🧠:<think>Okay, so I have this problem where an entrepreneur wants to expand their e-commerce business into three regions: A, B, and C. They've given me the mean and standard deviation for the monthly revenue in each region, and they want to know two things. First, the probability that the combined monthly revenue from all three regions exceeds 160,000. Second, they want to model the combined revenue growth over the next 12 months using an exponential growth model. Hmm, let me tackle these one by one.Starting with the first part: calculating the probability that the combined revenue exceeds 160,000. Since the revenues from each region are independent and normally distributed, I remember that the sum of independent normal variables is also normal. So, I can find the mean and variance of the combined revenue by adding the means and variances of each region.Let me write down the given data:- Region A: μ_A = 50,000, σ_A = 8,000- Region B: μ_B = 45,000, σ_B = 10,000- Region C: μ_C = 60,000, σ_C = 12,000So, the combined mean revenue, μ_total, should be μ_A + μ_B + μ_C. Let me calculate that:μ_total = 50,000 + 45,000 + 60,000 = 155,000.Wait, that's interesting. The mean combined revenue is 155,000. The entrepreneur wants the probability that this combined revenue exceeds 160,000. So, it's 5,000 above the mean. Now, to find this probability, I need the standard deviation of the combined revenue.Since the revenues are independent, the variance of the sum is the sum of the variances. So, variance_total = σ_A² + σ_B² + σ_C².Calculating each variance:σ_A² = (8,000)^2 = 64,000,000σ_B² = (10,000)^2 = 100,000,000σ_C² = (12,000)^2 = 144,000,000Adding them up:variance_total = 64,000,000 + 100,000,000 + 144,000,000 = 308,000,000So, the standard deviation, σ_total, is the square root of 308,000,000. Let me compute that:σ_total = sqrt(308,000,000) ≈ sqrt(3.08 * 10^8) ≈ 17,549.34 approximately.Wait, let me check that calculation again. 308,000,000 is 3.08 * 10^8. The square root of 10^8 is 10,000. So, sqrt(3.08) is approximately 1.7549, so 1.7549 * 10,000 = 17,549. So, yes, approximately 17,549.So, the combined revenue follows a normal distribution with μ = 155,000 and σ ≈ 17,549.Now, to find the probability that the combined revenue exceeds 160,000, which is P(X > 160,000). Since the normal distribution is symmetric, I can standardize this value to a Z-score.Z = (X - μ) / σ = (160,000 - 155,000) / 17,549 ≈ 5,000 / 17,549 ≈ 0.285.So, Z ≈ 0.285. Now, I need to find the probability that Z > 0.285. I can use a Z-table or a calculator for this. Looking at the standard normal distribution table, the area to the left of Z=0.285 is approximately 0.612. Therefore, the area to the right is 1 - 0.612 = 0.388.Wait, let me double-check that. If Z is 0.28, the cumulative probability is about 0.6103, and for Z=0.29, it's about 0.6141. Since 0.285 is halfway, maybe around 0.6122. So, 1 - 0.6122 ≈ 0.3878, which is approximately 38.78%.So, the probability that the combined revenue exceeds 160,000 is roughly 38.78%.But wait, let me make sure I didn't make a mistake in calculating the Z-score. 5,000 divided by 17,549 is indeed approximately 0.285. Yes, that seems right.Alternatively, I can use a calculator to compute the exact value. The Z-score is approximately 0.285, so the probability is about 0.387, or 38.7%.So, that's the first part.Now, moving on to the second part: modeling the combined revenue growth over the next 12 months using an exponential growth model. The growth rate g(t) is given by g(t) = a * e^(b*t), where a = 0.03 and b = 0.015. The initial combined monthly revenue is 155,000, and we need to find the expected total revenue at the end of 12 months.Hmm, so I need to model the growth over 12 months. The growth rate is given as a function of time, g(t) = 0.03 * e^(0.015*t). I think this is a continuous growth model, so the formula for the revenue at time t would be:Revenue(t) = Initial_Revenue * e^(∫g(t) dt from 0 to t)Wait, because in continuous growth, the growth rate is the derivative of the logarithm of the revenue. So, if g(t) is the instantaneous growth rate, then:dR/dt = g(t) * R(t)Which is a differential equation. The solution to this is R(t) = R0 * e^(∫g(t) dt from 0 to t)So, let's compute the integral of g(t) from 0 to 12.Given g(t) = 0.03 * e^(0.015*t)So, ∫g(t) dt = ∫0.03 * e^(0.015*t) dtLet me compute this integral.The integral of e^(kt) dt is (1/k) e^(kt) + C. So, here, k = 0.015.So, ∫0.03 * e^(0.015*t) dt = 0.03 / 0.015 * e^(0.015*t) + C = 2 * e^(0.015*t) + CTherefore, the integral from 0 to 12 is:[2 * e^(0.015*12)] - [2 * e^(0.015*0)] = 2 * e^(0.18) - 2 * e^0 = 2*(e^0.18 - 1)Calculating e^0.18: e^0.18 ≈ 1.1972 (since e^0.1 ≈ 1.1052, e^0.2 ≈ 1.2214, so 0.18 is closer to 0.2, so maybe around 1.197)So, e^0.18 ≈ 1.1972Therefore, 2*(1.1972 - 1) = 2*(0.1972) = 0.3944So, the integral from 0 to 12 is approximately 0.3944.Therefore, the revenue at t=12 is:R(12) = R0 * e^(0.3944)Given R0 = 155,000So, R(12) = 155,000 * e^0.3944Calculating e^0.3944: e^0.3944 ≈ e^0.4 ≈ 1.4918, but since 0.3944 is slightly less than 0.4, maybe around 1.483.Alternatively, let's compute it more accurately.We know that ln(1.483) ≈ 0.3944. Let me check:ln(1.483) = ln(1 + 0.483). Using Taylor series or calculator approximation.Alternatively, using a calculator:e^0.3944 ≈ 1.483.So, R(12) ≈ 155,000 * 1.483 ≈ Let's compute that.155,000 * 1.483:First, 155,000 * 1 = 155,000155,000 * 0.4 = 62,000155,000 * 0.08 = 12,400155,000 * 0.003 = 465Adding them up:155,000 + 62,000 = 217,000217,000 + 12,400 = 229,400229,400 + 465 = 229,865So, approximately 229,865.Wait, but let me check if I did the multiplication correctly.Alternatively, 155,000 * 1.483:Multiply 155,000 by 1.483:155,000 * 1 = 155,000155,000 * 0.4 = 62,000155,000 * 0.08 = 12,400155,000 * 0.003 = 465Adding: 155,000 + 62,000 = 217,000217,000 + 12,400 = 229,400229,400 + 465 = 229,865Yes, that seems correct.Alternatively, using another method:155,000 * 1.483 = 155,000 * (1 + 0.4 + 0.08 + 0.003) = same as above.So, approximately 229,865.But let me verify the integral calculation again because that's crucial.We had g(t) = 0.03 * e^(0.015*t)The integral from 0 to 12 is ∫0.03 * e^(0.015*t) dt from 0 to 12.As I did before, the integral is 2 * (e^(0.015*12) - 1) = 2*(e^0.18 - 1)e^0.18 ≈ 1.1972, so 2*(0.1972) ≈ 0.3944So, the exponent is 0.3944, so e^0.3944 ≈ 1.483Thus, R(12) = 155,000 * 1.483 ≈ 229,865.So, the expected total revenue at the end of 12 months is approximately 229,865.Wait, but let me make sure about the model. The problem says the growth rate g(t) follows an exponential growth model given by g(t) = a e^{bt}, where a=0.03 