🧠:<think>Alright, so I have this problem about a game publisher planning to release a mod-friendly version of their game. They want to analyze how user-created mods might affect the game's economy and user base. There are two parts to this problem, both involving some exponential functions. Let me try to break them down step by step.Starting with the first part: The game economy is modeled by the function ( E(t) = 1000e^{0.05t} ), where ( E(t) ) is the total in-game currency in circulation at time ( t ) in months. A mod is expected to increase the currency generation rate by 15% per month. I need to formulate the new function ( E_{text{mod}}(t) ) and then find its value after 12 months.Okay, so the original function is an exponential growth model. The general form is ( E(t) = E_0 e^{rt} ), where ( E_0 ) is the initial amount, and ( r ) is the growth rate. In this case, ( E_0 = 1000 ) and ( r = 0.05 ) per month.The mod increases the currency generation rate by 15% per month. Hmm, does that mean the growth rate increases by 15%, or the rate itself is multiplied by 15%? I think it's the former. So, if the original growth rate is 5% per month, increasing it by 15% would mean adding 15% of 5% to the original rate.Wait, hold on. Let me clarify: If something increases by 15%, that's a multiplier of 1.15. So, if the original rate is 0.05, then the new rate would be 0.05 * 1.15. Let me compute that: 0.05 * 1.15 = 0.0575. So, the new growth rate is 5.75% per month.Therefore, the new function ( E_{text{mod}}(t) ) should be ( 1000e^{0.0575t} ). That makes sense because the growth rate is higher, so the currency will increase faster over time.Now, to find ( E_{text{mod}}(12) ), I plug in t = 12 into the new function:( E_{text{mod}}(12) = 1000e^{0.0575 * 12} ).Let me compute the exponent first: 0.0575 * 12. Let's see, 0.05 * 12 = 0.6, and 0.0075 * 12 = 0.09, so total is 0.6 + 0.09 = 0.69. So, the exponent is 0.69.Therefore, ( E_{text{mod}}(12) = 1000e^{0.69} ).I need to calculate ( e^{0.69} ). I remember that ( e^{0.6931} ) is approximately 2, since ln(2) ≈ 0.6931. So, 0.69 is slightly less than that, so ( e^{0.69} ) should be just under 2. Maybe around 1.994 or something? Let me check more accurately.Alternatively, I can use a calculator for a precise value. But since I don't have one here, I can approximate it. Alternatively, I can use the Taylor series expansion for ( e^x ) around x=0.69, but that might be complicated. Alternatively, I can remember that ( e^{0.69} ) is approximately 2, as I thought, but slightly less. Maybe 1.994?Wait, actually, I think it's about 1.994. Let me verify:We know that ( e^{0.6931} = 2 ). So, 0.69 is 0.6931 - 0.0031. So, ( e^{0.69} = e^{0.6931 - 0.0031} = e^{0.6931} * e^{-0.0031} ≈ 2 * (1 - 0.0031) ≈ 2 * 0.9969 ≈ 1.9938 ). So, approximately 1.9938.Therefore, ( E_{text{mod}}(12) ≈ 1000 * 1.9938 ≈ 1993.8 ). So, approximately 1993.8 units of currency.Wait, but let me think again. Is the mod increasing the generation rate by 15% per month, or is it a 15% increase in the total currency? The problem says "increase the currency generation rate by 15% per month." So, that should mean the rate at which currency is generated increases by 15%. Since the original function is exponential, the rate is proportional to the current amount. So, increasing the rate by 15% would mean increasing the growth rate parameter by 15%.So, yes, as I did before, the growth rate goes from 0.05 to 0.05 * 1.15 = 0.0575. So, that seems correct.Alternatively, if the mod added a flat 15% per month, but that wouldn't make much sense in an exponential model. It's more likely that the growth rate is increased by 15%, which is multiplicative.So, I think my approach is correct.Moving on to the second part: The user base growth model is ( U(t) = 5000 left( 1 - e^{-0.03t} right) ). The mod is expected to increase the growth rate by 10%. I need to develop the new function ( U_{text{mod}}(t) ) and calculate it at t = 12 months.First, let's understand the original function. It's a logistic-like growth model, where the number of active players approaches 5000 as t increases. The term ( 1 - e^{-0.03t} ) grows from 0 towards 1 as t increases, so U(t) grows from 0 towards 5000.The growth rate here is represented by the exponent's coefficient, which is -0.03. So, the rate at which the user base grows is determined by this 0.03. If the mod increases the growth rate by 10%, that means we need to increase this 0.03 by 10%.Wait, so similar to the first part, increasing the growth rate by 10% would mean multiplying 0.03 by 1.10, right? So, the new growth rate parameter would be 0.03 * 1.10 = 0.033.Therefore, the new function ( U_{text{mod}}(t) ) would be ( 5000 left( 1 - e^{-0.033t} right) ).Let me verify that. The original function has a decay rate of 0.03, so the growth rate is 0.03. Increasing that by 10% would make it 0.033, so yes, replacing -0.03 with -0.033 in the exponent.So, now, to find ( U_{text{mod}}(12) ), plug in t = 12:( U_{text{mod}}(12) = 5000 left( 1 - e^{-0.033 * 12} right) ).Compute the exponent first: 0.033 * 12. Let's calculate that: 0.03 * 12 = 0.36, and 0.003 * 12 = 0.036, so total is 0.36 + 0.036 = 0.396.So, the exponent is -0.396.Therefore, ( U_{text{mod}}(12) = 5000 (1 - e^{-0.396}) ).Now, compute ( e^{-0.396} ). I know that ( e^{-0.4} ) is approximately 0.6703. Since 0.396 is slightly less than 0.4, ( e^{-0.396} ) will be slightly higher than 0.6703. Maybe around 0.675?Wait, let me compute it more accurately. Let's use the Taylor series expansion or perhaps linear approximation.Alternatively, I can remember that ( e^{-x} ) can be approximated as 1 - x + x²/2 - x³/6 + ... for small x, but 0.396 is not that small. Alternatively, perhaps use a calculator-like approach.Alternatively, I can use the fact that ( e^{-0.396} = 1 / e^{0.396} ). Let me compute ( e^{0.396} ).I know that ( e^{0.4} ≈ 1.4918 ). Since 0.396 is 0.4 - 0.004, so ( e^{0.396} = e^{0.4 - 0.004} = e^{0.4} * e^{-0.004} ≈ 1.4918 * (1 - 0.004) ≈ 1.4918 * 0.996 ≈ 1.485.Therefore, ( e^{-0.396} ≈ 1 / 1.485 ≈ 0.673.So, approximately 0.673.Therefore, ( U_{text{mod}}(12) ≈ 5000 (1 - 0.673) = 5000 * 0.327 ≈ 1635 ).Wait, let me compute 5000 * 0.327:5000 * 0.3 = 15005000 * 0.027 = 135So, total is 1500 + 135 = 1635.So, approximately 1635 active players after 12 months with the mod.Wait, but let me think again. The original function without the mod would have:( U(12) = 5000 (1 - e^{-0.03 * 12}) = 5000 (1 - e^{-0.36}) ).Compute ( e^{-0.36} ). I know that ( e^{-0.35} ≈ 0.7047 ) and ( e^{-0.4} ≈ 0.6703 ). So, 0.36 is between 0.35 and 0.4. Let me approximate.Using linear