🧠:<think>Alright, so I have these two problems to solve related to a software development project that uses DevOps practices. The first one is about deployment frequency following a Poisson distribution, and the second is about mean time to recovery modeled by an exponential distribution. Let me tackle them one by one.Starting with the first problem: 1. Deployment Frequency (DF) as a Poisson DistributionThe problem states that DF follows a Poisson distribution with a mean rate of λ deployments per week. The project manager wants an average deployment frequency of 10 deployments per week, so λ is 10. We need to find the probability that in any given week, the team will perform exactly 8 deployments. I remember that the Poisson probability mass function is given by:P(X = k) = (λ^k * e^(-λ)) / k!Where:- P(X = k) is the probability of k occurrences,- λ is the average rate (mean),- e is the base of the natural logarithm,- k! is the factorial of k.So, plugging in the values we have:λ = 10k = 8Therefore, the probability should be:P(X = 8) = (10^8 * e^(-10)) / 8!I can compute this numerically, but the question says to express the answer in terms of λ. Since λ is given as 10, maybe they just want the formula with λ instead of substituting 10? Let me check the question again.It says, "calculate the probability... Express your answer in terms of λ." Hmm, so perhaps they want the formula expressed with λ, not substituting 10. But wait, the project manager sets λ to 10, so maybe they just want the expression in terms of λ, which is 10. Hmm, maybe I should write it as (λ^8 * e^(-λ)) / 8! since λ is 10. Alternatively, maybe they just want the formula, but given the context, since λ is set to 10, perhaps substituting it is acceptable.Wait, the question says "the project manager sets a goal to achieve an average deployment frequency of 10 deployments per week," so λ is 10. So, the probability is (10^8 * e^(-10)) / 8!.But the question says "express your answer in terms of λ." So maybe they don't want the numerical value but the expression with λ. So, perhaps just leave it as (λ^8 * e^(-λ)) / 8!.But I'm a bit confused because λ is given as 10, so maybe they expect the numerical value. Let me check both.First, the formula in terms of λ is (λ^8 * e^(-λ)) / 8!.If I compute it numerically, let's see:10^8 is 100,000,000.e^(-10) is approximately 0.00004539993.8! is 40320.So, 100,000,000 * 0.00004539993 = approximately 4539.993.Then, 4539.993 / 40320 ≈ 0.1125.So, approximately 11.25% chance.But since the question says to express in terms of λ, maybe just leave it as the formula. Hmm.Wait, the question says "calculate the probability... Express your answer in terms of λ." So, perhaps they just want the formula, not the numerical value. So, I think the answer is (λ^8 * e^(-λ)) / 8!.But let me think again. If λ is 10, then the probability is (10^8 * e^(-10)) / 8! which is approximately 0.1125 or 11.25%. But since they specify to express in terms of λ, maybe they just want the formula. Alternatively, perhaps they accept either, but since λ is given, maybe it's better to compute the numerical value.Wait, the question says "the project manager sets a goal to achieve an average deployment frequency of 10 deployments per week," so λ is 10. So, the probability is (10^8 * e^(-10)) / 8!.But the question says "express your answer in terms of λ," so maybe they want the formula with λ, not substituting 10. So, perhaps the answer is (λ^8 * e^(-λ)) / 8!.But I'm not sure. Maybe I should compute it numerically as well, just in case.Alternatively, perhaps the question expects the formula, so I'll go with that.2. Mean Time to Recovery (MTTR) as an Exponential DistributionThe second problem states that MTTR is modeled by an exponential distribution with a rate parameter θ = 1/μ, where μ is the average recovery time. The project manager wants the probability of recovering from any failure within 2 hours to be at least 90%. We need to determine the maximum allowable average recovery time μ.I recall that the exponential distribution is memoryless, and the cumulative distribution function (CDF) is P(X ≤ x) = 1 - e^(-θx), where θ is the rate parameter.Given that θ = 1/μ, so θ = 1/μ.We want P(X ≤ 2) ≥ 0.90.So, 1 - e^(-θ*2) ≥ 0.90.Let me write that down:1 - e^(-2θ) ≥ 0.90Subtract 1 from both sides:-e^(-2θ) ≥ -0.10Multiply both sides by -1 (which reverses the inequality):e^(-2θ) ≤ 0.10Take the natural logarithm of both sides:ln(e^(-2θ)) ≤ ln(0.10)Simplify left side:-2θ ≤ ln(0.10)Multiply both sides by -1 (which reverses the inequality again):2θ ≥ -ln(0.10)Compute -ln(0.10):ln(0.10) is approximately -2.302585, so -ln(0.10) is approximately 2.302585.So, 2θ ≥ 2.302585Divide both sides by 2:θ ≥ 2.302585 / 2 ≈ 1.1512925But θ = 1/μ, so:1/μ ≥ 1.1512925Take reciprocal (remembering that inequality direction reverses when taking reciprocals because both sides are positive):μ ≤ 1 / 1.1512925 ≈ 0.868589So, μ ≤ approximately 0.868589 hours.But let's compute it more accurately.We have:θ ≥ ln(10)/2, because ln(10) is approximately 2.302585, so θ ≥ 2.302585 / 2 ≈ 1.1512925.Therefore, μ = 1/θ ≤ 1 / 1.1512925 ≈ 0.868589 hours.To express this as a fraction, since 0.868589 is approximately 13/15 hours, but let me check:13/15 = 0.866666..., which is close but not exact. Alternatively, 0.868589 is approximately 52.1153 minutes, but since the question asks for μ in hours, we can express it as approximately 0.8686 hours.But perhaps we can express it exactly in terms of ln(10). Let's see:From earlier steps:θ ≥ (ln(10))/2Thus, μ = 1/θ ≤ 2 / ln(10)Compute 2 / ln(10):ln(10) ≈ 2.302585093So, 2 / 2.302585093 ≈ 0.8685889637So, μ ≤ 2 / ln(10) hours.Thus, the maximum allowable average recovery time μ is 2 / ln(10) hours, which is approximately 0.8686 hours.To convert 0.8686 hours into minutes, multiply by 60: 0.8686 * 60 ≈ 52.116 minutes, so about 52 minutes and 7 seconds.But since the question asks for μ in hours, we can leave it as 2 / ln(10) hours or approximately 0.8686 hours.So, summarizing:1. The probability is (λ^8 * e^(-λ)) / 8! which, with λ=10, is approximately 0.1125 or 11.25%.2. The maximum allowable μ is 2 / ln(10) hours, approximately 0.8686 hours.Wait, but let me double-check the second problem.We have P(X ≤ 2) ≥ 0.90, which translates to 1 - e^(-2θ) ≥ 0.90.So, e^(-2θ) ≤ 0.10.Taking natural logs: -2θ ≤ ln(0.10).Multiply both sides by -1: 2θ ≥ -ln(0.10).Since ln(0.10) is negative, -ln(0.10) is positive, so 2θ ≥ positive number.Thus, θ ≥ (ln(10))/2, because ln(1/0.10) = ln(10).Wait, ln(0.10) = ln(1/10) = -ln(10), so -ln(0.10) = ln(10).Therefore, 2θ ≥ ln(10), so θ ≥ ln(10)/2.Thus, μ = 1/θ ≤ 2 / ln(10).Yes, that's correct.So, the maximum μ is 2 / ln(10) hours.I think that's accurate.So, to recap:1. For the Poisson distribution, the probability is (λ^8 * e^(-λ)) / 8! which with λ=10 is approximately 0.1125.2. For the exponential distribution, the maximum μ is 2 / ln(10) hours, approximately 0.8686 hours.I think that's it.</think>

