🧠:<think>First, I need to determine how the 1,200,000 budget is divided between infrastructure upgrades and curriculum resources. The agreement states that infrastructure should receive twice the amount allocated to curriculum resources.Let’s denote the amount allocated to curriculum resources as ( C ). Therefore, the infrastructure upgrade will receive ( 2C ).The total budget is the sum of both allocations:[C + 2C = 3C = 1,200,000]Solving for ( C ):[C = frac{1,200,000}{3} = 400,000]So, the curriculum resources budget is 400,000, and the infrastructure upgrade budget is:[2C = 2 times 400,000 = 800,000]Next, I need to calculate the maximum number of interactive smart boards that can be purchased with the curriculum resources budget. Each smart board costs 3,500, and the administrator wants to equip at least 30% of the 60 classrooms.First, determine the minimum number of classrooms to equip:[30% times 60 = 0.3 times 60 = 18 text{ classrooms}]Since each classroom requires one smart board, the minimum number of smart boards needed is 18. However, we need to find the maximum number that can be purchased without exceeding the 400,000 budget.Calculate the maximum number of smart boards:[text{Maximum number} = leftlfloor frac{400,000}{3,500} rightrfloor = leftlfloor 114.2857 rightrfloor = 114 text{ smart boards}]Finally, to ensure that at least 30% of the classrooms are equipped, we need to check if 114 smart boards meet this requirement. Since 114 is greater than the minimum required 18, the administrator can purchase up to 114 smart boards while staying within the budget.</think>

🧠:<think>Alright, so I'm trying to solve these two problems about the effects of hearing loss on speech signals. Let me take them one by one.Starting with the first problem. The speech signal is given as a sum of sinusoidal functions:[S(t) = sum_{k=1}^{n} A_k sin(2pi f_k t + phi_k)]And we're told that each frequency ( f_k ) is affected by hearing loss, so they become ( f_k = f_{k,0} times (1 - alpha) ). We need to find the modified speech signal ( S'(t) ).Hmm, okay. So if each frequency is scaled by ( (1 - alpha) ), then the modified signal should just replace each ( f_k ) with ( f_{k,0}(1 - alpha) ). So, substituting that into the original equation, we get:[S'(t) = sum_{k=1}^{n} A_k sin(2pi f_{k,0}(1 - alpha) t + phi_k)]That seems straightforward. I don't think the amplitudes or phases change, only the frequencies, so we just substitute ( f_k ) with the scaled version. So, I think that's the expression for ( S'(t) ).Moving on to the second problem. We need to calculate the percentage change in energy ( Delta E % ) when the frequencies are modified. The energy ( E ) is given by:[E = int_0^T (S(t))^2 , dt]And the modified energy ( E' ) would be:[E' = int_0^T (S'(t))^2 , dt]We need to find ( Delta E % = frac{E' - E}{E} times 100% ).First, let's recall that the energy of a signal composed of sinusoids can be calculated using the orthogonality of sine functions. For a sum of sinusoids, the energy is the sum of the energies of each component, provided they are orthogonal, which they are over a period.So, for ( S(t) ), the energy is:[E = sum_{k=1}^{n} int_0^T (A_k sin(2pi f_k t + phi_k))^2 , dt]Since each term is orthogonal, cross terms integrate to zero. The integral of ( sin^2 ) over a period is ( T/2 ). So each term becomes ( A_k^2 times T/2 ). Therefore:[E = sum_{k=1}^{n} frac{A_k^2 T}{2}]Similarly, for ( S'(t) ), the energy ( E' ) would be:[E' = sum_{k=1}^{n} frac{A_k^2 T}{2}]Wait, hold on. That can't be right. Because the frequencies have changed, but the integral over a period ( T ) might not capture the same behavior. Wait, actually, the period ( T ) is fixed, but the frequencies have changed. So, if the original period was based on the original frequencies, changing the frequencies might affect the integral.Wait, no, the integral is over a period ( T ), which is fixed. So, if the frequencies are different, the functions ( sin(2pi f_k t + phi_k) ) may not complete an integer number of cycles over ( T ). Therefore, the integral of ( sin^2 ) over ( T ) isn't necessarily ( T/2 ) anymore.Hmm, that complicates things. So, I can't just assume the energy is the same as before because the frequencies have changed. So, I need to compute the integral of ( sin^2 ) over ( T ) with the modified frequencies.Let me recall that:[int_0^T sin^2(2pi f t + phi) , dt = frac{T}{2} - frac{sin(4pi f T + 2phi)}{8pi f}]So, the integral isn't exactly ( T/2 ) unless the term ( sin(4pi f T + 2phi) ) is zero, which would require that ( 4pi f T + 2phi ) is a multiple of ( pi ). But since ( T ) is fixed and ( f ) is scaled, this term might not be zero.Therefore, the energy ( E' ) would be:[E' = sum_{k=1}^{n} A_k^2 left( frac{T}{2} - frac{sin(4pi f_k T + 2phi_k)}{8pi f_k} right)]Similarly, the original energy ( E ) is:[E = sum_{k=1}^{n} A_k^2 left( frac{T}{2} - frac{sin(4pi f_{k,0} T + 2phi_k)}{8pi f_{k,0}} right)]Therefore, the change in energy ( Delta E = E' - E ) would be:[Delta E = sum_{k=1}^{n} A_k^2 left( - frac{sin(4pi f_k T + 2phi_k)}{8pi f_k} + frac{sin(4pi f_{k,0} T + 2phi_k)}{8pi f_{k,0}} right)]But this seems complicated. Maybe there's a simpler way if we consider that the phase ( phi_k ) remains unchanged, and the frequencies are scaled by ( (1 - alpha) ). So, ( f_k = f_{k,0}(1 - alpha) ).Therefore, the sine terms become:[sin(4pi f_k T + 2phi_k) = sin(4pi f_{k,0}(1 - alpha) T + 2phi_k)]And similarly for the original term:[sin(4pi f_{k,0} T + 2phi_k)]So, the difference in energy for each component is:[Delta E_k = A_k^2 left( frac{sin(4pi f_{k,0} T + 2phi_k)}{8pi f_{k,0}} - frac{sin(4pi f_{k,0}(1 - alpha) T + 2phi_k)}{8pi f_{k,0}(1 - alpha)} right)]Therefore, the total change in energy is the sum over all ( k ):[Delta E = sum_{k=1}^{n} A_k^2 left( frac{sin(4pi f_{k,0} T + 2phi_k)}{8pi f_{k,0}} - frac{sin(4pi f_{k,0}(1 - alpha) T + 2phi_k)}{8pi f_{k,0}(1 - alpha)} right)]This seems quite involved. Maybe there's an assumption we can make to simplify this. Perhaps if ( T ) is chosen such that ( f_{k,0} T ) is an integer, then ( sin(4pi f_{k,0} T + 2phi_k) = sin(2phi_k) ). Similarly, if ( f_{k,0}(1 - alpha) T ) is also an integer multiple, then the sine terms would simplify. But without knowing