and b=0.015. So, g(t) is the growth rate at time t, which is in percentage per month.But in continuous growth models, the growth rate is usually expressed as a decimal, not a percentage. So, 0.03 is 3%, and 0.015 is 1.5%. So, that seems correct.But let me think again: if g(t) is the instantaneous growth rate, then the differential equation is dR/dt = g(t) * R(t). So, solving that gives R(t) = R0 * e^{∫g(t) dt from 0 to t}.Yes, that's correct.So, the integral of g(t) from 0 to 12 is 0.3944, so the multiplier is e^{0.3944} ≈ 1.483.Therefore, R(12) ≈ 155,000 * 1.483 ≈ 229,865.So, that's the expected total revenue at the end of 12 months.Wait, but let me check if the units are correct. The initial revenue is monthly, but the growth is over 12 months. So, is the expected total revenue the monthly revenue at month 12, or the total revenue over 12 months?The problem says "expected total revenue at the end of 12 months", so I think it refers to the monthly revenue at the end of 12 months, not the sum over 12 months. Because if it were the sum, it would be different. So, I think it's the monthly revenue after 12 months, which is R(12) ≈ 229,865.Alternatively, if it were the total revenue over 12 months, we would have to integrate R(t) from 0 to 12, but the problem says "expected total revenue at the end of 12 months", which I think refers to the revenue at that point, not the cumulative.So, I think 229,865 is the correct answer.Wait, but let me make sure. The initial combined monthly revenue is 155,000. So, after 12 months, the monthly revenue would have grown to approximately 229,865. So, that's the expected monthly revenue at the end of 12 months.Alternatively, if we were to compute the total revenue over 12 months, it would be the integral of R(t) from 0 to 12, but that's more complicated. But the problem says "expected total revenue at the end of 12 months", which I think is just R(12), not the sum. So, I think 229,865 is correct.So, to recap:1. The probability that the combined monthly revenue exceeds 160,000 is approximately 38.78%.2. The expected total revenue at the end of 12 months is approximately 229,865.I think that's it. Let me just check if I made any calculation errors.For the first part:μ_total = 155,000σ_total ≈ 17,549Z = (160,000 - 155,000)/17,549 ≈ 0.285P(Z > 0.285) ≈ 0.387, so 38.7%.Yes, that seems correct.For the second part:g(t) = 0.03 e^{0.015 t}Integral from 0 to 12: 2*(e^{0.18} - 1) ≈ 0.3944e^{0.3944} ≈ 1.483155,000 * 1.483 ≈ 229,865Yes, that seems correct.So, I think I'm confident with these answers.</think>

🧠:<think>Alright, so I have this problem about the Soviet Union's industrial output and population during the Khrushchev era. It's divided into two parts. Let me try to figure out each step carefully.First, the functions given are:- Industrial output: ( I(t) = 500e^{0.05t} )- Population: ( P(t) = 200 + 10t - 0.1t^2 )Where ( t ) is the number of years since 1960. So, 1960 corresponds to ( t = 0 ), 1961 is ( t = 1 ), and so on.Problem 1: Calculate the rate of change of the industrial output and the population in 1970.Okay, so 1970 is 10 years after 1960, so ( t = 10 ).I need to find the rate of change, which means I need to compute the derivatives of both ( I(t) ) and ( P(t) ) with respect to ( t ), and then evaluate them at ( t = 10 ).Starting with the industrial output ( I(t) ):The function is ( I(t) = 500e^{0.05t} ). The derivative of ( e^{kt} ) with respect to ( t ) is ( ke^{kt} ). So, the derivative ( I'(t) ) should be:( I'(t) = 500 * 0.05e^{0.05t} = 25e^{0.05t} )Now, evaluate this at ( t = 10 ):( I'(10) = 25e^{0.05*10} = 25e^{0.5} )I can calculate ( e^{0.5} ) approximately. I remember that ( e ) is about 2.71828, so ( e^{0.5} ) is roughly 1.6487.So, ( I'(10) ≈ 25 * 1.6487 ≈ 41.2175 ). So, the rate of change of industrial output in 1970 is approximately 41.22 units per year.Now, moving on to the population ( P(t) ):The function is ( P(t) = 200 + 10t - 0.1t^2 ). The derivative ( P'(t) ) is straightforward:( P'(t) = 10 - 0.2t )Evaluate this at ( t = 10 ):( P'(10) = 10 - 0.2*10 = 10 - 2 = 8 )So, the rate of change of the population in 1970 is 8 million per year.Wait, hold on. The population is given in millions, so the rate of change is 8 million per year? That seems a bit high, but let's check the units. The function ( P(t) ) is in millions, so the derivative ( P'(t) ) is in millions per year. So, yes, 8 million per year is correct.So, part 1 is done. The rate of change of industrial output is approximately 41.22 units per year, and the population is increasing by 8 million per year in 1970.Problem 2: Determine the year when the industrial output per capita was at its maximum during the Khrushchev era (1953-1964).Hmm, okay. So, the Khrushchev era is from 1953 to 1964. But our functions are defined starting from 1960 (( t = 0 )). So, we need to consider ( t ) from -10 (1953) to 4 (1964). Wait, actually, 1964 is 4 years after 1960, so ( t = 4 ). But 1953 is 7 years before 1960, so ( t = -7 ).But the functions are given for ( t geq 0 ), right? Because they are defined starting in 1960. So, maybe the problem is considering the Khrushchev era from 1960 to 1964? Or perhaps we need to extend the functions back to 1953? Hmm, the problem statement says "during the Khrushchev era (1953-1964)", so we need to consider the entire period.But the functions are given starting from 1960. So, perhaps we can assume that the functions are valid for ( t ) from -10 to 4? Or maybe we need to adjust the functions for the years before 1960.Wait, that might complicate things. Alternatively, maybe the functions are only given for ( t geq 0 ), so we can only compute from 1960 onwards. But the Khrushchev era started in 1953, so perhaps the question is only concerned with the period from 1960 to 1964? Or maybe it's expecting us to consider the entire era, but with the functions starting in 1960.This is a bit confusing. Let me read the problem again."During the Khrushchev era, the Soviet Union experienced significant changes in industrial output and population growth. Suppose in 1960, the industrial output... was represented by the function ( I(t) = 500e^{0.05t} ), where ( t ) is the number of years since 1960. Simultaneously, the population ( P(t) ) in millions was modeled by the function ( P(t) = 200 + 10t - 0.1t^2 )."So, the functions are defined starting in 1960, with ( t = 0 ) in 1960. So, the Khrushchev era is 1953-1964, but the functions are only given from 1960 onwards. So, perhaps the question is only concerned with the period from 1960 to 1964? Because before 1960, the functions aren't defined.Alternatively, maybe we can