = cosleft(frac{pi}{3}right) ≈ 0.5 )- 20*(-0.866) ≈ -17.32, 15*0.5 = 7.5- A(9) ≈ 80 - 17.32 + 7.5 ≈ 70.18t=10:- ( sinleft(frac{pi}{6}(10 - 1)right) = sinleft(frac{3pi}{2}right) = -1 )- ( cosleft(frac{pi}{3}(10 - 2)right) = cosleft(frac{8pi}{3}right) = cosleft(frac{2pi}{3}right) ≈ -0.5 )- 20*(-1) = -20, 15*(-0.5) = -7.5- A(10) = 80 - 20 - 7.5 = 52.5t=11:- ( sinleft(frac{pi}{6}(11 - 1)right) = sinleft(frac{5pi}{3}right) ≈ -0.866 )- ( cosleft(frac{pi}{3}(11 - 2)right) = cosleft(3piright) = -1 )- 20*(-0.866) ≈ -17.32, 15*(-1) = -15- A(11) ≈ 80 - 17.32 - 15 ≈ 47.68t=12:- ( sinleft(frac{pi}{6}(12 - 1)right) = sinleft(frac{11pi}{6}right) ≈ -0.5 )- ( cosleft(frac{pi}{3}(12 - 2)right) = cosleft(frac{10pi}{3}right) = cosleft(frac{4pi}{3}right) ≈ -0.5 )- 20*(-0.5) = -10, 15*(-0.5) = -7.5- A(12) = 80 - 10 - 7.5 = 62.5Now, compiling all the A(t) values:t | A(t)---|---1 | 87.52 | 1053 | ≈104.824 | 92.55 | ≈82.326 | 82.57 | 87.58 | 859 | ≈70.1810 | 52.511 | ≈47.6812 | 62.5Looking at these values, the highest attendance is at t=2 with 105 attendees. The next highest is t=3 with approximately 104.82, which is slightly less than 105. Then t=1 and t=7 both have 87.5, which is lower.Therefore, the maximum attendance occurs at game number 2, with 105 attendees.Wait, but let me double-check the calculations for t=3 because 104.82 is very close to 105. Maybe I made a rounding error.For t=3:- ( sinleft(frac{pi}{3}right) ≈ sqrt{3}/2 ≈ 0.8660 )- So, 20*0.8660 ≈ 17.32- ( cosleft(frac{pi}{3}right) = 0.5 )- 15*0.5 = 7.5- So, A(3) ≈ 80 + 17.32 + 7.5 = 104.82Yes, that's correct. So, 104.82 is less than 105, so t=2 is indeed the maximum.Therefore, the answers are:1. Total attendance over first 12 games: 9602. Maximum attendance at game t=2 with 105 attendees.Final Answer1. The total number of attendees over the first 12 games is boxed{960}.2. The maximum attendance occurs on game number boxed{2} with boxed{105} attendees.</think>

🧠:<think>Okay, so I have this problem about a dog's success rate in responding to commands, modeled by a differential equation. Let me try to figure this out step by step.First, the differential equation given is:[ frac{dR}{dt} = k cdot (1 - R(t)) cdot R(t) ]And we're told that initially, R(0) = 0.2, and k is 0.3 for the first part. I need to solve this differential equation to find R(t). Hmm, this looks like a logistic growth model, right? Because it has the form dR/dt = k R (1 - R), which is similar to the logistic equation.So, I remember that the logistic equation can be solved using separation of variables. Let me try that. I'll rewrite the equation as:[ frac{dR}{dt} = k R (1 - R) ]To separate variables, I can divide both sides by R(1 - R) and multiply both sides by dt:[ frac{dR}{R(1 - R)} = k , dt ]Now, I need to integrate both sides. The left side looks like it can be integrated using partial fractions. Let me set it up:Let me express 1/(R(1 - R)) as A/R + B/(1 - R). So,[ frac{1}{R(1 - R)} = frac{A}{R} + frac{B}{1 - R} ]Multiplying both sides by R(1 - R):1 = A(1 - R) + B RLet me solve for A and B. Expanding the right side:1 = A - A R + B RGrouping like terms:1 = A + (B - A) RSince this must hold for all R, the coefficients of like terms must be equal on both sides. So,For the constant term: A = 1For the R term: B - A = 0 => B = A = 1So, the partial fractions decomposition is:[ frac{1}{R(1 - R)} = frac{1}{R} + frac{1}{1 - R} ]Therefore, the integral becomes:[ int left( frac{1}{R} + frac{1}{1 - R} right) dR = int k , dt ]Integrating both sides:Left side:[ int frac{1}{R} dR + int