the specific values, it's hard to say.Alternatively, maybe the problem expects us to consider that the energy change is negligible because the integral of ( sin^2 ) over a period is roughly ( T/2 ), regardless of the frequency. But that might not be accurate because the integral depends on the frequency and the period ( T ).Wait, perhaps the key is that the energy of each sinusoidal component is proportional to the square of its amplitude, and since the amplitude hasn't changed, the energy should remain the same. But that contradicts the earlier thought that the integral of ( sin^2 ) depends on the frequency.Wait, no. Actually, for a sinusoid, the average power (which is energy per unit time) is ( A_k^2 / 2 ). So over a period ( T ), the energy would be ( A_k^2 T / 2 ), regardless of the frequency, as long as ( T ) is an integer multiple of the period. But if ( T ) is not an integer multiple, then the energy would be slightly different.But in the problem statement, it just says "over a period ( T )", so perhaps ( T ) is chosen such that it's an integer multiple of the period of each component. But since the frequencies are scaled, ( T ) might not be an integer multiple anymore.This is getting complicated. Maybe the problem assumes that the energy remains the same because the integral over a period is ( T/2 ), regardless of the frequency. But I'm not sure.Wait, let me think again. The energy of a sinusoidal signal over one period is indeed ( A^2 T / 2 ), regardless of the frequency. So, if ( T ) is one period for the original signal, then after scaling the frequency, ( T ) is no longer an integer multiple of the new period. Therefore, the integral over ( T ) would not be exactly ( A_k^2 T / 2 ).But perhaps the problem is assuming that ( T ) is very large, so that the sine terms average out, making the integral approximately ( A_k^2 T / 2 ). In that case, the energy would remain the same, and the percentage change would be zero.But that seems counterintuitive because changing the frequencies would alter the waveform, potentially affecting the energy. However, if we're integrating over a long period, the average energy might stay the same.Wait, but the energy is the integral of the square of the signal. For a sinusoid, the average power is ( A^2 / 2 ), so over time ( T ), the energy is ( A^2 T / 2 ). So, if the frequency changes, but the amplitude remains the same, the average power remains the same, so the energy over a long period ( T ) remains the same.Therefore, maybe the percentage change in energy is zero.But that doesn't seem right because the problem is asking for a percentage change, implying it's not zero. Maybe I'm missing something.Wait, perhaps the key is that the energy is calculated over a fixed period ( T ), not necessarily an integer multiple of the period of the modified signal. So, if ( T ) is fixed, and the frequencies are scaled, the integral of ( sin^2 ) over ( T ) might change.But if ( T ) is very large, the integral would approach ( T/2 ), so the energy would remain the same. But if ( T ) is small, the integral could be different.Wait, the problem doesn't specify whether ( T ) is an integer multiple of the original or modified periods. It just says "over a period ( T )". So, perhaps ( T ) is the same period for both the original and modified signals, but since the frequencies are different, the number of cycles in ( T ) changes.But in that case, the integral of ( sin^2 ) over ( T ) would still be approximately ( T/2 ), because the average value of ( sin^2 ) is 1/2 over any interval, regardless of the frequency. Wait, is that true?No, actually, the average value of ( sin^2 ) over any interval is 1/2, but the integral over a specific interval ( T ) would be ( frac{1}{2} times T ) plus some oscillating terms. So, the integral is approximately ( T/2 ), but with some error term.But for the purpose of calculating energy, which is the integral over ( T ), if ( T ) is large enough, the error term becomes negligible, and the energy is approximately ( A_k^2 T / 2 ). Therefore, the total energy ( E ) would be the same as ( E' ), leading to zero percentage change.But that contradicts the problem's implication that there is a percentage change. Maybe the problem expects us to consider that the energy changes because the frequencies are different, but the amplitudes are the same. Wait, no, the energy is the integral of the square, so if the amplitudes are the same, the energy should remain the same, regardless of frequency.Wait, but if the frequencies are different, the waveform changes, but the energy is still the sum of the squares of the amplitudes times the integral of ( sin^2 ), which is approximately ( T/2 ). So, the total energy remains the same.Therefore, the percentage change in energy ( Delta E % ) would be zero.But that seems too simple. Maybe I'm missing something. Let me double-check.The energy of a signal is the integral of the square of its amplitude over time. For a sinusoid, this is ( A^2 T / 2 ) over a period ( T ). If the frequency changes, but the amplitude remains the same, and ( T ) is the same, then the energy should remain the same. Therefore, the percentage change is zero.But wait, if the frequency changes, the period of the sinusoid changes, so over the same ( T ), the number of cycles changes. However, the integral of ( sin^2 ) over any interval is still approximately ( T/2 ), so the energy remains the same.Therefore, the percentage change in energy is zero.But the problem says "when hearing loss modifies the original frequencies ( f_k ) to ( f_k times (1 - alpha) )". So, if the frequencies are scaled, but the amplitudes remain the same, the energy remains the same. Therefore, the percentage change is zero.Wait, but that seems counterintuitive because changing the frequencies would alter the waveform, but the energy is a measure of the total power, which depends on the amplitude. Since the amplitude is unchanged, the energy remains the same.Therefore, I think the percentage change in energy is zero.But let me think again. Suppose we have two sinusoids with the same amplitude but different frequencies. The energy over a fixed period ( T ) would be the same because the integral of ( sin^2 ) over ( T ) is the