🧠:<think>Alright, so I have this problem about an actress's energy levels during a performance. The energy is modeled by the function ( E(t) = A sin(omega t + phi) + C ). There are two parts: first, I need to find the constants ( A ), ( omega ), ( phi ), and ( C ) given that she reaches her peak energy of 150 units at ( t = 5 ) minutes and her minimum energy of 50 units at ( t = 15 ) minutes. Second, I need to calculate the total energy expended over a 20-minute performance by integrating ( E(t) ) from 0 to 20.Starting with part 1. Let me recall that the general sine function is ( E(t) = A sin(omega t + phi) + C ). Here, ( A ) is the amplitude, ( omega ) is the angular frequency, ( phi ) is the phase shift, and ( C ) is the vertical shift or the average value.Given that she reaches a peak of 150 at ( t = 5 ) and a minimum of 50 at ( t = 15 ). So, the maximum value of ( E(t) ) is 150, and the minimum is 50. Since the sine function oscillates between -1 and 1, the amplitude ( A ) is half the difference between the maximum and minimum values. So, ( A = frac{150 - 50}{2} = 50 ). That seems straightforward.Next, the vertical shift ( C ) is the average of the maximum and minimum values. So, ( C = frac{150 + 50}{2} = 100 ). So, now we have ( A = 50 ) and ( C = 100 ).Now, we need to find ( omega ) and ( phi ). Let's think about the period of the sine function. The time between the peak and the minimum is 10 minutes (from t=5 to t=15). In a sine wave, the time between a peak and a trough is half the period. So, if the time between peak and trough is 10 minutes, the full period ( T ) is 20 minutes. Therefore, the angular frequency ( omega ) is ( frac{2pi}{T} = frac{2pi}{20} = frac{pi}{10} ) radians per minute.So, ( omega = frac{pi}{10} ).Now, we need to find the phase shift ( phi ). We know that at ( t = 5 ), the energy is at its peak, which corresponds to the sine function reaching its maximum. The sine function reaches its maximum at ( frac{pi}{2} ) radians. So, we can set up the equation:( omega t + phi = frac{pi}{2} ) when ( t = 5 ).Plugging in ( omega = frac{pi}{10} ) and ( t = 5 ):( frac{pi}{10} times 5 + phi = frac{pi}{2} )Simplify:( frac{pi}{2} + phi = frac{pi}{2} )Subtract ( frac{pi}{2} ) from both sides:( phi = 0 )Wait, that seems too simple. Let me double-check. If ( phi = 0 ), then the function is ( E(t) = 50 sinleft(frac{pi}{10} tright) + 100 ). Let's test this at ( t = 5 ):( E(5) = 50 sinleft(frac{pi}{10} times 5right) + 100 = 50 sinleft(frac{pi}{2}right) + 100 = 50 times 1 + 100 = 150 ). That works.Now, let's check at ( t = 15 ):( E(15) = 50 sinleft(frac{pi}{10} times 15right) + 100 = 50 sinleft(frac{3pi}{2}right) + 100 = 50 times (-1) + 100 = -50 + 100 = 50 ). That also works.So, it seems ( phi = 0 ) is correct. Therefore, the function is ( E(t) = 50 sinleft(frac{pi}{10} tright) + 100 ).Wait a second, but let me think about the period again. If the period is 20 minutes, then the function completes one full cycle every 20 minutes. So, starting at t=0, the sine function would start at 0, go up to 150 at t=5, back down to 100 at t=10, down to 50 at t=15, and back to 100 at t=20. That seems consistent.But let me also consider the phase shift. If ( phi = 0 ), the sine function starts at 0 when t=0. So, at t=0, ( E(0) = 50 sin(0) + 100 = 100 ). That's the average energy. So, she starts at average energy, goes up to peak at t=5, down to minimum at t=15, and back to average at t=20. That seems reasonable.So, I think my calculations are correct. So, ( A = 50 ), ( omega = frac{pi}{10} ), ( phi = 0 ), and ( C = 100 ).Moving on to part 2: calculating the total energy expended over a 20-minute performance by integrating ( E(t) ) from 0 to 20.So, the integral of ( E(t) ) from 0 to 20 is:( int_{0}^{20} [50 sinleft(frac{pi}{10} tright) + 100] dt )I can split this integral into two parts:( 50 int_{0}^{20} sinleft(frac{pi}{10} tright) dt + 100 int_{0}^{20} dt )Let's compute each integral separately.First integral: ( 50 int_{0}^{20} sinleft(frac{pi}{10} tright) dt )Let me make a substitution to solve this integral. Let ( u = frac{pi}{10} t ), so ( du = frac{pi}{10} dt ), which means ( dt = frac{10}{pi} du ).Changing the limits of integration: when t=0, u=0; when t=20, u= ( frac{pi}{10} times 20 = 2pi ).So, the integral becomes:( 50 times frac{10}{pi} int_{0}^{2pi} sin(u) du )Compute the integral:( 50 times frac{10}{pi} [ -cos(u) ]_{0}^{2pi} )Calculate the antiderivative at the limits:( 50 times frac{10}{pi} [ -cos(2pi) + cos(0) ] )We know that ( cos(2pi) = 1 ) and ( cos(0) = 1 ), so:( 50 times frac{10}{pi} [ -1 + 1 ] = 50 times frac{10}{pi} times 0 = 0 )So, the first integral is 0. That makes sense because the sine function is symmetric over its period, so the area above the x-axis cancels out the area below.Now, the second integral: ( 100 int_{0}^{20} dt )This is straightforward:( 100 [ t ]_{0}^{20} = 100 (20 - 0) = 2000 )Therefore, the total integral is 0 + 2000 = 2000 units.So, the total energy expended over the 20-minute performance is 2000 units.Wait, let me just make sure I didn't make a mistake in the substitution. The substitution was correct: ( u = frac{pi}{10} t ), so ( du = frac{pi}{10} dt ), hence ( dt = frac{10}{pi} du ). The limits from t=0 to t=20 become u=0 to u=2π. The integral of sin(u) over 0 to 2π is indeed 0 because it's a full period. So, that part checks out.And the second integral is just the integral of a constant, which is the constant times the interval length. So, 100 over 20 minutes is 2000. That seems correct.So, putting it all together, the total energy expended is 2000 units.Final Answer1. The constants are ( A = boxed{50} ), ( omega = boxed{dfrac{pi}{10}} ), ( phi = boxed{0} ), and ( C = boxed{100} ).2. The total energy expended over the 20-minute performance is ( boxed{2000} ) units.</think>

🧠:<think>Okay, so I have this problem about Alex revisiting his favorite RTS game, Galactic Conquest. He needs to figure out the resources over time, and there are two sub-problems. Let me tackle them one by one.Starting with Sub-problem 1: Determine the total amount of each resource accumulated over a 24-hour period using integral calculus. The resources are minerals, gas, and credits, each given by their respective functions.First, let me write down the functions again to make sure I have them right.1. ( M(t) = 500 sinleft(frac{pi t}{12}right) + 1000t )2. ( G(t) = 300 cosleft(frac{pi t}{12}right) + 800t )3. ( C(t) = 2000e^{0.05t} )So, for each resource, I need to compute the integral from t=0 to t=24. That will give me the total accumulated resources over 24 hours.Let me start with minerals, M(t). The integral of M(t) from 0 to 24 is:( int_{0}^{24} left[500 sinleft(frac{pi t}{12}right) + 1000tright] dt )I can split this integral into two parts:( 500 int_{0}^{24} sinleft(frac{pi t}{12}right) dt + 1000 int_{0}^{24} t dt )Let me compute each part separately.First integral: ( 500 int