🧠:<think>Okay, so I have this problem about a coffee shop owner using an AI system to predict demand and optimize pricing. There are two parts to the problem. Let me try to tackle them step by step.Starting with part 1: The demand function is given as D(t) = 50 + 30 sin(π/12 t), where t is the time in hours from 6 AM (t=0) to 6 PM (t=12). I need to calculate the total predicted demand over the 12-hour period and determine the time intervals when demand exceeds 70 cups per hour.First, total demand over the 12-hour period. Since demand is a continuous function, I think I need to integrate D(t) from t=0 to t=12. That should give me the total number of cups demanded over the day.So, the integral of D(t) dt from 0 to 12. Let me write that down:Total Demand = ∫₀¹² [50 + 30 sin(π/12 t)] dtI can split this integral into two parts: the integral of 50 dt and the integral of 30 sin(π/12 t) dt.Calculating the first part: ∫₀¹² 50 dt = 50t evaluated from 0 to 12. That would be 50*12 - 50*0 = 600.Now the second part: ∫₀¹² 30 sin(π/12 t) dtLet me make a substitution to solve this integral. Let u = π/12 t, so du/dt = π/12, which means dt = (12/π) du.Changing the limits: when t=0, u=0; when t=12, u=π.So the integral becomes 30 ∫₀^π sin(u) * (12/π) duWhich is (30 * 12 / π) ∫₀^π sin(u) duCalculating the integral of sin(u) is -cos(u), so:(360 / π) [ -cos(u) ] from 0 to π= (360 / π) [ -cos(π) + cos(0) ]We know that cos(π) = -1 and cos(0) = 1, so:= (360 / π) [ -(-1) + 1 ] = (360 / π) [1 + 1] = (360 / π) * 2 = 720 / πSo the second integral is 720 / π.Therefore, the total demand is 600 + 720 / π.Let me compute that numerically. π is approximately 3.1416, so 720 / π ≈ 720 / 3.1416 ≈ 229.183.So total demand ≈ 600 + 229.183 ≈ 829.183 cups.Hmm, so approximately 829.18 cups over 12 hours.Wait, but is that correct? Let me double-check my substitution.Original integral: ∫₀¹² 30 sin(π/12 t) dtLet u = π/12 t, so du = π/12 dt, so dt = (12/π) du.So, 30 * (12/π) ∫₀^π sin(u) duWhich is (360 / π) ∫₀^π sin(u) duYes, that's correct. And the integral of sin(u) from 0 to π is 2, so 360/π * 2 = 720/π. So that part is correct.So total demand is 600 + 720/π ≈ 600 + 229.183 ≈ 829.183. So approximately 829.18 cups.Okay, moving on to the second part of question 1: Determine the time intervals during which demand exceeds 70 cups per hour.So, we need to solve D(t) > 70.Given D(t) = 50 + 30 sin(π/12 t) > 70Subtract 50: 30 sin(π/12 t) > 20Divide both sides by 30: sin(π/12 t) > 20/30 = 2/3 ≈ 0.6667So, sin(θ) > 2/3, where θ = π/12 t.We need to find all t in [0,12] such that sin(θ) > 2/3.First, let's find the values of θ where sin(θ) = 2/3.The general solution for sin(θ) = k is θ = arcsin(k) + 2πn and θ = π - arcsin(k) + 2πn, for integer n.So, θ = arcsin(2/3) and θ = π - arcsin(2/3).Compute arcsin(2/3). Let me calculate that.arcsin(2/3) is approximately, since sin(π/6)=0.5, sin(π/4)=√2/2≈0.707, so 2/3≈0.6667 is between π/6 and π/4.Using a calculator, arcsin(2/3) ≈ 0.7297 radians.So, θ ≈ 0.7297 and θ ≈ π - 0.7297 ≈ 2.4119 radians.So, sin(θ) > 2/3 when θ is in (0.7297, 2.4119) + 2πn.But since θ = π/12 t and t is between 0 and 12, θ ranges from 0 to π.So, in the interval θ ∈ [0, π], sin(θ) > 2/3 occurs when θ ∈ (0.7297, 2.4119).Therefore, θ ∈ (0.7297, 2.4119) corresponds to t ∈ (θ * 12 / π, θ * 12 / π).Wait, hold on. θ = π/12 t, so t = (12 / π) θ.So, when θ is between 0.7297 and 2.4119, t is between:Lower bound: (12 / π) * 0.7297 ≈ (12 / 3.1416) * 0.7297 ≈ (3.8197) * 0.7297 ≈ 2.785 hoursUpper bound: (12 / π) * 2.4119 ≈ 3.8197 * 2.4119 ≈ 9.216 hoursSo, t is between approximately 2.785 hours and 9.216 hours.But t is measured from 6 AM. So, 2.785 hours after 6 AM is approximately 6 AM + 2 hours 47 minutes, which is around 8:47 AM.Similarly, 9.216 hours after 6 AM is approximately 6 AM + 9 hours 13 minutes, which is around 3:13 PM.Therefore, the demand exceeds 70 cups per hour from approximately 8:47 AM to 3:13 PM.Wait, let me verify that.Wait, 2.785 hours is 2 hours and 0.785*60 minutes ≈ 47 minutes. So, 6 AM + 2h47m = 8:47 AM.Similarly, 9.216 hours is 9 hours and 0.216*60 ≈ 13 minutes, so 6 AM + 9h13m = 3:13 PM.Yes, that seems correct.But let me check if the sine function is above 2/3 in that interval.Since sin(theta) is increasing from 0 to pi/2 and decreasing from pi/2 to pi.So, between theta = arcsin(2/3) ≈ 0.7297 and theta = pi - arcsin(2/3) ≈ 2.4119, sin(theta) is above 2/3.Therefore, the corresponding t is between 2.785 and 9.216 hours.So, the time intervals when demand exceeds 70 cups per hour are approximately from 8:47 AM to 3:13 PM.Wait, but let me think again. The sine function is symmetric around pi/2, so the duration when it's above 2/3 is symmetric around pi/2.But in terms of t, it's symmetric around t=6 hours, which is 12 PM.Wait, let me compute the exact times.Compute t1 = (12 / π) * arcsin(2/3) ≈ (12 / 3.1416) * 0.7297 ≈ 3.8197 * 0.7297 ≈ 2.785 hours.Similarly, t2 = (12 / π) * (π - arcsin(2/3)) ≈ (12 / 3.1416) * (3.1416 - 0.7297) ≈ 3.8197 * 2.4119 ≈ 9.216 hours.So, t1 ≈ 2.785 hours, which is 2 hours and 47 minutes, and t2 ≈ 9.216 hours, which is 9 hours and 13 minutes.Therefore, the demand exceeds 70 cups per hour from approximately 8:47 AM to 3:13 PM.Wait, but let me check if the sine function is indeed above 2/3 in that interval.At t=0, D(t)=50 + 30 sin(0)=50. At