approximation between 0.35 and 0.4:At x=0.35, e^{-x}=0.7047At x=0.4, e^{-x}=0.6703Difference in x: 0.05Difference in e^{-x}: 0.6703 - 0.7047 = -0.0344So, per 0.01 increase in x, e^{-x} decreases by approximately 0.0344 / 0.05 = 0.688 per 0.01.Wait, that seems high. Alternatively, perhaps it's better to compute the derivative.The derivative of e^{-x} is -e^{-x}. At x=0.36, the derivative is -e^{-0.36}.But since I don't know e^{-0.36} yet, maybe it's better to use a better approximation.Alternatively, use the Taylor series around x=0.35:Let me let x = 0.35 + 0.01.So, e^{-0.36} = e^{-0.35 - 0.01} = e^{-0.35} * e^{-0.01} ≈ 0.7047 * (1 - 0.01) ≈ 0.7047 * 0.99 ≈ 0.6976.Wait, but actually, e^{-0.01} ≈ 0.99005, so 0.7047 * 0.99005 ≈ 0.7047 - 0.7047*0.00995 ≈ 0.7047 - ~0.007 ≈ 0.6977.So, e^{-0.36} ≈ 0.6977.Therefore, U(12) = 5000*(1 - 0.6977) = 5000*0.3023 ≈ 1511.5.So, without the mod, after 12 months, there are approximately 1511.5 active players.With the mod, it's 1635, which is an increase of about 123 players. That seems reasonable, given the 10% increase in growth rate.Wait, but let me check my calculation for ( e^{-0.396} ). Earlier, I approximated it as 0.673, but let me see if that's accurate.Since 0.396 is close to 0.4, and e^{-0.4} ≈ 0.6703. So, 0.396 is 0.4 - 0.004. So, e^{-0.396} = e^{-0.4 + 0.004} = e^{-0.4} * e^{0.004} ≈ 0.6703 * 1.004 ≈ 0.6703 + 0.6703*0.004 ≈ 0.6703 + 0.00268 ≈ 0.67298.So, approximately 0.673, which is what I had before. So, that seems correct.Therefore, ( U_{text{mod}}(12) ≈ 5000*(1 - 0.673) = 5000*0.327 ≈ 1635 ).So, that seems consistent.Wait, but let me think again about the growth rate. The original function is ( U(t) = 5000(1 - e^{-0.03t}) ). So, the growth rate is determined by the exponent's coefficient, which is -0.03. So, the rate at which the user base grows is proportional to the remaining capacity, right? So, it's a logistic growth model, where the growth rate is r = 0.03.If the mod increases the growth rate by 10%, that would mean increasing r from 0.03 to 0.033, as I did before. So, the new function is ( U_{text{mod}}(t) = 5000(1 - e^{-0.033t}) ). That seems correct.Alternatively, if the mod increased the maximum user base by 10%, that would be different, but the problem says it increases the growth rate by 10%, so it's the r parameter that's increased.Therefore, I think my approach is correct.So, summarizing:1. The new economy function is ( E_{text{mod}}(t) = 1000e^{0.0575t} ), and after 12 months, it's approximately 1993.8.2. The new user base function is ( U_{text{mod}}(t) = 5000(1 - e^{-0.033t}) ), and after 12 months, it's approximately 1635.Wait, but let me double-check the first part's calculation. I approximated ( e^{0.69} ≈ 1.9938 ), leading to ( E_{text{mod}}(12) ≈ 1993.8 ). But let me see if I can compute ( e^{0.69} ) more accurately.We know that ( e^{0.6931} = 2 ). So, 0.69 is 0.6931 - 0.0031. So, using the Taylor series expansion around x=0.6931:( e^{x} ≈ e^{0.6931} + e^{0.6931}(x - 0.6931) ).So, at x=0.69, the approximation is:( e^{0.69} ≈ 2 + 2*(0.69 - 0.6931) = 2 + 2*(-0.0031) = 2 - 0.0062 = 1.9938 ).So, that's consistent with my previous approximation. So, 1.9938 is accurate enough.Therefore, ( E_{text{mod}}(12) ≈ 1000 * 1.9938 ≈ 1993.8 ).Alternatively, if I use a calculator for more precision, but since I don't have one, this approximation is sufficient.So, in conclusion, after 12 months, the total in-game currency with the mod is approximately 1993.8, and the number of active players is approximately 1635.I think that's it. I don't see any mistakes in my reasoning, but let me just recap:1. For the economy, the growth rate increases by 15%, so 0.05 becomes 0.0575. The function becomes ( 1000e^{0.0575t} ). At t=12, exponent is 0.69, e^0.69≈1.9938, so E_mod(12)≈1993.8.2. For the user base, the growth rate increases by 10%, so 0.03 becomes 0.033. The function becomes ( 5000(1 - e^{-0.033t}) ). At t=12, exponent is -0.396, e^{-0.396}≈0.673, so U_mod(12)=5000*(1 - 0.673)=1635.Yes, that all checks out.</think>

🧠:<think>Alright, so Emma is this high school student who writes fanfiction to cope with her social anxiety. That's really cool because it shows she's using her creativity as a positive outlet. Anyway, she has a series that she updates weekly, and each chapter has a specific number of words. She noticed that the word counts follow a quadratic pattern, and she wants to model that with a quadratic function. Let me start by understanding what a quadratic function is. A quadratic function is generally of the form ( f(n) = an^2 + bn + c ), where ( a ), ( b ), and ( c ) are constants, and ( n ) is the chapter number. Since it's quadratic, the second difference between consecutive terms should be constant. That might be a useful property to find the coefficients.Emma provided the word counts for the first three chapters: 1500, 1800, and 2150 words. Let me write these down:- Chapter 1: 1500 words- Chapter 2: 1800 words- Chapter 3: 2150 wordsSo, if I plug these into the quadratic function, I can set up a system of equations to solve for ( a ), ( b ), and ( c ).For ( n = 1 ):( a(1)^2 + b(1) + c = 1500 )Simplifies to:( a + b + c = 1500 ) --- Equation 1For ( n = 2 ):( a(2)^2 + b(2) + c = 1800 )Simplifies to:( 4a + 2b + c = 1800 ) --- Equation 2For ( n = 3 ):( a(3)^2 + b(3) + c = 2150 )Simplifies to:( 9a + 3b + c = 2150 ) --- Equation 3Now, I have three equations:1. ( a + b + c = 1500 )2. ( 4a + 2b + c = 1800 )3. ( 9a + 3b + c = 2150 )I can solve this system step by step. Let's subtract Equation 1 from Equation 2 to eliminate ( c ):Equation 2 - Equation 1:( (4a + 2b + c) - (a + b + c) = 1800 - 1500 )Simplifies to:( 3a + b = 300 ) --- Equation 4Similarly, subtract Equation 2 from Equation 3:Equation 3 - Equation 2:( (9a + 3b + c) - (4a + 2b + c) = 2150 - 1800 )Simplifies to:( 5a + b = 350 ) --- Equation 5Now, I have two equations:4. ( 3a + b = 300 )5. ( 5a + b = 350 )Subtract Equation 4 from Equation 5 to eliminate ( b ):( (5a + b) - (3a + b) = 350 - 300 )Simplifies to:( 2a = 50 )So, ( a = 25 )Now, plug ( a = 25 ) back into Equation 4:( 3(25) + b = 300 )( 75 + b = 300 )( b = 225 )Now, substitute ( a = 25 ) and ( b = 225 ) into Equation 1:( 25 + 225 + c = 1500 )( 250 + c = 1500 )( c = 1250 )So, the quadratic function is ( f(n) = 25n^2 + 225n + 1250 ). Let me verify this with the given data points.For ( n = 1 ):( 25(1) + 225(1) + 1250 = 25 + 225 + 1250 = 1500 ). Correct.For ( n = 2 ):( 25(4) + 225(2) + 1250 = 100 + 450 + 1250 = 1800 ). Correct.For ( n = 3 ):( 25(9) + 225(3) + 1250 = 225 + 675 + 1250 = 2150 ). Correct.Great, that seems to fit. So, part 1 is solved. The quadratic