sinleft(frac{pi t}{12}right) dt )The integral of sin(ax) dx is -(1/a)cos(ax) + C. So, here, a = π/12.So, the integral becomes:500 * [ -12/π cos(π t /12) ] evaluated from 0 to 24.Let me compute this:At t=24:-12/π cos(π*24/12) = -12/π cos(2π) = -12/π * 1 = -12/πAt t=0:-12/π cos(0) = -12/π *1 = -12/πSo, subtracting:[ -12/π ] - [ -12/π ] = 0Wait, that's interesting. So the integral of the sine function over one full period is zero? Because the sine function is symmetric and over a full period, the positive and negative areas cancel out.So, the first integral is zero.Now, the second integral: 1000 ∫ t dt from 0 to24.The integral of t dt is (1/2)t².So, 1000 * [ (1/2)(24)^2 - (1/2)(0)^2 ] = 1000 * (1/2)(576) = 1000 * 288 = 288,000.So, the total minerals accumulated over 24 hours is 288,000.Wait, but let me double-check the first integral. The function is 500 sin(π t /12). The period of this sine function is 2π / (π/12) = 24 hours. So, over 24 hours, it completes exactly one full cycle. Therefore, the integral over one period is indeed zero because the area above the x-axis cancels the area below. So, yes, the first part is zero.Okay, moving on to gas, G(t):( G(t) = 300 cosleft(frac{pi t}{12}right) + 800t )Similarly, the integral from 0 to24 is:( 300 int_{0}^{24} cosleft(frac{pi t}{12}right) dt + 800 int_{0}^{24} t dt )Again, split into two integrals.First integral: 300 ∫ cos(π t /12) dtThe integral of cos(ax) dx is (1/a) sin(ax) + C.So, 300 * [12/π sin(π t /12)] evaluated from 0 to24.Compute at t=24:12/π sin(2π) = 12/π * 0 = 0At t=0:12/π sin(0) = 0So, the integral is 300*(0 - 0) = 0.Again, similar to the sine function, the integral over one full period is zero because the positive and negative areas cancel out.Second integral: 800 ∫ t dt from 0 to24.Same as before, integral of t is (1/2)t².So, 800 * [ (1/2)(24)^2 - 0 ] = 800 * 288 = 230,400.So, total gas accumulated is 230,400.Wait, let me confirm. The period of the cosine function is also 24 hours, so over 24 hours, the integral is zero. Correct.Now, moving on to credits, C(t):( C(t) = 2000e^{0.05t} )Integral from 0 to24:2000 ∫ e^{0.05t} dt from 0 to24.The integral of e^{kt} dt is (1/k)e^{kt} + C.So, 2000 * [ (1/0.05) e^{0.05t} ] evaluated from 0 to24.Compute:2000 * (1/0.05) [ e^{0.05*24} - e^{0} ] = 2000 * 20 [ e^{1.2} - 1 ]Calculate e^{1.2}:I know that e^1 is approximately 2.71828, e^0.2 is approximately 1.2214.So, e^{1.2} ≈ e^1 * e^0.2 ≈ 2.71828 * 1.2214 ≈ let's compute that.2.71828 * 1.2 = 3.2619362.71828 * 0.0214 ≈ 0.0581So, total ≈ 3.261936 + 0.0581 ≈ 3.3200So, e^{1.2} ≈ 3.3200Therefore, 2000 * 20 [3.3200 -1] = 40,000 [2.3200] = 40,000 * 2.32 = 92,800.Wait, let me compute 40,000 * 2.32:40,000 * 2 = 80,00040,000 * 0.32 = 12,800So, total is 80,000 + 12,800 = 92,800.Therefore, the total credits accumulated over 24 hours is 92,800.Wait, but let me check the integral again.Yes, integral of e^{0.05t} is (1/0.05)e^{0.05t} = 20 e^{0.05t}.So, 2000 * 20 [e^{1.2} -1] = 40,000*(e^{1.2}-1). Since e^{1.2}≈3.3201, so 3.3201 -1=2.3201, so 40,000*2.3201≈92,804.So, approximately 92,804. But since the problem says to compute exact values, maybe I should keep it in terms of e^{1.2}.Wait, the problem says to use integral calculus to compute the exact values. So, perhaps I should leave it in terms of exponentials rather than approximate.So, let me write it as:2000 * (1/0.05) [e^{1.2} -1] = 40,000 (e^{1.2} -1)So, exact value is 40,000(e^{1.2} -1). If needed, we can compute it numerically, but since the problem says exact, maybe that's acceptable.But let me check if it's necessary to compute it numerically. The problem says "exact values," so perhaps leaving it in terms of e^{1.2} is acceptable. Alternatively, maybe they expect a numerical value. Hmm.But given that the other resources had exact values (288,000 and 230,400), which are integers, maybe for credits, we can compute it exactly as 40,000(e^{1.2} -1). But let me compute e^{1.2} more accurately.Using a calculator, e^{1.2} is approximately 3.3201169227766016.So, 40,000*(3.3201169227766016 -1) = 40,000*2.3201169227766016 ≈ 92,804.67691106406.So, approximately 92,804.68. Since resources are likely in whole numbers, maybe 92,805.But the problem says "exact values," so perhaps we can write it as 40,000(e^{1.2} -1). Alternatively, if they accept decimal, 92,804.68.But let me see if I can compute it more precisely.Alternatively, maybe I can compute e^{1.2} using Taylor series for more precision, but that might be overkill. Alternatively, just use a calculator value.But perhaps the problem expects the exact integral expression, so 40,000(e^{1.2} -1). Alternatively, if they want a numerical value, then 92,804.68.Wait, looking back at the problem statement: "Use integral calculus to compute the exact values." So, exact values would mean not approximate, so perhaps leaving it in terms of e^{1.2} is acceptable. But sometimes, in calculus, exact can mean evaluated as much as possible, so maybe it's acceptable to write 40,000(e^{1.2} -1). Alternatively, if they expect a numerical value, perhaps we can compute it as 92,804.68.But let me check the other resources. For minerals, it was 288,000, gas 230,400, both integers. So, maybe for credits, they expect a numerical value as well, so 92,805.Alternatively, perhaps the problem expects all three resources to be computed as exact integrals, so maybe for credits, it's 40,000(e^{1.2} -1). But let me see if I can write it as a numerical value.Alternatively, maybe I can compute it more accurately. Let me compute e^{1.2} using a calculator:e^{1.2} ≈ 3.3201169227766016So, 3.3201169227766016 -1 = 2.3201169227766016Multiply by 40,000: 2.3201169227766016 * 40,000 = 92,804.67691106406So, approximately 92,804.68. So, I can write it as 92,804.68, but since resources are in whole numbers, maybe 92,805.Alternatively, perhaps the problem expects the exact expression, so 40,000(e^{1.2} -1).But let me check the problem statement again: "Use integral calculus to compute the exact values." So, exact values would mean not approximate, so perhaps the answer is 40,000(e^{1.2} -1). Alternatively, if they accept decimal, 92,804.68.Wait, but in the other resources, the integrals resulted in exact integers, so maybe for credits, it's acceptable to write it as 40,000(e^{1.2} -1). Alternatively, maybe they expect a numerical value, so 92,804.68.But I think, given that the other resources are exact, perhaps for credits, it's acceptable to write it as 40,000(e^{1.2} -1). But let me see if I can compute it more precisely.Alternatively, perhaps I can write it as 40,000(e^{1.2} -1) ≈ 92,804.68.But let me check if I can write it as an exact value. Since e^{1.2} is a transcendental number, it can't be expressed exactly as a finite decimal or fraction, so the exact value would be 40,000(e^{1.2} -1). So, perhaps that's the answer they expect.So, summarizing:Minerals: 288,000Gas: 230,400Credits: 40,000(e^{1.2} -1) ≈ 92,804.68But let me check if I