t=6, D(t)=50 + 30 sin(pi/2)=80. At t=12, D(t)=50 + 30 sin(pi)=50.So, the demand peaks at 80 cups at 12 PM, which is t=6.So, the demand goes up from 50 at 6 AM to 80 at 12 PM, then back down to 50 at 6 PM.So, the demand crosses 70 on the way up and on the way down.So, the times when D(t)=70 are t1 and t2, which are symmetric around t=6.So, the duration when D(t) >70 is from t1 to t2, which is approximately 8:47 AM to 3:13 PM.So, that seems correct.So, summarizing part 1: total demand is approximately 829.18 cups, and the demand exceeds 70 cups per hour from approximately 8:47 AM to 3:13 PM.Moving on to part 2: The AI suggests a dynamic pricing model where P(t) = k / D(t), and the owner wants the average price over 12 hours to be 3. We need to find k.So, average price over 12 hours is given by (1/12) ∫₀¹² P(t) dt = 3.So, (1/12) ∫₀¹² [k / D(t)] dt = 3.Therefore, ∫₀¹² [k / D(t)] dt = 36.So, k ∫₀¹² [1 / D(t)] dt = 36.Therefore, k = 36 / ∫₀¹² [1 / D(t)] dt.So, I need to compute ∫₀¹² [1 / (50 + 30 sin(π/12 t))] dt.Hmm, that integral might be a bit tricky. Let me think about how to approach it.The integral of 1 / (A + B sin(theta)) d(theta) is a standard integral, but I need to recall the formula.I remember that ∫ 1 / (A + B sin(theta)) d(theta) can be solved using the Weierstrass substitution or other methods.Alternatively, there is a formula:∫ 1 / (A + B sin(theta)) d(theta) = (2 / sqrt(A² - B²)) arctan( tan(theta/2) + sqrt((A - B)/(A + B)) ) + C, when A > B.In our case, A=50, B=30, so A > B, so we can use this formula.But let me verify.Wait, the integral is ∫ [1 / (50 + 30 sin(π/12 t))] dt from 0 to 12.Let me make a substitution to make it easier.Let me set theta = π/12 t, so d(theta) = π/12 dt, so dt = (12/π) d(theta).When t=0, theta=0; when t=12, theta=pi.So, the integral becomes:∫₀^π [1 / (50 + 30 sin(theta))] * (12/π) d(theta)= (12 / π) ∫₀^π [1 / (50 + 30 sin(theta))] d(theta)So, now I need to compute ∫₀^π [1 / (50 + 30 sin(theta))] d(theta).Using the standard integral formula:∫₀^π [1 / (A + B sin(theta))] d(theta) = π / sqrt(A² - B²)Wait, is that correct?Wait, let me recall.The integral over 0 to pi of 1 / (A + B sin(theta)) d(theta) is equal to pi / sqrt(A² - B²), provided that A > |B|.Yes, that seems familiar.Let me check with A=50, B=30.So, sqrt(A² - B²) = sqrt(2500 - 900) = sqrt(1600) = 40.Therefore, ∫₀^π [1 / (50 + 30 sin(theta))] d(theta) = pi / 40.Therefore, the integral becomes (12 / pi) * (pi / 40) = 12 / 40 = 3/10.Wait, that seems too straightforward.Wait, let me double-check.Yes, the formula is ∫₀^π [1 / (A + B sin(theta))] d(theta) = pi / sqrt(A² - B²).So, with A=50, B=30, sqrt(50² - 30²)=sqrt(2500-900)=sqrt(1600)=40.So, the integral is pi / 40.Therefore, ∫₀¹² [1 / D(t)] dt = (12 / pi) * (pi / 40) = 12 / 40 = 3/10.So, ∫₀¹² [1 / D(t)] dt = 3/10.Therefore, k = 36 / (3/10) = 36 * (10/3) = 120.So, k=120.Wait, let me confirm that.Given that the average price is 3 dollars, so:Average price = (1/12) ∫₀¹² P(t) dt = 3.Which implies ∫₀¹² P(t) dt = 36.But P(t) = k / D(t), so ∫₀¹² [k / D(t)] dt = 36.We found that ∫₀¹² [1 / D(t)] dt = 3/10.Therefore, k * (3/10) = 36.So, k = 36 / (3/10) = 36 * (10/3) = 120.Yes, that seems correct.Therefore, k=120.So, summarizing part 2: The value of k is 120.Let me just recap:1. Total demand is approximately 829.18 cups, and demand exceeds 70 cups per hour from approximately 8:47 AM to 3:13 PM.2. The constant k is 120.I think that's all.Final Answer1. The total predicted demand is boxed{829.18} cups, and demand exceeds 70 cups per hour from approximately boxed{8:47 text{AM}} to boxed{3:13 text{PM}}.2. The value of ( k ) is boxed{120}.</think>

🧠:<think>Alright, so I've got these two problems about Dilworth, Oklahoma. Let me try to tackle them one by one. I'm a bit nervous because calculus can be tricky, but I'll take it step by step.Starting with the first problem about the population. The function given is ( P(t) = 500 cdot e^{0.02t} ). I need to find the population in 1951 and the year when the population reaches 2000.First, let's figure out what ( t ) is for the year 1951. Since the model starts in 1901, ( t ) is the number of years since 1901. So, 1951 minus 1901 is 50 years. Therefore, ( t = 50 ).Plugging that into the population function: ( P(50) = 500 cdot e^{0.02 times 50} ). Let me compute the exponent first: 0.02 times 50 is 1. So, it's ( 500 cdot e^{1} ). I remember that ( e ) is approximately 2.71828. So, multiplying that by 500: 500 * 2.71828. Let me do that calculation.500 * 2 is 1000, 500 * 0.71828 is approximately 500 * 0.7 = 350, and 500 * 0.01828 is about 9.14. So adding those together: 1000 + 350 + 9.14 = 1359.14. So, approximately 1359 people in 1951. Hmm, that seems reasonable.Now, the second part is to find the year when the population reaches 2000. So, we need to solve ( 500 cdot e^{0.02t} = 2000 ).Let me write that equation down:( 500 cdot e^{0.02t} = 2000 )First, divide both sides by 500:( e^{0.02t} = 4 )Now, take the natural logarithm of both sides to solve for ( t ):( ln(e^{0.02t}) = ln(4) )Simplifying the left side:( 0.02t = ln(4) )So, ( t = frac{ln(4)}{0.02} )Calculating ( ln(4) ). I remember that ( ln(4) ) is approximately 1.386294.So, ( t = frac{1.386294}{0.02} ). Let me compute that: 1.386294 divided by 0.02 is the same as multiplying by 50, so 1.386294 * 50 = 69.3147.So, ( t ) is approximately 69.3147 years. Since ( t ) is the number of years since 1901, adding that to 1901 gives the year. 1901 + 69 is 1970, and 0.3147 of a year is roughly 0.3147 * 365 ≈ 115 days. So, approximately mid-May 1970. But since we're talking about population, it's probably rounded to the nearest year, so 1970.Wait, let me double-check my calculations. 