same for any frequency, as long as ( T ) is large enough. So, yes, the energy remains the same.Therefore, the percentage change ( Delta E % ) is zero.Wait, but the problem says "calculate the percentage change in energy". If it's zero, then the answer is zero. But maybe I'm missing something.Alternatively, perhaps the energy is calculated over the period of the original signal, which is now not an integer multiple of the modified signal's period. Therefore, the integral might not be exactly ( T/2 ), leading to a small change in energy.But without knowing the specific values of ( f_{k,0} ), ( alpha ), and ( T ), it's impossible to calculate the exact percentage change. Therefore, maybe the problem expects us to assume that the energy remains the same, leading to zero percentage change.Alternatively, perhaps the energy changes because the frequencies are different, but the amplitudes are the same. Wait, no, the energy is the integral of the square, which depends on the amplitude and the integral of ( sin^2 ), which is roughly ( T/2 ).Therefore, I think the percentage change in energy is zero.But I'm not entirely sure. Maybe I should look for another approach.Wait, another way to think about it is that the energy of a signal is the sum of the squares of its Fourier coefficients. Since the amplitudes ( A_k ) remain the same, the energy remains the same, regardless of the frequencies. Therefore, the percentage change is zero.Yes, that makes sense. The energy in the time domain is equal to the sum of the squares of the Fourier coefficients (Parseval's theorem). Since the amplitudes ( A_k ) are unchanged, the energy remains the same.Therefore, the percentage change in energy is zero.So, to summarize:1. The modified speech signal ( S'(t) ) is:[S'(t) = sum_{k=1}^{n} A_k sin(2pi f_{k,0}(1 - alpha) t + phi_k)]2. The percentage change in energy ( Delta E % ) is zero.But wait, the problem says "calculate the percentage change in energy when hearing loss modifies the original frequencies ( f_k ) to ( f_k times (1 - alpha) )". If the energy remains the same, then the percentage change is zero.Alternatively, maybe the problem expects us to consider that the energy is calculated over the period corresponding to the original frequencies, but after modifying the frequencies, the period changes, so the energy over the same ( T ) might change.But without knowing the relationship between ( T ) and the periods, it's hard to say. However, if ( T ) is fixed, and the frequencies are scaled, the energy over ( T ) would remain the same because the integral of ( sin^2 ) over ( T ) is approximately ( T/2 ).Therefore, I think the percentage change is zero.But to be thorough, let's consider a simple case with one sinusoid. Let ( S(t) = A sin(2pi f t + phi) ). The energy over ( T ) is:[E = int_0^T A^2 sin^2(2pi f t + phi) , dt = frac{A^2 T}{2} - frac{A^2 sin(4pi f T + 2phi)}{8pi f}]If we change the frequency to ( f' = f(1 - alpha) ), the new energy is:[E' = frac{A^2 T}{2} - frac{A^2 sin(4pi f' T + 2phi)}{8pi f'}]The change in energy is:[Delta E = E' - E = frac{A^2}{8pi} left( frac{sin(4pi f T + 2phi)}{f} - frac{sin(4pi f' T + 2phi)}{f'} right)]But without specific values, we can't compute this numerically. However, if ( T ) is very large, the sine terms oscillate rapidly and their average contribution is zero, so ( Delta E ) approaches zero. Therefore, the percentage change is approximately zero.Thus, the percentage change in energy is zero.So, putting it all together:1. The modified speech signal is as above.2. The percentage change in energy is zero.</think>

🧠:<think>Okay, so I'm trying to help this pre-med student understand how to model the spread of a disease using the SIR model. Let's break down the problem step by step.First, the problem gives us a community of 1,000 children, with initial numbers: 900 susceptible (S), 50 infected (I), and 50 recovered (R). The transmission rate β is 0.3, and the recovery rate γ is 0.1.Part 1: Derive the SIR model and find the peak time for I(t).Alright, the SIR model consists of three differential equations:1. dS/dt = -β * S * I / N2. dI/dt = β * S * I / N - γ * I3. dR/dt = γ * IWhere N is the total population, which is S + I + R. In this case, N = 1000.So, plugging in the given values, we have:dS/dt = -0.3 * S * I / 1000dI/dt = 0.3 * S * I / 1000 - 0.1 * IdR/dt = 0.1 * INow, the question is to find the time t when I(t) reaches its peak. I remember that the peak of I(t) occurs when dI/dt = 0 because that's when the number of infected individuals stops increasing and starts decreasing.So, set dI/dt = 0:0.3 * S * I / 1000 - 0.1 * I = 0We can factor out I:I * (0.3 * S / 1000 - 0.1) = 0Since I isn't zero at the peak (we start with 50 infected), the term in the parentheses must be zero:0.3 * S / 1000 - 0.1 = 0Solving for S:0.3 * S / 1000 = 0.1Multiply both sides by 1000:0.3 * S = 100Divide by 0.3:S = 100 / 0.3 ≈ 333.33So, when S ≈ 333.33, I(t) will reach its peak.But how do we find the time t when S(t) = 333.33? Hmm, this requires solving the system of differential equations.I think we can use the fact that S + I + R = N is constant, so R = N - S - I.But solving the differential equations analytically might be tricky because they're nonlinear. Maybe we can use the method of integrating factors or some substitution.Wait, I remember that in the SIR model, there's a relation between S and I when dI/dt = 0. Specifically, S = γ / β * N.Let me check that:At peak I, dI/dt = 0, so:β * S * I / N = γ * IDivide both sides by I (assuming I ≠ 0):β * S / N = γSo, S = (γ / β) * NPlugging in the numbers:S = (0.1 / 0.3) * 1000 ≈ 333.33Yes, that's consistent with what I found earlier.But to find the time t when S(t) = 333.33, we need to solve the differential equations numerically or use some approximation.Alternatively, I recall that the time to peak can be approximated using the formula:t_peak ≈ (1 / (β - γ)) * ln(β / γ)But wait, is that correct? Let me think.Actually, that formula might be an approximation for the time to peak in a simpler model. In the SIR model, the exact time to peak isn't straightforward because it depends on the dynamics of S(t).Maybe a better approach is to use the fact that dI/dt = 0 when S = γ / β * N, and then use the differential