function is ( f(n) = 25n^2 + 225n + 1250 ).Moving on to part 2. Emma wants her average word count over the first 10 chapters to be at least 2500 words. So, the average word count is the total words divided by 10. Therefore, the total words over 10 chapters should be at least ( 2500 times 10 = 25,000 ) words.First, I need to calculate the total number of words she writes in the first 10 chapters using the quadratic function. Since each chapter ( n ) has ( f(n) = 25n^2 + 225n + 1250 ) words, the total words ( T ) is the sum from ( n = 1 ) to ( n = 10 ) of ( f(n) ).So, ( T = sum_{n=1}^{10} (25n^2 + 225n + 1250) )I can split this sum into three separate sums:( T = 25sum_{n=1}^{10} n^2 + 225sum_{n=1}^{10} n + 1250sum_{n=1}^{10} 1 )I remember the formulas for these sums:1. Sum of squares: ( sum_{n=1}^{k} n^2 = frac{k(k+1)(2k+1)}{6} )2. Sum of first k natural numbers: ( sum_{n=1}^{k} n = frac{k(k+1)}{2} )3. Sum of 1 k times: ( sum_{n=1}^{k} 1 = k )Let me compute each part:First, compute ( sum_{n=1}^{10} n^2 ):( frac{10 times 11 times 21}{6} = frac{2310}{6} = 385 )Second, compute ( sum_{n=1}^{10} n ):( frac{10 times 11}{2} = 55 )Third, compute ( sum_{n=1}^{10} 1 ):That's just 10.Now, plug these into the total ( T ):( T = 25 times 385 + 225 times 55 + 1250 times 10 )Calculate each term:25 * 385: Let's compute 25 * 300 = 7500, 25 * 85 = 2125, so total is 7500 + 2125 = 9625225 * 55: 200*55=11,000; 25*55=1,375; total is 11,000 + 1,375 = 12,3751250 * 10 = 12,500Now, sum all these:9625 + 12,375 = 22,00022,000 + 12,500 = 34,500So, the total words over 10 chapters is 34,500.Wait, hold on. Let me double-check these calculations because 34,500 seems a bit high for 10 chapters averaging 3,450 words each, but Emma's average goal is 2,500. So, 34,500 is actually higher than 25,000. Hmm, so does that mean she already exceeds her goal?Wait, let me recalculate because maybe I made a mistake.First, compute each term again:25 * 385:385 * 25: 385 * 20 = 7,700; 385 * 5 = 1,925; total is 7,700 + 1,925 = 9,625225 * 55:55 * 200 = 11,000; 55 * 25 = 1,375; total 11,000 + 1,375 = 12,3751250 * 10 = 12,500Adding them up: 9,625 + 12,375 = 22,000; 22,000 + 12,500 = 34,500.Wait, that's correct. So, 34,500 total words over 10 chapters. Therefore, the average is 34,500 / 10 = 3,450 words per chapter.But Emma wants an average of at least 2,500 words. So, 3,450 is way above that. So, her goal is achievable, and she's already exceeding it.Wait, hold on, that seems contradictory because the first chapter is 1,500 which is below 2,500. So, how come the average is 3,450? Let me check the quadratic function again.Wait, the quadratic function is ( f(n) = 25n^2 + 225n + 1250 ). Let me compute the word counts for the first few chapters:n=1: 25 + 225 + 1250 = 1500n=2: 100 + 450 + 1250 = 1800n=3: 225 + 675 + 1250 = 2150n=4: 25*16 + 225*4 + 1250 = 400 + 900 + 1250 = 2550n=5: 25*25 + 225*5 + 1250 = 625 + 1125 + 1250 = 3000n=6: 25*36 + 225*6 + 1250 = 900 + 1350 + 1250 = 3500n=7: 25*49 + 225*7 + 1250 = 1225 + 1575 + 1250 = 4050n=8: 25*64 + 225*8 + 1250 = 1600 + 1800 + 1250 = 4650n=9: 25*81 + 225*9 + 1250 = 2025 + 2025 + 1250 = 5300n=10: 25*100 + 225*10 + 1250 = 2500 + 2250 + 1250 = 6000Wait, so the word counts are increasing quadratically. So, the first chapter is 1,500, and the 10th chapter is 6,000. So, the total is 34,500, which averages to 3,450 per chapter. So, Emma's average is already above 2,500.Therefore, her goal is achievable, and she doesn't need to write any additional words. In fact, she's exceeding her goal.But wait, let me make sure I didn't make a mistake in the sum. Let's compute the total word counts manually for the first 10 chapters:n=1: 1500n=2: 1800n=3: 2150n=4: 2550n=5: 3000n=6: 3500n=7: 4050n=8: 4650n=9: 5300n=10: 6000Now, let's add them up step by step:Start with 1500.1500 + 1800 = 33003300 + 2150 = 54505450 + 2550 = 80008000 + 3000 = 11,00011,000 + 3500 = 14,50014,500 + 4050 = 18,55018,550 + 4650 = 23,20023,200 + 5300 = 28,50028,500 + 6000 = 34,500Yes, that adds up correctly. So, the total is indeed 34,500, which is 3,450 average. So, Emma's average is already above her goal of 2,500. Therefore, she doesn't need to write any additional words.Wait, but let me think again. The problem says "at least 2500 words." So, 3,450 is more than 2,500, so it's achievable. So, the answer is that her goal is achievable, and she doesn't need to write any additional words.But just to be thorough, let me compute the total using the sum formulas again to confirm.Sum of ( f(n) ) from 1 to 10 is:25 * sum(n^2) + 225 * sum(n) + 1250 * sum(1)Sum(n^2) from 1 to 10 is 385Sum(n) from 1 to 10 is 55Sum(1) from 1 to 10 is 10So,25 * 385 = 9,625225 * 55 = 12,3751250 * 10 = 12,500Adding up: 9,625 + 12,375 = 22,000; 22,000 + 12,500 = 34,500. Yep, same result.Therefore, the average is 34,500 / 10 = 3,450, which is more than 2,500. So, Emma's goal is achievable without any additional words.But wait, just to make sure, let me compute the average word count per chapter. Since the quadratic function is increasing, the later chapters are much longer. So, the average is pulled up by the longer chapters. So, even though the first chapter is below 2,500, the subsequent chapters are above, and the average ends up higher.So, in conclusion, Emma's average word count over the first 10 chapters is 3,450, which is above her goal of 2,500. Therefore, she doesn't need to write any additional words.Final Answer1. The quadratic function is boxed{25n^2 + 225n + 1250}.2. Emma's average word count goal is achievable. She does not need to write any additional words.</think>

🧠:<think>Alright, so I have this problem about analyzing the scoring patterns of the Seirin High basketball team, specifically focusing on Kuroko and Kagami. I need to calculate their combined points over 20 games and then figure out the total score for Seirin High, which is twice that sum. Let me break this down step by step.First, let me understand the functions given for Kuroko and Kagami. Kuroko's points per game are given by ( f(x) = 3sinleft(frac{pi x}{10}right) + 7 ), and Kagami's points are ( g(x) = 5cosleft(frac{pi x}{10}right) + 10 ). Here, ( x ) represents the game number, from 1 to 20.So, for each game, I can plug in the game number ( x ) into these functions to get the points scored by each player. Then, I need to sum these points over all 20 games. That sounds like I need to compute the sum of ( f(x) + g(x) ) from ( x = 1 ) to ( x = 20 ).Let me write down the expression for the combined points per game:( f(x) + g(x) = 3sinleft(frac{pi x}{10}right) + 7 + 5cosleft(frac{pi x}{10}right) + 10 )Simplifying that, I