🧠:<think>Alright, so I have this problem where a Ukrainian human rights organization needs to allocate a budget of 1,000,000 across five regions: A, B, C, D, and E. The goal is to maximize the total impact on human rights conditions. Each region has a different impact factor, which is the improvement per dollar allocated. First, I need to understand what exactly is being asked. It's a linear programming problem, right? So, I have to formulate the problem with variables, an objective function, and constraints. Then, solve it to find the optimal allocation. After that, there's a second part where an additional constraint is introduced, and I have to re-solve it.Starting with part 1. The impact factors are given as:- I_A = 0.5- I_B = 0.3- I_C = 0.4- I_D = 0.6- I_E = 0.2So, the higher the impact factor, the more effective each dollar is in that region. That makes sense. So, intuitively, to maximize the total impact, we should allocate as much as possible to the region with the highest impact factor, which is region D with I_D = 0.6. Then, if there's remaining budget, allocate to the next highest, which is region A with 0.5, followed by region C with 0.4, then B with 0.3, and finally E with 0.2.But since this is a linear programming problem, I should set it up formally.Let me define the variables:Let x_A, x_B, x_C, x_D, x_E be the amounts allocated to regions A, B, C, D, E respectively in dollars.Our objective is to maximize the total impact, which is the sum of (impact factor * allocation) for each region. So, the objective function is:Maximize Z = 0.5x_A + 0.3x_B + 0.4x_C + 0.6x_D + 0.2x_ESubject to the constraints:1. The total allocation cannot exceed the budget: x_A + x_B + x_C + x_D + x_E ≤ 1,000,0002. All allocations must be non-negative: x_A, x_B, x_C, x_D, x_E ≥ 0That's the standard setup for a linear programming problem where we want to maximize a linear objective function subject to linear constraints.Now, to solve this, since it's a maximization problem with a single constraint (apart from non-negativity), the optimal solution will allocate as much as possible to the variable with the highest coefficient in the objective function, then the next highest, and so on until the budget is exhausted.Looking at the coefficients:- x_D: 0.6 (highest)- x_A: 0.5- x_C: 0.4- x_B: 0.3- x_E: 0.2 (lowest)So, the optimal allocation should be to put as much as possible into D, then A, then C, then B, then E.Since the budget is 1,000,000, we allocate all of it to D first. But wait, can we? There's no upper limit on each region except the total budget. So, if we put all 1,000,000 into D, that would give the maximum impact.But let me verify that. If we allocate all to D, the total impact would be 0.6 * 1,000,000 = 600,000.Alternatively, if we allocate some to A, which has the next highest impact factor, but since 0.6 is higher than 0.5, it's better to allocate all to D.So, the optimal solution is x_D = 1,000,000 and x_A = x_B = x_C = x_E = 0.Wait, but is there any constraint that prevents allocating all to D? The problem doesn't specify any other constraints except the total budget and non-negativity. So, yes, this should be the optimal solution.But let me think again. Sometimes, in real-world scenarios, you might have constraints like minimum allocations or other considerations, but in this case, the problem only mentions the total budget and non-negativity. So, I think it's correct.Moving on to part 2. Now, an additional constraint is introduced: Region D must receive at least 200,000. So, we have to ensure that x_D ≥ 200,000.So, we need to reformulate the linear programming problem with this new constraint.The new constraints are:1. x_A + x_B + x_C + x_D + x_E ≤ 1,000,0002. x_D ≥ 200,0003. x_A, x_B, x_C, x_D, x_E ≥ 0So, now, we have to allocate at least 200,000 to D, and the rest can be allocated as per the impact factors.Since D still has the highest impact factor, after allocating the minimum required to D, we should allocate the remaining budget to the next highest impact regions.So, the remaining budget after allocating 200,000 to D is 1,000,000 - 200,000 = 800,000.Now, we need to allocate this 800,000 to the remaining regions in order of their impact factors.The impact factors are:- A: 0.5- C: 0.4- B: 0.3- E: 0.2So, we should allocate as much as possible to A first, then C, then B, then E.But since there's no upper limit on the allocations except the total budget, we can allocate all 800,000 to A, which has the next highest impact factor.Wait, but let me check: If we allocate all 800,000 to A, then the total allocation would be x_D = 200,000 and x_A = 800,000, with