extrapolate the functions back to 1953? But that might not be accurate, as the functions are given for ( t geq 0 ).Wait, the problem says "during the Khrushchev era (1953-1964)", so it's likely that the functions are meant to model the entire era, but starting from 1960. So, perhaps the functions are only valid from 1960 to 1964, which is the latter part of Khrushchev's era. Alternatively, maybe the functions are supposed to be extended back to 1953, but that would require knowing the values before 1960.Hmm, this is a bit unclear. Let me think. Since the functions are given starting in 1960, perhaps the question is only considering the period from 1960 to 1964, which is the last part of Khrushchev's era. So, ( t ) would range from 0 to 4.Alternatively, maybe the functions are valid for the entire Khrushchev era, so we can use them for ( t ) from -7 (1953) to 4 (1964). But since the functions are given as ( I(t) = 500e^{0.05t} ) and ( P(t) = 200 + 10t - 0.1t^2 ), which are defined for all real numbers, perhaps we can use them for ( t ) from -7 to 4.But I need to check if that makes sense. Let's see, in 1953, which is ( t = -7 ), what would the population be?( P(-7) = 200 + 10*(-7) - 0.1*(-7)^2 = 200 - 70 - 0.1*49 = 200 - 70 - 4.9 = 125.1 ) million.Is that a reasonable population? The Soviet Union's population in 1953 was around 190 million, I think. So, 125 million is way too low. So, perhaps the functions are only valid from 1960 onwards, and we can't use them before that.Therefore, maybe the question is only concerned with the period from 1960 to 1964, which is ( t = 0 ) to ( t = 4 ).So, the industrial output per capita is ( frac{I(t)}{P(t)} ). We need to find the value of ( t ) in [0,4] that maximizes this function.So, let's define the function ( Q(t) = frac{I(t)}{P(t)} = frac{500e^{0.05t}}{200 + 10t - 0.1t^2} )To find its maximum, we can take the derivative ( Q'(t) ) and set it equal to zero.First, let me write ( Q(t) = frac{500e^{0.05t}}{200 + 10t - 0.1t^2} )Let me denote the denominator as ( D(t) = 200 + 10t - 0.1t^2 )So, ( Q(t) = frac{500e^{0.05t}}{D(t)} )To find ( Q'(t) ), we can use the quotient rule:( Q'(t) = frac{D(t) * frac{d}{dt}[500e^{0.05t}] - 500e^{0.05t} * D'(t)}{[D(t)]^2} )Compute each part:First, ( frac{d}{dt}[500e^{0.05t}] = 500 * 0.05e^{0.05t} = 25e^{0.05t} )Second, ( D'(t) = 10 - 0.2t )So, putting it together:( Q'(t) = frac{D(t) * 25e^{0.05t} - 500e^{0.05t} * (10 - 0.2t)}{[D(t)]^2} )We can factor out ( 25e^{0.05t} ) from the numerator:( Q'(t) = frac{25e^{0.05t} [D(t) - 20(10 - 0.2t)]}{[D(t)]^2} )Wait, let me check that:Wait, 500e^{0.05t} is 20 * 25e^{0.05t}, right? Because 25 * 20 = 500.So, 500e^{0.05t} = 20 * 25e^{0.05t}So, the numerator is:25e^{0.05t} * D(t) - 20 * 25e^{0.05t} * (10 - 0.2t)Factor out 25e^{0.05t}:25e^{0.05t} [D(t) - 20*(10 - 0.2t)]So, yes, that's correct.Therefore, ( Q'(t) = frac{25e^{0.05t} [D(t) - 20*(10 - 0.2t)]}{[D(t)]^2} )Since ( 25e^{0.05t} ) and ( [D(t)]^2 ) are always positive (as exponential and square functions are positive), the sign of ( Q'(t) ) depends on the term in the brackets:( D(t) - 20*(10 - 0.2t) )Let me compute this:First, ( D(t) = 200 + 10t - 0.1t^2 )Second, ( 20*(10 - 0.2t) = 200 - 4t )So, subtracting:( D(t) - 20*(10 - 0.2t) = (200 + 10t - 0.1t^2) - (200 - 4t) )Simplify:200 - 200 + 10t + 4t - 0.1t^2 = 14t - 0.1t^2So, the numerator simplifies to ( 14t - 0.1t^2 )Therefore, ( Q'(t) = frac{25e^{0.05t} (14t - 0.1t^2)}{[D(t)]^2} )So, to find critical points, set numerator equal to zero:( 14t - 0.1t^2 = 0 )Factor:t(14 - 0.1t) = 0So, solutions are t = 0 and 14 - 0.1t = 0 => 0.1t = 14 => t = 140But our domain is t in [0,4], so t = 140 is outside the domain. Therefore, the only critical point in [0,4] is at t = 0.But we need to check if this is a maximum or not. Since t = 0 is the left endpoint, we should check the behavior of Q(t) at t = 0 and t = 4, and also see if Q'(t) changes sign in the interval.Wait, but the derivative Q'(t) is proportional to (14t - 0.1t^2). Let's analyze the sign of this expression in [0,4].At t = 0: 0At t approaching 0 from the right: 14t is positive, so the expression is positive.At t = 4: 14*4 - 0.1*(4)^2 = 56 - 1.6 = 54.4, which is positive.So, in the interval [0,4], the expression 14t - 0.1t^2 is always positive (except at t = 0 where it's zero). Therefore, Q'(t) is positive throughout [0,4], meaning that Q(t) is increasing on [0,4].Therefore, the maximum occurs at t = 4, which is 1964.Wait, but let me double-check. If Q(t) is increasing on [0,4], then yes, the maximum is at t=4.But let me compute Q(t) at t=0 and t=4 to confirm.At t=0:I(0) = 500e^0 = 500P(0) = 200 + 0 - 0 = 200So, Q(0) = 500 / 200 = 2.5At t=4:I(4) = 500e^{0.2} ≈ 500 * 1.2214 ≈ 610.7P(4) = 200 + 40 - 0.1*16 = 200 + 40 - 1.6 = 238.4So, Q(4) ≈ 610.7 / 238.4 ≈ 2.56So, indeed, Q(t) increased from 2.5 to approximately 2.56 over the interval. So, it's increasing, so maximum at t=4, which is 1964.But wait, the Khrushchev era is from 1953 to 1964. So, if we consider the entire era, including before 1960, but our functions don't extend back. So, if the functions are only valid from 1960, then the maximum per capita output in the Khrushchev era would be in 1964.But just to be thorough, let's check if maybe before 1960, the per capita output was higher. But since we don't have the functions defined before 1960, we can't compute that. So, perhaps the question is only considering the period from 1960 to 1964, in which case, the maximum is in 1964.Alternatively, if we assume that the functions can be extended back to 1953, even though they don't make sense (as population was lower than reality), then we can check if the maximum occurs before 1960.But given that the population function P(t) is a quadratic opening downward (since the coefficient of t^2 is negative), it has a maximum at t = -b/(2a) = -10/(2*(-0.1)) = -10 / (-0.2) = 50. So, the population would peak at t=50, which is way beyond our interval. So, in our interval from t=-7 to t=4, the population is increasing because the vertex is at t=50, so from t=-7 to t=4, it's on the increasing side of the parabola.Wait, let me compute P(t) at t=-7 and t=4.At t=-7:P(-7) = 200 + 10*(-7) - 0.1*(-7)^2 = 200 -70 -4.9 = 125.1 millionAt t=4:P(4) = 200 + 40 - 1.6 = 238.4 millionSo, population is increasing from 125.1 million in 1953 to 238.4 million in 1964.Similarly, the industrial output is growing exponentially, so it's increasing as well.Therefore, the per capita output is I(t)/P(t). Since both I(t) and P(t) are increasing, but I(t) is exponential and P(t) is quadratic, it's possible that per capita output could have a maximum somewhere.But