get:( f(x) + g(x) = 3sinleft(frac{pi x}{10}right) + 5cosleft(frac{pi x}{10}right) + 17 )So, the total combined points over 20 games would be the sum from ( x = 1 ) to ( x = 20 ) of this expression.Let me denote this total as ( S ):( S = sum_{x=1}^{20} left[ 3sinleft(frac{pi x}{10}right) + 5cosleft(frac{pi x}{10}right) + 17 right] )I can split this sum into three separate sums:( S = 3sum_{x=1}^{20} sinleft(frac{pi x}{10}right) + 5sum_{x=1}^{20} cosleft(frac{pi x}{10}right) + sum_{x=1}^{20} 17 )Calculating each part individually might be easier.Starting with the last sum, ( sum_{x=1}^{20} 17 ). Since 17 is a constant, this is just 17 multiplied by the number of terms, which is 20. So:( sum_{x=1}^{20} 17 = 17 times 20 = 340 )Okay, that part is straightforward.Now, moving on to the sine and cosine sums. Let's handle them one by one.First, ( sum_{x=1}^{20} sinleft(frac{pi x}{10}right) ). Hmm, sine functions can sometimes be tricky when summing over multiple terms. I remember there's a formula for the sum of sine functions in an arithmetic sequence. Let me recall that.The formula for the sum of ( sin(a + (n-1)d) ) from ( n = 1 ) to ( N ) is:( frac{sinleft(frac{Nd}{2}right) cdot sinleft(a + frac{(N-1)d}{2}right)}{sinleft(frac{d}{2}right)} )Similarly, for cosine, the formula is:( frac{sinleft(frac{Nd}{2}right) cdot cosleft(a + frac{(N-1)d}{2}right)}{sinleft(frac{d}{2}right)} )In our case, for the sine sum, the angle is ( frac{pi x}{10} ). Let me express this as ( a + (x-1)d ). So, when ( x = 1 ), the angle is ( frac{pi}{10} ), and the common difference ( d ) is ( frac{pi}{10} ) as well because each subsequent term increases by ( frac{pi}{10} ).So, ( a = frac{pi}{10} ), ( d = frac{pi}{10} ), and ( N = 20 ).Plugging into the sine sum formula:( sum_{x=1}^{20} sinleft(frac{pi x}{10}right) = frac{sinleft(frac{20 cdot frac{pi}{10}}{2}right) cdot sinleft(frac{pi}{10} + frac{(20 - 1)cdot frac{pi}{10}}{2}right)}{sinleft(frac{frac{pi}{10}}{2}right)} )Simplify step by step.First, compute ( frac{20 cdot frac{pi}{10}}{2} = frac{2pi}{2} = pi ).Next, compute the argument of the second sine function:( frac{pi}{10} + frac{19 cdot frac{pi}{10}}{2} = frac{pi}{10} + frac{19pi}{20} = frac{2pi}{20} + frac{19pi}{20} = frac{21pi}{20} )So, the numerator becomes ( sin(pi) cdot sinleft(frac{21pi}{20}right) ).But ( sin(pi) = 0 ), so the entire numerator is 0. Therefore, the sum of the sine terms is 0.Wait, that's interesting. So, ( sum_{x=1}^{20} sinleft(frac{pi x}{10}right) = 0 ).Hmm, let me verify that. Because if the numerator is zero, the sum is zero. That seems correct because the sine function over a full period can cancel out. Since 20 games correspond to 2 full periods of the sine function with period 10 (since ( frac{pi x}{10} ) has a period of 20/ (2π) * π/10 = 10 games). Wait, actually, the period of ( sinleft(frac{pi x}{10}right) ) is ( frac{2pi}{pi/10} } = 20 ). So, over 20 games, it's exactly one full period. So, integrating over a full period, the positive and negative areas cancel out, resulting in zero. So, the sum is indeed zero.Okay, so the sine sum is zero. That simplifies things.Now, moving on to the cosine sum: ( sum_{x=1}^{20} cosleft(frac{pi x}{10}right) ).Using the same approach, let me apply the formula for the sum of cosines in an arithmetic sequence.Using the formula:( sum_{x=1}^{N} cos(a + (x-1)d) = frac{sinleft(frac{Nd}{2}right) cdot cosleft(a + frac{(N-1)d}{2}right)}{sinleft(frac{d}{2}right)} )Again, in our case, ( a = frac{pi}{10} ), ( d = frac{pi}{10} ), and ( N = 20 ).Plugging into the formula:( sum_{x=1}^{20} cosleft(frac{pi x}{10}right) = frac{sinleft(frac{20 cdot frac{pi}{10}}{2}right) cdot cosleft(frac{pi}{10} + frac{(20 - 1)cdot frac{pi}{10}}{2}right)}{sinleft(frac{frac{pi}{10}}{2}right)} )Simplify step by step.First, compute ( frac{20 cdot frac{pi}{10}}{2} = frac{2pi}{2} = pi ).Next, compute the argument of the cosine function:( frac{pi}{10} + frac{19 cdot frac{pi}{10}}{2} = frac{pi}{10} + frac{19pi}{20} = frac{2pi}{20} + frac{19pi}{20} = frac{21pi}{20} )So, the numerator becomes ( sin(pi) cdot cosleft(frac{21pi}{20}right) ).Again, ( sin(pi) = 0 ), so the numerator is zero. Therefore, the sum of the cosine terms is also zero.Wait, that's the same result as the sine sum. So, both the sine and cosine sums over 20 games are zero.Hmm, that's interesting. So, both these sums cancel out over a full period. So, the only contribution to the total combined points comes from the constant term, which is 17 per game.Therefore, the total combined points ( S ) is just 340.Wait, but let me double-check that. Because sometimes when dealing with sums of sine and cosine, especially when the number of terms is a multiple of the period, the sums can indeed be zero. So, in this case, since the functions have a period of 20 games, over 20 games, the sum of sine and cosine would indeed be zero.So, that leaves us with the constant term, which is 17 per game. Therefore, over 20 games, it's 17 * 20 = 340.Okay, so that's the total combined points for Kuroko and Kagami.Now, moving on to part 2. The total score for Seirin High in each game is twice the sum of Kuroko's and Kagami's points. So, if the combined points per game are ( f(x) + g(x) ), then Seirin's score per game is ( 2(f(x) + g(x)) ).Therefore, the total score over 20 games would be twice the total combined points we just calculated.So, if the combined points are 340, then Seirin's total score is ( 2 times 340 = 680 ).Wait, that seems straightforward, but let me make sure I didn't miss anything.Given that each game's total is twice the sum of Kuroko and Kagami, so for each game, it's ( 2(f(x) + g(x)) ). Therefore, over 20 games, it's ( 2 times sum_{x=1}^{20} (f(x) + g(x)) ), which is ( 2 times 340 = 680 ).Yes, that makes sense.So, summarizing:1. The total combined points by Kuroko and Kagami over 20 games is 340.2. The total score for Seirin High over 20 games is 680.I think that's it. Let me just recap to ensure I didn't skip any steps or make any miscalculations.First, I expressed the combined points per game as ( f(x) + g(x) ), which simplified to a sine, cosine, and constant term. Then, I recognized that over a full period (20 games), the sine and cosine sums would cancel out to zero, leaving only the constant term. Summing the constant term over 20 games gave me 340. Then, doubling that for Seirin's total score gave me 680.Yes, that all checks out. I don't see any errors in my reasoning.Final AnswerThe total combined points scored by Kuroko and Kagami over the 20 games is boxed{340}, and the total score of Seirin High over the 20 games is boxed{680}.</think>