the rest zero. That would give a total impact of 0.6*200,000 + 0.5*800,000 = 120,000 + 400,000 = 520,000.Alternatively, if we allocate some to A and some to C, would that give a higher impact? Let's see.Suppose we allocate x_A = 800,000, then impact is 0.5*800,000 = 400,000.If we allocate x_A = 700,000 and x_C = 100,000, the impact would be 0.5*700,000 + 0.4*100,000 = 350,000 + 40,000 = 390,000, which is less than 400,000.Similarly, any allocation to C instead of A would decrease the total impact because 0.5 > 0.4.Therefore, the optimal allocation after the constraint is x_D = 200,000 and x_A = 800,000, with x_B = x_C = x_E = 0.Wait, but let me think again. Is there a possibility that allocating some to C could allow more allocation to A? No, because the total remaining budget is fixed at 800,000. So, allocating more to A means less to others, but since A has a higher impact factor, it's better to allocate as much as possible to A.Alternatively, if the impact factors were such that a combination could yield a higher total, but in this case, since 0.5 is higher than 0.4, it's better to maximize A.So, the optimal solution is x_D = 200,000, x_A = 800,000, and the rest zero.But let me verify this with the linear programming approach.In linear programming, when you have a constraint that a variable must be at least a certain value, you can think of it as fixing that variable at the minimum and then optimizing the rest.So, in this case, we fix x_D = 200,000, and then the remaining budget is 800,000, which we allocate to the next highest impact regions, which is A, then C, etc.Since A has the highest impact factor among the remaining, we allocate all 800,000 to A.Therefore, the new optimal allocations are:x_D = 200,000x_A = 800,000x_B = x_C = x_E = 0This should give the maximum total impact under the new constraint.I think that's correct. Let me just recap:Without the constraint, all goes to D.With the constraint that D must get at least 200,000, we allocate 200,000 to D, and the rest to the next highest impact region, which is A.Yes, that makes sense.So, summarizing:1. Optimal allocation without additional constraints: All 1,000,000 to D.2. Optimal allocation with D needing at least 200,000: 200,000 to D, and the remaining 800,000 to A.I think that's the solution.</think>

🧠:<think>Okay, so I'm trying to solve this problem about the harmonic series generated by early electronic synthesizers. Let me take it step by step.First, part 1 asks me to derive the general equation for the nth harmonic frequency ( f_n ) when the fundamental frequency ( f_0 ) is 440 Hz. Hmm, I remember that harmonics are integer multiples of the fundamental frequency. So, the first harmonic is ( f_0 ), the second is ( 2f_0 ), the third is ( 3f_0 ), and so on. Therefore, the nth harmonic should be ( f_n = n f_0 ). Let me write that down: ( f_n = n times f_0 ). Since ( f_0 ) is given as 440 Hz, then for the 12th harmonic, n is 12. So, ( f_{12} = 12 times 440 ) Hz. Let me calculate that: 12 times 400 is 4800, and 12 times 40 is 480, so adding them together gives 5280 Hz. So, the 12th harmonic is 5280 Hz.Moving on to part 2. The synthesizer produces a waveform modeled as a sum of its harmonics: ( S(t) = sum_{n=1}^{N} frac{1}{n} sin(2pi n f_0 t) ). It says the amplitude of each harmonic decreases inversely with its harmonic number, which makes sense because the coefficient is ( frac{1}{n} ). So, the Fourier series representation is already given here, right? It's a sum of sine terms with decreasing amplitudes.But wait, the question says to derive the Fourier series representation. Hmm, maybe I need to recognize that this is a Fourier series of a specific function. I recall that the sum ( sum_{n=1}^{infty} frac{sin(ntheta)}{n} ) is a known Fourier series. Isn't that related to the sawtooth wave or something similar?Let me think. The Fourier series for a sawtooth wave is indeed ( sum_{n=1}^{infty} frac{sin(ntheta)}{n} ), which converges to ( frac{pi - theta}{2} ) for ( 0 < theta < 2pi ). So, in this case, ( theta = 2pi f_0 t ), so the function ( S(t) ) would be similar to a sawtooth wave but scaled and shifted.But since we're only summing up to N=5, it's a partial sum of the Fourier series. So, the Fourier series representation is as given, and it's a partial sum approximation of a sawtooth wave.Now, the second part of question 2 asks me to determine the value of ( S(t) ) at ( t = frac{1}{8 f_0} ) for ( N = 5 ). Let me plug in the values.First, ( t = frac{1}{8 f_0} ). So, let's compute each term in the sum for n from 1 to 5.The general term is ( frac{1}{n} sin(2pi n f_0 t) ). Plugging in