0.02t = ln(4). So, t = ln(4)/0.02. Yes, that's correct. And ln(4) is about 1.386, so 1.386 / 0.02 is indeed 69.3. So, 1901 + 69 is 1970, as I thought. So, the population reaches 2000 around 1970.Alright, moving on to the second problem about the number of historical sites. The function is given as ( H(t) = int_{1901}^{t} 2 + cos(pi x / 50) , dx ). I need to find the number of historical sites in 1951 and the rate of increase in 2001.First, let's understand what ( H(t) ) represents. It's the integral from 1901 to ( t ) of the function ( 2 + cos(pi x / 50) ) with respect to ( x ). So, ( H(t) ) is the accumulation of historical sites over time, starting from 1901.To find the number of historical sites in 1951, I need to compute ( H(1951) ). But wait, the integral is from 1901 to ( t ), so ( t ) is the year. But in calculus, when we have integrals with variable limits, it's often expressed in terms of a variable, say ( t ), which is the number of years since 1901. Wait, hold on, the problem says ( H(t) ) is the number of historical sites at time ( t ) years since 1901. So, actually, ( t ) is the number of years since 1901, not the year itself.Wait, that's a crucial point. So, in the function ( H(t) = int_{1901}^{t} 2 + cos(pi x / 50) , dx ), is ( t ) the year or the number of years since 1901? The problem says ( t ) is the time in years since 1901. So, actually, the integral should be from 1901 to ( 1901 + t ). Hmm, that might complicate things.Wait, let me read the problem again: "the number of historical sites in Dilworth has been modeled by the function ( H(t) = int_{1901}^{t} 2 + cos(pi x / 50) , dx ), where ( H(t) ) is the number of historical sites at time ( t ) years since 1901."Hmm, that seems contradictory because if ( t ) is years since 1901, then the integral should be from 0 to ( t ), not from 1901 to ( t ). Maybe it's a typo or misinterpretation. Alternatively, perhaps ( x ) is the year, so the integral is over the years from 1901 to ( t ), where ( t ) is the year. But then ( t ) is the year, not the number of years since 1901.Wait, the problem says ( t ) is the time in years since 1901. So, ( t ) is 0 in 1901, 1 in 1902, etc. So, the integral is from 1901 to ( 1901 + t ). Hmm, that would make sense because ( t ) is the number of years since 1901. So, the upper limit is 1901 + t.But in the function, it's written as ( int_{1901}^{t} ). So, if ( t ) is in years since 1901, then the upper limit is 1901 + t. So, perhaps the integral should be ( int_{1901}^{1901 + t} ). Maybe that's a misprint in the problem.Alternatively, maybe ( x ) is the number of years since 1901, so ( x ) goes from 0 to ( t ). Let me think.Wait, the integrand is ( 2 + cos(pi x / 50) ). If ( x ) is the year, then ( pi x / 50 ) would have units of year, which doesn't make much sense inside a cosine function, because cosine expects a dimensionless argument. So, that suggests that ( x ) is the number of years since 1901, so that ( pi x / 50 ) is dimensionless.Therefore, the integral should be from 0 to ( t ), where ( t ) is the number of years since 1901. So, perhaps the problem has a typo, and it should be ( int_{0}^{t} ) instead of ( int_{1901}^{t} ). Alternatively, maybe the integral is expressed in terms of the year, so ( x ) is the year, and ( t ) is the year. But then ( t ) would be the year, not the number of years since 1901.This is a bit confusing. Let me try to clarify.Given that ( H(t) ) is the number of historical sites at time ( t ) years since 1901, so ( t = 0 ) corresponds to 1901, ( t = 50 ) corresponds to 1951, etc.Therefore, the integral should be from 0 to ( t ), because we're integrating over the number of years since 1901.But the problem says ( int_{1901}^{t} ). That suggests that ( t ) is the year, not the number of years since 1901. So, perhaps the problem is using ( t ) as the year, not the number of years since 1901. That would make more sense with the integral limits.Wait, let me check the problem statement again:"The number of historical sites in Dilworth has been modeled by the function ( H(t) = int_{1901}^{t} 2 + cos(pi x / 50) , dx ), where ( H(t) ) is the number of historical sites at time ( t ) years since 1901."Hmm, so ( t ) is the time in years since 1901, so ( t ) is 0 in 1901, 1 in 1902, etc. Therefore, the integral is from 1901 to ( t ), but ( t ) is the number of years since 1901. That doesn't make sense because 1901 is a year, and ( t ) is a number of years. So, the integral limits are inconsistent.Alternatively, maybe the integral is supposed to be from 0 to ( t ), where ( t ) is the number of years since 1901, and ( x ) is also the number of years since 1901. So, the integrand is ( 2 + cos(pi x / 50) ), which would make sense because ( x ) is in years, so ( pi x / 50 ) is dimensionless.Therefore, perhaps the integral should be ( int_{0}^{t} 2 + cos(pi x / 50) , dx ). Maybe the problem had a typo, and it's supposed to be from 0 to ( t ). Alternatively, maybe ( x ) is the year, and ( t ) is the year, so the integral is from 1901 to ( t ), where ( t ) is the year. But then ( H(t) ) is the number of historical sites at time ( t ) years since 1901, which would mean ( t ) is both the year and the number of years since 1901, which is confusing.Wait, maybe it's better to proceed with the assumption that ( t ) is the year, so ( H(t) ) is the number of historical sites in the year ( t ). So, for example, ( H(1901) = 0 ), since the integral from 1901 to 1901 is zero. Then, ( H(1951) ) would be the integral from 1901 to 1951.But the problem says ( t ) is the time in years since 1901, so ( t = 0 ) in 1901, ( t = 50 ) in 1951, etc. Therefore, the integral should be from 0 to ( t ), where ( x ) is the number of years since 1901. So, perhaps the problem has a typo, and it's supposed to be ( int_{0}^{t} ).Alternatively, maybe ( x ) is the year, and ( t ) is the year. So, for example, ( H(1951) = int_{1901}^{1951} 2 + cos(pi x / 50) , dx ). That would make sense because ( x ) is the year, so ( pi x / 50 ) would be in terms of year, but cosine is unitless, so that doesn't make sense. Wait, no, ( x ) is the year, so ( pi x / 50 ) would have units of year, which is not acceptable inside a cosine function.Therefore, the only way this makes sense is if ( x ) is the number of years since 1901, so ( x ) is unitless, and ( pi x / 50 ) is also unitless. Therefore, the integral should be from 0 to ( t ), where ( t ) is the number of years since 1901. So, I think the problem has a typo, and it should be ( int_{0}^{t} ) instead of ( int_{1901}^{t} ).Given that, I'll proceed with the assumption that the integral is from 0 to ( t ), where ( t ) is the number of years since 1901. Therefore, ( H(t) = int_{0}^{t} 2 + cos(pi x / 50) , dx ).So, for the first part, calculating the number of historical sites in 1951. Since 1951 is 50 years after 1901, ( t = 50 ).Therefore, ( H(50) = int_{0}^{50} 2 + cos(pi x / 50) , dx ).Let me compute this integral. The integral of 2 with respect to x is 2x. The integral of ( cos(pi x / 50) ) with respect to x is ( frac{50}{pi} sin(pi x / 50) ). So, putting it together:( H(50) = [2x + frac{50}{pi} sin(pi x / 50)] ) evaluated from 0 to 50.Calculating at x = 50:2*50 = 100( frac{50}{pi} sin(pi * 50 / 50) = frac{50}{pi} sin(pi) ). But ( sin(pi) = 0 ), so that term is 0.At x = 0:2*0 = 0( frac{50}{pi} sin(0) = 0 )Therefore, ( H(50) = 100 + 0 - (0 + 0) = 100 ).So, there are 100 historical sites in 1951.Wait, that seems straightforward. Let me double-check the integral.Yes, integral of 2 is 2x, integral of cosine is sine with the coefficient adjusted. So, ( int cos(ax) dx = frac{1}{a} sin(ax) ). So, here, a = ( pi / 50 ), so the integral is ( frac{50}{pi} sin(pi x / 50) ). Correct.Evaluating from 0 to 50:At 50: 2*50 = 100, ( sin(pi) = 0 )At 0: 0, ( sin(0) = 0 )So, yes, H(50) = 100.Now, the second part is to find the rate of increase in the number of historical sites in the year 2001. So, the rate of increase is the derivative of H(t) with respect to t, evaluated at the appropriate t.But wait, H(t) is defined as the integral from 0 to t of 2 + cos(πx/50) dx. Therefore, by the Fundamental Theorem of Calculus, the derivative H’(t) is just the integrand evaluated at t. So, H’(t) = 2 + cos(πt / 50).Therefore, the rate of increase in 2001 is H’(t) where t is the number of years since 1901. 2001 - 1901 = 100 years. So, t = 100.Therefore, H’(100) = 2 + cos(π * 100 / 50) = 2 + cos(2π). Since cos(2π) is 1, this becomes 2 + 1 = 3.So, the rate of increase in 2001 is 3 historical sites per year.Wait, let me make sure. The derivative of H(t) is indeed the integrand, so H’(t) = 2 + cos(πt / 50). So, at t = 100, it's 2 + cos(2π) = 2 + 1 = 3. Yes, that seems correct.But hold on, earlier I assumed that the integral was from 0 to t, but the problem says it's from 1901 to t. If t is the year, then in 2001, t = 2001, and the integral is from 1901 to 2001. But then, the derivative would be the integrand evaluated at t, which is 2 + cos(π * 2001 / 50). Wait, that would be a different result.Wait, this is confusing again. Let me clarify.If H(t) is the integral from 1901 to t (where t is the year) of 2 + cos(πx / 50) dx, then H(t) is a function of the year t. Therefore, the derivative H’(t) with respect to t is the integrand evaluated at t, so H’(t) = 2 + cos(πt / 50). Therefore, in 2001, the rate is 2 + cos(π*2001 / 50).But 2001 / 50 is 40.02. So, π*40.02 is approximately 125.84 radians. But cosine is periodic with period 2π, so 125.84 / (2π) is approximately 20.01. So, 125.84 radians is equivalent to 125.84 - 20*2π = 125.84 - 125.66 = 0.18 radians. So, cos(0.18) ≈ 0.983.Therefore, H’(2001) ≈ 2 + 0.983 ≈ 2.983, approximately 3.Wait, but earlier, when I assumed t was the number of years since 1901, I got exactly 3. So, both interpretations lead to approximately the same result, but one is exact and the other is approximate.Wait, if t is the year, then in 2001, t = 2001, so H’(2001) = 2 + cos(π*2001 / 50). Let me compute π*2001 / 50.2001 / 50 = 40.02. So, π*40.02 ≈ 3.1416 * 40.02 ≈ 125.84 radians.Now, 125.84 radians is a large angle. To find cos(125.84), we can subtract multiples of 2π until we get an angle between 0 and 2π.2π ≈ 6.2832. So, 125.84 / 6.2832 ≈ 20.01. So, 20 full rotations, which is 20*2π ≈ 125.66 radians. So, 125.84 - 125.66 ≈ 0.18 radians.Therefore, cos(125.84) = cos(0.18) ≈ 0.983.So, H’(2001) ≈ 2 + 0.983 ≈ 2.983, which is approximately 3.Alternatively, if t is the number of years since 1901, then in 2001, t = 100. So, H’(100) = 2 + cos(π*100 / 50) = 2 + cos(2π) = 2 + 1 = 3.So, depending on the interpretation, we get either exactly 3 or approximately 3. Given that, I think the problem expects us to interpret t as the number of years since 1901, so the derivative is exactly 3.Therefore, the rate of increase in 2001 is 3 historical sites per year.Wait, but let me think again. If t is the year, then the integral is from 1901 to t, and the derivative is 2 + cos(πt / 50). So, in 2001, t = 2001, so cos(π*2001 / 50) is approximately 0.983, so the rate is approximately 2.983, which is roughly 3.But if t is the number of years since 1901, then t = 100 in 2001, and cos(π*100 / 50) = cos(2π) = 1, so the rate is exactly 3.Given that the problem says "where H(t) is the number of historical sites at time t years since 1901," it's more consistent to interpret t as the number of years since 1901. Therefore, the integral should be from 0 to t, and the derivative is exactly 3.Therefore, the rate of increase in 2001 is 3 historical sites per year.So, summarizing:1. Population in 1951: approximately 1359.2. Population reaches 2000 around 1970.3. Number of historical sites in 1951: 100.4. Rate of increase in 2001: 3 per year.I think that's it. I hope I didn't make any mistakes in interpreting the problem, especially with the integral limits. But given the problem statement, I think this is the correct approach.</think>

🧠:<think>Alright, so I've got this problem about designing a water filtration system for a growing city. It's divided into two main parts: water flow optimization and environmental impact assessment. Let me try to break it down step by step.Starting with the first part, water flow optimization. The city currently uses Q0 = 10,000 cubic meters per day. The population is growing at 2.5% per year, and I need to find the required flow rate Q(t) after t years. Hmm, population growth is typically modeled with exponential growth, right? So the formula should be something like Q(t) = Q0 * (1 + r)^t, where r is the growth rate. Plugging in the numbers, that would be Q(t) = 10,000 * (1.025)^t. That makes sense because each year the flow rate increases by 2.5%.Next, I need to design a filtration system that can handle this future flow rate while maintaining a filtration efficiency η of at least 95%. The efficiency is given by η(Q) = 100 - Q/a, where a is a constant. So, I need to find a such that when Q is the future flow rate, η is still 95%.Wait, but hold on. The efficiency function is given as η(Q) = 100 - Q/a. So, when Q is 10,000, what is η? Let me see. At current conditions, I guess η should be 100 - 10,000/a. But the problem says the filtration efficiency needs to be at least 95%, so 100 - Q(t)/a >= 95. That simplifies to Q(t)/a <= 5, so a >= Q(t)/5.But wait, a is a constant, so I need to determine a based on initial conditions. At t=0, Q(0) = 10,000. So, plugging into η(Q), η(10,000) = 100 - 10,000/a. The initial efficiency isn't specified, but I think we can assume that at t=0, the system is designed to have 95% efficiency. So, η(10,000) = 95. Therefore, 95 = 100 - 10,000/a. Solving for a, 10,000/a = 5, so a = 10,000 / 5 = 2,000. So, a is 2,000.Wait, but hold on. If a is 2,000, then η(Q) = 100 - Q/2000. So, as Q increases, η decreases. So, to maintain η >= 95%, Q must be <= 5*2000 = 10,000? That can't be, because Q is increasing over time. So, maybe I misunderstood.Wait, no. If a is 2,000, then η(Q) = 100 - Q/2000. So, when Q is 10,000, η would be 100 - 10,000/2000 = 100 - 5 = 95%. So, at t=0, the efficiency is 95%. But as Q increases, η decreases. So, to maintain η >= 95%, Q must not exceed 10,000. But the population is growing, so Q will exceed 10,000 in the future. Therefore, the filtration system's efficiency will drop below 95% unless we adjust a.Wait, but the problem says to design a system that can handle the future flow rate while maintaining η of at least 95%. So, perhaps we need to adjust a so that even when Q(t) increases, η remains 95%. So, η(Q(t)) = 95 = 100 - Q(t)/a. Therefore, Q(t)/a = 5, so a = Q(t)/5. But a is a constant, so we need to set a such that even at the maximum Q(t) we expect, η remains 95%. But since the population is projected to grow indefinitely, Q(t) will keep increasing, so a would have to be set to infinity, which isn't practical.Wait, maybe I'm overcomplicating. Perhaps the efficiency is 95% at the current flow rate, and as the flow rate increases, the efficiency decreases, but we need to ensure that even at the future flow rate, the efficiency is still 95%. But that would require a to be adjusted for each t, which isn't possible since a is a constant.Alternatively, maybe the filtration system can be scaled up as needed. So, perhaps we need to determine a such that at any t, η(Q(t)) = 95. But that would require a = Q(t)/5 for all t, which is impossible because a is a constant. Therefore, perhaps the system needs to be designed with a higher a so that even as Q increases, η remains above 95%.Wait, but if a is larger, then η(Q) = 100 - Q/a would be higher for the same Q. So, if we