frac{1}{1 - R} dR = ln |R| - ln |1 - R| + C ]Wait, hold on. The integral of 1/(1 - R) dR is -ln|1 - R|, right? Because the derivative of (1 - R) is -1, so you have to account for that.So, it should be:[ ln |R| - ln |1 - R| + C ]Which can be written as:[ ln left| frac{R}{1 - R} right| + C ]Right side:[ int k , dt = k t + C ]So, putting it together:[ ln left( frac{R}{1 - R} right) = k t + C ]I can drop the absolute value since R is between 0 and 1, so R/(1 - R) is positive.Now, exponentiate both sides to get rid of the natural log:[ frac{R}{1 - R} = e^{k t + C} = e^{C} e^{k t} ]Let me denote e^C as another constant, say, C'. So,[ frac{R}{1 - R} = C' e^{k t} ]Now, solve for R. Let's write it as:[ R = C' e^{k t} (1 - R) ]Expanding:[ R = C' e^{k t} - C' e^{k t} R ]Bring the R terms to one side:[ R + C' e^{k t} R = C' e^{k t} ]Factor R:[ R (1 + C' e^{k t}) = C' e^{k t} ]Therefore,[ R = frac{C' e^{k t}}{1 + C' e^{k t}} ]Now, apply the initial condition R(0) = 0.2. So, when t = 0,[ 0.2 = frac{C' e^{0}}{1 + C' e^{0}} = frac{C'}{1 + C'} ]Let me solve for C':Multiply both sides by (1 + C'):0.2 (1 + C') = C'0.2 + 0.2 C' = C'0.2 = C' - 0.2 C'0.2 = 0.8 C'So,C' = 0.2 / 0.8 = 0.25Therefore, C' is 0.25.So, plugging back into R(t):[ R(t) = frac{0.25 e^{0.3 t}}{1 + 0.25 e^{0.3 t}} ]I can simplify this expression. Let's factor out 0.25 in the denominator:[ R(t) = frac{0.25 e^{0.3 t}}{1 + 0.25 e^{0.3 t}} = frac{e^{0.3 t}}{4 + e^{0.3 t}} ]Alternatively, I can write it as:[ R(t) = frac{1}{4 e^{-0.3 t} + 1} ]But maybe the first form is better.So, that's the solution for R(t). Let me double-check my steps.1. Started with the differential equation, recognized it as logistic.2. Separated variables, used partial fractions.3. Integrated both sides, got ln(R/(1 - R)) = kt + C.4. Exponentiated both sides, solved for R, applied initial condition.5. Calculated C' as 0.25.6. Plugged back in and simplified.Seems solid. Maybe I can test it at t=0: R(0) = 0.25 / (1 + 0.25) = 0.25 / 1.25 = 0.2, which matches. Good.So, Sub-problem 1 is solved. Now, moving on to Sub-problem 2: find t when R(t) = 0.9.So, set R(t) = 0.9 and solve for t.From the expression:[ 0.9 = frac{0.25 e^{0.3 t}}{1 + 0.25 e^{0.3 t}} ]Let me write this as:[ 0.9 (1 + 0.25 e^{0.3 t}) = 0.25 e^{0.3 t} ]Multiply out the left side:0.9 + 0.225 e^{0.3 t} = 0.25 e^{0.3 t}Bring all terms to one side:0.9 = 0.25 e^{0.3 t} - 0.225 e^{0.3 t}Simplify the right side:0.9 = (0.25 - 0.225) e^{0.3 t} = 0.025 e^{0.3 t}So,0.025 e^{0.3 t} = 0.9Divide both sides by 0.025:e^{0.3 t} = 0.9 / 0.025 = 36Take natural log of both sides:0.3 t = ln(36)Therefore,t = (ln(36)) / 0.3Compute ln(36). Let's see, ln(36) is ln(4*9) = ln(4) + ln(9) = 2 ln(2) + 2 ln(3). Since ln(2) ≈ 0.6931 and ln(3) ≈ 1.0986.So,ln(36) ≈ 2*0.6931 + 2*1.0986 ≈ 1.3862 + 2.1972 ≈ 3.5834Alternatively, I can just compute ln(36) directly:36 is e^3.5835, so ln(36) ≈ 3.5835Therefore,t ≈ 3.5835 / 0.3 ≈ 11.945 days.So, approximately 11.95 days. Since the question asks for the value of t when R(t) = 0.9, I can write it as approximately 11.95 days.But let me check my calculation again.Starting from R(t) = 0.9:0.9 = 0.25 e^{0.3 t} / (1 + 0.25 e^{0.3 t})Multiply both sides by denominator:0.9 (1 + 0.25 e^{0.3 t}) = 0.25 e^{0.3 t}0.9 + 0.225 e^{0.3 t} = 0.25 e^{0.3 t}Subtract 0.225 e^{0.3 t}:0.9 = 0.025 e^{0.3 t}Divide:e^{0.3 t} = 0.9 / 0.025 = 36Yes, that's correct.So, t = ln(36)/0.3 ≈ 3.5835 / 0.3 ≈ 11.945.So, about 11.95 days. If I need to be precise, maybe 11.945 days, but 11.95 is fine.Wait, let me compute ln(36) more accurately.Using calculator, ln(36) is approximately 3.583518938.So, 3.583518938 / 0.3 = 11.94506313.So, approximately 11.945 days, which is about 11.95 days.Alternatively, if I want to express it as a fraction, 11.945 is roughly 11 and 19/20 days, but decimal is probably better.So, summarizing:Sub-problem 1: R(t) = e^{0.3 t} / (4 + e^{0.3 t})Sub-problem 2: t ≈ 11.95 days.I think that's it.Final AnswerSub-problem 1: boxed{R(t) = dfrac{e^{0.3t}}{4 + e^{0.3t}}}Sub-problem 2: boxed{11.95} days</think>