t:( frac{1}{n} sinleft(2pi n f_0 times frac{1}{8 f_0}right) )Simplify the argument of the sine function:( 2pi n f_0 times frac{1}{8 f_0} = frac{2pi n}{8} = frac{pi n}{4} )So, each term becomes ( frac{1}{n} sinleft(frac{pi n}{4}right) ).Now, let's compute this for n = 1 to 5.For n=1:( frac{1}{1} sinleft(frac{pi}{4}right) = sinleft(frac{pi}{4}right) = frac{sqrt{2}}{2} approx 0.7071 )For n=2:( frac{1}{2} sinleft(frac{2pi}{4}right) = frac{1}{2} sinleft(frac{pi}{2}right) = frac{1}{2} times 1 = 0.5 )For n=3:( frac{1}{3} sinleft(frac{3pi}{4}right) = frac{1}{3} times frac{sqrt{2}}{2} approx frac{1}{3} times 0.7071 approx 0.2357 )For n=4:( frac{1}{4} sinleft(frac{4pi}{4}right) = frac{1}{4} sin(pi) = frac{1}{4} times 0 = 0 )For n=5:( frac{1}{5} sinleft(frac{5pi}{4}right) = frac{1}{5} times (-frac{sqrt{2}}{2}) approx frac{1}{5} times (-0.7071) approx -0.1414 )Now, summing all these up:n=1: ~0.7071n=2: +0.5 → total ~1.2071n=3: +0.2357 → total ~1.4428n=4: +0 → total remains ~1.4428n=5: -0.1414 → total ~1.3014So, approximately 1.3014.But let me compute it more accurately.Compute each term precisely:n=1: sin(π/4) = √2/2 ≈ 0.7071067812n=2: sin(π/2) = 1, so 1/2 = 0.5n=3: sin(3π/4) = √2/2 ≈ 0.7071067812, so 1/3 * 0.7071067812 ≈ 0.2357022604n=4: sin(π) = 0, so 0n=5: sin(5π/4) = -√2/2 ≈ -0.7071067812, so 1/5 * (-0.7071067812) ≈ -0.1414213562Now, adding them up:0.7071067812 + 0.5 = 1.20710678121.2071067812 + 0.2357022604 = 1.44280904161.4428090416 + 0 = 1.44280904161.4428090416 - 0.1414213562 ≈ 1.3013876854So, approximately 1.3013876854. Let me round this to, say, four decimal places: 1.3014.But let me check if I did everything correctly.Wait, the time t is 1/(8f0). So, plugging into the sine function, each term is sin(2πn f0 t) = sin(2πn f0 * 1/(8f0)) = sin(2πn /8) = sin(πn/4). That's correct.So, for each n, we have sin(πn/4). So, n=1: π/4, n=2: π/2, n=3: 3π/4, n=4: π, n=5:5π/4. Correct.So, the sine values are:n=1: √2/2 ≈0.7071n=2:1n=3:√2/2n=4:0n=5:-√2/2So, the coefficients are 1/n times these values.So, adding them up:n=1: 0.7071n=2: 0.5n=3: 0.2357n=4:0n=5:-0.1414Total: 0.7071 + 0.5 = 1.2071; +0.2357 = 1.4428; -0.1414 = 1.3014.Yes, that seems correct.So, the value of S(t) at t=1/(8f0) is approximately 1.3014.But let me express it more precisely. Since √2 ≈1.41421356, so √2/2≈0.70710678.So, n=1: 0.70710678n=2:0.5n=3:0.70710678/3≈0.23570226n=5: -0.70710678/5≈-0.14142136Adding them:0.70710678 + 0.5 = 1.207106781.20710678 + 0.23570226 = 1.442809041.44280904 - 0.14142136 ≈1.30138768So, approximately 1.30138768.I can write it as approximately 1.3014.But maybe we can express it in terms of √2. Let's see:Compute each term:n=1: √2/2n=2:1/2n=3:√2/6n=5:-√2/10So, total S(t) = √2/2 + 1/2 + √2/6 - √2/10Combine the √2 terms:√2/2 + √2/6 - √2/10 = √2*(1/2 + 1/6 -1/10)Compute the coefficients:1/2 = 15/301/6 =5/301/10=3/30So, 15/30 +5/30 -3/30=17/30Thus, √2*(17/30)=17√2/30Then, add the 1/2 term:Total S(t)=17√2/30 +1/2Convert 1/2 to 15/30:So, 17√2/30 +15/30= (15 +17√2)/30So, exact value is (15 +17√2)/30. Let me compute that:17√2≈17*1.41421356≈24.0416305215 +24.04163052≈39.04163052Divide by 30:≈1.30138768So, exactly, it's (15 +17√2)/30, which is approximately 1.3014.So, that's the exact value.Therefore, the value of S(t) at t=1/(8f0) for N=5 is (15 +17√2)/30, approximately 1.3014.Now, exploring the mathematical and musical relationship.Mathematically, the waveform is a sum of sine waves with decreasing amplitudes. This is a Fourier series, specifically a partial sum of the Fourier series for a sawtooth wave. The sawtooth wave has a sharp rising edge and a gradual fall, and its Fourier series includes all integer harmonics with amplitudes decreasing as 1/n. So, as N increases, the approximation gets closer to a sawtooth wave.Musically, the presence of multiple harmonics creates a timbre that is rich and complex. The fundamental frequency determines the pitch, while the harmonics contribute to the tone color. In this case, since the amplitudes decrease with each harmonic, the higher harmonics are softer, which might give the sound a smoother quality compared to a square wave or a sawtooth wave with equal amplitude harmonics.The waveform at t=1/(8f0) is a specific point in time. Since t=1/(8f0), the period T=1/f0, so 1/(8f0)=T/8. So, we're evaluating the waveform at 1/8th of the period. For a sawtooth wave, at t=T/8, the value would be linearly increasing from 0 to some maximum over the period. However, since we're using a partial sum, the approximation might not be perfect, but it should still show a rising trend.Looking at the calculated value of approximately 1.3014, which is positive, it suggests that at this