set a higher, then even as Q increases, η doesn't drop too much. But how high should a be?Wait, maybe the question is that the filtration efficiency is a function of flow rate, given by η(Q) = 100 - Q/a, and we need to determine a based on initial conditions, which is when Q=10,000, η=95. So, as I did before, a=2000. Therefore, the filtration efficiency will decrease as Q increases. So, for future Q(t), η will be less than 95%. But the problem says the system must maintain η of at least 95%. Therefore, perhaps we need to adjust the system so that even as Q increases, η remains 95%.But since η(Q) = 100 - Q/a, to have η=95 for any Q, we need a = Q/5. But a is a constant, so unless we can change a, which we can't, perhaps the system needs to be designed with a higher a. Wait, but a is determined by initial conditions. So, maybe the filtration system's a is fixed at 2000, and as Q increases, η decreases. Therefore, to maintain η >=95, Q must not exceed 10,000. But since the population is growing, Q will exceed 10,000, so η will drop below 95%. Therefore, the system as designed cannot handle future growth while maintaining 95% efficiency. So, perhaps we need to redesign the system with a higher a.Wait, but how? Because a is determined by the initial condition. If we set a higher, say a=4000, then at Q=10,000, η=100 - 10,000/4000=75%, which is too low. So, we can't do that. Alternatively, maybe the filtration system can be scaled up by adding more filters, effectively increasing a.Wait, perhaps the system can be designed with multiple filtration units, each contributing to a. So, if we have n units, each with a_i, then total a = sum(a_i). So, as Q increases, we can add more units, increasing a, thus keeping η(Q) = 100 - Q/a above 95%.But the problem doesn't specify that we can scale up the system. It just says to design a filtration system that can handle the future flow rate while maintaining η of at least 95%. So, perhaps we need to find the maximum t such that Q(t) <= a*5, where a is determined by initial conditions.Wait, but a is fixed at 2000, so Q(t) = 10,000*(1.025)^t <= 2000*5=10,000. So, Q(t) <=10,000. But since Q(t) starts at 10,000 and grows, this would mean t=0 is the maximum. That doesn't make sense.I think I'm missing something here. Maybe the filtration efficiency is given as η(Q) = 100 - Q/a, and we need to find a such that at the future Q(t), η is still 95%. So, setting η(Q(t)) =95, which gives Q(t)/a =5, so a= Q(t)/5. But a is a constant, so we need to set a such that for the maximum Q(t) we expect, a= Q(t)/5. But since the population is growing indefinitely, Q(t) will go to infinity, so a would have to be infinity, which isn't practical.Alternatively, perhaps the problem is that the filtration system's efficiency is 95% at the current flow rate, and as the flow rate increases, the efficiency decreases, but we need to ensure that even at the future flow rate, the efficiency is still 95%. But that would require a to be set such that a= Q(t)/5 for all t, which is impossible because a is a constant.Wait, maybe I'm overcomplicating. Let's go back. The problem says: "design a filtration system that can handle this future flow rate while maintaining a filtration efficiency η of at least 95%." So, perhaps we need to find the required a such that η(Q(t)) >=95 for all t. Given that a is determined based on initial conditions, which is when t=0, Q=10,000, η=95. So, a=2000. Therefore, for any t, η(Q(t))=100 - Q(t)/2000. We need to ensure that 100 - Q(t)/2000 >=95, which simplifies to Q(t) <=10,000. But Q(t) is growing, so this is only true at t=0. Therefore, the system as designed cannot handle future growth while maintaining 95% efficiency. Therefore, perhaps the system needs to be designed with a higher a.But how? Because a is determined by initial conditions. If we set a higher, say a=4000, then at t=0, η=100 -10,000/4000=75%, which is too low. So, perhaps the system needs to be designed with multiple units, each with a_i, so that a can be increased as needed. For example, if we have n units, each with a_i=2000, then total a=2000n. So, as Q increases, we can add more units, increasing a, thus keeping η above 95%.But the problem doesn't specify that we can add units. It just says to design a system. So, perhaps the answer is that the system cannot maintain 95% efficiency as the population grows, unless it's scaled up, which would require increasing a. Therefore, the filtration system needs to be designed with a higher a, but since a is determined by initial conditions, perhaps we need to set a higher a initially.Wait, but if we set a higher a initially, η at t=0 would be lower. For example, if a=4000, η=75% at t=0, which is too low. So, perhaps the only way is to have a system that can be scaled, adding more filtration units as needed. Therefore, the system should be designed with the ability to increase a as Q increases, ensuring η remains above 95%.But the problem doesn't mention scaling, so maybe I'm supposed to assume that a is fixed, and thus the system can only handle up to Q=10,000 with η=95%. Beyond that, η drops. Therefore, the system cannot handle future growth while maintaining 95% efficiency. So, perhaps the answer is that the system needs to be designed with a higher a, but since a is fixed by initial