since our functions are only defined from t=0 onwards, and in that interval, per capita output is increasing, the maximum would be at t=4 (1964).But just to be thorough, let's consider the possibility that before 1960, per capita output might have been higher.But since we don't have the functions defined before 1960, we can't compute it. So, perhaps the question is only concerned with the period from 1960 to 1964.Alternatively, if we assume that the functions can be extended back, even though they don't make sense, let's see.Compute Q(t) at t=-7:I(-7) = 500e^{-0.35} ≈ 500 * 0.7047 ≈ 352.35P(-7) = 125.1So, Q(-7) ≈ 352.35 / 125.1 ≈ 2.816At t=0:Q(0) = 2.5So, Q(t) was higher in 1953 (t=-7) than in 1960 (t=0). So, if we consider the entire Khrushchev era, the per capita output was higher in 1953 than in 1960, but since our functions don't model before 1960, we can't be sure.But the problem says "during the Khrushchev era (1953-1964)", but the functions are given starting in 1960. So, perhaps the question is only considering the period from 1960 to 1964, in which case, the maximum is at 1964.Alternatively, maybe the functions are meant to be used for the entire era, even though they don't fit perfectly. In that case, we can compute Q(t) from t=-7 to t=4.But since we don't have the functions before 1960, it's unclear. Maybe the question expects us to consider the period from 1960 to 1964.Given that, and since Q(t) is increasing on [0,4], the maximum is at t=4, which is 1964.But just to be thorough, let's check the derivative before t=0, assuming the functions are extended.Wait, but the derivative expression we had was:( Q'(t) = frac{25e^{0.05t} (14t - 0.1t^2)}{[D(t)]^2} )So, for t < 0, say t = -7:Compute 14t - 0.1t^2:14*(-7) - 0.1*(-7)^2 = -98 - 4.9 = -102.9So, the numerator is negative, meaning Q'(t) is negative for t < 0. So, Q(t) is decreasing before t=0.Therefore, if we consider the entire interval from t=-7 to t=4, Q(t) is decreasing from t=-7 to t=0, and increasing from t=0 to t=4.Therefore, the minimum occurs at t=0, and the maximums are at the endpoints t=-7 and t=4.But since Q(-7) ≈ 2.816 and Q(4) ≈ 2.56, so Q(-7) > Q(4). Therefore, the maximum per capita output occurs at t=-7, which is 1953.But wait, that contradicts our earlier conclusion. So, if we consider the entire Khrushchev era, the per capita output was highest in 1953, then decreased until 1960, and then started increasing again, but didn't surpass the 1953 level by 1964.But the problem is that the functions don't accurately model the population before 1960, as we saw earlier, because in reality, the population was higher in 1953 than what the function suggests.So, perhaps the question is only concerned with the period from 1960 to 1964, in which case, the maximum per capita output is in 1964.Alternatively, if we take the functions at face value, even though they don't align with real data before 1960, then the maximum per capita output is in 1953.But the problem statement says "during the Khrushchev era (1953-1964)", so it's likely that they expect us to consider the entire era, even though the functions are defined from 1960. So, perhaps we need to consider t from -7 to 4.But since the functions are given starting from 1960, it's unclear. Maybe the question is only considering the period from 1960 to 1964.Given that, and since Q(t) is increasing on [0,4], the maximum is at t=4, which is 1964.But let me think again. If we consider the entire era, the per capita output was highest in 1953, then decreased until 1960, and then started increasing again, but didn't reach the 1953 level by 1964.But since the functions don't model reality before 1960, perhaps the question is only considering the period from 1960 to 1964, in which case, the maximum is at 1964.Alternatively, maybe the functions are valid for the entire era, and we can use them to find the maximum.But given that the population function is a quadratic that peaks at t=50, which is way beyond our interval, and in our interval, it's increasing, so the per capita output is influenced by both I(t) and P(t) increasing.But since I(t) is exponential and P(t) is quadratic, the per capita output might have a maximum somewhere.Wait, but in our derivative, we found that Q'(t) is positive throughout [0,4], meaning it's increasing. So, the maximum is at t=4.But if we consider t < 0, Q(t) is decreasing, so the maximum in the entire era would be at t=-7, which is 1953.But again, the functions don't model reality before 1960, so it's unclear.Given the ambiguity, perhaps the question expects us to consider the period from 1960 to 1964, in which case, the maximum is at 1964.Alternatively, maybe the question is considering the entire era, and the maximum occurs at t=0, but that doesn't make sense because Q(t) is increasing after t=0.Wait, no, Q(t) is decreasing before t=0 and increasing after t=0, so the minimum is at t=0, and the maximums are at the endpoints.But since the functions are only given from t=0 onwards, perhaps the question is only concerned with t >=0.Given that, and since Q(t) is increasing on [0,4], the maximum is at t=4, which is 1964.Therefore, the year when the industrial output per capita was at its maximum during the Khrushchev era is 1964.But wait, let me check the derivative again.We had:Q'(t) = [25e^{0.05t} (14t - 0.1t^2)] / [D(t)]^2So, for t in [0,4], 14t - 0.1t^2 is positive, as 14*4 - 0.1*16 = 56 - 1.6 = 54.4 >0Therefore, Q'(t) >0 in [0,4], so Q(t) is increasing, so maximum at t=4.Therefore, the answer is 1964.But let me compute Q(t) at t=4 and t=5 to see if it's still increasing beyond 1964, but since the Khrushchev era ends in 1964, we don't need to go beyond.So, yes, the maximum is in 1964.Final Answer1. The rate of change of industrial output in 1970 was approximately boxed{41.22} units per year, and the population was increasing at a rate of boxed{8} million per year.2. The industrial output per capita was at its maximum in the year boxed{1964}.</think>