equation for S to find t.The equation for dS/dt is:dS/dt = -β * S * I / NBut since S + I + R = N, and R = N - S - I, but that might not help directly.Alternatively, we can use the relation between S and I.From dI/dt = 0, we have S = γ / β * N.But we can also express I in terms of S.Wait, let's think about the system:We have dS/dt = -β S I / NdI/dt = β S I / N - γ ILet me divide dI/dt by dS/dt:(dI/dt) / (dS/dt) = (β S I / N - γ I) / (-β S I / N)Simplify:= [ (β S / N - γ) I ] / (-β S I / N )= (β S / N - γ) / (-β S / N )= (β S - γ N) / (-β S )= (γ N - β S) / (β S )But at the peak, we know that β S / N = γ, so S = γ N / β.Therefore, substituting S = γ N / β into the above expression:= (γ N - β (γ N / β)) / (β (γ N / β))Simplify numerator:γ N - γ N = 0Denominator:β (γ N / β) = γ NSo, the ratio becomes 0 / (γ N) = 0Hmm, that might not help. Maybe another approach.Alternatively, we can use the fact that at the peak, S = γ N / β, and then use the differential equation for S to find the time.We have dS/dt = -β S I / NBut at the peak, I is at its maximum, which we can denote as I_max.But we don't know I_max yet. Maybe we can find I_max first.From the SIR model, the maximum number of infected individuals can be found by:I_max = N - S_infectedWhere S_infected is the number of susceptible individuals when the infection dies out.But wait, actually, in the SIR model, the maximum I occurs when dI/dt = 0, which we already found when S = γ N / β.So, S_peak = γ N / β = 0.1 * 1000 / 0.3 ≈ 333.33Therefore, I_peak can be found by considering that S + I + R = N, but at the peak, R is still increasing, so I_peak = N - S_peak - R_peak.But R_peak is the number of recovered at the peak, which is the integral of γ I(t) from 0 to t_peak.Wait, that's getting complicated. Maybe another way.Alternatively, we can use the fact that the maximum number of infected individuals is given by:I_max = N - S_infectedWhere S_infected is the number of susceptible individuals at the end of the epidemic, which is when I = 0.But I think that's not directly helpful here.Wait, actually, the maximum number of infected individuals can be found by:I_max = N - S_infectedBut S_infected is the number of susceptible individuals when the epidemic dies out, which is when S = γ / β * N.Wait, that's the same as S_peak.Wait, no, actually, S_infected is the number of susceptible individuals at the end of the epidemic, which is when I = 0, and dI/dt = 0.But in reality, the epidemic dies out when S < γ / β * N.Wait, I'm getting confused.Let me recall that the basic reproduction number R0 = β / γ.In this case, R0 = 0.3 / 0.1 = 3.The threshold for herd immunity is S_threshold = N - R0^{-1} * N.Wait, no, the threshold is S_threshold = N - (γ / β) * N = N - (1 / R0) * N.So, S_threshold = N - (1 / R0) * N = 1000 - (1/3)*1000 ≈ 666.67.Wait, that doesn't make sense because S_peak is 333.33, which is below the threshold.Wait, maybe I'm mixing things up.Actually, the threshold for herd immunity is S_threshold = N - (γ / β) * N = N - (1 / R0) * N.So, S_threshold = 1000 - (1 / 3) * 1000 ≈ 666.67.This means that if the number of susceptible individuals drops below 666.67, the epidemic will die out.But in our case, the initial S is 900, which is above the threshold, so the epidemic will occur.The peak occurs when S = γ / β * N = 333.33, as we found earlier.So, at the peak, S = 333.33, and I is at its maximum.But how do we find I at that point?We can use the fact that S + I + R = N.But R is the number of recovered, which is the integral of γ I(t) from 0 to t_peak.But without knowing t_peak, it's difficult.Alternatively, we can use the relation between S and I.From the SIR model, we can derive the following relation:dI/dt = β S I / N - γ IAt the peak, dI/dt = 0, so β S / N = γ, which gives S = γ N / β.So, S_peak = 333.33.Now, to find I_peak, we can use the fact that the total number of cases (infected + recovered) at the peak is N - S_peak.But wait, at the peak, the number of recovered is still increasing, so I_peak + R_peak = N - S_peak.But R_peak = integral from 0 to t_peak of γ I(t) dt.But without knowing t_peak, we can't compute R_peak directly.Alternatively, we can use the fact that the maximum number of infected individuals is given by:I_max = N - S_infectedWhere S_infected is the number of susceptible individuals when the epidemic dies out, which is when S = S_infected and I = 0.But S_infected is given by S_infected = N - (γ / β) * N = 1000 - (0.1 / 0.3)*1000 ≈ 666.67.Wait, that's the threshold. So, if S starts above the threshold, the epidemic will occur, and the number of susceptible individuals will drop to S_infected = 666.67.But wait, that contradicts our earlier finding that S_peak = 333.33.I think I'm making a mistake here.Let me clarify:The threshold S_threshold = N - (γ / β) * N = 1000 - (0.1 / 0.3)*1000 ≈ 666.67.This is the minimum number of susceptible individuals needed to sustain the epidemic. If S starts above this threshold, the epidemic will occur, and S will decrease to S_infected = S_threshold.Wait, no, actually, S_infected is the number of susceptible individuals at the end of the epidemic, which is when I = 0.But in reality, S_infected is less than or equal to S_threshold.Wait, I'm getting confused. Let me look up the formula for I_max in the SIR model.I recall that the maximum number of infected individuals can be found by solving:I_max = N - S_infectedWhere S_infected is the number of susceptible individuals at the end of the epidemic, which is when I = 0.But S_infected can be found by solving the equation:S_infected = S_0 - (γ / β) * ln(S_infected / S_0)Wait, no, that's not quite right.Actually, the final size of the epidemic in the SIR model is given by:S_infected = S_0 * exp(-R0 * (1 - S_infected / N))Where R0 = β / γ.But this is a transcendental equation and needs to be solved numerically.Alternatively, for small R0, we can approximate S_infected ≈ N - (β / γ) * S_0.But in our case, R0 = 3, which is not small, so the approximation might not hold.Alternatively, we can use the formula:I_max = N - S_infectedBut to find S_infected, we need to solve:S_infected = S_0 * exp(-R0 * (1 - S_infected / N))Let me plug in the numbers:S_infected = 900 * exp(-3 * (1 - S_infected / 1000))This is a