made any mistakes in the integrals.For minerals:Integral of 500 sin(π t /12) from 0 to24 is zero, as it's over one period.Integral of 1000t from 0 to24 is 1000*(24^2)/2 = 1000*288 = 288,000. Correct.Gas:Integral of 300 cos(π t /12) from 0 to24 is zero, over one period.Integral of 800t from 0 to24 is 800*(24^2)/2 = 800*288 = 230,400. Correct.Credits:Integral of 2000 e^{0.05t} from 0 to24 is 2000*(1/0.05)(e^{1.2} -1) = 40,000(e^{1.2} -1). Correct.So, that's Sub-problem 1 done.Now, moving on to Sub-problem 2: Determine the exact time t when Alex will have enough of each resource to build a unit requiring 15,000 minerals, 12,000 gas, and 25,000 credits.So, we need to find the smallest t such that:M(t) ≥ 15,000G(t) ≥ 12,000C(t) ≥ 25,000And t must satisfy all three inequalities simultaneously.So, we need to solve for t in each inequality and find the maximum of the minimal t's.So, let's solve each inequality separately.Starting with minerals:M(t) = 500 sin(π t /12) + 1000t ≥ 15,000Similarly for gas:G(t) = 300 cos(π t /12) + 800t ≥ 12,000And for credits:C(t) = 2000 e^{0.05t} ≥ 25,000Let me start with credits because it's likely the easiest, as it's a single exponential function without oscillations.Credits: 2000 e^{0.05t} ≥ 25,000Divide both sides by 2000:e^{0.05t} ≥ 12.5Take natural logarithm on both sides:0.05t ≥ ln(12.5)Compute ln(12.5):ln(12.5) = ln(25/2) = ln(25) - ln(2) = 3.2189 - 0.6931 ≈ 2.5258So, 0.05t ≥ 2.5258t ≥ 2.5258 / 0.05 ≈ 50.516 hours.So, t must be at least approximately 50.516 hours to have enough credits.Now, let's check gas:G(t) = 300 cos(π t /12) + 800t ≥ 12,000We need to solve for t in this inequality.Let me rearrange:800t + 300 cos(π t /12) ≥ 12,000Let me subtract 800t:300 cos(π t /12) ≥ 12,000 - 800tBut this seems a bit messy. Alternatively, perhaps I can write it as:800t ≥ 12,000 - 300 cos(π t /12)But since cos(π t /12) oscillates between -1 and 1, the term 300 cos(...) oscillates between -300 and 300.Therefore, 12,000 - 300 cos(...) oscillates between 12,000 - 300 = 11,700 and 12,000 + 300 = 12,300.So, 800t must be ≥ 11,700 - 300 cos(...). But since cos(...) can be as low as -1, the right-hand side can be as low as 11,700 - (-300) = 12,000.Wait, that might not be the right approach. Let me think differently.Alternatively, perhaps I can write:800t ≥ 12,000 - 300 cos(π t /12)But since cos(...) ≤ 1, the right-hand side is ≥ 12,000 - 300*1 = 11,700.Therefore, 800t ≥ 11,700t ≥ 11,700 / 800 = 14.625 hours.But this is a lower bound. However, since cos(...) can be negative, the actual t needed might be higher.Wait, but perhaps it's better to solve 800t + 300 cos(π t /12) = 12,000 numerically.Similarly for minerals, we'll have to solve 500 sin(π t /12) + 1000t = 15,000 numerically.But let's first see what t we get from credits: approximately 50.516 hours.So, let's check if at t=50.516, gas and minerals are sufficient.But before that, let's see if gas and minerals can be satisfied at t=50.516.But perhaps it's better to find the minimal t for each resource and then take the maximum.So, for gas:We need to solve 800t + 300 cos(π t /12) ≥ 12,000Let me try to find t such that 800t + 300 cos(π t /12) = 12,000This is a transcendental equation, so we'll need to use numerical methods.Similarly for minerals:500 sin(π t /12) + 1000t = 15,000Again, transcendental, so numerical methods needed.But let's see.First, let's solve for gas.We can write:800t + 300 cos(π t /12) = 12,000Let me rearrange:800t = 12,000 - 300 cos(π t /12)t = (12,000 - 300 cos(π t /12)) / 800t = 15 - (300 / 800) cos(π t /12)t = 15 - 0.375 cos(π t /12)This is a fixed-point equation. We can use iterative methods like the Newton-Raphson method or fixed-point iteration.Alternatively, we can use the Newton-Raphson method.Let me define f(t) = 800t + 300 cos(π t /12) - 12,000We need to find t such that f(t) = 0.Compute f(t):f(t) = 800t + 300 cos(π t /12) - 12,000Compute f'(t):f'(t) = 800 - 300*(π/12) sin(π t /12) = 800 - 25π sin(π t /12)We can use Newton-Raphson:t_{n+1} = t_n - f(t_n)/f'(t_n)We need an initial guess. Let's see.At t=15:f(15) = 800*15 + 300 cos(π*15/12) - 12,000= 12,000 + 300 cos(5π/4) - 12,000cos(5π/4) = -√2/2 ≈ -0.7071So, f(15) ≈ 0 + 300*(-0.7071) ≈ -212.13So, f(15) ≈ -212.13At t=16:f(16) = 800*16 + 300 cos(π*16/12) - 12,000= 12,800 + 300 cos(4π/3) - 12,000cos(4π/3) = -0.5So, f(16) = 12,800 - 12,000 + 300*(-0.5) = 800 - 150 = 650So, f(16) = 650So, between t=15 and t=16, f(t) crosses zero.At t=15, f(t)≈-212.13At t=16, f(t)=650So, let's use t0=15.5 as initial guess.Compute f(15.5):f(15.5) = 800*15.5 + 300 cos(π*15.5/12) - 12,000= 12,400 + 300 cos(15.5π/12) - 12,00015.5π/12 ≈ 15.5*0.2618 ≈ 4.0627 radianscos(4.0627) ≈ cos(π + 1.0627) ≈ -cos(1.0627) ≈ -0.4818So, f(15.5) ≈ 12,400 - 12,000 + 300*(-0.4818) ≈ 400 - 144.54 ≈ 255.46So, f(15.5)≈255.46We can see that f(15)= -212.13, f(15.5)=255.46So, the root is between 15 and 15.5.Let me use t0=15.25f(15.25)=800*15.25 +300 cos(π*15.25/12) -12,000= 12,200 + 300 cos(15.25π/12) -12,00015.25π/12≈15.25*0.2618≈3.990 radianscos(3.990)≈cos(π + 0.848)≈-cos(0.848)≈-0.660So, f(15.25)=12,200 -12,000 +300*(-0.660)=200 -198=2So, f(15.25)=2That's very close to zero.Compute f(15.25)=2Compute f'(15.25)=800 -25π sin(π*15.25/12)π*15.25/12≈3.990 radianssin(3.990)≈sin(π + 0.848)≈-sin(0.848)≈-0.750So, f'(15.25)=800 -25π*(-0.750)=800 +25π*0.750≈800 +25*3.1416*0.750≈800 +25*2.356≈800 +58.9≈858.9So, Newton-Raphson step:t1 = t0 - f(t0)/f'(t0)=15.25 - 2/858.9≈15.25 -0.0023≈15.2477Compute f(15.2477):f(t)=800*15.2477 +300 cos(π*15.2477/12) -12,000Compute 800*15.2477≈12,198.16Compute π*15.2477/12≈3.990 radianscos(3.990)≈-0.660So, 300*(-0.660)= -198So, f(t)=12,198.16 -198 -12,000≈12,198.16 -12,198≈0.16So, f(t)=0.16Compute f'(t)=800 -25π sin(3.990)≈800 -25π*(-0.750)=800 +58.9≈858.9So, t2=15.2477 -0.16/858.9≈15.2477 -0.000186≈15.2475Compute f(t2)=800*15.2475 +300 cos(π*15.2475/12) -12,000800*15.2475≈12,198cos(π*15.2475/12)=cos(3.990)≈-0.660So, 300*(-0.660)= -198Thus, f(t2)=12,198 -198 -12,000=0So, t≈15.2475 hours.So, approximately 15.2475 hours.So, gas is satisfied at t≈15.25 hours.Now, let's solve for minerals:M(t)=500 sin(π t /12) +1000t=15,000So, 500 sin(π t /12) +1000t=15,000Let me rearrange:1000t=15,000 -500 sin(π t /12)t=15 -0.5 sin(π t /12)Again, a transcendental equation. Let's use Newton-Raphson.Define f(t)=500 sin(π t /12) +1000t -15,000f(t)=0Compute f(t):f(t)=500 sin(π t /12) +1000t -15,000f'(t)=500*(π/12) cos(π t /12) +1000≈130.8997 cos(π t /12) +1000We need to find t such that f(t)=0.Let's estimate t.At t=15:f(15)=500 sin(π*15/12) +1000*15 -15,000=500 sin(5π/4) +15,000 -15,000sin(5π/4)= -√2/2≈-0.7071So, f(15)=500*(-0.7071)= -353.55At t=15.5:f(15.5)=500 sin(π*15.5/12) +1000*15.5 -15,000=500 sin(15.5π/12) +15,500 -15,00015.5π/12≈3.990 radianssin(3.990)≈sin(π +0.848)= -sin(0.848)≈-0.750So, f(15.5)=500*(-0.750)= -375 +500= -375 +500=125Wait, wait, 1000*15.5=15,500, so 15,500 -15,000=500So, f(15.5)=500*(-0.750) +500= -375 +500=125So, f(15.5)=125So, between t=15 and t=15.5, f(t) goes from -353.55 to 125.Let's try t=15.25f(15.25)=500 sin(π*15.25/12) +1000*15.25 -15,000=500 sin(15.25π/12) +15,250 -15,00015.25π/12≈3.990 radianssin(3.990)≈-0.750So, f(15.25)=500*(-0.750)= -375 +250= -125Wait, 1000*15.25=15,250, so 15,250 -15,000=250So, f(15.25)=500*(-0.750) +250= -375 +250= -125So, f(15.25)= -125So, f(15.25)= -125, f(15.5)=125So, the root is between 15.25 and 15.5.Let me try t=15.375f(15.375)=500 sin(π*15.375/12) +1000*15.375 -15,000=500 sin(15.375π/12) +15,375 -15,00015.375π/12≈3.990 + (0.375π/12)=3.990 +0.09817≈4.088 radianssin(4.088)=sin(π +1.088)= -sin(1.088)≈-0.877So, f(15.375)=500*(-0.877)= -438.5 +375= -63.5Wait, 1000*15.375=15,375, so 15,375 -15,000=375So, f(15.375)=500*(-0.877) +375≈-438.5 +375≈-63.5So, f(15.375)= -63.5Now, let's try t=15.4375f(15.4375)=500 sin(π*15.4375/12) +1000*15.4375 -15,000=500 sin(15.4375π/12) +15,437.5 -15,00015.4375π/12≈3.990 + (0.4375π/12)=3.990 +0.1134≈4.1034 radianssin(4.1034)=sin(π +1.1034)= -sin(1.1034)≈-0.896So, f(15.4375)=500*(-0.896)= -448 +437.5≈-10.5Wait, 1000*15.4375=15,437.5, so 15,437.5 -15,000=437.5So, f(15.4375)=500*(-0.896) +437.5≈-448 +437.5≈-10.5So, f(15.4375)= -10.5Now, let's try t=15.46875f(15.46875)=500 sin(π*15.46875/12) +1000*15.46875 -15,000=500 sin(15.46875π/12) +15,468.75 -15,00015.46875π/12≈3.990 + (0.46875π/12)=3.990 +0.1207≈4.1107 radianssin(4.1107)=sin(π +1.1107)= -sin(1.1107)≈-0.898So, f(15.46875)=500*(-0.898)= -449 +468.75≈19.75So, f(15.46875)=19.75So, between t=15.4375 and t=15.46875, f(t) goes from -10.5 to 19.75Let me try t=15.453125f(15.453125)=500 sin(π*15.453125/12) +1000*15.453125 -15,000=500 sin(15.453125π/12) +15,453.125 -15,00015.453125π/12≈3.990 + (0.453125π/12)=3.990 +0.1178≈4.1078 radianssin(4.1078)=sin(π +1.1078)= -sin(1.1078)≈-0.896So, f(t)=500*(-0.896)= -448 +453.125≈5.125So, f(15.453125)=5.125Now, let's try t=15.446875f(15.446875)=500 sin(π*15.446875/12) +1000*15.446875 -15,000=500 sin(15.446875π/12) +15,446.875 -15,00015.446875π/12≈3.990 + (0.446875π/12)=3.990 +0.116≈4.106 radianssin(4.106)=sin(π +1.106)= -sin(1.106)≈-0.895So, f(t)=500*(-0.895)= -447.5 +446.875≈-0.625So, f(15.446875)= -0.625So, between t=15.446875 and t=15.453125, f(t) goes from -0.625 to 5.125Let me use linear approximation.The change in t is 15.453125 -15.446875=0.00625The change in f(t) is 5.125 - (-0.625)=5.75We need to find t where f(t)=0.From t=15.446875, f(t)= -0.625To reach f(t)=0, need to cover 0.625 over a total change of 5.75 over 0.00625 t.So, delta t= (0.625 /5.75)*0.00625≈(0.1087)*0.00625≈0.000679So, t≈15.446875 +0.000679≈15.447554So, t≈15.4476 hours.Let me check f(15.4476):f(t)=500 sin(π*15.4476/12) +1000*15.4476 -15,000Compute π*15.4476/12≈3.990 + (0.4476π/12)=3.990 +0.116≈4.106 radianssin(4.106)=sin(π +1.106)= -sin(1.106)≈-0.895So, f(t)=500*(-0.895)= -447.5 +15,447.6 -15,000≈-447.5 +447.6≈0.1So, f(t)=0.1, very close to zero.Compute f'(t)=130.8997 cos(π t /12) +1000At t=15.4476, π t /12≈4.106 radianscos(4.106)=cos(π +1.106)= -cos(1.106)≈-0.443So, f'(t)=130.8997*(-0.443) +1000≈-58.1 +1000≈941.9So, Newton-Raphson step:t1=15.4476 -0.1/941.9≈15.4476 -0.000106≈15.4475So, t≈15.4475 hours.So, minerals are satisfied at t≈15.4475 hours.Now, we have:Credits needed at t≈50.516 hoursGas needed at t≈15.2475 hoursMinerals needed at t≈15.4475 hoursSo, the latest time is t≈50.516 hours, which is when credits are satisfied. But we need to check if at t=50.516, gas and minerals are also satisfied.Wait, but let me check if gas and minerals are satisfied at t=50.516.Compute M(50.516)=500 sin(π*50.516/12) +1000*50.516Compute π*50.516/12≈50.516*0.2618≈13.23 radians13.23 radians is equivalent to 13.23 - 4π≈13.23 -12.566≈0.664 radiansSo, sin(13.23)=sin(0.664)≈0.618So, M(50.516)=500*0.618 +1000*50.516≈309 +50,516≈50,825Which is way more than 15,000.Similarly, G(50.516)=300 cos(π*50.516/12) +800*50.516Compute π*50.516/12≈13.23 radianscos(13.23)=cos(0.664)≈0.786So, G(50.516)=300*0.786 +800*50.516≈235.8 +40,412.8≈40,648.6Which is way more than 12,000.So, at t≈50.516 hours, all resources are satisfied.But wait, the problem says "the exact time t when Alex will have enough of each resource to build this unit." So, the earliest time when all three resources are satisfied is the maximum of the minimal t's for each resource. Since credits require t≈50.516, which is later than gas and minerals, the answer is t≈50.516 hours.But let me check if at t=50.516, the resources are indeed above the required amounts.As computed, M(t)=50,825 ≥15,000G(t)=40,648.6 ≥12,000C(t)=2000 e^{0.05*50.516}=2000 e^{2.5258}=2000*12.5≈25,000Wait, exactly 25,000.Wait, because we set C(t)=25,000, so at t≈50.516, C(t)=25,000.But let me compute it more accurately.C(t)=2000 e^{0.05t}=25,000So, e^{0.05t}=12.5So, 0.05t=ln(12.5)=2.5258So, t=2.5258/0.05≈50.516 hours.So, at t=50.516, C(t)=25,000 exactly.But we need to check if at t=50.516, M(t) and G(t) are ≥15,000 and 12,000 respectively.As computed, M(t)=50,825 ≥15,000G(t)=40,648.6 ≥12,000So, yes, at t≈50.516 hours, all resources are satisfied.But wait, let me check if there's a time before 50.516 where all resources are satisfied.Wait, no, because credits require t≈50.516, which is the latest. So, the earliest time when all resources are satisfied is t≈50.516 hours.But let me confirm if at t=50.516, M(t) and G(t) are indeed above their required amounts.As computed, M(t)=50,825, which is more than 15,000.G(t)=40,648.6, which is more than 12,000.So, yes, t≈50.516 hours is the earliest time when all resources are satisfied.But let me check if there's a time before 50.516 where all resources are satisfied.Wait, no, because credits are only satisfied at t≈50.516, so before that, credits are insufficient.Therefore, the exact time is t≈50.516 hours.But let me compute it more accurately.We had t=ln(12.5)/0.05=ln(12.5)/0.05≈2.5258/0.05≈50.516 hours.So, t≈50.516 hours.But let me compute it more precisely.ln(12.5)=2.525884798156828So, t=2.525884798156828 /0.05=50.51769596313656So, t≈50.5177 hours.So, approximately 50.5177 hours.But let me check if at t=50.5177, M(t) and G(t) are indeed above their thresholds.Compute M(t)=500 sin(π*50.5177/12) +1000*50.5177π*50.5177/12≈50.5177*0.2617993878≈13.23 radians13.23 radians is equivalent to 13.23 - 4π≈13.23 -12.566≈0.664 radianssin(0.664)≈0.618So, M(t)=500*0.618 +1000*50.5177≈309 +50,517.7≈50,826.7Which is above 15,000.Similarly, G(t)=300 cos(π*50.5177/12) +800*50.5177cos(0.664)≈0.786So, G(t)=300*0.786 +800*50.5177≈235.8 +40,414.16≈40,649.96Which is above 12,000.So, yes, at t≈50.5177 hours, all resources are satisfied.Therefore, the exact time is t≈50.5177 hours.But let me check if the problem expects an exact expression or a numerical value.The problem says "Use numerical methods if necessary to find the precise time."So, we can write it as t≈50.5177 hours, or more precisely, t=ln(12.5)/0.05≈50.5177 hours.Alternatively, since ln(12.5)=ln(25/2)=ln(25)-ln(2)=3.2189 -0.6931≈2.5258, so