🧠:<think>Okay, so I have this problem about a homeowner who wants to build a custom modular shelving system. It's divided into two parts. Let me try to tackle them one by one.Starting with the first part: Each unit has dimensions x, y, z meters for width, height, and depth respectively. The volume of each unit is exactly 0.5 cubic meters. So, that gives me the equation:x * y * z = 0.5That's straightforward. Now, the total surface area of all the units should not exceed 30 square meters. Hmm, surface area for a rectangular unit is 2(xy + yz + xz). Since there are n units, the total surface area would be n times that. So:n * 2(xy + yz + xz) ≤ 30So, putting it together, the relationship between x, y, z, and n is:x * y * z = 0.5and2n(xy + yz + xz) ≤ 30I think that's the first part. Now, moving on to the second part.The homeowner wants to arrange these units along a wall that's 10 meters long. They want exactly 0.5 meters of space between each unit. So, if there are n units, how much space will they take up?Each unit has a width of x meters, and between each unit, there's 0.5 meters. So, for n units, there will be (n - 1) spaces between them. Therefore, the total length occupied by the units and the spaces is:n * x + (n - 1) * 0.5 ≤ 10We need to find the maximum number of units n that can fit along the wall. So, we need to solve for n in this inequality.But wait, we don't know x yet. Hmm. Maybe we can express x in terms of y and z from the first part? Since x * y * z = 0.5, x = 0.5 / (y * z). But that might complicate things because we don't have values for y and z.Alternatively, maybe we can express the surface area constraint in terms of x, y, z, and n, and then find a relationship that allows us to find n.Wait, but in the second part, we're only concerned with the width x and the spacing. Maybe we can find the maximum n by assuming that x is as small as possible? Or perhaps we need to find x in terms of n?Let me think. Since each unit has a volume of 0.5, if we minimize x, that would allow more units along the wall. But minimizing x would require maximizing y and z, but we don't have constraints on y and z except for the surface area.Alternatively, maybe we can express the surface area in terms of x, given that x * y * z = 0.5.Let me try that. Let's solve for y * z:y * z = 0.5 / xSo, the surface area per unit is 2(xy + yz + xz). Let's substitute yz with 0.5 / x:Surface area per unit = 2(xy + 0.5/x + xz)But we still have y and z in there. Maybe we can express y and z in terms of x? Or perhaps use some inequality to find the minimum surface area.Wait, maybe using the AM-GM inequality? Since we have terms like xy and xz, maybe we can find a lower bound for the surface area.Let me recall that for positive numbers a, b, c, the AM-GM inequality states that (a + b + c)/3 ≥ (abc)^(1/3). But I'm not sure if that directly applies here.Alternatively, for the surface area, maybe we can express it as 2(xy + yz + xz) and try to find its minimum value given that xyz = 0.5.Yes, that sounds like a calculus optimization problem. Let's try that.Let me denote S = 2(xy + yz + xz). We need to minimize S given that xyz = 0.5.Using Lagrange multipliers, perhaps. Let me set up the function:f(x, y, z) = 2(xy + yz + xz)subject to the constraint g(x, y, z) = xyz - 0.5 = 0The Lagrangian would be:L = 2(xy + yz + xz) - λ(xyz - 0.5)Taking partial derivatives:∂L/∂x = 2(y + z) - λ(yz) = 0∂L/∂y = 2(x + z) - λ(xz) = 0∂L/∂z = 2(x + y) - λ(xy) = 0And the constraint:xyz = 0.5So, from the partial derivatives:2(y + z) = λ(yz) ...(1)2(x + z) = λ(xz) ...(2)2(x + y) = λ(xy) ...(3)Let me try to solve these equations.From equation (1) and (2):2(y + z)/ (yz) = λ = 2(x + z)/(xz)So,2(y + z)/ (yz) = 2(x + z)/(xz)Simplify:(y + z)/(yz) = (x + z)/(xz)Multiply both sides by yz * xz:x(y + z) = y(x + z)Expand:xy + xz = xy + yzSubtract xy from both sides:xz = yzAssuming z ≠ 0, we can divide both sides by z:x = ySimilarly, from equations (1) and (3):2(y + z)/(yz) = 2(x + y)/(xy)But since x = y, let's substitute:2(x + z)/(xz) = 2(x + x)/(x*x) = 2(2x)/(x²) = 4/xLeft side: 2(x + z)/(xz) = 2(1/z + x/xz) = 2(1/z + 1/z) = 4/zSo, 4/z = 4/xThus, z = xSo, x = y = zTherefore, the unit is a cube? Wait, but the volume is 0.5, so x³ = 0.5, so x = (0.5)^(1/3) ≈ 0.7937 meters.But wait, if x = y = z, then the surface area per unit is 2(3x²) = 6x².So, total surface area for n units is 6n x².Given that x³ = 0.5, x = (0.5)^(1/3), so x² = (0.5)^(2/3).Thus, total surface area is 6n * (0.5)^(2/3).We need this to be ≤ 30.So,6n * (0.5)^(2/3) ≤ 30Divide both sides by 6:n * (0.5)^(2/3) ≤ 5Calculate (0.5)^(2/3):0.5^(2/3) = (2^(-1))^(2/3) = 2^(-2/3) ≈ 0.63So,n * 0.63 ≤ 5Thus,n ≤ 5 / 0.63 ≈ 7.936Since n must be an integer, n ≤ 7.But wait, this is under the assumption that the units are cubes, which might not necessarily be the case. Maybe the minimal surface area occurs when x = y = z, but perhaps the actual x is different.But in the second part, we need to find the maximum n based on the wall length, which depends on x. So, if we can find the minimal possible x, that would allow the maximum n.But from the first part, the surface area constraint might limit how small x can be.Wait, perhaps I need to approach this differently.From the first part, we have:x * y * z = 0.5and2n(xy + yz + xz) ≤ 30We need to find the relationship between x, y, z, and n.But for the second part, we need to find the maximum n such that:n * x + (n - 1) * 0.5 ≤ 10So, we can write:n * x ≤ 10 - 0.5(n - 1)n * x ≤ 10 - 0.5n + 0.5n * x ≤ 10.5 - 0.5nSo,x ≤ (10.5 - 0.5n)/nx ≤ 10.5/n - 0.5But from the volume constraint, x = 0.5/(y * z)So, to minimize x, we need to maximize y * z.But y * z is related to the surface area.Wait, maybe we can express y * z in terms of x, which is 0.5/(y * z) = x, so y * z = 0.5/x.Then, the surface area per unit is 2(xy + yz + xz) = 2(x(y + z) + yz)But yz = 0.5/x, so:Surface area per unit = 2(x(y + z) + 0.5/x)But we don't know y + z. Hmm.Alternatively, maybe we can express y + z in terms of x.Wait, from the volume, y * z = 0.5/x.If we consider y and z as variables with a fixed product, the sum y + z is minimized when y = z, by AM-GM.So, y + z ≥ 2√(yz) = 2√(0.5/x) = 2*(0.5/x)^(1/2) = 2*(√(0.5)/√x) = √(2)/√xTherefore, the surface area per unit is:2(x(y + z) + 0.5/x) ≥ 2(x*(√(2)/√x) + 0.5/x) = 2(√(2x) + 0.5/x)So, the minimal surface area per unit is 2(√(2x) + 0.5/x)Therefore, total surface area is n * 2(√(2x) + 0.5/x) ≤ 30So, we have:2n(√(2x) + 0.5/x) ≤ 30Simplify:n(√(2x) + 0.5/x) ≤ 15But we also have from the wall length:x ≤ (10.5 - 0.5n)/nSo, x ≤ 10.5/n - 0.5Let me denote x = 10.5/n - 0.5 - ε, where ε ≥ 0, but since we are looking for maximum n, we can assume ε = 0 for the minimal x.Wait, actually, to maximize n, we need to minimize x. So, x should be as small as possible, but subject to the surface area constraint.So, substituting x from the wall constraint into the surface area constraint:n(√(2x) + 0.5/x) ≤ 15But x = (10.5 - 0.5n)/nSo,n(√(2*(10.5 - 0.5n)/n) + 0.5/((10.5 - 0.5n)/n)) ≤ 15Simplify:n(√(2*(10.5 - 0.5n)/n) + 0.5n/(10.5 - 0.5n)) ≤ 15Let me compute each term:First term inside the brackets:√(2*(10.5 - 0.5n)/n) = √(21 - n)/√nSecond term:0.5n/(10.5 - 0.5n) = (0.5n)/(10.5 - 0.5n) = (n)/(21 - n)So, the inequality becomes:n(√(21 - n)/√n + n/(21 - n)) ≤ 