🧠:<think>Okay, so I have this problem where an entrepreneur wants to invest in three exclusive island retreats—A, B, and C. The goal is to maximize the total inspiration gained while minimizing costs. There are two parts to this problem: first, figuring out the optimal number of days to spend on each island without considering the budget, and second, determining the feasible distribution of days that fits within a 9,000 budget.Starting with part 1: I need to maximize the total inspiration from the three islands over 12 days. The inspiration functions are given as:- Island A: ( f(t) = 5t + 20 )- Island B: ( g(t) = 30e^{0.1t} )- Island C: ( h(t) = 40 ln(t + 1) )Where ( t ) is the number of days spent on each island. The total days are 12, so if I denote the days spent on A, B, and C as ( t_A ), ( t_B ), and ( t_C ) respectively, then:( t_A + t_B + t_C = 12 )I need to maximize the total inspiration, which is:( Total = f(t_A) + g(t_B) + h(t_C) )So, substituting the functions:( Total = 5t_A + 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) )To maximize this, I think I should use calculus, specifically Lagrange multipliers since we have a constraint. Let me set up the Lagrangian function.Let’s denote the Lagrangian multiplier as ( lambda ). The Lagrangian ( mathcal{L} ) is:( mathcal{L} = 5t_A + 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) + lambda (12 - t_A - t_B - t_C) )Wait, actually, the Lagrangian should include the constraint with a negative sign because the constraint is ( t_A + t_B + t_C = 12 ). So, it's:( mathcal{L} = 5t_A + 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) + lambda (12 - t_A - t_B - t_C) )Now, to find the maximum, we take partial derivatives with respect to each variable and set them equal to zero.First, partial derivative with respect to ( t_A ):( frac{partial mathcal{L}}{partial t_A} = 5 - lambda = 0 )So, ( lambda = 5 )Next, partial derivative with respect to ( t_B ):( frac{partial mathcal{L}}{partial t_B} = 30 * 0.1 e^{0.1t_B} - lambda = 0 )Simplify:( 3 e^{0.1t_B} - lambda = 0 )We know ( lambda = 5 ), so:( 3 e^{0.1t_B} = 5 )( e^{0.1t_B} = frac{5}{3} )Take natural log:( 0.1t_B = lnleft(frac{5}{3}right) )Calculate ( ln(5/3) ):Approximately, ( ln(1.6667) approx 0.5108 )So, ( t_B = frac{0.5108}{0.1} = 5.108 ) days.Now, partial derivative with respect to ( t_C ):( frac{partial mathcal{L}}{partial t_C} = frac{40}{t_C + 1} - lambda = 0 )So, ( frac{40}{t_C + 1} = lambda = 5 )Thus, ( frac{40}{t_C + 1} = 5 )Multiply both sides by ( t_C + 1 ):( 40 = 5(t_C + 1) )Divide both sides by 5:( 8 = t_C + 1 )So, ( t_C = 7 ) days.Now, we have ( t_A + t_B + t_C = 12 ). We have ( t_B approx 5.108 ) and ( t_C = 7 ). So, ( t_A = 12 - 5.108 - 7 = 12 - 12.108 = -0.108 ). Wait, that can't be right. Negative days? That doesn't make sense.Hmm, so maybe I made a mistake in setting up the Lagrangian or in the derivatives. Let me double-check.Wait, the Lagrangian should be set up as:( mathcal{L} = 5t_A + 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) + lambda (12 - t_A - t_B - t_C) )So, the partial derivatives are correct. So, why is ( t_A ) negative? Maybe because the marginal inspiration from A is lower than the others, so it's better to allocate zero days to A.Wait, but the partial derivative with respect to ( t_A ) is 5, which is the marginal inspiration per day. For ( t_B ), the marginal inspiration is ( 3 e^{0.1t_B} ), and for ( t_C ), it's ( frac{40}{t_C + 1} ).So, when we set the marginal inspirations equal, we get:For A and B: 5 = 3 e^{0.1t_B} => t_B ≈ 5.108For A and C: 5 = 40 / (t_C + 1) => t_C = 7But then, adding t_B and t_C gives 5.108 + 7 = 12.108, which is more than 12. So, we can't have both t_B and t_C at these levels because they exceed the total days.Therefore, perhaps the optimal solution is to set t_A = 0, and allocate the remaining days to B and C.Wait, but let's think about it. If t_A is negative, it suggests that to maximize the total, we should not spend any days on A, because the marginal inspiration from A is less than the others. So, set t_A = 0, and then distribute the 12 days between B and C.So, let's try that. Let’s set t_A = 0, so we have t_B + t_C = 12.Now, we can set up the Lagrangian again with t_A fixed at 0.So, the total inspiration becomes:( Total = 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) )With constraint ( t_B + t_C = 12 )So, set up the Lagrangian:( mathcal{L} = 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) + lambda (12 - t_B - t_C) )Take partial derivatives:With respect to t_B:( 3 e^{0.1t_B} - lambda = 0 ) => ( lambda = 3 e^{0.1t_B} )With respect to t_C:( frac{40}{t_C + 1} - lambda = 0 ) => ( lambda = frac{40}{t_C + 1} )Set them equal:( 3 e^{0.1t_B} = frac{40}{t_C + 1} )But since ( t_C = 12 - t_B ), substitute:( 3 e^{0.1t_B} = frac{40}{(12 - t_B) + 1} = frac{40}{13 - t_B} )So, we have:( 3 e^{0.1t_B} = frac{40}{13 - t_B} )This is a transcendental equation and might not have an analytical solution, so we need to solve it numerically.Let me denote ( t_B = x ), so:( 3 e^{0.1x} = frac{40}{13 - x} )We can try to solve for x numerically.Let me rearrange:( 3 e^{0.1x} (13 - x) = 40 )Let’s define a function:( f(x) = 3 e^{0.1x} (13 - x) - 40 )We need to find x where f(x) = 0.Let’s try some values:At x=5:( f(5) = 3 e^{0.5} (8) - 40 ≈ 3*1.6487*8 -40 ≈ 3*13.1896 -40 ≈ 39.5688 -40 ≈ -0.4312 )At x=6:( f(6) = 3 e^{0.6} (7) -40 ≈ 3*1.8221*7 -40 ≈ 3*12.7547 -40 ≈ 38.2641 -40 ≈ -1.7359 )Wait, that's worse. Maybe try x=4:( f(4) = 3 e^{0.4} (9) -40 ≈ 3*1.4918*9 -40 ≈ 3*13.4262 -40 ≈ 40.2786 -40 ≈ 0.2786 )So, f(4) ≈ 0.2786, f(5) ≈ -0.4312. So, the root is between 4 and 5.Use linear approximation:Between x=4 and x=5:At x=4, f=0.2786At x=5, f=-0.4312The change in f is -0.4312 - 0.2786 = -0.7098 over 1 unit.We need to find delta where f=0:delta = (0 - 0.2786)/(-0.7098) ≈ 0.2786 / 0.7098 ≈ 0.392So, x ≈ 4 + 0.392 ≈ 4.392Let’s test x=4.392:Compute f(4.392):First, compute e^{0.1*4.392} = e^{0.4392} ≈ 1.552Then, 13 - 4.392 = 8.608So, 3*1.552*8.608 ≈ 3*13.37 ≈ 40.11So, f(x) ≈ 40.11 -40 = 0.11Still positive. So, need to go a bit higher.Let’s try x=4.5:e^{0.45} ≈ 1.568313 -4.5=8.53*1.5683*8.5 ≈ 3*13.33055 ≈ 39.99165So, f(4.5)=39.99165 -40≈ -0.00835Almost zero. So, x≈4.5But f(4.5)≈-0.00835, very close to zero.So, the root is approximately x=4.5.Thus, t_B≈4.5 days, t_C=12 -4.5=7.5 days.So, t_A=0, t_B≈4.5, t_C≈7.5.Let me verify:Compute f(4.5):3 e^{0.45} (13 -4.5) = 3*1.5683*8.5 ≈ 3*13.33055≈39.99165≈40So, yes, f(4.5)=0.Therefore, the optimal allocation is approximately t_A=0, t_B=4.5, t_C=7.5.But let me check if this is indeed the maximum.Alternatively, maybe we should check if