point in time, the waveform is in the rising phase. The exact value depends on the contributions of each harmonic, with the lower harmonics having a more significant impact due to their higher amplitudes.An interesting pattern is that the waveform is constructed from an infinite series of sine waves, each contributing to the overall shape. Even with just five harmonics, we can see the beginnings of a sawtooth-like shape, though it's still somewhat jagged. As more harmonics are added, the waveform becomes smoother and more like a true sawtooth wave.Another observation is the Gibbs phenomenon, where the Fourier series overshoots near discontinuities. However, since we're only summing up to N=5, the overshoot isn't very pronounced yet. As N increases, the overshoot becomes more noticeable near the discontinuity, but the overall shape still converges to the sawtooth wave except near the jumps.In terms of music, this waveform would produce a sound with a certain brightness or richness due to the presence of higher harmonics, even though their amplitudes are decreasing. The fundamental frequency sets the pitch, and the harmonics add complexity to the sound, making it more interesting than a pure sine wave.So, in summary, the nth harmonic frequency is ( f_n = n f_0 ), and the 12th harmonic is 5280 Hz. The Fourier series is a sum of sine terms with decreasing amplitudes, and at t=1/(8f0), the value is approximately 1.3014, which is a point in the rising phase of the waveform, illustrating how the harmonics contribute to the overall shape and timbre of the sound.</think>

🧠:<think>Alright, so I have this problem about calculating the trajectory of a spacecraft using a Hohmann transfer orbit. I'm supposed to find the semi-major axis, the period of the transfer orbit, and the total change in velocity required. Hmm, okay, let me try to remember what I know about Hohmann transfers.First, I recall that a Hohmann transfer is an elliptical orbit that allows a spacecraft to move from one circular orbit to another. It's the most efficient way in terms of fuel usage, so that makes sense why it's being used here. The transfer orbit touches both the Earth's orbit and Mars' orbit, so the perihelion is at Earth's orbit and the aphelion is at Mars' orbit.Given data:- Earth's orbital radius, ( r_1 = 1 ) AU- Mars' orbital radius, ( r_2 = 1.524 ) AU- Gravitational constant times the mass of the Sun, ( mu = 1.327 times 10^{20} ) m³/s²I need to calculate the semi-major axis ( a ) of the Hohmann transfer orbit. I think the semi-major axis is the average of the perihelion and aphelion distances. So, since the transfer orbit goes from Earth's orbit to Mars' orbit, the perihelion is ( r_1 ) and the aphelion is ( r_2 ). Therefore, the semi-major axis should be:( a = frac{r_1 + r_2}{2} )Let me plug in the numbers:( a = frac{1 + 1.524}{2} ) AUCalculating that:( a = frac{2.524}{2} = 1.262 ) AUOkay, so the semi-major axis is 1.262 AU. That seems straightforward.Next, I need to find the period ( T ) of the Hohmann transfer orbit. I remember Kepler's third law, which relates the period of an orbit to its semi-major axis. The formula is:( T = 2pi sqrt{frac{a^3}{mu}} )But wait, the units here are a bit mixed. The semi-major axis is given in AU, and ( mu ) is in m³/s². I need to make sure the units are consistent. Let me convert AU to meters because ( mu ) is already in meters.1 AU is approximately ( 1.496 times 10^{11} ) meters. So, 1.262 AU is:( a = 1.262 times 1.496 times 10^{11} ) mCalculating that:First, multiply 1.262 by 1.496:1.262 * 1.496 ≈ 1.889So, ( a ≈ 1.889 times 10^{11} ) meters.Now, plug this into Kepler's third law:( T = 2pi sqrt{frac{(1.889 times 10^{11})^3}{1.327 times 10^{20}}} )First, calculate the cube of ( a ):( (1.889 times 10^{11})^3 = (1.889)^3 times 10^{33} )Calculating ( 1.889^3 ):1.889 * 1.889 ≈ 3.5683.568 * 1.889 ≈ 6.735So, ( a^3 ≈ 6.735 times 10^{33} ) m³Now, divide that by ( mu ):( frac{6.735 times 10^{33}}{1.327 times 10^{20}} ≈ 5.076 times 10^{13} )Taking the square root:( sqrt{5.076 times 10^{13}} ≈ 7.125 times 10^{6} ) secondsMultiply by ( 2pi ):( T ≈ 2 times 3.1416 times 7.125 times 10^{6} ≈ 44.78 times 10^{6} ) secondsConvert seconds to years to make it more understandable. There are approximately ( 3.154 times 10^{7} ) seconds in a year.So, ( T ≈ frac{44.78 times 10^{6}}{3.154 times 10^{7}} ≈ 1.42 ) yearsWait, that seems a bit long. Let me double-check my calculations.Wait, maybe I messed up the cube calculation. Let me recalculate ( a^3 ):( a = 1.889 times 10^{11} ) mSo, ( a^3 = (1.889)^3 times (10^{11})^3 = 6.735 times 10^{33} ) m³. That seems correct.Then, ( frac{6.735 times 10^{33}}{1.327 times 10^{20}} ≈ 5.076 times 10^{13} ). Correct.Square root of that is ( sqrt{5.076 times 10^{13}} ). Let me compute that more accurately.