t≈2.5258/0.05≈50.516 hours.But for precision, let's use more decimal places.ln(12.5)=2.525884798156828So, t=2.525884798156828 /0.05=50.51769596313656So, t≈50.5177 hours.Alternatively, we can write it as t= (ln(12.5))/0.05, which is the exact expression.But since the problem says "use numerical methods if necessary," I think they expect a numerical value.So, t≈50.5177 hours.But let me check if I can write it as t= (ln(25/2))/0.05= (ln(25)-ln(2))/0.05= (3.2189 -0.6931)/0.05≈2.5258/0.05≈50.516 hours.So, t≈50.516 hours.But to be precise, t≈50.5177 hours.So, rounding to four decimal places, t≈50.5177 hours.Alternatively, to three decimal places, t≈50.518 hours.But let me check the problem statement again: "Determine the exact time t when Alex will have enough of each resource to build this unit. Use numerical methods if necessary to find the precise time."So, "exact time" might mean the exact expression, but since it's a transcendental equation, the exact time can't be expressed in a closed form, so we have to use numerical methods to find the precise time, which is approximately 50.5177 hours.Therefore, the answer is t≈50.5177 hours.But let me check if I can write it as t=ln(12.5)/0.05, which is exact, but it's not a numerical value.Alternatively, perhaps the problem expects the answer in terms of ln(12.5)/0.05, but since it's a numerical value, I think they expect the decimal.So, t≈50.5177 hours.But let me check if I can compute it more accurately.Using a calculator, ln(12.5)=2.525884798156828So, t=2.525884798156828 /0.05=50.51769596313656So, t≈50.5177 hours.Alternatively, t≈50.5177 hours.So, that's the answer.But let me check if I can write it as t= (ln(25/2))/0.05, which is exact.But since the problem says "use numerical methods if necessary," I think they expect a numerical value.So, t≈50.5177 hours.But let me check if I can write it as t=50.5177 hours.Alternatively, perhaps they expect it in hours and minutes.50.5177 hours is 50 hours and 0.5177*60≈31.06 minutes.So, approximately 50 hours and 31 minutes.But the problem doesn't specify the format, so probably decimal hours is fine.So, summarizing:Sub-problem 1:Minerals: 288,000Gas: 230,400Credits: 40,000(e^{1.2} -1)≈92,804.68Sub-problem 2:t≈50.5177 hoursBut let me check if I can write it as t= (ln(12.5))/0.05, which is exact.Alternatively, if they accept decimal, t≈50.5177 hours.But let me check if I can write it as t=50.5177 hours.Alternatively, perhaps they expect it in terms of ln(12.5)/0.05, but since it's a numerical value, I think decimal is fine.So, final answers:Sub-problem 1:Minerals: 288,000Gas: 230,400Credits: 40,000(e^{1.2} -1)≈92,804.68Sub-problem 2:t≈50.5177 hoursBut let me check if I can write it as t=50.5177 hours.Alternatively, perhaps they expect it in terms of ln(12.5)/0.05, but since it's a numerical value, I think decimal is fine.So, I think that's it.</think>

🧠:<think>Okay, so I have these two sub-problems to solve about the arithmetic mean and harmonic mean of movie lengths. Let me take them one at a time.Starting with Sub-problem 1: I need to show that the arithmetic mean (AM) is greater than or equal to the harmonic mean (HM). Hmm, I remember that there's an inequality called the AM-HM inequality which states exactly that. But I should probably prove it from scratch to make sure I understand it.First, let me recall the definitions. The arithmetic mean of a set of positive real numbers ( {L_1, L_2, ldots, L_n} ) is given by:[bar{L} = frac{1}{n} sum_{i=1}^{n} L_i]And the harmonic mean is:[H = frac{n}{sum_{i=1}^{n} frac{1}{L_i}}]I need to prove that ( bar{L} geq H ). I think the Cauchy-Schwarz inequality might be useful here. The Cauchy-Schwarz inequality states that for any real numbers ( a_i ) and ( b_i ):[left( sum_{i=1}^{n} a_i b_i right)^2 leq left( sum_{i=1}^{n} a_i^2 right) left( sum_{i=1}^{n} b_i^2 right)]Maybe I can apply this inequality to specific sequences ( a_i ) and ( b_i ) related to the movie lengths.Let me set ( a_i = sqrt{L_i} ) and ( b_i = frac{1}{sqrt{L_i}} ). Then, the left-hand side of Cauchy-Schwarz becomes:[left( sum_{i=1}^{n} sqrt{L_i} cdot frac{1}{sqrt{L_i}} right)^2 = left( sum_{i=1}^{n} 1 right)^2 = n^2]The right-hand side becomes:[left( sum_{i=1}^{n} L_i right) left( sum_{i=1}^{n} frac{1}{L_i} right)]Putting it together, we have:[n^2 leq left( sum_{i=1}^{n} L_i right) left( sum_{i=1}^{n} frac{1}{L_i} right)]Dividing both sides by ( n sum_{i=1}^{n} frac{1}{L_i} ), we get:[frac{n}{sum_{i=1}^{n} frac{1}{L_i}} leq frac{1}{n} sum_{i=1}^{n} L_i]Which simplifies to:[H leq bar{L}]So, that proves the first part. The arithmetic mean is indeed greater than or equal to the harmonic mean.Moving on to Sub-problem 2: Win Wu YanHong releases a new movie that's ( k ) minutes longer than the average length of his previous ( n ) movies. The new movie's length is ( L_{n+1} = bar{L} + k ). I need to find the new arithmetic mean ( bar{L}_{new} ) in terms of ( bar{L} ) and ( k ), and then find the value of ( k ) such that the new harmonic mean ( H_{new} ) equals the new arithmetic mean.First, let's find ( bar{L}_{new} ). The arithmetic mean is the total sum divided by the number of elements. The original total sum is ( n bar{L} ). Adding the new movie, the total sum becomes ( n bar{L} + L_{n+1} = n bar{L} + (bar{L} + k) ).So, the new total sum is ( n bar{L} + bar{L} + k = (n + 1) bar{L} + k ). Therefore, the new arithmetic mean is:[bar{L}_{new} = frac{(n + 1) bar{L} + k}{n + 1}]Simplifying that, we can write:[bar{L}_{new} = bar{L} + frac{k}{n + 1}]Okay, that's the expression for the new arithmetic mean.Now, I need to find ( k ) such that the new harmonic mean ( H_{new} ) equals ( bar{L}_{new} ). Let's recall the formula for the harmonic mean:[H_{new} = frac{n + 1}{sum_{i=1}^{n+1} frac{1}{L_i}}]We know that the original harmonic mean is ( H = frac{n}{sum_{i=1}^{n} frac{1}{L_i}} ). Let me denote ( S = sum_{i=1}^{n} frac{1}{L_i} ), so ( H = frac{n}{S} ).Then, the new harmonic mean is:[H_{new} = frac{n + 1}{S + frac{1}{L_{n+1}}} = frac{n + 1}{S + frac{1}{bar{L} + k}}]We want this to be equal to ( bar{L}_{new} ), which is ( bar{L} + frac{k}{n + 1} ). So, setting them equal:[frac{n + 1}{S + frac{1}{bar{L} + k}} = bar{L} + frac{k}{n + 1}]Let me substitute ( S = frac{n}{H} ), but since we don't know ( H ), maybe it's better to express everything in terms of ( bar{L} ) and ( k ). Alternatively, perhaps express ( S ) in terms of ( bar{L} ) and ( H ). Wait, from the first sub-problem, we know that ( bar{L} geq H ), but I don't know if that helps directly.Alternatively, maybe express ( S ) in terms of the original arithmetic mean. Let me think.Wait, actually, ( S = sum_{i=1}^{n} frac{1}{L_i} ). We can express this in terms of ( H ) since ( H = frac{n}{S} ), so ( S = frac{n}{H} ). But since ( H leq bar{L} ), ( S geq frac{n}{bar{L}} ). Hmm, not sure if that helps.Alternatively, perhaps express ( S ) in terms of ( bar{L} ) and the individual ( L_i ). But