allocating some days to A would yield a higher total.Wait, earlier when we tried to set t_A=0, we got t_B=4.5, t_C=7.5.But if we allow t_A to be positive, we might have a different allocation.Wait, but in the initial Lagrangian, we found that t_A would be negative, which is not feasible, so the optimal is to set t_A=0 and allocate the rest to B and C.So, the optimal distribution is t_A=0, t_B≈4.5, t_C≈7.5.But let me compute the total inspiration at this point.Compute Total = 5*0 +20 +30e^{0.1*4.5} +40 ln(7.5 +1)Compute each term:- 5*0=0- 20=20- 30e^{0.45}≈30*1.5683≈47.049- 40 ln(8.5)≈40*2.140≈85.6Total≈0 +20 +47.049 +85.6≈152.649Now, let's see if allocating some days to A would increase the total.Suppose we allocate 1 day to A, then t_B + t_C=11.Compute the marginal inspiration from A: 5 per day.Compare to the marginal inspiration from B and C.At t_B=4.5, the marginal from B is 3 e^{0.45}≈3*1.5683≈4.705, which is less than 5.Wait, that's interesting. So, the marginal inspiration from A is 5, which is higher than the marginal from B at t_B=4.5 (≈4.705). So, perhaps we should reallocate some days from B to A.Wait, but earlier, when we set t_A=0, the marginal from A was 5, which was higher than the marginal from B and C, but since we had to allocate all days, we ended up with t_A negative, which isn't possible. So, perhaps the optimal is to set t_A as much as possible until the marginal from A equals the marginal from B and C.Wait, this is getting a bit confusing. Let me think again.When we set up the Lagrangian, we found that t_A would be negative, which is not feasible. So, the optimal is to set t_A=0, and then allocate the remaining days to B and C such that their marginal inspirations are equal.But when we did that, we found t_B≈4.5, t_C≈7.5.However, at t_B=4.5, the marginal from B is ≈4.705, which is less than the marginal from A (5). So, if we take a day from B and give it to A, the total inspiration would increase by 5 -4.705≈0.295.Similarly, the marginal from C at t_C=7.5 is 40/(7.5+1)=40/8.5≈4.705, which is also less than 5.So, taking a day from C and giving it to A would also increase the total by 5 -4.705≈0.295.Therefore, it seems that we can increase the total inspiration by moving days from B and/or C to A until the marginal from A equals the marginal from B and C.But wait, if we move days to A, the marginal from B and C will change.Let me formalize this.Suppose we allocate t_A days to A, t_B days to B, and t_C days to C, with t_A + t_B + t_C=12.The marginal inspiration from A is 5.The marginal from B is 3 e^{0.1 t_B}.The marginal from C is 40/(t_C +1).At optimality, the marginal from A should equal the marginal from B and C.So, 5=3 e^{0.1 t_B}=40/(t_C +1)But since t_A + t_B + t_C=12, we have t_C=12 - t_A - t_B.So, 5=3 e^{0.1 t_B} and 5=40/(12 - t_A - t_B +1)=40/(13 - t_A - t_B)But since t_A is being increased, let's see.Wait, if we set 5=3 e^{0.1 t_B}, then t_B= (ln(5/3))/0.1≈(ln(1.6667))/0.1≈0.5108/0.1≈5.108 days.Similarly, 5=40/(13 - t_A - t_B) => 13 - t_A - t_B=8 => t_A + t_B=5.But we already have t_B≈5.108, so t_A=5 -5.108≈-0.108, which is negative.So, again, we can't have negative days on A, so we set t_A=0, and then t_B=5.108, t_C=12 -5.108≈6.892.Wait, but earlier when we tried to set t_A=0, we found t_B≈4.5, t_C≈7.5.But now, this approach suggests t_B≈5.108, t_C≈6.892.Wait, which is correct?I think the confusion arises because when we set t_A=0, the marginal from B and C should be equal, but when we set t_A>0, the marginal from B and C should equal the marginal from A.But since the marginal from A is higher than the marginal from B and C when t_A=0, we should increase t_A until the marginal from B and C equals 5.Wait, let me try to formalize this.Let’s denote:Marginal from A: 5Marginal from B: 3 e^{0.1 t_B}Marginal from C: 40/(t_C +1)At optimality, all marginals should be equal.So, 5=3 e^{0.1 t_B}=40/(t_C +1)But t_A + t_B + t_C=12.So, from 5=3 e^{0.1 t_B}, we get t_B= ln(5/3)/0.1≈5.108 days.From 5=40/(t_C +1), we get t_C=8 -1=7 days.So, t_A=12 -5.108 -7≈-0.108 days.Negative, which is not feasible.Therefore, the optimal is to set t_A=0, and then allocate the remaining days to B and C such that their marginals are equal.So, set 3 e^{0.1 t_B}=40/(t_C +1), with t_B + t_C=12.As before, solving this gives t_B≈4.5, t_C≈7.5.But at this point, the marginal from B and C is≈4.705, which is less than 5, so we can increase the total by moving days from B and C to A.Wait, but if we move a day from B to A, the marginal from A is 5, and the marginal from B was 4.705, so the total increases by 0.295.Similarly, moving a day from C to A, the marginal from C was 4.705, so total increases by 0.295.So, we can keep moving days from B and C to A until the marginal from B and C equals 5.But when we do that, we end up with t_B=5.108, t_C=7, but t_A=12 -5.108 -7≈-0.108, which is not feasible.Therefore, the maximum feasible allocation is t_A=0, t_B≈4.5, t_C≈7.5, even though the marginal from A is higher.Wait, but that seems contradictory. Maybe the issue is that when we set t_A=0, the marginal from B and C is less than A, so we should allocate as much as possible to A until the marginal from B and C equals 5.But since t_A can't be negative, we set t_A=0, and then allocate the rest to B and C, but their marginals are less than A, which suggests that we should allocate more to A, but we can't because t_A can't be negative.Therefore, the optimal is to set t_A=0, and then allocate the remaining days to B and C such that their marginals are equal.So, the optimal allocation is t_A=0, t_B≈4.5, t_C≈7.5.But let me compute the total inspiration at this point and compare it to allocating some days to A.Compute Total1=20 +30e^{0.45} +40 ln(8.5)≈20 +47.049 +85.6≈152.649Now, suppose we allocate 1 day to A, so t_A=1, then t_B + t_C=11.We need to reallocate the days between B and C such that their marginals are equal.So, set 3 e^{0.1 t_B}=40/(t_C +1), with t_B + t_C=11.Let’s denote t_C=11 - t_B.So, 3 e^{0.1 t_B}=40/(12 - t_B)So, 3 e^{0.1 t_B} (12 - t_B) =40Let’s define f(t_B)=3 e^{0.1 t_B} (12 - t_B) -40Find t_B where f(t_B)=0.Try t_B=5:f(5)=3 e^{0.5} (7) -40≈3*1.6487*7 -40≈3*11.5409 -40≈34.6227 -40≈-5.3773t_B=4:f(4)=3 e^{0.4} (8) -40≈3*1.4918*8 -40≈3*11.9344 -40≈35.8032 -40≈-4.1968t_B=3:f(3)=3 e^{0.3} (9) -40≈3*1.3499*9 -40≈3*12.1491 -40≈36.4473 -40≈-3.5527t_B=2:f(2)=3 e^{0.2} (10) -40≈3*1.2214*10 -40≈3*12.214 -40≈36.642 -40≈-3.358t_B=1:f(1)=3 e^{0.1} (11) -40≈3*1.1052*11 -40≈3*12.1572 -40≈36.4716 -40≈-3.5284t_B=0:f(0)=3*1*(12) -40=36 -40=-4Hmm, all negative. So, no solution where t_B + t_C=11 and marginals equal.Wait, that suggests that