( sqrt{5.076 times 10^{13}} = sqrt{5.076} times 10^{6.5} )( sqrt{5.076} ≈ 2.253 )( 10^{6.5} = 10^{6} times 10^{0.5} ≈ 3.162 times 10^{6} )So, multiplying together: 2.253 * 3.162 ≈ 7.125, so 7.125 * 10^6 seconds. That's correct.Then, 7.125 * 10^6 seconds * 2π ≈ 44.78 * 10^6 seconds.Convert to years: 44.78e6 / 3.154e7 ≈ 1.42 years. Hmm, okay, so about 1.42 years. That seems plausible because a Hohmann transfer from Earth to Mars takes roughly 8-9 months, but wait, 1.42 years is about 17 months, which is longer than that. Hmm, maybe I made a mistake in unit conversion.Wait, no, actually, the period of the transfer orbit is longer than Earth's year, which is 1 year, so 1.42 years is correct. But the time taken for the transfer is half the period because the spacecraft only goes from perihelion to aphelion, which is half the orbit. So, the transfer time is ( T/2 ≈ 0.71 ) years, which is about 8.5 months. That aligns with what I know. So, the period is 1.42 years, but the transfer time is half that.But the question specifically asks for the period of the Hohmann transfer orbit, so 1.42 years is correct.Wait, but let me check if I used the right formula. The formula ( T = 2pi sqrt{a^3/mu} ) is correct for the period, yes. So, 1.42 years is the period, and half of that is the transfer time.Okay, moving on to part 2: calculating the total change in velocity ( Delta v ) required for the Hohmann transfer.I remember that the total ( Delta v ) is the sum of two burns: one to leave Earth's orbit into the transfer orbit, and another to match Mars' orbit.The formula for the velocity change is:( Delta v = Delta v_1 + Delta v_2 )Where:- ( Delta v_1 ) is the velocity change to enter the transfer orbit from Earth's orbit.- ( Delta v_2 ) is the velocity change to leave the transfer orbit and enter Mars' orbit.First, I need to find the velocities at Earth's orbit, in the transfer orbit at Earth's radius, and at Mars' orbit in the transfer orbit.The velocity in a circular orbit is given by:( v = sqrt{frac{mu}{r}} )So, Earth's orbital velocity ( v_1 ) is:( v_1 = sqrt{frac{mu}{r_1}} )Similarly, Mars' orbital velocity ( v_2 ) is:( v_2 = sqrt{frac{mu}{r_2}} )But for the transfer orbit, the velocities at Earth's and Mars' radii are different. The velocity at perihelion (Earth's radius) in the transfer orbit is higher than Earth's circular velocity, and the velocity at aphelion (Mars' radius) is lower than Mars' circular velocity.The velocities in the transfer orbit can be found using the vis-viva equation:( v = sqrt{mu left( frac{2}{r} - frac{1}{a} right)} )Where ( r ) is the current radius, and ( a ) is the semi-major axis of the transfer orbit.So, at Earth's radius ( r_1 ), the velocity in the transfer orbit ( v_t1 ) is:( v_t1 = sqrt{mu left( frac{2}{r_1} - frac{1}{a} right)} )Similarly, at Mars' radius ( r_2 ), the velocity in the transfer orbit ( v_t2 ) is:( v_t2 = sqrt{mu left( frac{2}{r_2} - frac{1}{a} right)} )Therefore, the velocity changes are:( Delta v_1 = v_t1 - v_1 )( Delta v_2 = v_2 - v_t2 )So, total ( Delta v = Delta v_1 + Delta v_2 )Let me compute each step.First, compute Earth's orbital velocity ( v_1 ):( v_1 = sqrt{frac{mu}{r_1}} )But wait, ( r_1 ) is given in AU, and ( mu ) is in m³/s². So, I need to convert ( r_1 ) to meters.1 AU = ( 1.496 times 10^{11} ) m, so ( r_1 = 1.496 times 10^{11} ) m.So,( v_1 = sqrt{frac{1.327 times 10^{20}}{1.496 times 10^{11}}} )Calculate the division:( frac{1.327 times 10^{20}}{1.496 times 10^{11}} ≈ 8.87 times 10^{8} )So, ( v_1 = sqrt{8.87 times 10^{8}} ≈ 29780 ) m/sWait, that seems high. Wait, Earth's orbital velocity is approximately 29.78 km/s, so 29780 m/s is correct.Similarly, compute Mars' orbital velocity ( v_2 ):( v_2 = sqrt{frac{mu}{r_2}} )( r_2 = 1.524 ) AU = ( 1.524 times 1.496 times 10^{11} ) m ≈ ( 2.279 times 10^{11} ) mSo,( v_2 = sqrt{frac{1.327 times 10^{20}}{2.279 times 10^{11}}} )Calculate the division:( frac{1.327 times 10^{20}}{2.279 times 10^{11}} ≈ 5.825 times 10^{8} )So, ( v_2 = sqrt{5.825 times 10^{8}} ≈ 24135 ) m/sOkay, so Earth's velocity is ~29.78 km/s, Mars' is ~24.14 km/s.Now, compute the velocities in the transfer orbit at Earth's and Mars' radii.First, ( v_t1 ):( v_t1 = sqrt{mu left( frac{2}{r_1} - frac{1}{a} right)} )We have ( a = 1.262 ) AU = ( 1.262 times 1.496 times 10^{11} ) m ≈ ( 1.889 times 10^{11} ) mSo,( frac{2}{r_1} = frac{2}{1.496 times 10^{11}} ≈ 1.337 times 10^{-11} ) 1/m( frac{1}{a} = frac{1}{1.889 times 10^{11}} ≈ 5.297 times 