15Simplify:n*(√(21 - n)/√n) + n*(n/(21 - n)) ≤ 15Which is:√n * √(21 - n) + n²/(21 - n) ≤ 15Let me denote k = √n, so n = k²Then, the inequality becomes:k * √(21 - k²) + k⁴/(21 - k²) ≤ 15This seems complicated, but maybe we can try integer values of n to find the maximum n that satisfies the inequality.Let me try n=7:First term: √7 * √(21 - 7) = √7 * √14 ≈ 2.6458 * 3.7417 ≈ 10.000Second term: 7²/(21 -7) = 49/14 ≈ 3.5Total: 10 + 3.5 = 13.5 ≤15. So, n=7 is possible.Try n=8:First term: √8 * √(21 -8)= √8 * √13 ≈ 2.828 * 3.6055 ≈ 10.19Second term: 64/(21 -8)=64/13≈4.923Total≈10.19 +4.923≈15.113>15. So, n=8 is too much.Wait, but 15.113 is just over 15, so maybe n=8 is possible if we adjust x slightly? Or perhaps the minimal surface area is when x is as small as possible, but maybe the actual surface area is higher, so n=8 might not be possible.Alternatively, maybe n=7 is the maximum.But let's check n=7:x = (10.5 -0.5*7)/7 = (10.5 -3.5)/7=7/7=1So, x=1 meter.Then, y*z=0.5/1=0.5Surface area per unit: 2(xy + yz +xz)=2(1*y +0.5 +1*z)=2(y + z +0.5)But since y*z=0.5, the minimal y+z is √(4*0.5)=√2≈1.414, by AM-GM.So, minimal surface area per unit is 2(1.414 +0.5)=2(1.914)=3.828Total surface area for 7 units:7*3.828≈26.8 ≤30. So, it's okay.If we try n=8:x=(10.5 -0.5*8)/8=(10.5 -4)/8=6.5/8≈0.8125Then, y*z=0.5/0.8125≈0.6154Surface area per unit:2(xy + yz +xz)=2(0.8125y +0.6154 +0.8125z)=2(0.8125(y+z)+0.6154)Again, y+z ≥√(4*0.6154)=√2.4616≈1.569So, minimal surface area per unit≈2(0.8125*1.569 +0.6154)=2(1.274 +0.6154)=2(1.8894)=3.7788Total surface area for 8 units≈8*3.7788≈30.23>30. So, it's over the limit.Therefore, n=8 is not possible because it exceeds the surface area constraint.Thus, the maximum n is 7.Wait, but earlier when I assumed x=1 for n=7, the surface area was 26.8, which is under 30. So, maybe we can actually have a larger x, which would allow more units? Wait, no, because x is determined by the wall length.Wait, no, x is fixed by the wall constraint for n=7 as x=1. So, we can't have a larger x because that would require fewer units.Wait, perhaps if we don't have x=1, but a smaller x, which would allow more units, but then the surface area might go up.Wait, but for n=7, x=1 is the minimal x to fit on the wall. If we make x smaller, we could fit more units, but the surface area would increase.But in the surface area constraint, we have a maximum of 30. So, maybe with n=7, x=1, surface area is 26.8, which is under 30. So, maybe we can have a larger x, but that would allow fewer units, which is not what we want.Alternatively, maybe n=7 is the maximum because n=8 exceeds the surface area.So, the maximum number of units is 7.But let me double-check.For n=7:x=1y*z=0.5Surface area per unit:2(xy + yz +xz)=2(1*y +0.5 +1*z)=2(y + z +0.5)To minimize surface area, y=z=√(0.5)≈0.707So, y+z≈1.414Thus, surface area per unit≈2(1.414 +0.5)=2*1.914≈3.828Total surface area≈7*3.828≈26.8Which is under 30.If we try to make x smaller, say x=0.9, then n would be:From the wall constraint:n *0.9 +0.5*(n-1) ≤100.9n +0.5n -0.5 ≤101.4n ≤10.5n≤7.5So, n=7But x=0.9, then y*z=0.5/0.9≈0.5556Surface area per unit:2(0.9y +0.5556 +0.9z)=2(0.9(y+z)+0.5556)y+z≥√(4*0.5556)=√2.222≈1.491So, surface area per unit≈2(0.9*1.491 +0.5556)=2(1.342 +0.5556)=2*1.8976≈3.795Total surface area≈7*3.795≈26.565, still under 30.So, even with x=0.9, n=7 is possible.But if we try n=8:x=(10.5 -0.5*8)/8=6.5/8≈0.8125y*z=0.5/0.8125≈0.6154Surface area per unit≈2(0.8125(y+z)+0.6154)y+z≥√(4*0.6154)=√2.4616≈1.569So, surface area per unit≈2(0.8125*1.569 +0.6154)=2(1.274 +0.6154)=2*1.8894≈3.7788Total surface area≈8*3.7788≈30.23>30Thus, n=8 is not allowed.Therefore, the maximum number of units is 7.</think>

🧠:<think>Alright, so I have this problem about modeling the weight gain of cows and sheep using differential equations. It's a system of two equations, right? Let me try to understand what each part means.First, for Group A (cows), the differential equation is:[ frac{dW_A}{dt} = aW_A - bW_A^2 + cW_B ]And for Group B (sheep):[ frac{dW_B}{dt} = dW_B - eW_B^2 + fW_A ]Okay, so both groups have their own growth terms, which are linear (aW_A and dW_B) and quadratic terms (-bW_A² and -eW_B²). These quadratic terms probably represent some kind of saturation effect, meaning that as the weight increases, the growth rate slows down because of limited resources or something like that. Then, there are cross terms: cW_B for cows and fW_A for sheep. So, the weight gain of cows is influenced by the sheep's weight and vice versa. Interesting, so they interact in some way.The first part asks to determine the equilibrium points and classify their stability using the Jacobian matrix. Hmm, equilibrium points are where the derivatives are zero, so set dW_A/dt = 0 and dW_B/dt = 0.Let me write down the equations for equilibrium:1. ( aW_A - bW_A^2 + cW_B = 0 )2. ( dW_B - eW_B^2 + fW_A = 0 )So, we have a system of two nonlinear equations. To find equilibrium points, I need to solve these simultaneously.Let me denote equation 1 as:( aW_A - bW_A^2 + cW_B = 0 ) --> equation (1)And equation 2 as:( dW_B - eW_B^2 + fW_A = 0 ) --> equation (2)I can try to solve for one variable in terms of the other. Let's solve equation (1) for W_B:From equation (1):( cW_B = -aW_A + bW_A^2 )So,( W_B = frac{-aW_A + bW_A^2}{c} )Similarly, from equation (2):( dW_B - eW_B^2 = -fW_A )So,( W_B(d - eW_B) = -fW_A )But since I have W_B expressed in terms of W_A from equation (1), I can substitute that into equation (2). Let me do that.Substitute W_B = ( -aW_A + bW_A² ) / c into equation (2):( d left( frac{-aW_A + bW_A^2}{c} right) - e left( frac{-aW_A + bW_A^2}{c} right)^2 + fW_A = 0 )Wow, that looks complicated. Let me simplify step by step.First, compute each term:1. ( d times frac{-aW_A + bW_A^2}{c} = frac{-a d W_A + b d W_A^2}{c} )2. ( e times left( frac{-aW_A + bW_A^2}{c} right)^2 = e times frac{( -aW_A + bW_A^2 )^2}{c^2} )Let me expand the squared term:( (-aW_A + bW_A^2)^2 = a²W_A² - 2abW_A³ + b²W_A⁴ )So, the second term becomes:( frac{e(a²W_A² - 2abW_A³ + b²W_A⁴)}{c²} )Putting it all together, equation (2) becomes:( frac{-a d W_A + b d W_A^2}{c} - frac{e(a²W_A² - 2abW_A³ + b²W_A⁴)}{c²} + fW_A = 0 )To make this manageable, let's multiply both sides by c² to eliminate denominators:( (-a d W_A + b d W_A^2)c - e(a²W_A² - 2abW_A³ + b²W_A⁴) + fW_A c² = 0 )Expanding each term:1. ( (-a d W_A)c = -a d c W_A )2. ( (b d W_A²)c = b d c W_A² )3. ( -e(a²W_A²) = -e a² W_A² )4. ( -e(-2abW_A³) = +2 a b e W_A³ )5. ( -e(b²W_A⁴) = -e b² W_A⁴ )6. ( fW_A c² = f c² W_A )So, combining all these:- a d c W_A + b d c W_A² - e a² W_A² + 2 a b e W_A³ - e b² W_A⁴ + f c² W_A = 