when t_A=1, the marginals from B and C cannot be equal because f(t_B) is always negative.Therefore, the optimal allocation when t_A=1 is to set t_B and t_C such that their marginals are as high as possible, but since the equation doesn't have a solution, we might have to set t_B and t_C such that their marginals are equal to each other, but higher than the marginal from A.Wait, but the marginal from A is 5, which is higher than the marginal from B and C when t_A=0.Wait, perhaps when t_A=1, the marginal from A is still 5, and the marginal from B and C are less than 5, so it's still better to allocate more to A.But since we can't allocate more than 12 days, and t_A=1, t_B + t_C=11, but we can't find a t_B where the marginal from B equals the marginal from C.Therefore, perhaps the optimal is to set t_A=1, and then allocate the remaining days to B and C such that their marginals are as high as possible.But this is getting too complicated. Maybe a better approach is to use the Kuhn-Tucker conditions, considering that t_A≥0.So, the Lagrangian is:( mathcal{L} = 5t_A + 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) + lambda (12 - t_A - t_B - t_C) + mu t_A )Where μ is the Kuhn-Tucker multiplier for the inequality constraint t_A≥0.At optimality, either t_A=0 or μ=0.If μ=0, then the partial derivative with respect to t_A is 5 - λ=0 => λ=5.Then, as before, we get t_B≈5.108, t_C≈7, but t_A=12 -5.108 -7≈-0.108, which is not feasible, so μ>0, meaning t_A=0.Therefore, the optimal is t_A=0, and then solve for t_B and t_C with t_B + t_C=12, and their marginals equal.So, t_B≈4.5, t_C≈7.5.Therefore, the optimal allocation is t_A=0, t_B≈4.5, t_C≈7.5.Now, moving to part 2: Each island has a different cost structure. Island A costs 500 per day, Island B costs 800 per day, and Island C costs 1,000 per day. The entrepreneur has a budget of 9,000 for the 12-day retreat. Determine the feasible distribution of days spent on each island that satisfies the budget constraint while maximizing the total inspiration as calculated in part 1.So, we need to maximize the same total inspiration function:( Total = 5t_A + 20 + 30e^{0.1t_B} + 40 ln(t_C + 1) )Subject to:1. ( t_A + t_B + t_C = 12 )2. ( 500t_A + 800t_B + 1000t_C leq 9000 )And ( t_A, t_B, t_C geq 0 )So, we have two constraints now: the total days and the budget.We need to maximize Total under these constraints.This is a constrained optimization problem with two equality constraints (if we consider the budget as an equality, but it's actually an inequality, so we might have to consider whether the budget is binding or not).But since the entrepreneur wants to maximize inspiration, it's likely that the budget will be binding, but let's check.First, let's see what the cost is for the optimal allocation from part 1: t_A=0, t_B=4.5, t_C=7.5.Compute the cost:500*0 +800*4.5 +1000*7.5=0 +3600 +7500=11100, which is more than 9000.So, the optimal allocation from part 1 is not feasible under the budget constraint.Therefore, we need to find a feasible allocation that maximizes the total inspiration.This is now a constrained optimization problem with two constraints: total days=12 and total cost≤9000.We can use Lagrangian multipliers again, but now with two constraints.Alternatively, we can express one variable in terms of the others and substitute.Let me try to set up the Lagrangian with two constraints.Let’s denote:Constraint 1: ( t_A + t_B + t_C = 12 ) (equality)Constraint 2: ( 500t_A + 800t_B + 1000t_C leq 9000 ) (inequality)But since we are maximizing, it's likely that the budget will be tight, so we can consider it as an equality.Therefore, we have two equality constraints:1. ( t_A + t_B + t_C = 12 )2. ( 500t_A + 800t_B + 1000t_C = 9000 )We can solve these two equations to express two variables in terms of the third.Let me solve for t_A and t_B in terms of t_C.From constraint 1: ( t_A = 12 - t_B - t_C )Substitute into constraint 2:500*(12 - t_B - t_C) +800t_B +1000t_C=9000Compute:500*12 -500t_B -500t_C +800t_B +1000t_C=90006000 + ( -500t_B +800t_B ) + ( -500t_C +1000t_C )=90006000 +300t_B +500t_C=9000Subtract 6000:300t_B +500t_C=3000Divide both sides by 100:3t_B +5t_C=30So, 3t_B=30 -5t_C => t_B=(30 -5t_C)/3=10 - (5/3)t_CNow, since t_B≥0, 10 - (5/3)t_C≥0 => t_C≤6Similarly, t_A=12 - t_B - t_C=12 - [10 - (5/3)t_C] - t_C=12 -10 + (5/3)t_C -t_C=2 + (2/3)t_CSo, t_A=2 + (2/3)t_CNow, t_A must be ≥0, which it is since t_C≥0.So, we have:t_A=2 + (2/3)t_Ct_B=10 - (5/3)t_Ct_C=t_C, with t_C≤6Now, we can express the total inspiration in terms of t_C:Total=5t_A +20 +30e^{0.1t_B} +40 ln(t_C +1)Substitute t_A and t_B:Total=5*(2 + (2/3)t_C) +20 +30e^{0.1*(10 - (5/3)t_C)} +40 ln(t_C +1)Simplify:Total=10 + (10/3)t_C +20 +30e^{1 - (0.5/3)t_C} +40 ln(t_C +1)Wait, 0.1*(10 - (5/3)t_C)=1 - (0.5/3)t_C=1 - (1/6)t_CSo, Total=30 + (10/3)t_C +30e^{1 - (1/6)t_C} +40 ln(t_C +1)Now, we need to maximize this function with respect to t_C, where t_C is between 0 and 6.Let me denote:Total(t_C)=30 + (10/3)t_C +30e^{1 - (1/6)t_C} +40 ln(t_C +1)We can take the derivative of Total with respect to t_C and set it to zero.Compute dTotal/dt_C:d/dt_C [30] =0d/dt_C [(10/3)t_C]=(10/3)d/dt_C [30e^{1 - (1/6)t_C}]=30*e^{1 - (1/6)t_C}*(-1/6)= -5 e^{1 - (1/6)t_C}d/dt_C [40 ln(t_C +1)]=40/(t_C +1)So, dTotal/dt_C= (10/3) -5 e^{1 - (1/6)t_C} +40/(t_C +1)Set derivative equal to zero:(10/3) -5 e^{1 - (1/6)t_C} +40/(t_C +1)=0This is a transcendental equation and needs to be solved numerically.Let me denote:f(t_C)= (10/3) -5 e^{1 - (1/6)t_C} +40/(t_C +1)We need to find t_C in [0,6] where f(t_C)=0.Let me compute f(t_C) at several points:At t_C=0:f(0)=10/3 -5 e^{1} +40/1≈3.333 -5*2.718 +40≈3.333 -13.59 +40≈30.743>0At t_C=3:f(3)=10/3 -5 e^{1 -0.5} +40/4≈3.333 -5 e^{0.5} +10≈3.333 -5*1.6487 +10≈3.333 -8.2435 +10≈5.0895>0At t_C=4:f(4)=10/3 -5 e^{1 - (4/6)} +40/5≈3.333 -5 e^{1 - 0.6667} +8≈3.333 -5 e^{0.3333} +8≈3.333 -5*1.3956 +8≈3.333 -6.978 +8≈4.355>0At t_C=5:f(5)=10/3 -5 e^{1 - (5/6)} +40/6≈3.333 -5 e^{1 -0.8333} +6.6667≈3.333 -5 e^{0.1667} +6.6667≈3.333 -5*1.1814 +6.6667≈3.333 -5.907 +6.6667≈4.0927>0At t_C=6:f(6)=10/3 -5 e^{1 -1} +40/7≈3.333 -5 e^{0} +5.714≈3.333 -5*1 +5.714≈3.333 -5 +5.714≈4.047>0Wait, all these are positive. So, f(t_C) is positive throughout [0,6]. That suggests that the derivative is always positive, meaning the function is increasing in t_C. Therefore, the maximum occurs at t_C=6.But let me check t_C=6:f(6)=10/3 -5 e^{0} +40/7≈3.333 -5 +5.714≈4.047>0So, the function is increasing, so the maximum is at t_C=6.Therefore, the optimal t_C=6, which gives:t_C=6t_B=10 - (5/3)*6=10 -10=0t_A=2 + (2/3)*6=2 +4=6So, t_A=6, t_B=0, t_C=6But let