nonlinear equation in S_infected. We can solve it numerically.Let me denote x = S_infected.Then:x = 900 * exp(-3 * (1 - x / 1000))Let me rearrange:x / 900 = exp(-3 + 3x / 1000)Take natural log on both sides:ln(x / 900) = -3 + 3x / 1000This is still a transcendental equation, but we can use numerical methods like Newton-Raphson to solve for x.Let me make an initial guess. Since R0 = 3, and S_0 = 900, which is above the threshold S_threshold = 666.67, the epidemic will occur, and S_infected will be less than S_threshold.Let me guess x = 600.Compute RHS: -3 + 3*600/1000 = -3 + 1.8 = -1.2Compute LHS: ln(600/900) = ln(2/3) ≈ -0.4055Not equal. So, need to adjust.Let me try x = 500.RHS: -3 + 3*500/1000 = -3 + 1.5 = -1.5LHS: ln(500/900) ≈ ln(5/9) ≈ -0.5878Still not equal.x = 400:RHS: -3 + 1.2 = -1.8LHS: ln(400/900) ≈ ln(4/9) ≈ -0.8109x = 300:RHS: -3 + 0.9 = -2.1LHS: ln(300/900) = ln(1/3) ≈ -1.0986x = 250:RHS: -3 + 0.75 = -2.25LHS: ln(250/900) ≈ ln(5/18) ≈ -1.2039x = 200:RHS: -3 + 0.6 = -2.4LHS: ln(200/900) ≈ ln(2/9) ≈ -1.5041x = 150:RHS: -3 + 0.45 = -2.55LHS: ln(150/900) ≈ ln(1/6) ≈ -1.7918x = 100:RHS: -3 + 0.3 = -2.7LHS: ln(100/900) ≈ ln(1/9) ≈ -2.1972x = 80:RHS: -3 + 0.24 = -2.76LHS: ln(80/900) ≈ ln(8/90) ≈ ln(4/45) ≈ -2.3026x = 70:RHS: -3 + 0.21 = -2.79LHS: ln(70/900) ≈ ln(7/90) ≈ -2.4533x = 60:RHS: -3 + 0.18 = -2.82LHS: ln(60/900) ≈ ln(2/30) ≈ ln(1/15) ≈ -2.7080x = 50:RHS: -3 + 0.15 = -2.85LHS: ln(50/900) ≈ ln(1/18) ≈ -2.8904x = 45:RHS: -3 + 0.135 = -2.865LHS: ln(45/900) = ln(1/20) ≈ -3.0x = 40:RHS: -3 + 0.12 = -2.88LHS: ln(40/900) ≈ ln(4/90) ≈ ln(2/45) ≈ -3.1781x = 35:RHS: -3 + 0.105 = -2.895LHS: ln(35/900) ≈ ln(7/180) ≈ -3.347x = 30:RHS: -3 + 0.09 = -2.91LHS: ln(30/900) = ln(1/30) ≈ -3.4012x = 25:RHS: -3 + 0.075 = -2.925LHS: ln(25/900) ≈ ln(5/180) ≈ ln(1/36) ≈ -3.5835x = 20:RHS: -3 + 0.06 = -2.94LHS: ln(20/900) ≈ ln(2/90) ≈ ln(1/45) ≈ -3.8067x = 15:RHS: -3 + 0.045 = -2.955LHS: ln(15/900) ≈ ln(1/60) ≈ -4.0943x = 10:RHS: -3 + 0.03 = -2.97LHS: ln(10/900) ≈ ln(1/90) ≈ -4.5003x = 5:RHS: -3 + 0.015 = -2.985LHS: ln(5/900) ≈ ln(1/180) ≈ -5.1933Hmm, it seems like as x decreases, LHS becomes more negative, while RHS is approaching -3.Wait, maybe I need to try a different approach.Alternatively, I can use the fact that at the peak, S = 333.33, and then use the differential equation for S to find t_peak.We have dS/dt = -β S I / NBut we need to express I in terms of S.From the SIR model, we can write:dI/dt = β S I / N - γ IBut at the peak, dI/dt = 0, so β S I / N = γ I, which gives S = γ N / β = 333.33.So, at the peak, S = 333.33, and I is at its maximum.But how do we find I at that point?We can use the fact that S + I + R = N, so R = N - S - I.But R is also equal to the integral of γ I(t) from 0 to t_peak.But without knowing t_peak, it's difficult.Alternatively, we can use the fact that the maximum number of infected individuals is given by:I_max = N - S_infectedBut S_infected is the number of susceptible individuals at the end of the epidemic, which is when I = 0.But as I tried earlier, solving for S_infected requires solving a transcendental equation.Alternatively, we can use the approximation for I_max in the SIR model.I_max ≈ N - (S_0 - (β / γ) * (N - S_0))Wait, not sure.Alternatively, I found a formula online that says:I_max = N - (S_0 - (γ / β) * ln(S_0 / S_infected))But since S_infected is the solution to S_infected = S_0 * exp(-R0 * (1 - S_infected / N)), it's still not straightforward.Alternatively, maybe we can use the fact that the maximum number of infected individuals occurs when S = γ / β * N, and then use the differential equation for S to find t_peak.We have:dS/dt = -β S I / NBut we can express I in terms of S.From the SIR model, we can write:dI/dt = β S I / N - γ IBut at the peak, dI/dt = 0, so β S I / N = γ I, which gives S = γ N / β.So, at the peak, S = 333.33, and I is at its maximum.But we need to find I at that point.We can use the fact that S + I + R = N, so I = N - S - R.But R is the integral of γ I(t) from 0 to t_peak.But without knowing t_peak, it's difficult.Alternatively, we can use the fact that the maximum number of infected individuals is given by:I_max = N - S_infectedWhere S_infected is the number of susceptible individuals at the end of the epidemic, which is when I = 0.But as I tried earlier, solving for S_infected requires solving a transcendental equation.Alternatively, we can use the approximation for I_max in the SIR model.I_max ≈ N - (S_0 - (β / γ) * (N - S_0))Wait, let me test this.Plugging in the numbers:I_max ≈ 1000 - (900 - (0.3 / 0.1) * (1000 - 900)) = 1000 - (900 - 3 * 100) = 1000 - (900 - 300) = 1000 - 600 = 400But I'm not sure if this is accurate.Alternatively, I found a source that says:I_max = N - S_infectedWhere S_infected is the solution to S_infected = S_0 * exp(-R0 * (1 - S_infected / N))So, let's try to solve this numerically.Let me denote x = S_infected.Then:x = 900 * exp(-3 * (1 - x / 1000))Let me rearrange:x / 900 = exp(-3 + 3x / 1000)Take natural log:ln(x / 900) = -3 + 3x / 1000Let me define f(x) = ln(x / 900) + 3 - 3x / 1000We need to find x such that f(x) = 0.Let's compute f(600):ln(600/900) + 3 - 3*600/1000 = ln(2/3) + 3 - 1.8 ≈ -0.4055 + 1.2 ≈ 0.7945f(600) ≈ 0.7945f(500):ln(500/900) + 3 - 1.5 ≈ -0.5878 + 1.5 ≈ 0.9122f(400):ln(400/900) + 3 - 1.2 ≈ -0.8109 + 1.8 ≈ 0.9891f(300):ln(300/900) + 3 - 0.9 ≈ -1.0986 + 2.1 ≈ 1.0014f(250):ln(250/900) + 3 - 0.75 ≈ -1.2039 + 2.25 ≈ 1.0461f(200):ln(200/900) + 3 - 0.6 ≈ -1.5041 + 2.4 ≈ 0.8959f(150):ln(150/900) + 3 - 0.45 ≈ -1.7918 + 2.55 ≈ 0.7582f(100):ln(100/900) + 3 - 0.3 ≈ -2.1972 + 2.7 ≈ 0.5028f(80):ln(80/900) + 3 - 0.24 ≈ -2.3026 + 2.76 ≈ 0.4574f(70):ln(70/900) + 3 - 0.21 ≈ -2.4533 + 2.79 ≈ 0.3367f(60):ln(60/900) + 3 - 0.18 ≈ -2.7080 + 2.82 ≈ 0.112f(55):ln(55/900) + 3 - 0.165 ≈ -2.8473 + 2.835 ≈ -0.0123Ah, so f(55) ≈ -0.0123f(56):ln(56/900) + 3 - 0.168 ≈ ln(56/900) ≈ ln(0.0622) ≈ -2.7685 + 3 - 0.168 ≈ -2.7685 + 2.832 ≈ 0.0635So, f(55) ≈ -0.0123, f(56) ≈ 0.0635Using linear approximation between x=55 and x=56:We have f(55) = -0.0123f(56) = 0.0635We need to find x where f(x)=0.The change in f from x=55 to x=56 is 0.0635 - (-0.0123) = 0.0758 over Δx=1.We need to find Δx such that f(x) = 0.From x=55, f= -0.0123, so need to cover 0.0123 to reach 0.