0Now, let's collect like terms:- Terms with W_A:(-a d c + f c²) W_A- Terms with W_A²:(b d c - e a²) W_A²- Terms with W_A³:2 a b e W_A³- Terms with W_A⁴:- e b² W_A⁴So, the equation becomes:(-a d c + f c²) W_A + (b d c - e a²) W_A² + 2 a b e W_A³ - e b² W_A⁴ = 0This is a quartic equation in W_A. Solving quartic equations analytically is complicated, but maybe we can factor out W_A:W_A [ (-a d c + f c²) + (b d c - e a²) W_A + 2 a b e W_A² - e b² W_A³ ] = 0So, one solution is W_A = 0. Then, substituting back into equation (1):If W_A = 0, then equation (1) becomes c W_B = 0, so W_B = 0.Therefore, (0, 0) is an equilibrium point.Now, the other solutions come from the cubic equation inside the brackets:(-a d c + f c²) + (b d c - e a²) W_A + 2 a b e W_A² - e b² W_A³ = 0Let me denote this as:A + B W_A + C W_A² + D W_A³ = 0Where:A = -a d c + f c²B = b d c - e a²C = 2 a b eD = -e b²So, the equation is D W_A³ + C W_A² + B W_A + A = 0Given that a, b, c, d, e, f are positive constants, let's see the signs of A, B, C, D:A = -a d c + f c² = c(-a d + f c)Since f and c are positive, if f c > a d, A is positive; otherwise, negative.Similarly, B = b d c - e a², which is positive if b d c > e a², else negative.C = 2 a b e > 0D = -e b² < 0So, the cubic equation is:- e b² W_A³ + 2 a b e W_A² + (b d c - e a²) W_A + ( -a d c + f c² ) = 0This is a cubic equation, which can have up to three real roots. Depending on the constants, we might have multiple equilibrium points.But since the problem gives specific constants, maybe I can compute the roots numerically?Wait, actually, in part 2, they give specific constants: a=0.1, b=0.01, c=0.05, d=0.2, e=0.02, f=0.03.So, maybe I can compute the coefficients A, B, C, D with these values.Let me compute A, B, C, D:First, compute A:A = -a d c + f c²Plugging in the values:A = -0.1 * 0.2 * 0.05 + 0.03 * (0.05)^2Compute each term:-0.1 * 0.2 = -0.02; -0.02 * 0.05 = -0.0010.03 * 0.0025 = 0.000075So, A = -0.001 + 0.000075 = -0.000925B = b d c - e a²Compute:b d c = 0.01 * 0.2 * 0.05 = 0.0001e a² = 0.02 * (0.1)^2 = 0.02 * 0.01 = 0.0002So, B = 0.0001 - 0.0002 = -0.0001C = 2 a b e = 2 * 0.1 * 0.01 * 0.02 = 2 * 0.00002 = 0.00004D = -e b² = -0.02 * (0.01)^2 = -0.02 * 0.0001 = -0.000002So, the cubic equation is:D W_A³ + C W_A² + B W_A + A = 0Plugging in:-0.000002 W_A³ + 0.00004 W_A² - 0.0001 W_A - 0.000925 = 0Hmm, that's a very small coefficients. Maybe I can multiply both sides by 1,000,000 to make it manageable:-2 W_A³ + 40 W_A² - 100 W_A - 925 = 0Wait, let me check:-0.000002 * 1,000,000 = -20.00004 * 1,000,000 = 40-0.0001 * 1,000,000 = -100-0.000925 * 1,000,000 = -925Yes, so the equation becomes:-2 W_A³ + 40 W_A² - 100 W_A - 925 = 0Multiply both sides by -1:2 W_A³ - 40 W_A² + 100 W_A + 925 = 0So, 2 W_A³ - 40 W_A² + 100 W_A + 925 = 0Hmm, solving this cubic equation. Maybe I can try rational roots. The possible rational roots are factors of 925 over factors of 2.Factors of 925: 1, 5, 25, 37, 185, 925Possible roots: ±1, ±5, ±25, ±37, ±185, ±925, ±1/2, ±5/2, etc.Let me test W_A = 5:2*(125) - 40*(25) + 100*5 + 925 = 250 - 1000 + 500 + 925 = (250 + 500 + 925) - 1000 = 1675 - 1000 = 675 ≠ 0W_A = -5:2*(-125) - 40*(25) + 100*(-5) + 925 = -250 - 1000 - 500 + 925 = (-250 - 1000 - 500) + 925 = -1750 + 925 = -825 ≠ 0W_A = 25:2*(15625) - 40*(625) + 100*25 + 925 = 31250 - 25000 + 2500 + 925 = (31250 + 2500 + 925) - 25000 = 34675 - 25000 = 9675 ≠ 0Too big. Maybe W_A = 10:2*1000 - 40*100 + 100*10 + 925 = 2000 - 4000 + 1000 + 925 = (2000 + 1000 + 925) - 4000 = 3925 - 4000 = -75 ≠ 0Close. W_A = 10 gives -75.W_A = 11:2*1331 - 40*121 + 100*11 + 925 = 2662 - 4840 + 1100 + 925 = (2662 + 1100 + 925) - 4840 = 4687 - 4840 = -153 ≠ 0W_A = 12:2*1728 - 40*144 + 100*12 + 925 = 3456 - 5760 + 1200 + 925 = (3456 + 1200 + 925) - 5760 = 5581 - 5760 = -179 ≠ 0Hmm, not helpful. Maybe W_A = 15:2*3375 - 40*225 + 100*15 + 925 = 6750 - 9000 + 1500 + 925 = (6750 + 1500 + 925) - 9000 = 9175 - 9000 = 175 ≠ 0Wait, positive now. So between 12 and 15, the function goes from -179 to +175, so there is a root between 12 and 15.Similarly, let's try W_A = 13:2*2197 - 40*169 + 100*13 + 925 = 4394 - 6760 + 1300 + 925 = (4394 + 1300 + 925) - 6760 = 6619 - 6760 = -141 ≠ 0W_A = 14:2*2744 - 40*196 + 100*14 + 925 = 5488 - 7840 + 1400 + 925 = (5488 + 1400 + 925) - 7840 = 7813 - 7840 = -27 ≠ 0W_A = 14.5:2*(14.5)^3 - 40*(14.5)^2 + 100*14.5 + 925Compute 14.5³: 14.5*14.5=210.25; 210.25*14.5≈3048.625So, 2*3048.625≈6097.2514.5²=210.25; 40*210.25=8410100*14.5=1450So, total: 6097.25 - 8410 + 1450 + 925 ≈ (6097.25 + 1450 + 925) - 8410 ≈ 8472.25 - 8410 ≈ 62.25 > 0So, at W_A=14.5, f(W_A)=62.25At W_A=14, f= -27; at 14.5, f=62.25. So, root between 14 and 14.5.Let me use linear approximation.Between W_A=14 (-27) and W_A=14.5 (62.25). The difference in f is 62.25 - (-27)=89.25 over 0.5 interval.We need to find delta where f=0:delta = (0 - (-27)) / 89.25 * 0.5 ≈ (27 / 89.25)*0.5 ≈ (0.3025)*0.5 ≈ 0.15125So, approximate root at 14 + 0.15125 ≈14.15125Similarly, check W_A=14.15:Compute f(W_A)=2*(14.15)^3 -40*(14.15)^2 +100*14.15 +925Compute 14.15³:14.15²=200.2225; 200.2225*14.15≈200.2225*14 +200.2225*0.15≈2803.115 +30.033≈2833.148So, 2*2833.148≈5666.29614.15²=200.2225; 40*200.2225≈8008.9100*14.15=1415So, f=5666.296 -8008.9 +1415 +925≈(5666.296 +1415 +925) -8008.9≈8006.296 -8008.9≈-2.604Close to zero. So, f(14.15)≈-2.604Similarly, at W_A=14.15, f≈-2.6; at W_A=14.15 + delta, f=0.Compute derivative at W_A=14.15:f'(W_A)=6 W_A² -80 W_A +100At W_A=14.15:6*(14.15)^2 -80*14.15 +10014.15²=200.2225; 6*200.2225≈1201.33580*14.15=1132So, f'≈1201.335 -1132 +100≈169.335So, using Newton-Raphson:Next approximation: W_A =14.15 - f(W_A)/f'(W_A)=14.15 - (-2.604)/169.335≈14.15 +0.0154≈14.1654Compute f(14.1654):14.1654³≈(14.1654)^3. Let me compute 14.1654² first:14.1654²≈200.666 (since 14²=196, 0.1654²≈0.027, cross term 2*14*0.1654≈4.631; total≈196 +4.631 +0.027≈200.658)Then, 14.1654³≈14.1654*200.658≈14*200.658 +0.1654*200.658≈2809.212 +33.19≈2842.402So, 2*2842.402≈5684.80414.1654²≈200.658; 40*200.658≈8026.32100*14.1654≈1416.54So, f≈5684.804 -8026.32 +1416.54 +925≈(5684.804 +1416.54 +925) -8026.32≈8026.344 -8026.32≈0.024Almost zero. So, f≈0.024 at W_A≈14.1654So, approximately, W_A≈14.165So, one equilibrium point is around W_A≈14.165 kgThen, substituting back into equation (1):W_B = (-a W_A + b W_A²)/cPlugging in the values:a=0.1, b=0.01, c=0.05, W_A≈14.165Compute numerator:-0.1*14.165 +0.01*(14.165)^2Compute each term:-0.1*14.165≈-1.41650.01*(14.165)^2≈0.01*200.658≈2.00658So, numerator≈-1.4165 +2.00658≈0.59008Thus, W_B≈0.59008 /0.05≈11.8016 kgSo, another equilibrium point is approximately (14.165, 11.802)Wait, but we had a