10^{-12} ) 1/mSo,( frac{2}{r_1} - frac{1}{a} ≈ 1.337 times 10^{-11} - 5.297 times 10^{-12} ≈ 8.073 times 10^{-12} ) 1/mTherefore,( v_t1 = sqrt{1.327 times 10^{20} times 8.073 times 10^{-12}} )Multiply inside the square root:( 1.327 times 10^{20} times 8.073 times 10^{-12} ≈ 1.071 times 10^{9} )So,( v_t1 = sqrt{1.071 times 10^{9}} ≈ 32730 ) m/sSo, ( v_t1 ≈ 32.73 ) km/sTherefore, ( Delta v_1 = v_t1 - v_1 = 32730 - 29780 ≈ 2950 ) m/sNow, compute ( v_t2 ):( v_t2 = sqrt{mu left( frac{2}{r_2} - frac{1}{a} right)} )We have ( r_2 = 2.279 times 10^{11} ) mSo,( frac{2}{r_2} = frac{2}{2.279 times 10^{11}} ≈ 8.77 times 10^{-12} ) 1/m( frac{1}{a} = 5.297 times 10^{-12} ) 1/m (from earlier)So,( frac{2}{r_2} - frac{1}{a} ≈ 8.77 times 10^{-12} - 5.297 times 10^{-12} ≈ 3.473 times 10^{-12} ) 1/mTherefore,( v_t2 = sqrt{1.327 times 10^{20} times 3.473 times 10^{-12}} )Multiply inside the square root:( 1.327 times 10^{20} times 3.473 times 10^{-12} ≈ 4.603 times 10^{8} )So,( v_t2 = sqrt{4.603 times 10^{8}} ≈ 21450 ) m/sSo, ( v_t2 ≈ 21.45 ) km/sTherefore, ( Delta v_2 = v_2 - v_t2 = 24135 - 21450 ≈ 2685 ) m/sAdding both ( Delta v ):Total ( Delta v = 2950 + 2685 ≈ 5635 ) m/sSo, approximately 5.635 km/s.Wait, let me double-check these calculations because the numbers seem a bit high, but I think they are correct.Earth's velocity: ~29.78 km/sTransfer orbit at Earth: ~32.73 km/s, so burn needed is ~2.95 km/sTransfer orbit at Mars: ~21.45 km/sMars' velocity: ~24.14 km/s, so burn needed is ~2.685 km/sTotal: ~5.635 km/sYes, that seems correct. I remember that typical Hohmann transfer requires about 5.5 km/s, so this is in the right ballpark.So, summarizing:1. Semi-major axis ( a = 1.262 ) AU2. Period ( T ≈ 1.42 ) years3. Total ( Delta v ≈ 5.635 ) km/sI think that's it. Let me just make sure I didn't make any calculation errors.Wait, when I calculated ( v_t1 ), I had:( frac{2}{r_1} - frac{1}{a} ≈ 8.073 times 10^{-12} ) 1/mThen, ( mu times that = 1.327e20 * 8.073e-12 ≈ 1.071e9 ), square root is ~32730 m/s. Correct.Similarly, ( v_t2 ):( frac{2}{r_2} - frac{1}{a} ≈ 3.473e-12 )Multiply by ( mu ): 1.327e20 * 3.473e-12 ≈ 4.603e8, square root is ~21450 m/s. Correct.So, the calculations seem right.Final Answer1. The semi-major axis of the Hohmann transfer orbit is (boxed{1.262}) AU and the period is (boxed{1.42}) years.2. The total change in velocity required is (boxed{5.64}) km/s.</think>

🧠:<think>Alright, so I have this problem where a business owner is comparing two competitors, Company A and Company B, in terms of their customer acquisition and retention strategies. The problem has two parts: first, figuring out which company acquires more customers over the first 12 months, and second, determining which company retains more customers over the same period. Let me start with the first sub-problem. I need to determine the total number of customers acquired by each company over the first 12 months. For Company A, the acquisition rate is given by ( A(t) = 100e^{0.05t} ) customers per month, and for Company B, it's ( B(t) = 120e^{0.04t} ). Hmm, so both companies have exponential growth in their acquisition rates, which makes sense because exponential functions grow over time. The question is, which one grows faster? Company A has a higher coefficient (100 vs. 120), but Company B's exponent is smaller (0.04 vs. 0.05). Wait, actually, Company A's exponent is higher, which would mean it grows faster over time. But Company B starts with a higher base. I need to calculate the total over 12 months.To find the total number of customers acquired, I should integrate the acquisition rate over the time period from 0 to 12 months. That is, for Company A, the total would be the integral of ( A(t) ) from 0 to 12, and similarly for Company B.So, let me write that down:For Company A:Total acquired, ( T_A = int_{0}^{12} 100e^{0.05t} dt )For Company B:Total acquired, ( T_B = int_{0}^{12} 120e^{0.04t} dt )I need to compute these integrals. Let's start with Company A.The integral of ( e^{kt} ) is ( frac{1}{k}e^{kt} ), so applying that:( T_A = 100 times left[ frac{1}{0.05} e^{0.05t} right]_0^{12} )Simplify:( T_A = 100 times 20 times (e^{0.6} - 1) )Because ( 1/0.05 = 20 ), and evaluating from 0 to 12 gives ( e^{0.05*12} - e^{0} = e^{0.6} - 1 ).Calculating ( e^{0.6} ): I remember that ( e^{0.6} ) is approximately 1.8221.So, ( T_A = 2000 times (1.8221 - 1) = 2000 times 0.8221 = 1644.2 ) customers.Wait, hold on, that seems a bit high. Let me check my calculations again.Wait, no, actually, 100 * 20 is 2000, and then multiplied by (1.8221 - 1) which is 0.8221. So, 2000 * 0.8221 is indeed 1644.2. So, approximately 1644 