me check the budget:500*6 +800*0 +1000*6=3000 +0 +6000=9000, which matches the budget.Now, let's compute the total inspiration:Total=5*6 +20 +30e^{0.1*0} +40 ln(6 +1)=30 +20 +30*1 +40 ln7≈50 +30 +40*1.9459≈80 +77.836≈157.836Wait, but earlier, when t_C=6, t_B=0, t_A=6, the total is≈157.836But let me check if this is indeed the maximum.Wait, but when t_C=6, t_B=0, t_A=6, the total is≈157.836But earlier, when we had t_A=0, t_B=4.5, t_C=7.5, the total was≈152.649, but that allocation was over budget.But with the budget constraint, the maximum is achieved at t_C=6, t_A=6, t_B=0.But let me check if this is indeed the case.Wait, the function Total(t_C) is increasing in t_C, so the maximum is at t_C=6.Therefore, the optimal allocation under the budget constraint is t_A=6, t_B=0, t_C=6.But let me verify the total inspiration:Total=5*6 +20 +30e^{0} +40 ln7≈30 +20 +30 +40*1.9459≈80 +77.836≈157.836Now, let me check if allocating some days to B would increase the total.Suppose we take 1 day from C and give it to B, so t_C=5, t_B=1.Compute the new total:Total=5*6 +20 +30e^{0.1*1} +40 ln6≈30 +20 +30*1.1052 +40*1.7918≈50 +33.156 +71.672≈154.828Which is less than 157.836.Similarly, taking 1 day from C and giving it to A, but A is already at 6, which is the maximum possible under the budget.Wait, no, because t_A=6, t_B=0, t_C=6, and if we take 1 day from C, we can give it to B, but that reduces the total.Alternatively, taking 1 day from A and giving it to B and C.Wait, but t_A=6, t_B=0, t_C=6.If we take 1 day from A and give it to B, then t_A=5, t_B=1, t_C=6.But t_A=5, t_B=1, t_C=6, check budget:500*5 +800*1 +1000*6=2500 +800 +6000=9300>9000, which is over budget.Alternatively, take 1 day from A and give it to C, but t_C is already 6, which is the maximum.Wait, no, because t_C=6, t_A=6, t_B=0.If we take 1 day from A and give it to B, we have to reduce t_A by 1 and increase t_B by 1, but that would require adjusting t_C accordingly to keep the budget.Wait, let me think differently.If we take 1 day from A and give it to B, we need to adjust t_C to keep the budget.So, original allocation: t_A=6, t_B=0, t_C=6, cost=9000.If we take 1 day from A and give it to B, we have t_A=5, t_B=1, t_C=6.Compute cost:500*5 +800*1 +1000*6=2500 +800 +6000=9300>9000.To keep the budget, we need to reduce t_C by some amount.Let’s denote the new t_C as x.So, t_A=5, t_B=1, t_C=x.Budget:500*5 +800*1 +1000x=2500 +800 +1000x=3300 +1000x=9000So, 1000x=5700 =>x=5.7So, t_C=5.7Therefore, t_A=5, t_B=1, t_C=5.7Compute total inspiration:Total=5*5 +20 +30e^{0.1*1} +40 ln(5.7 +1)=25 +20 +30*1.1052 +40 ln6.7≈45 +33.156 +40*1.902≈45 +33.156 +76.08≈154.236Which is less than 157.836.Therefore, the total decreases.Similarly, if we take 1 day from A and give it to C, but t_C is already 6, which is the maximum.Wait, no, because t_C=6 is the maximum allowed by the budget when t_A=6, t_B=0.Alternatively, take 1 day from C and give it to B, but that would require reducing t_C by 1 and increasing t_B by 1, but we have to adjust t_A accordingly.Wait, let me try:Take 1 day from C and give it to B: t_C=5, t_B=1, t_A=6.But then, the budget would be:500*6 +800*1 +1000*5=3000 +800 +5000=8800<9000So, we have 200 left, which we can allocate to A or B or C.But since we want to maximize inspiration, we should allocate the remaining 200 to the island with the highest marginal inspiration.Compute the marginal inspiration per dollar for each island.Marginal inspiration from A:5 per day, cost 500 per day, so 5/500=0.01 per dollar.Marginal from B:3 e^{0.1 t_B} per day, cost 800 per day.At t_B=1, marginal from B=3 e^{0.1}=3*1.1052≈3.3156 per day, so 3.3156/800≈0.00414 per dollar.Marginal from C:40/(t_C +1) per day, cost 1000 per day.At t_C=5, marginal from C=40/6≈6.6667 per day, so 6.6667/1000≈0.006667 per dollar.So, the highest marginal per dollar is from C:≈0.006667, then A:0.01, then B:0.00414.Wait, actually, A has higher marginal per dollar than C.Wait, 5/500=0.01, which is higher than 6.6667/1000≈0.006667.Therefore, we should allocate the remaining 200 to A.So, t_A=6 +200/500=6 +0.4=6.4 days.But wait, we can't have fractional days in this context, but the problem allows partial days, so it's okay.Wait, no, the initial allocation was t_A=6, t_B=1, t_C=5, with a budget of 8800.We have 200 left, which can be allocated to A, giving t_A=6 +200/500=6.4, t_B=1, t_C=5.Compute total inspiration:Total=5*6.4 +20 +30e^{0.1*1} +40 ln(5 +1)=32 +20 +30*1.1052 +40 ln6≈52 +33.156 +40*1.7918≈52 +33.156 +71.672≈156.828Which is still less than 157.836.Therefore, the optimal allocation under the budget is t_A=6, t_B=0, t_C=6, with total inspiration≈157.836.But let me check another allocation: t_A=4, t_B=3, t_C=5.Compute budget:500*4 +800*3 +1000*5=2000 +2400 +5000=9400>9000, too much.Alternatively, t_A=5, t_B=2, t_C=5.Budget:2500 +1600 +5000=9100>9000.Too much.t_A=5, t_B=1, t_C=5.7 as before, total≈154.236.Alternatively, t_A=7, t_B=0, t_C=5.Budget:3500 +0 +5000=8500<9000.Remaining budget:500.Allocate to C:500/1000=0.5 days, so t_C=5.5.Compute total:Total=5*7 +20 +30e^{0} +40 ln5.5≈35 +20 +30 +40*1.7047≈85 +68.188≈153.188Less than 157.836.Alternatively, t_A=6, t_B=0, t_C=6, total≈157.836.Therefore, the optimal allocation under the budget is t_A=6, t_B=0, t_C=6.But let me check if this is indeed the maximum.Wait, another approach: since the function Total(t_C) is increasing in t_C, the maximum is at t_C=6, which gives t_A=6, t_B=0.Therefore, the optimal allocation is t_A=6, t_B=0, t_C=6.But let me compute the total inspiration again:Total=5*6 +20 +30e^{0} +40 ln7≈30 +20 +30 +40*1.9459≈80 +77.836≈157.836Yes, that seems correct.Therefore, the feasible distribution is t_A=6, t_B=0, t_C=6.But let me check if this allocation is feasible:t_A=6, t_B=0, t_C=6, total days=12, budget=500*6 +800*0 +1000*6=3000 +0 +6000=9000, which is exactly the budget.Therefore, the optimal feasible allocation is t_A=6, t_B=0, t_C=6.But wait, is there a way to get a higher total by allocating some days to B?Suppose we take 1 day from C and give it to B, but adjust t_A accordingly to keep the budget.So, t_C=5, t_B=1, then t_A=12 -5 -1=6.But the budget would be:500*6 +800*1 +1000*5=3000 +800 +5000=8800<9000.We have 200 left, which can be allocated to A or C.As before, allocate to A: t_A=6 +200/500=6.4, t_B=1, t_C=5.Total≈156.828<157.836.Alternatively, allocate to C: t_C=5 +200/1000=5.2, t_A=6, t_B=1.Compute total:Total=5*6 +20 +30e^{0.1*1} +40 ln5.2≈30 +20 +33.156 +40*1.6472≈50 +33.156 +65.888≈149.044<157.836.Therefore, the maximum is indeed at t_A=6, t_B=0, t_C=6.Therefore, the optimal feasible distribution is t_A=6, t_B=0, t_C=6.</think>