quartic equation earlier, so maybe there are more equilibrium points?Wait, no, because after factoring out W_A, the remaining equation was a cubic, which can have up to three real roots. But in our case, with the given constants, we found one positive real root. The other roots might be negative or complex.But since weight can't be negative, we only consider positive roots. So, in this case, we have two equilibrium points: (0,0) and approximately (14.165, 11.802)Wait, but let me check if the cubic equation has more positive roots.We found one at around 14.165. Let me check behavior as W_A approaches infinity:The leading term is -e b² W_A³, which is negative, so as W_A→∞, f(W_A)→-∞Wait, but in our scaled equation, it was 2 W_A³ -40 W_A² +100 W_A +925, which as W_A→∞, tends to +∞. Wait, no, because we multiplied by -1 earlier.Wait, original equation after scaling was:-2 W_A³ +40 W_A² -100 W_A -925=0So, as W_A→∞, -2 W_A³ dominates, so f(W_A)→-∞At W_A=0, f(0)= -925At W_A=14.165, f≈0As W_A increases beyond 14.165, f becomes negative again. So, only one positive real root.Therefore, only two equilibrium points: (0,0) and (14.165, 11.802)Wait, but let me confirm with the original equations.At (0,0): both W_A and W_B are zero. That's trivial.At (14.165, 11.802): positive weights.So, these are the only two equilibrium points.Now, to classify their stability, we need to compute the Jacobian matrix at each equilibrium point and find the eigenvalues.The Jacobian matrix J is given by:[ ∂(dW_A/dt)/∂W_A , ∂(dW_A/dt)/∂W_B ][ ∂(dW_B/dt)/∂W_A , ∂(dW_B/dt)/∂W_B ]Compute the partial derivatives:From dW_A/dt = a W_A - b W_A² + c W_B∂(dW_A/dt)/∂W_A = a - 2b W_A∂(dW_A/dt)/∂W_B = cFrom dW_B/dt = d W_B - e W_B² + f W_A∂(dW_B/dt)/∂W_A = f∂(dW_B/dt)/∂W_B = d - 2e W_BSo, Jacobian matrix J is:[ a - 2b W_A , c ][ f , d - 2e W_B ]Now, evaluate J at each equilibrium point.First, at (0,0):J = [ a , c ] [ f , d ]So, J = [0.1, 0.05; 0.03, 0.2]Compute eigenvalues. The eigenvalues λ satisfy:det(J - λ I) = 0So,|0.1 - λ 0.05 ||0.03 0.2 - λ | = 0Compute determinant:(0.1 - λ)(0.2 - λ) - (0.05)(0.03) = 0Expand:0.02 - 0.1λ - 0.2λ + λ² - 0.0015 = 0Simplify:λ² - 0.3λ + 0.02 - 0.0015 = λ² - 0.3λ + 0.0185 = 0Solve quadratic equation:λ = [0.3 ± sqrt(0.09 - 4*1*0.0185)] / 2Compute discriminant:0.09 - 0.074 = 0.016So,λ = [0.3 ± sqrt(0.016)] / 2sqrt(0.016)≈0.12649Thus,λ1≈(0.3 +0.12649)/2≈0.42649/2≈0.213245λ2≈(0.3 -0.12649)/2≈0.17351/2≈0.086755Both eigenvalues are positive, so the equilibrium point (0,0) is an unstable node.Now, evaluate J at (14.165, 11.802):Compute each entry:∂(dW_A/dt)/∂W_A = a - 2b W_A =0.1 - 2*0.01*14.165≈0.1 -0.2833≈-0.1833∂(dW_A/dt)/∂W_B = c=0.05∂(dW_B/dt)/∂W_A = f=0.03∂(dW_B/dt)/∂W_B = d - 2e W_B=0.2 - 2*0.02*11.802≈0.2 -0.47208≈-0.27208So, Jacobian matrix J is:[ -0.1833 , 0.05 ][ 0.03 , -0.27208 ]Compute eigenvalues. The characteristic equation is:det(J - λ I)=0So,| -0.1833 - λ 0.05 || 0.03 -0.27208 - λ | =0Compute determinant:(-0.1833 - λ)(-0.27208 - λ) - (0.05)(0.03)=0First, expand the product:(0.1833 + λ)(0.27208 + λ) -0.0015=0Compute (0.1833 + λ)(0.27208 + λ):=0.1833*0.27208 +0.1833λ +0.27208λ +λ²≈0.0500 +0.45538λ +λ²So, equation becomes:0.0500 +0.45538λ +λ² -0.0015=0Simplify:λ² +0.45538λ +0.0485=0Compute discriminant:(0.45538)^2 -4*1*0.0485≈0.2074 -0.194≈0.0134So,λ = [-0.45538 ± sqrt(0.0134)] /2sqrt(0.0134)≈0.1158Thus,λ1≈(-0.45538 +0.1158)/2≈(-0.33958)/2≈-0.1698λ2≈(-0.45538 -0.1158)/2≈(-0.57118)/2≈-0.2856Both eigenvalues are negative, so the equilibrium point (14.165, 11.802) is a stable node.Therefore, the system has two equilibrium points: (0,0) which is unstable, and approximately (14.165, 11.802) which is stable.Now, moving to part 2: simulate the system with initial conditions W_A(0)=50 kg, W_B(0)=30 kg, and constants a=0.1, b=0.01, c=0.05, d=0.2, e=0.02, f=0.03 over 100 days.But wait, the equilibrium points we found were around (14.165, 11.802). However, the initial conditions are much higher: 50 and 30. So, I wonder if the system will approach the stable equilibrium or diverge.But given that the stable equilibrium is around (14,12), and initial conditions are higher, perhaps the system will decrease towards the equilibrium.But let me think about the system dynamics.Given the equations:dW_A/dt =0.1 W_A -0.01 W_A² +0.05 W_BdW_B/dt =0.2 W_B -0.02 W_B² +0.03 W_ASo, both groups have positive growth terms, but also negative quadratic terms, and positive cross terms.Given that the initial weights are higher than the equilibrium, the quadratic terms will dominate, leading to negative growth rates, causing the weights to decrease towards the equilibrium.But let me confirm.At W_A=50, W_B=30:Compute dW_A/dt=0.1*50 -0.01*2500 +0.05*30=5 -25 +1.5= -18.5dW_B/dt=0.2*30 -0.02*900 +0.03*50=6 -18 +1.5= -10.5So, both derivatives are negative, meaning both weights will decrease initially.As they decrease, the quadratic terms will decrease, and the cross terms will become more significant.Eventually, they should approach the equilibrium point.To simulate this, I can use numerical methods like Euler's method or Runge-Kutta. Since this is a thought process, I can outline the steps.But since I can't actually compute the numerical simulation here, I can discuss the expected behavior.Given that the stable equilibrium is at around (14.165, 11.802), and the initial conditions are above that, the system should spiral or approach this equilibrium.But since both eigenvalues at the equilibrium are real and negative, it's a stable node, so the approach will be monotonic, not oscillatory.Therefore, over time, both W_A and W_B will decrease towards their equilibrium values.Implications for the holistic approach: The model suggests that the weight gains will stabilize at certain levels despite initial higher weights. The interaction between cows and sheep (positive cross terms) affects each other's weight gain. The stable equilibrium indicates that the system will reach a balance where both groups contribute to each other's weight gain without unbounded growth, which could be due to resource limitations (quadratic terms). This balance is crucial for a holistic farm management approach, ensuring sustainable growth without overexploitation of resources.So, in conclusion, the system will approach the stable equilibrium, indicating that the feed's impact leads to a balanced weight gain for both cows and sheep, supporting a holistic health approach.Final Answer1. The equilibrium points are (boxed{(0, 0)}) (unstable) and approximately (boxed{(14.17, 11.80)}) (stable).2. The system approaches the stable equilibrium, indicating balanced weight gain for cows and sheep over time.</think>