customers for Company A.Now, for Company B:( T_B = int_{0}^{12} 120e^{0.04t} dt )Again, integrating:( T_B = 120 times left[ frac{1}{0.04} e^{0.04t} right]_0^{12} )Simplify:( T_B = 120 times 25 times (e^{0.48} - 1) )Because ( 1/0.04 = 25 ), and ( 0.04*12 = 0.48 ).Calculating ( e^{0.48} ): I think that's approximately 1.6161.So, ( T_B = 3000 times (1.6161 - 1) = 3000 times 0.6161 = 1848.3 ) customers.So, Company B has acquired approximately 1848 customers over 12 months, compared to Company A's 1644. So, based on total customers acquired, Company B has a more effective acquisition strategy.Wait, but let me double-check the calculations because sometimes constants can be tricky.For Company A:Integral of 100e^{0.05t} from 0 to12:= 100 * (1/0.05)(e^{0.6} - 1)= 2000*(1.8221 - 1)= 2000*0.8221= 1644.2Yes, that's correct.For Company B:Integral of 120e^{0.04t} from 0 to12:= 120*(1/0.04)(e^{0.48} - 1)= 3000*(1.6161 -1)= 3000*0.6161= 1848.3Yes, that's correct as well.So, indeed, Company B has a higher total acquisition over 12 months.Moving on to the second sub-problem. Now, I need to find the total retained customer base for each company at the end of 12 months. The retention rates are given as functions of time: for Company A, ( R_A(t) = 0.9 + 0.05 sin(pi t /6) ), and for Company B, ( R_B(t) = 0.85 + 0.1 cos(pi t /6) ).The problem states that the retention rates apply continuously to the customers acquired in each month. So, this means that for each month, the number of customers acquired that month will be multiplied by the retention rate for each subsequent month until month 12.Wait, actually, it's a bit more precise. Since retention applies continuously, I think we need to model the number of customers retained over time. For each month t, the number of customers acquired in month t will be subject to the retention rate for each month from t to 12.But since retention rates vary with time, this might get a bit complicated. Let me think.I think the total retained customers at the end of 12 months can be calculated by integrating the acquisition rate multiplied by the retention rate over the remaining time. That is, for each month t, the number of customers acquired is A(t) or B(t), and each of these customers will be retained for the remaining (12 - t) months, but the retention rate varies each month.Wait, no, actually, retention rate is a monthly rate, so each month, a certain percentage of customers are retained. So, if a customer is acquired in month t, they will be subject to the retention rates in months t, t+1, ..., 12.But since the retention rates are functions of time, we can't just multiply by a constant. Instead, for each customer acquired in month t, the probability that they are still retained at month 12 is the product of the retention rates from month t to month 12.But integrating this over all t from 0 to 12 would give the total retained customers.Wait, that might be a bit involved. Let me formalize this.For each infinitesimal time dt, the number of customers acquired is A(t) dt (for Company A). Each of these customers will be subject to retention rates from time t to time 12. So, the number of customers retained from this acquisition is A(t) dt multiplied by the product of retention rates from t to 12.But since the retention rate varies continuously, we can model the survival probability as the exponential of the integral of the logarithm of the retention rate. Wait, that might be overcomplicating.Alternatively, since the retention rate is monthly, perhaps we can model the survival probability as the product of monthly retention rates. But since we're dealing with continuous time, maybe we can approximate it as a continuous decay process.Wait, perhaps it's better to model the retention as a continuous process. If the retention rate is R(t), then the fraction of customers retained each month is R(t). So, the number of customers remaining after each month is multiplied by R(t). Therefore, the survival function S(t) from time t to T is the product of R(s) from s = t to T.But since R(t) is a function, and we're dealing with continuous time, the survival function can be written as:( S(t) = expleft( int_{t}^{12} ln R(s) ds right) )But this might be a bit too complex. Alternatively, if the retention rate is applied continuously, perhaps we can model it as a differential equation.Let me think. Let’s denote C_A(t) as the number of customers acquired at time t. Then, the number of customers remaining at time T is C_A(t) multiplied by the product of retention rates from t to T.But since the retention rate is a function of time, this product is equivalent to the