without knowing the specific values, it's tricky.Wait, maybe it's better to express everything in terms of ( bar{L} ) and ( k ). Let me denote ( bar{L} ) as ( mu ) for simplicity.So, ( mu = bar{L} ), and ( L_{n+1} = mu + k ). The new arithmetic mean is ( mu_{new} = mu + frac{k}{n + 1} ).The harmonic mean is:[H_{new} = frac{n + 1}{S + frac{1}{mu + k}} = mu + frac{k}{n + 1}]So, we have:[frac{n + 1}{S + frac{1}{mu + k}} = mu + frac{k}{n + 1}]Let me solve for ( k ). First, let's denote ( S = sum_{i=1}^{n} frac{1}{L_i} ). From the original harmonic mean, ( H = frac{n}{S} ), so ( S = frac{n}{H} ). But since ( H leq mu ), ( S geq frac{n}{mu} ).But without knowing ( H ), maybe I can express ( S ) in terms of ( mu ) and the individual ( L_i ). Wait, perhaps I can relate ( S ) to ( mu ) using the Cauchy-Schwarz inequality or something else.Wait, another approach: Let's consider the relationship between the harmonic mean and the arithmetic mean. Since ( H leq mu ), and ( H_{new} = mu_{new} ), which is slightly larger than ( mu ) if ( k > 0 ).But maybe I can express ( S ) in terms of ( mu ) and the individual ( L_i ). Let me think about the sum ( S = sum_{i=1}^{n} frac{1}{L_i} ). If all ( L_i ) were equal to ( mu ), then ( S = frac{n}{mu} ). But since they are not necessarily equal, ( S ) could be larger or smaller? Wait, no. If all ( L_i ) are equal, then ( S = frac{n}{mu} ). If some ( L_i ) are larger and some smaller, then ( S ) would be larger because the reciprocals would be smaller for larger ( L_i ) and larger for smaller ( L_i ). Wait, actually, if some ( L_i ) are smaller, their reciprocals are larger, so ( S ) would be larger than ( frac{n}{mu} ). Therefore, ( S geq frac{n}{mu} ).But I'm not sure if that helps directly. Maybe I need to express ( S ) in terms of ( mu ) and some other variable.Alternatively, perhaps I can write the equation:[frac{n + 1}{S + frac{1}{mu + k}} = mu + frac{k}{n + 1}]Let me cross-multiply:[(n + 1) = left( S + frac{1}{mu + k} right) left( mu + frac{k}{n + 1} right)]Expanding the right-hand side:[(n + 1) = S mu + S cdot frac{k}{n + 1} + frac{mu}{mu + k} + frac{k}{(n + 1)(mu + k)}]This seems complicated. Maybe I can rearrange terms or find a substitution.Wait, let me denote ( k = t mu ), so ( k ) is a fraction of ( mu ). Maybe that can simplify the equation.Let ( k = t mu ), where ( t ) is a scalar. Then, ( L_{n+1} = mu + t mu = mu(1 + t) ).The new arithmetic mean becomes:[mu_{new} = mu + frac{t mu}{n + 1} = mu left( 1 + frac{t}{n + 1} right)]The harmonic mean equation becomes:[frac{n + 1}{S + frac{1}{mu(1 + t)}} = mu left( 1 + frac{t}{n + 1} right)]Let me substitute ( S = frac{n}{H} ), but since ( H leq mu ), ( S geq frac{n}{mu} ). Let me express ( S ) as ( frac{n}{mu} + d ), where ( d geq 0 ). Wait, that might not be helpful.Alternatively, let me express ( S ) in terms of ( mu ). From the original harmonic mean, ( H = frac{n}{S} leq mu ), so ( S geq frac{n}{mu} ). Let me denote ( S = frac{n}{mu} + c ), where ( c geq 0 ). Then, substituting into the equation:[frac{n + 1}{frac{n}{mu} + c + frac{1}{mu(1 + t)}} = mu left( 1 + frac{t}{n + 1} right)]Simplify the denominator:[frac{n}{mu} + c + frac{1}{mu(1 + t)} = frac{n(1 + t) + 1}{mu(1 + t)} + c]Wait, that might not be the best approach. Let me instead try to express everything in terms of ( mu ) and ( t ).Let me rewrite the equation:[frac{n + 1}{S + frac{1}{mu(1 + t)}} = mu left( 1 + frac{t}{n + 1} right)]Taking reciprocal on both sides:[frac{S + frac{1}{mu(1 + t)}}{n + 1} = frac{1}{mu left( 1 + frac{t}{n + 1} right)}]Multiply both sides by ( n + 1 ):[S + frac{1}{mu(1 + t)} = frac{n + 1}{mu left( 1 + frac{t}{n + 1} right)}]Simplify the right-hand side:[frac{n + 1}{mu left( frac{n + 1 + t}{n + 1} right)} = frac{(n + 1)^2}{mu(n + 1 + t)}]So, the equation becomes:[S + frac{1}{mu(1 + t)} = frac{(n + 1)^2}{mu(n + 1 + t)}]Let me isolate ( S ):[S = frac{(n + 1)^2}{mu(n + 1 + t)} - frac{1}{mu(1 + t)}]Factor out ( frac{1}{mu} ):[S = frac{1}{mu} left( frac{(n + 1)^2}{n + 1 + t} - frac{1}{1 + t} right)]Let me combine the terms inside the parentheses:Find a common denominator for the two fractions, which is ( (n + 1 + t)(1 + t) ).So,[frac{(n + 1)^2 (1 + t) - (n + 1 + t)}{(n + 1 + t)(1 + t)}]Expanding the numerator:[(n + 1)^2 (1 + t) - (n + 1 + t) = (n + 1)^2 + (n + 1)^2 t - n - 1 - t]Simplify term by term:First term: ( (n + 1)^2 = n^2 + 2n + 1 )Second term: ( (n + 1)^2 t = (n^2 + 2n + 1) t )Third term: ( -n - 1 )Fourth term: ( -t )So, combining:[n^2 + 2n + 1 + (n^2 + 2n + 1) t - n - 1 - t]Simplify the constants:( n^2 + 2n + 1 - n - 1 = n^2 + n )Simplify the terms with ( t ):( (n^2 + 2n + 1) t - t = (n^2 + 2n + 1 - 1) t = (n^2 + 2n) t )So, the numerator becomes:[n^2 + n + (n^2 + 2n) t]Factor out ( n ):[n(n + 1) + n(n + 2) t]So, putting it back into the expression for ( S ):[S = frac{1}{mu} cdot frac{n(n + 1) + n(n + 2) t}{(n + 1 + t)(1 + t)}]Factor ( n ) in the numerator:[S = frac{n}{mu} cdot frac{(n + 1) + (n + 2) t}{(n + 1 + t)(1 + t)}]But from earlier, we know that ( S = sum_{i=1}^{n} frac{1}{L_i} ). However, without knowing the specific values of ( L_i ), it's hard to relate this further. Maybe I need to make an assumption or find a way to express ( S ) in terms of ( mu ).Wait, perhaps I can relate ( S ) to ( mu ) using the Cauchy-Schwarz inequality again. From the first sub-problem, we have:[left( sum_{i=1}^{n} L_i right) left( sum_{i=1}^{n} frac{1}{L_i} right) geq n^2]Which is:[n mu cdot S geq n^2 implies mu S geq n implies S geq frac{n}{mu}]So, ( S geq frac{n}{mu} ). Therefore, the expression we have for ( S ) must satisfy:[frac{n}{mu} leq frac{n}{mu} cdot frac{(n + 1) + (n + 2) t}{(n + 1 + t)(1 + t)}]Wait, that seems a bit circular. Let me see.From the equation:[S = frac{n}{mu} cdot frac{(n + 1) + (n + 2) t}{(n + 1 + t)(1 + t)}]But since ( S geq frac{n}{mu} ), we have:[frac{n}{mu} cdot frac{(n + 1) + (n + 2) t}{(n + 1 + t)(1 + t)} geq frac{n}{mu}]Divide both sides by ( frac{n}{mu} ):[frac{(n + 1) + (n + 2) t}{(n + 1 + t)(1 + t)} geq 1]Multiply both sides by the denominator (which is positive since all terms are positive):[(n + 1) + (n + 2) t geq (n + 1 + t)(1 + t)]Expand the right-hand side:[(n + 1)(1) + (n + 1)t + t(1) + t^2 = (n + 1) + (n + 1)t + t + t^2 = (n + 1) + (n + 2)t + t^2]So, the inequality becomes:[(n + 1) + (n + 2) t geq (n + 1) + (n + 2) t + t^2]Subtract ( (n + 1) + (n + 2) t ) from both sides:[0 geq t^2]Which implies ( t^2 leq 0 ). Since ( t^2 ) is always non-negative, this means ( t^2 = 0 ), so ( t = 0 ).Wait, that suggests that the only solution is ( t = 0 ), which would mean ( k = 0 ). But that can't be right because if ( k = 0 ), the new movie is exactly the average length, so the harmonic mean would remain the same as the arithmetic mean? Wait, no, actually, adding a movie equal to the average length might change the harmonic mean.Wait, let me test this with ( k = 0 ). If ( k = 0 ), then ( L_{n+1} = mu ). The new arithmetic mean is ( mu_{new} = frac{n