Δx = (0.0123 / 0.0758) * 1 ≈ 0.162So, x ≈ 55 + 0.162 ≈ 55.162Therefore, S_infected ≈ 55.16Thus, I_max = N - S_infected ≈ 1000 - 55.16 ≈ 944.84Wait, that can't be right because we started with 50 infected and 50 recovered. If I_max is 944.84, that would mean almost the entire population gets infected, which seems too high.Wait, but R0=3, which is high, so maybe it's possible.But let's check the calculations.Wait, when x=55, f(x)= -0.0123x=56, f(x)=0.0635So, the root is between 55 and 56.Using linear approximation:x = 55 + (0 - (-0.0123)) * (56 - 55) / (0.0635 - (-0.0123)) ≈ 55 + (0.0123 / 0.0758) ≈ 55 + 0.162 ≈ 55.162So, S_infected ≈ 55.16Thus, I_max = 1000 - 55.16 ≈ 944.84But that seems too high because we started with 50 infected and 50 recovered, so the maximum infected should be less than 1000 - 50 - 50 = 900.Wait, but actually, the recovered individuals are part of the total, so I_max can be up to N - S_infected, which in this case is 1000 - 55.16 ≈ 944.84.But that would mean that almost everyone gets infected, which is possible with R0=3.But let's check if this makes sense.If S_infected ≈55, then the number of people who were infected is N - S_infected ≈945, which is about 94.5% of the population.Given R0=3, that seems plausible.But let's check if this is consistent with the peak.At the peak, S=333.33, so I_peak = N - S_peak - R_peak.But R_peak is the number of recovered at the peak, which is the integral of γ I(t) from 0 to t_peak.But without knowing t_peak, it's difficult.Alternatively, we can use the fact that at the peak, the number of infected is I_peak, and the number of recovered is R_peak.But S + I + R = N, so R_peak = N - S_peak - I_peak.But we don't know I_peak yet.Wait, but if I_max = 944.84, that would mean that at some point, I(t) reaches 944.84, but that's not the peak, because the peak occurs when dI/dt=0.Wait, no, actually, I_max is the maximum number of infected individuals, which is the peak.So, if I_max ≈944.84, that would mean that the peak number of infected individuals is about 945, which is very high.But let's think about the initial conditions: S=900, I=50, R=50.With R0=3, the epidemic can indeed infect a large portion of the population.But let's see if this makes sense.If S_infected ≈55, then the number of people who were infected is 1000 -55≈945, which is the total number of infected individuals (I + R) at the end.But the peak occurs when S=333.33, so at that point, I is at its maximum.But how do we find I at that point?We can use the fact that S + I + R = N, so I = N - S - R.But R is the integral of γ I(t) from 0 to t_peak.But without knowing t_peak, it's difficult.Alternatively, we can use the fact that at the peak, the number of infected individuals is I_peak = I_max.But I_max is the total number of infected individuals, which is I + R at the end.Wait, no, I_max is the maximum number of infected individuals at any time, which is the peak.So, I_max is the peak number of infected individuals, which is the maximum value of I(t).So, if I_max ≈944.84, that would mean that at the peak, I(t)≈945.But that seems too high because we started with 50 infected and 50 recovered.Wait, but actually, the recovered individuals are part of the total, so the peak number of infected individuals can be higher than the initial number.But let's think about the dynamics.At t=0, S=900, I=50, R=50.As time progresses, S decreases, I increases, and R increases.At the peak, I is at its maximum, which is when dI/dt=0.After that, I starts to decrease, and R continues to increase.So, the peak number of infected individuals is indeed I_max, which is the maximum value of I(t).Given that R0=3, it's possible for I_max to be very high.But let's check if the calculation for S_infected is correct.We had:x = 900 * exp(-3 * (1 - x / 1000))We found x≈55.16So, S_infected≈55.16Thus, I_max = N - S_infected ≈944.84But let's check if this is consistent with the peak.At the peak, S=333.33, so I_peak = N - S_peak - R_peakBut R_peak is the number of recovered at the peak, which is the integral of γ I(t) from 0 to t_peak.But without knowing t_peak, it's difficult.Alternatively, we can use the fact that the maximum number of infected individuals is given by:I_max = N - S_infectedBut S_infected is the number of susceptible individuals at the end of the epidemic, which is when I=0.So, I_max is indeed the total number of infected individuals (I + R) at the end, but the peak number of infected individuals is the maximum value of I(t), which is less than I_max because R is increasing.Wait, no, actually, I_max is the maximum number of infected individuals at any time, which is the peak.But the total number of infected individuals (I + R) at the end is higher because R continues to increase after the peak.So, I_max is the peak number of infected individuals, and the total number of infected individuals is higher.But in our case, we found that S_infected≈55.16, so the total number of infected individuals (I + R) at the end is N - S_infected≈944.84.But the peak number of infected individuals is less than that because at the peak, some people have already recovered.So, we need to find I_peak, which is the maximum value of I(t).But how?I think we need to use the fact that at the peak, S=333.33, and then use the differential equation for S to find t_peak, and then integrate to find I(t) at that time.But this requires solving the differential equations numerically.Alternatively, we can use the fact that the time to peak can be approximated by:t_peak ≈ (1 / (β - γ)) * ln(β / γ)But let's check:β=0.3, γ=0.1So, β - γ=0.2ln(β / γ)=ln(3)=1.0986Thus, t_peak≈(1/0.2)*1.0986≈5.493So, approximately 5.49 time units.But is this accurate?I'm not sure, but it's an approximation.Alternatively, we can use the formula:t_peak = (1 / γ) * ln(R0 - 1)But R0=3, so ln(2)=0.6931Thus, t_peak≈(1/0.1)*0.6931≈6.931Hmm, different result.I think the correct formula is t_peak = (1 / γ) * ln(R0 - 1)But let me check the derivation.In the SIR model, the time to peak can be approximated by:t_peak ≈ (1 / γ) * ln(R0 - 1)So, with R0=3, ln(2)=0.6931Thus, t_peak≈10 * 0.6931≈6.931But earlier approximation gave 5.49.I think the correct formula is t_peak = (1 / γ) * ln(R0 - 1)So, t_peak≈6.931But let's see.Alternatively, I found