conditions, it's impossible unless we scale the system.Wait, but the problem says to design the system, so perhaps we need to set a such that even at the maximum Q(t) we expect, η=95. But since the population grows indefinitely, we can't set a finite a. Therefore, perhaps the system needs to be designed with a variable a, which isn't practical. Alternatively, maybe the problem expects us to find a as 2000, and then note that as Q increases, η decreases, so the system will not maintain 95% efficiency in the future.But the problem says to design a system that can handle the future flow rate while maintaining η of at least 95%. So, perhaps the answer is that a must be set to Q(t)/5 for the maximum t we're considering. But since t can be any number, we need to set a to infinity, which isn't possible. Therefore, perhaps the system cannot maintain 95% efficiency as the population grows, and thus, the filtration system needs to be upgraded or scaled up as the population increases.But since the problem is asking to design the system, perhaps the answer is that a=2000, and the system will have decreasing efficiency as Q increases, but to maintain 95%, the system must be upgraded when Q exceeds 10,000. But the problem doesn't specify a time frame, so maybe we need to calculate when Q(t) exceeds 10,000, which is at t=0, so it's already at 10,000. Therefore, the system is designed to handle 10,000 with 95% efficiency, and as Q increases, efficiency drops.Wait, but the problem says to handle both current and projected future demand. So, perhaps the system needs to be designed with a higher a so that even as Q increases, η remains above 95%. Therefore, let's find a such that for the maximum Q(t) we expect, η=95. But since the population is growing indefinitely, we can't set a finite a. Therefore, perhaps the problem expects us to find a based on initial conditions, which is a=2000, and then note that as Q increases, η decreases, so the system will not maintain 95% efficiency in the future.Alternatively, maybe the problem is expecting us to set a such that η=95 at the future Q(t). But since a is a constant, we need to set a= Q(t)/5 for the future Q(t). But without knowing t, we can't set a. Therefore, perhaps the problem is expecting us to express a in terms of t, but a is a constant, so that's not possible.Wait, maybe I'm overcomplicating. Let's go back to the first part. The problem says: "Determine the required flow rate Q(t) in cubic meters per day after t years, accounting for the population growth." So, Q(t)=10,000*(1.025)^t.Then, design a filtration system that can handle this future flow rate while maintaining η of at least 95%. The efficiency is given by η(Q)=100 - Q/a. So, to maintain η>=95, we need Q/a <=5, so a>=Q/5.But a is a constant, so we need to set a such that a>=Q(t)/5 for all t. Since Q(t) increases exponentially, a would have to be infinite, which isn't possible. Therefore, the only way is to have a system that can be scaled, increasing a as Q increases. Therefore, the filtration system should be designed with the ability to add more units, each contributing to a, so that a can be increased as needed to maintain η>=95%.But since the problem doesn't mention scaling, perhaps the answer is that a=2000, and the system will have decreasing efficiency as Q increases, but to maintain 95%, the system must be upgraded when Q exceeds 10,000. But since Q is already 10,000 at t=0, that doesn't make sense.Wait, perhaps the problem is expecting us to find a such that η=95 at the future Q(t). So, if we know t, we can set a=Q(t)/5. But since t is variable, perhaps the problem is expecting us to express a in terms of t, but a is a constant, so that's not possible.Alternatively, maybe the problem is expecting us to find a such that η=95 at the current Q=10,000, which gives a=2000, and then note that as Q increases, η decreases, so the system will not maintain 95% efficiency in the future. Therefore, the filtration system needs to be upgraded or scaled up as the population grows.But since the problem says to design the system, perhaps the answer is that a=2000, and the system will have decreasing efficiency as Q increases, but to maintain 95%, the system must be upgraded when Q exceeds 10,000. But since Q is already 10,000 at t=0, that doesn't make sense.Wait, maybe I'm overcomplicating. Let's just proceed with a=2000, as determined by initial conditions, and then note that as Q increases, η decreases. So, the filtration system will have decreasing efficiency as the population grows, which may impact the environmental assessment.Moving on to the second part, environmental impact assessment. The filtration system releases water back into a river, and the pollutant concentration decreases over time according to dC/dt = -k(C - C_f), where C_f=0.5 mg/L. We're told that after 3 years, C(t)=2 mg/L, and we need to find k, then estimate C after 10 years.This is a first-order linear differential equation. The general solution is C(t) = C_f + (C0 - C_f)e^{-kt}. Plugging in the values, C(t)=0.5 + (5 -0.5)e^{-kt}=0.5 +4.5e^{-kt}.We're told that at t=3, C(3)=2. So, 2=0.5 +4.5e^{-3k}. Subtracting 0.5, 1.5=4.5e^{-3k}. Dividing both sides by 4.5, 1.5/4.5= e^{-3k}=1/3. Taking natural log, ln(1/3)= -3k, so k= -ln(1/3)/3= ln(3)/3≈1.0986/3≈0.3662 per year.Now, to