mu + mu}{n + 1} = mu ). The harmonic mean becomes:[H_{new} = frac{n + 1}{S + frac{1}{mu}}]But ( S = sum_{i=1}^{n} frac{1}{L_i} ). If all ( L_i = mu ), then ( S = frac{n}{mu} ), so ( H_{new} = frac{n + 1}{frac{n}{mu} + frac{1}{mu}} = frac{n + 1}{frac{n + 1}{mu}} = mu ). So, in that case, ( H_{new} = mu = mu_{new} ).But in our case, the original harmonic mean ( H ) is less than or equal to ( mu ). So, unless all ( L_i = mu ), ( H < mu ). Therefore, if we add a movie equal to ( mu ), the new harmonic mean becomes ( mu ), which equals the new arithmetic mean. So, in that case, ( k = 0 ) is a solution.But the problem says "the new movie is ( k ) minutes longer than the average length of his previous ( n ) movies", so ( k ) is positive. So, in that case, ( k = 0 ) is not a valid solution because the new movie is exactly the average, not longer.Wait, but from our earlier derivation, the only solution is ( t = 0 ), which implies ( k = 0 ). So, does that mean that the only way for the new harmonic mean to equal the new arithmetic mean is if ( k = 0 )? That seems counterintuitive because if we add a movie longer than the average, maybe the harmonic mean could increase to meet the arithmetic mean.Wait, but in reality, adding a longer movie would make the harmonic mean increase, but since the harmonic mean is always less than or equal to the arithmetic mean, the only way they can be equal is if all the terms are equal. So, if we add a movie equal to the average, then all terms become equal, making both means equal. If we add a longer movie, the harmonic mean would still be less than the arithmetic mean because the reciprocals would be smaller.Wait, let me test with an example. Suppose we have two movies with lengths 1 and 1. So, ( n = 2 ), ( mu = 1 ), ( H = 1 ). Now, add a new movie that's ( k ) minutes longer than the average, so ( L_3 = 1 + k ).New arithmetic mean: ( mu_{new} = frac{1 + 1 + (1 + k)}{3} = frac{3 + k}{3} = 1 + frac{k}{3} ).New harmonic mean: ( H_{new} = frac{3}{1 + 1 + frac{1}{1 + k}} = frac{3}{2 + frac{1}{1 + k}} ).Set ( H_{new} = mu_{new} ):[frac{3}{2 + frac{1}{1 + k}} = 1 + frac{k}{3}]Multiply both sides by denominator:[3 = left(1 + frac{k}{3}right) left(2 + frac{1}{1 + k}right)]Expand the right-hand side:[3 = 2 left(1 + frac{k}{3}right) + frac{1}{1 + k} left(1 + frac{k}{3}right)]Simplify:[3 = 2 + frac{2k}{3} + frac{1}{1 + k} + frac{k}{3(1 + k)}]Combine like terms:[3 = 2 + frac{2k}{3} + frac{1}{1 + k} left(1 + frac{k}{3}right)]Simplify the fraction:[frac{1}{1 + k} left(1 + frac{k}{3}right) = frac{1 + frac{k}{3}}{1 + k} = frac{3 + k}{3(1 + k)}]So, the equation becomes:[3 = 2 + frac{2k}{3} + frac{3 + k}{3(1 + k)}]Subtract 2 from both sides:[1 = frac{2k}{3} + frac{3 + k}{3(1 + k)}]Multiply both sides by 3(1 + k) to eliminate denominators:[3(1 + k) = 2k(1 + k) + (3 + k)]Expand:Left-hand side: ( 3 + 3k )Right-hand side: ( 2k + 2k^2 + 3 + k = 2k^2 + 3k + 3 )So, equation:[3 + 3k = 2k^2 + 3k + 3]Subtract ( 3 + 3k ) from both sides:[0 = 2k^2]Which implies ( k = 0 ). So, again, the only solution is ( k = 0 ). Therefore, in this specific case, the only way for the harmonic mean to equal the arithmetic mean after adding a new movie is if ( k = 0 ).This suggests that in general, the only solution is ( k = 0 ). Therefore, the value of ( k ) that makes ( H_{new} = bar{L}_{new} ) is ( k = 0 ).But wait, in the problem statement, it says "the new movie is ( k ) minutes longer than the average length of his previous ( n ) movies". So, ( k ) is positive. Therefore, there is no positive ( k ) that satisfies ( H_{new} = bar{L}_{new} ). The only solution is ( k = 0 ), which is not longer, just equal.Therefore, perhaps the answer is that no such positive ( k ) exists, except ( k = 0 ). But let me think again.Wait, in the general case, when we added the new movie, we derived that ( t = 0 ), which implies ( k = 0 ). So, regardless of ( n ), the only solution is ( k = 0 ). Therefore, the value of ( k ) is 0.But the problem says "the new movie is ( k ) minutes longer than the average length", implying ( k > 0 ). So, maybe there's no solution for ( k > 0 ). Alternatively, perhaps I made a mistake in the derivation.Wait, let me go back to the general case. We had:[frac{n + 1}{S + frac{1}{mu + k}} = mu + frac{k}{n + 1}]And after some manipulations, we arrived at ( t = 0 ), implying ( k = 0 ). So, unless ( S = frac{n}{mu} ), which would mean all ( L_i = mu ), then adding another ( mu ) would keep both means equal. But if the original set doesn't have all ( L_i = mu ), then ( S > frac{n}{mu} ), and adding a longer movie would not make ( H_{new} = mu_{new} ).Therefore, the only way for ( H_{new} = mu_{new} ) is if all the original movies are equal to ( mu ), and the new movie is also equal to ( mu ), meaning ( k = 0 ).So, in conclusion, the value of ( k ) is 0.But let me check with another example. Suppose we have three movies: 1, 1, and 1. So, ( n = 3 ), ( mu = 1 ), ( H = 1 ). Add a new movie ( L_4 = 1 + k ).New arithmetic mean: ( mu_{new} = frac{3 + 1 + k}{4} = 1 + frac{k}{4} ).New harmonic mean: ( H_{new} = frac{4}{1 + 1 + 1 + frac{1}{1 + k}} = frac{4}{3 + frac{1}{1 + k}} ).Set equal:[frac{4}{3 + frac{1}{1 + k}} = 1 + frac{k}{4}]Multiply both sides by denominator:[4 = left(1 + frac{k}{4}right) left(3 + frac{1}{1 + k}right)]Expand:[4 = 3 left(1 + frac{k}{4}right) + frac{1}{1 + k} left(1 + frac{k}{4}right)]Simplify:[4 = 3 + frac{3k}{4} + frac{1}{1 + k} + frac{k}{4(1 + k)}]Combine like terms:[4 = 3 + frac{3k}{4} + frac{1}{1 + k} left(1 + frac{k}{4}right)]Simplify the fraction:[frac{1}{1 + k} left(1 + frac{k}{4}right) = frac{4 + k}{4(1 + k)}]So, the equation becomes:[4 = 3 + frac{3k}{4} + frac{4 + k}{4(1 + k)}]Subtract 3 from both sides:[1 = frac{3k}{4} + frac{4 + k}{4(1 + k)}]Multiply both sides by 4(1 + k):[4(1 + k) = 3k(1 + k) + (4 + k)]Expand:Left-hand side: ( 4 + 4k )Right-hand side: ( 3k + 3k^2 + 4 + k = 3k^2 + 4k + 4 )So, equation:[4 + 4k = 3k^2 + 4k + 4]Subtract ( 4 + 4k ) from both sides:[0 = 3k^2]Which implies ( k = 0 ). Again, the only solution is ( k = 0 ).Therefore, it seems that regardless of ( n ), the only solution is ( k = 0 ). So, the value of ( k ) is 0.But wait, the problem says "the new movie is ( k ) minutes longer than the average length of his previous ( n ) movies". So, ( k ) is positive, but the only solution is ( k = 0 ). Therefore, there is no positive ( k ) that satisfies the condition. Hence, the answer is ( k = 0 ).But let me think again. Maybe I made a mistake in assuming ( S ) is related to ( mu ) in a certain way. Alternatively, perhaps the problem expects ( k ) in terms of ( n ) and ( mu ), but from the derivation, it seems ( k ) must be zero.Alternatively, maybe I need to consider that when adding a new term, the harmonic mean can sometimes increase enough to meet the arithmetic mean. But from the examples, it seems that it's only possible when ( k = 0 ).Therefore, the conclusion is that ( k = 0 ).</think>