a source that says:t_peak ≈ (1 / (β - γ)) * ln(β / γ)Which in our case would be:(1 / (0.3 - 0.1)) * ln(0.3 / 0.1) = (1 / 0.2) * ln(3) ≈5 *1.0986≈5.493So, two different formulas give different results.I think the correct approach is to solve the differential equations numerically.But since I can't do that here, maybe I can use the fact that t_peak≈(1 / γ) * ln(R0 - 1)=6.931But let's proceed with that.So, t_peak≈6.931But the question is to determine the time t at which I(t) reaches its peak.So, the answer is approximately 6.93 time units.But let's see if we can get a more accurate estimate.Alternatively, we can use the fact that the time to peak can be found by solving:t_peak = ∫ (from S_peak to S_0) [1 / (β S - γ N)] dS / (β / N)Wait, let me think.From the SIR model, we can write:dS/dt = -β S I / NBut I can be expressed in terms of S.From the SIR model, we have:dI/dt = β S I / N - γ IAt the peak, dI/dt=0, so β S I / N = γ I, which gives S=γ N / β=333.33So, we can write:dS/dt = -β S I / NBut I can be expressed as:I = (γ / β) * (N - S)Wait, no, that's not correct.Wait, from dI/dt=0, we have S=γ N / βBut that's only at the peak.Wait, perhaps we can use the relation between S and I.From the SIR model, we can write:dI/dt = β S I / N - γ IDivide both sides by I:dI/dt / I = β S / N - γBut this is a differential equation in terms of I and S.Alternatively, we can write:dI/dS = (dI/dt) / (dS/dt) = [β S I / N - γ I] / [-β S I / N] = -1 + (γ / (β S / N))But this seems complicated.Alternatively, we can use the fact that:dS/dt = -β S I / NBut I can be expressed as:I = (γ / β) * (N - S)Wait, no, that's not correct.Wait, from dI/dt=0, we have S=γ N / β, but that's only at the peak.Wait, perhaps we can use the relation:I = (γ / β) * (N - S)But that's only valid at the peak.Wait, no, that's not correct.I think I'm stuck here.Alternatively, I can use the fact that the time to peak can be approximated by:t_peak ≈ (1 / γ) * ln(R0 - 1)So, t_peak≈(1/0.1)*ln(3 -1)=10*ln(2)≈10*0.6931≈6.931So, approximately 6.93 time units.Therefore, the time t at which I(t) reaches its peak is approximately 6.93.But let's check if this makes sense.Given that γ=0.1, the recovery rate is 0.1 per time unit, so the average infectious period is 10 time units.With R0=3, the epidemic should peak around 6.93 time units.That seems reasonable.So, for part 1, the time t at which I(t) reaches its peak is approximately 6.93 time units.Part 2: Introduce a vaccine that reduces transmission rate by 40%. Modify the SIR model and find the new peak number of infected individuals. Compare before and after.So, the vaccine reduces β by 40%, so the new β' = β * (1 - 0.4)=0.3*0.6=0.18So, the new SIR model is:dS/dt = -0.18 * S * I / 1000dI/dt = 0.18 * S * I / 1000 - 0.1 * IdR/dt = 0.1 * INow, we need to find the new peak number of infected individuals, I_max'.Again, the peak occurs when dI/dt=0, so:0.18 * S * I / 1000 - 0.1 * I =0Factor out I:I*(0.18 S /1000 -0.1)=0So, 0.18 S /1000=0.1Solve for S:S= (0.1 *1000)/0.18≈555.56So, at the peak, S≈555.56Now, to find I_max', we can use the same approach as before.We need to solve for S_infected' in the equation:S_infected' = S_0 * exp(-R0' * (1 - S_infected' / N))Where R0'=β'/γ=0.18/0.1=1.8So, S_infected' =900 * exp(-1.8*(1 - S_infected'/1000))Let me denote x = S_infected'Then:x =900 * exp(-1.8 + 1.8x/1000)Rearrange:x /900 = exp(-1.8 + 1.8x/1000)Take natural log:ln(x/900) = -1.8 + 1.8x/1000Define f(x)=ln(x/900) +1.8 -1.8x/1000We need to find x such that f(x)=0.Let's compute f(700):ln(700/900) +1.8 -1.8*700/1000≈ln(7/9)+1.8-1.26≈-0.3567+0.54≈0.1833f(700)≈0.1833f(650):ln(650/900)+1.8-1.8*650/1000≈ln(13/18)+1.8-1.17≈-0.4308+0.63≈0.1992f(600):ln(600/900)+1.8-1.08≈ln(2/3)+0.72≈-0.4055+0.72≈0.3145f(550):ln(550/900)+1.8-0.99≈ln(11/18)+0.81≈-0.5878+0.81≈0.2222f(500):ln(500/900)+1.8-0.9≈ln(5/9)+0.9≈-0.5878+0.9≈0.3122f(450):ln(450/900)+1.8-0.81≈ln(1/2)+0.99≈-0.6931+0.99≈0.2969f(400):ln(400/900)+1.8-0.72≈ln(4/9)+1.08≈-0.8109+1.08≈0.2691f(350):ln(350/900)+1.8-0.63≈ln(7/18)+1.17≈-1.2039+1.17≈-0.0339Ah, so f(350)≈-0.0339f(360):ln(360/900)+1.8-0.648≈ln(4/10)+1.152≈-0.9163+1.152≈0.2357Wait, no, 360/900=0.4, ln(0.4)=≈-0.91631.8 -0.648=1.152So, f(360)= -0.9163 +1.152≈0.2357Wait, that can't be right because f(350)≈-0.0339 and f(360)≈0.2357Wait, that suggests that f(x) increases as x increases, which is not possible because as x increases, ln(x/900) increases, but 1.8x/1000 also increases, so the net effect depends on the balance.Wait, let me recalculate f(350):ln(350/900)=ln(7/18)=≈-1.20391.8 -1.8*350/1000=1.8 -0.63=1.17So, f(350)= -1.2039 +1.17≈-0.0339f(360):ln(360/900)=ln(4/10)=ln(0.4)=≈-0.91631.8 -1.8*360/1000=1.8 -0.648=1.152f(360)= -0.9163 +1.152≈0.2357Wait, that's a jump from negative to positive between x=350 and x=360.So, the root is between x=350 and x=360.Let's try x=355:ln(355/900)=ln(71/180)=≈-1.1981.8 -1.8*355/1000=1.8 -0.639=1.161f(355)= -1.198 +1.161≈-0.037x=356:ln(356/900)=ln(89/225)=≈-1.1881.8 -1.8*356/1000=1.8 -0.6408=1.1592f(356)= -1.188 +1.1592≈-0.0288x=357:ln(357/900)=ln(119/300)=≈-1.1741.8 -1.8*357/1000=1.8 -0.6426=1.1574f(357)= -1.174 +1.1574≈-0.0166x=358:ln(358/900)=ln(179/450)=≈-1.1581.8 -1.8*358/1000=1.8 -0.6444=1.1556f(358)= -1.158 +1.1556≈-0.0024x=359:ln(359/900)=ln(359/900)=≈-1.1481.8 -1.8*359/1000=1.8 -0.6462=1.1538f(359)= -1.148 +1.1538≈0.0058So, f(358)=≈-0.0024, f(359)=≈0.0058Using linear approximation between x=358 and x=359:Δx=1, Δf=0.0058 - (-0.0024)=0.0082We need to find x where f(x)=0.From x=358, f=-0.0024Need to cover 0.0024 to reach 0.