find C(10), plug t=10 into the equation: C(10)=0.5 +4.5e^{-0.3662*10}=0.5 +4.5e^{-3.662}. Calculating e^{-3.662}≈0.0255. So, C(10)=0.5 +4.5*0.0255≈0.5 +0.1148≈0.6148 mg/L.So, after 10 years, the concentration is approximately 0.615 mg/L.Now, discussing sustainability and environmental impact. The filtration system reduces the pollutant concentration over time, approaching the equilibrium concentration of 0.5 mg/L. After 10 years, it's at ~0.615 mg/L, which is still above 0.5, but getting closer. The rate constant k is about 0.366 per year, which indicates a moderate rate of reduction.However, from the first part, the filtration system's efficiency decreases as the population grows, which might affect the rate of pollutant removal. If η decreases, the filtration system is less effective, which could slow down the reduction of pollutant concentration. Therefore, the system's ability to maintain high efficiency is crucial for the environmental impact. If the system can't maintain 95% efficiency, the pollutant concentration might not decrease as quickly, or could even increase if the system becomes overwhelmed.In conclusion, while the filtration system effectively reduces pollutant concentration in the short term, the long-term sustainability depends on maintaining high filtration efficiency as the population grows. If the system's efficiency drops, the environmental benefits may diminish, and the system may need upgrades to continue effectively reducing pollutants.</think>

🧠:<think>Alright, so I've got these two problems about Wanda and her socialite graph. Let me try to figure them out step by step.Starting with the first problem: Wanda has a planar graph G with n vertices and m edges. I need to find the maximum number of edges m in terms of n. Hmm, planar graphs... I remember there's a formula related to Euler's formula that gives the maximum number of edges a planar graph can have without being too dense. Euler's formula is V - E + F = 2, right? Where V is vertices, E is edges, and F is faces.For planar graphs, there's also a relation that comes from the fact that each face must be bounded by at least three edges. So, if we consider that each edge is shared by two faces, we can say that 3F ≤ 2E. Rearranging that, we get F ≤ (2/3)E. Plugging that into Euler's formula: V - E + (2/3)E ≥ 2. Simplifying that, V - (1/3)E ≥ 2, which leads to (1/3)E ≤ V - 2, so E ≤ 3V - 6.Wait, so that means for a planar graph, the maximum number of edges is 3n - 6. But hold on, is that always true? I think that applies to simple planar graphs without any loops or multiple edges. Since the problem says it's a simple graph, that should be fine. So, yeah, the maximum number of edges m is 3n - 6.Okay, moving on to the second problem. Wanda wants to calculate the chromatic polynomial P(G, k) for a complete subgraph K4 within G. I need to provide the general form of the chromatic polynomial and explain how it reflects the color assignments.First, the chromatic polynomial counts the number of ways to color the vertices of a graph with k colors such that no two adjacent vertices share the same color. For a complete graph K_n, the chromatic polynomial is known. Since K4 is a complete graph with 4 vertices, each connected to every other, the chromatic polynomial should be k(k - 1)(k - 2)(k - 3).Let me verify that. For a complete graph K_n, the chromatic polynomial is indeed k(k - 1)(k - 2)...(k - n + 1). So for n=4, it's k(k - 1)(k - 2)(k - 3). That makes sense because each vertex must have a unique color, and since they're all connected, you can't have any repeats. So the number of colorings is the number of permutations of k colors taken 4 at a time, which is k! / (k - 4)!.So the general form is P(K4, k) = k(k - 1)(k - 2)(k - 3). This reflects the fact that each vertex in K4 must be assigned a distinct color, and with k colors available, the number of ways to do this is the product of decreasing integers starting from k down to k - 3.Wait, let me think if there's another way to express this. It can also be written as P(K4, k) = k^4 - 6k^3 + 11k^2 - 6k. But the factored form is more insightful because it clearly shows the requirement for each vertex to have a unique color, hence the consecutive decreasing terms.So, in summary, the chromatic polynomial for K4 is k(k - 1)(k - 2)(k - 3), which represents the number of valid colorings where each of the four socialites in the clique gets a distinct color from the available k colors.I think that covers both problems. Let me just recap:1. For a planar graph, the maximum number of edges is 3n - 6.2. The chromatic polynomial for K4 is k(k - 1)(k - 2)(k - 3), which counts the number of ways to color the four vertices with k colors without any adjacent vertices sharing the same color.Yeah, that seems right. I don't think I made any mistakes here. The first part is a standard result from graph theory, and the second part is specific to complete graphs, which I remember correctly.Final Answer1. The maximum number of edges ( m ) is boxed{3n - 6}.2. The chromatic polynomial ( P(G, k) ) for ( K_4 ) is ( k(k - 1)(k - 2)(k - 3) ), which represents the number of ways to color the four vertices with ( k ) colors such that no two adjacent vertices share the same color. The general form is boxed{k(k - 1)(k - 2)(k - 3)}.</think>