🧠:<think>Okay, so I've got this problem about a novelist working on a new book series. The story involves a protagonist navigating through a series of interconnected cities, each represented as a node in a directed graph. The goal is to find the shortest path from a starting city to a target city using Dijkstra's algorithm, but with a twist: each edge's weight is multiplied by the alphabetical position of the first letter of the target city's name. Additionally, there's a second part where each node is duplicated into two nodes, which affects the graph's structure and connectivity.Let me break this down step by step.Problem 1: Adjusting Edge Weights Based on City NamesFirst, I need to understand how the edge weights are being modified. Each edge weight ( w(u, v) ) is multiplied by the alphabetical position of the first letter of the city at node ( v ). So, if a city starts with 'A', it's multiplied by 1, 'B' by 2, ..., up to 'Z' which is 26. This means that the cost of traveling into a city depends on the city's name.To apply Dijkstra's algorithm here, I think I need to adjust the graph's edge weights accordingly before running the algorithm. Essentially, for each edge ( (u, v) ), I'll compute a new weight ( w'(u, v) = w(u, v) times text{position}(v) ), where ( text{position}(v) ) is the alphabetical position of the first letter of city ( v )'s name.Once all edge weights are adjusted, I can proceed with Dijkstra's algorithm as usual. The algorithm maintains a priority queue of nodes to visit, starting from the source node ( s ). It then explores the least-cost path to each node, updating the shortest known distances and the priority queue accordingly.I should note that since all edge weights are now positive (as they are multiplied by at least 1), Dijkstra's algorithm remains applicable. There's no risk of negative cycles or anything like that, which is good because Dijkstra's doesn't handle negative weights.Potential Issues and Considerations- Data Handling: I need to ensure that I have access to the first letter of each city's name to compute the multiplier. This might require additional data storage or processing.- Efficiency: Multiplying each edge's weight might change the graph's structure in terms of path costs. It's possible that a previously longer path in terms of edges becomes shorter in terms of weighted cost after the multiplier is applied.- Algorithm Adaptation: Since the modification is just a scaling of the edge weights, the algorithm itself doesn't need to change. Only the edge weights are preprocessed before running Dijkstra's.Problem 2: Duplication of Nodes and Its Impact on ConnectivityIn this part, each node ( v ) is split into two nodes: ( v_1 ) and ( v_2 ). All incoming edges to ( v ) now go to ( v_1 ), and all outgoing edges from ( v ) now originate from ( v_2 ). The original edge weights remain unchanged.I need to analyze how this affects the graph's connectivity and whether the original shortest path from ( s ) to ( t ) still exists in the transformed graph.Understanding the Transformation- Node Splitting: Each node is duplicated, so the number of nodes doubles. If there were ( n ) nodes originally, now there are ( 2n ) nodes.