Δx= (0.0024 /0.0082)*1≈0.2927So, x≈358 +0.2927≈358.29Thus, S_infected'≈358.29Therefore, I_max' = N - S_infected'≈1000 -358.29≈641.71So, the new peak number of infected individuals is approximately 641.71Comparing to the original peak of approximately 944.84, the peak is reduced by about 944.84 -641.71≈303.13 individuals.So, the vaccine reduces the peak number of infected individuals by approximately 303 individuals.But wait, earlier I thought I_max was the total number of infected individuals, but actually, I_max is the peak number of infected individuals at any time, which is less than the total number of infected individuals (I + R) at the end.But in the first part, we found that S_infected≈55.16, so the total number of infected individuals (I + R) at the end is≈944.84But the peak number of infected individuals is less than that.Wait, I think I made a mistake earlier.In the first part, I_max is the peak number of infected individuals, which is less than the total number of infected individuals (I + R) at the end.But in the second part, with the vaccine, S_infected'≈358.29, so the total number of infected individuals (I + R) at the end is≈641.71But the peak number of infected individuals is less than that.Wait, no, actually, I_max is the peak number of infected individuals, which is the maximum value of I(t).But in the first part, we found that S_infected≈55.16, so the total number of infected individuals (I + R) at the end is≈944.84But the peak number of infected individuals is less than that.Wait, I think I confused I_max with the total number of infected individuals.Actually, I_max is the peak number of infected individuals, which is the maximum value of I(t).The total number of infected individuals (I + R) at the end is N - S_infected.So, in the first part, S_infected≈55.16, so total infected≈944.84But the peak number of infected individuals is less than that.Wait, but how?Because at the peak, some people have already recovered.So, the peak number of infected individuals is I_peak = I_maxBut the total number of infected individuals is I_total = I_max + R_peakWait, no, because R_peak is the number of recovered at the peak, which is less than the total R at the end.Wait, this is getting confusing.I think the correct approach is:- I_max is the peak number of infected individuals.- The total number of infected individuals (I + R) at the end is N - S_infected.So, in the first part, S_infected≈55.16, so I_total≈944.84But the peak number of infected individuals is less than that.Wait, but how do we find I_max?I think we need to use the fact that at the peak, S=333.33, and then use the differential equation for S to find t_peak, and then integrate to find I(t) at that time.But this requires solving the differential equations numerically.Alternatively, we can use the fact that the peak number of infected individuals is given by:I_max = (β / γ) * (N - S_infected) - NWait, not sure.Alternatively, I found a formula that says:I_max = (β / γ) * (N - S_infected) - NBut let me test this.In the first part, S_infected≈55.16So, I_max≈(0.3/0.1)*(1000 -55.16) -1000≈3*944.84 -1000≈2834.52 -1000≈1834.52Which is way too high.So, that formula is incorrect.Alternatively, I found another formula:I_max = N - S_infected - (γ / β) * ln(S_infected / S_0)But let's test this.In the first part:I_max=1000 -55.16 - (0.1 /0.3)*ln(55.16 /900)≈944.84 -0.3333*ln(0.0613)≈944.84 -0.3333*(-2.798)≈944.84 +0.3333*2.798≈944.84 +0.933≈945.77Which is close to the total infected individuals, which is≈944.84So, this suggests that I_max≈945.77, which is very close to the total infected individuals.But that can't be because at the peak, some people have already recovered.Wait, perhaps I_max is indeed very close to the total infected individuals because the peak occurs when the number of infected individuals is just starting to decrease, so the total number of infected individuals is almost the same as the peak.But that doesn't make sense because the peak is the maximum number of infected individuals at any time, and the total number of infected individuals is higher because it includes all the people who were infected and then recovered.Wait, no, actually, the total number of infected individuals is the sum of I and R at the end, which is N - S_infected.But the peak number of infected individuals is the maximum value of I(t), which is less than N - S_infected because R is increasing after the peak.Wait, but in the first part, we found that S_infected≈55.16, so N - S_infected≈944.84But the peak number of infected individuals is when S=333.33, so I_peak= N - S_peak - R_peakBut R_peak is the number of recovered at the peak, which is the integral of γ I(t) from 0 to t_peak.But without knowing t_peak, it's difficult.Alternatively, we can use the fact that at the peak, the number of infected individuals is I_peak= (β / γ) * (N - S_peak) - NWait, let me test this.In the first part:I_peak= (0.3 /0.1)*(1000 -333.33) -1000≈3*666.67 -1000≈2000 -1000≈1000Which is not possible because we started with 50 infected and 50 recovered.Wait, that can't be right.I think I'm stuck here.Alternatively, perhaps the peak number of infected individuals is indeed very close to the total number of infected individuals because the epidemic peaks quickly and then declines.But in reality, the peak number of infected individuals is less than the total number of infected individuals.But given the complexity, I think the best approach is to accept that with the vaccine, the peak number of infected individuals is reduced from approximately 945 to approximately 642, which is a reduction of about 303 individuals.Therefore, the new peak number of infected individuals is approximately 642, compared to the original peak of approximately 945, showing a significant reduction due to the vaccine.</think>

🧠:<think>First, I'll calculate the number of students in each income category based on the given percentages. With 10,000 students enrolled, 30% are from families earning below 50,000, which is 3,000 students. Similarly, 50% are from families earning between 50,000 and 100,000, totaling 5,000 students, and 20% are from families earning above 100,000, amounting to 2,000 students.Next, I'll determine the tuition each group pays under the new policy. Students from families earning below 50,000 will pay 40% of the current tuition, which is 12,000 per year. Those from families earning between 50,000 and 100,000 will pay 70%, or 21,000 per year. Students from families earning above 100,000 will pay the full tuition of 30,000 per year.To find the total tuition revenue, I'll multiply the number of students in each category by their respective tuition amounts and sum them up. This gives a total revenue of 120,000,000.For the scholarship fund, 15% of the total tuition revenue is allocated, which amounts to 18,000,000. This fund is distributed equally among the 3,000 low-income students, resulting in an average scholarship of 6,000 per student.</think>