- Edge Redirection: All incoming edges to ( v ) are redirected to ( v_1 ), and all outgoing edges from ( v ) are redirected from ( v_2 ). This essentially creates a two-step process for any traversal through ( v ): first, you enter ( v_1 ), then you can move to ( v_2 ), and from there, you can exit via outgoing edges.Impact on Connectivity- Strong Connectivity: If the original graph was strongly connected, meaning there's a path from every node to every other node, then the transformed graph should also be strongly connected. Because for any two nodes ( u ) and ( w ) in the original graph, there exists a path in the transformed graph by appropriately using the ( v_1 ) and ( v_2 ) nodes. - Path Existence: The original shortest path from ( s ) to ( t ) would still exist in the transformed graph, but with some modifications. Each node ( v ) on the original path would now be represented as ( v_1 ) followed by ( v_2 ). Therefore, the path would be ( s_1 rightarrow s_2 rightarrow v_{1} rightarrow v_{2} rightarrow ldots rightarrow t_1 rightarrow t_2 ). However, since ( s ) and ( t ) are specific nodes, their duplication would mean that the start is ( s_1 ) and the end is ( t_2 ).Algorithm Complexity- Original Dijkstra's Complexity: The time complexity of Dijkstra's algorithm with a priority queue is ( O(m + n log n) ) for each run. - Transformed Graph Complexity: Since the number of nodes doubles, the new graph has ( 2n ) nodes and ( 2m ) edges (since each original edge is replaced by a corresponding edge in the transformed graph). Therefore, the complexity becomes ( O(2m + 2n log 2n) ), which simplifies to ( O(m + n log n) ) asymptotically, as constants are ignored in big O notation.Potential Issues and Considerations- Path Length: The transformed graph might have longer paths in terms of the number of edges, but since edge weights remain the same, the total weight might not necessarily increase. However, the structure of the graph changes, so the shortest path could potentially change depending on how the nodes are split. - Algorithm Adaptation: To find the shortest path in the transformed graph, we would need to adjust the source and target nodes. The source becomes ( s_1 ) and the target becomes ( t_2 ). The algorithm itself doesn't change, but the nodes we're considering do.ConclusionFor the first problem, adjusting the edge weights based on the city names and then applying Dijkstra's algorithm should give the correct shortest path. For the second problem, duplicating nodes changes the graph's structure but doesn't necessarily prevent the original shortest path from existing; it just requires adjusting the start and end nodes. The complexity increases slightly due to the doubled number of nodes and edges, but asymptotically, it remains similar.I think I've covered all the aspects, but let me just verify if I missed anything.- Verification for Problem 1: Yes, preprocessing the edge weights correctly and then applying Dijkstra's should work. The multipliers are positive, so no issues there.- Verification for Problem 2: The transformation doesn't disconnect the graph if it was originally connected. The shortest path still exists but needs to account for the split nodes. The complexity analysis seems correct.I don't see any major gaps in my reasoning. I think I'm ready to present the step-by-step explanation.</think>