exponential of the integral of the logarithm of R(t). So,( text{Retained customers from acquisition at t} = C_A(t) times expleft( int_{t}^{12} ln R_A(s) ds right) )Similarly for Company B.Therefore, the total retained customers at time 12 is the integral from 0 to 12 of C_A(t) multiplied by the survival probability from t to 12.So, for Company A:( text{Total retained}_A = int_{0}^{12} A(t) expleft( int_{t}^{12} ln R_A(s) ds right) dt )Similarly for Company B:( text{Total retained}_B = int_{0}^{12} B(t) expleft( int_{t}^{12} ln R_B(s) ds right) dt )This seems like the correct approach, but it's quite involved. Let me see if I can simplify it.First, let's compute the inner integral for each company.Starting with Company A:( R_A(t) = 0.9 + 0.05 sinleft( frac{pi t}{6} right) )So, ( ln R_A(t) = lnleft( 0.9 + 0.05 sinleft( frac{pi t}{6} right) right) )Similarly, for Company B:( R_B(t) = 0.85 + 0.1 cosleft( frac{pi t}{6} right) )So, ( ln R_B(t) = lnleft( 0.85 + 0.1 cosleft( frac{pi t}{6} right) right) )These integrals don't have elementary antiderivatives, so I might need to approximate them numerically. But since I'm doing this by hand, perhaps I can find a pattern or simplify the expression.Wait, let me consider the periodicity of the sine and cosine functions. Both R_A(t) and R_B(t) have a period of 12 months because the argument is ( pi t /6 ), so the period is ( 2pi / (pi /6) ) = 12 ). Therefore, over the interval from 0 to 12, the functions complete one full period.This might help in evaluating the integrals, but I'm not sure. Alternatively, maybe I can compute the average value of ( ln R_A(t) ) and ( ln R_B(t) ) over the period and use that to approximate the integral.But that might not be accurate enough. Alternatively, I can use numerical integration techniques like Simpson's rule or the trapezoidal rule to approximate the inner integrals.But since this is a thought process, let me outline the steps:1. For each company, compute the inner integral ( int_{t}^{12} ln R(s) ds ) for each t from 0 to 12.2. Then, exponentiate the result to get the survival probability.3. Multiply by the acquisition rate A(t) or B(t) and integrate over t from 0 to 12.This seems quite involved, but let's try to tackle it step by step.First, let's consider Company A.Compute ( int_{t}^{12} ln R_A(s) ds ) where ( R_A(s) = 0.9 + 0.05 sin(pi s /6) ).Similarly, for Company B, ( R_B(s) = 0.85 + 0.1 cos(pi s /6) ).Let me note that ( sin(pi s /6) ) and ( cos(pi s /6) ) are periodic with period 12, so over the interval from t to 12, the integral might have some symmetry.But perhaps instead of trying to find an analytical solution, I can approximate the integral numerically.Let me consider using the trapezoidal rule for numerical integration. Since I'm doing this manually, I'll have to approximate the integrals at several points and then sum them up.But this will be time-consuming. Alternatively, maybe I can find the average value of ( ln R_A(t) ) and ( ln R_B(t) ) over the period and use that to approximate the integral.Wait, the average value of ( ln R(t) ) over one period multiplied by the period length would give the integral over one period. But since we're integrating from t to 12, which is less than a full period if t >0, this might not help directly.Alternatively, perhaps I can compute the integral over the entire period and then subtract the integral from 0 to t.Wait, that's a good point. Let me denote:For Company A:( int_{t}^{12} ln R_A(s) ds = int_{0}^{12} ln R_A(s) ds - int_{0}^{t} ln R_A(s) ds )Similarly for Company B.So, if I can compute ( int_{0}^{12} ln R_A(s) ds ) and ( int_{0}^{t} ln R_A(s) ds ), then I can find the integral from t to 12.But again, without numerical methods, it's difficult. Let me see if I can compute ( int_{0}^{12} ln R_A(s) ds ) approximately.Let me compute ( ln R_A(s) ) at several points and use the trapezoidal rule.For Company A:( R_A(s) = 0.9 + 0.05 sin(pi s /6) )Let me compute ( ln R_A(s) ) at s = 0, 3, 6, 9, 12.At s=0:( R_A(0) = 0.9 + 0.05 sin(0) = 0.9 )So, ( ln(0.9) ≈ -0.10536 )At s=3:( R_A(3) = 0.9 + 0.05 sin(pi *3 /6) = 0.9 + 0.05 sin(pi/2) = 0.9 + 0.05*1 = 0.95 )So, ( ln(0.95) ≈ -0.051293 )At s=6:( R_A(6) = 0.9 + 0.05 sin(pi *6 /6) = 0.9 + 0.05 sin(pi) = 0.9 + 0 = 0.9 )So, ( ln(0.9) ≈ -0.10536 )At s=9:( R_A(9) = 0.9 + 0.05 sin(pi *9 /6) = 0.9 + 0.05 sin(3pi/2) = 0.9 + 0.05*(-1) = 0.85 )So, ( ln(0.85) ≈ -0.1625 )At s=12:( R_A(12) = 0.9 + 0.05 sin(pi *12 /6) = 0.9 + 0.05 sin(2pi) = 0.9 + 0 = 0.9 )So, ( ln(0.9) ≈ -0.10536 )Now, using the trapezoidal rule with these 5 points (s=0,3,6,9,12), each interval is 3 months.The trapezoidal rule formula is:( int_{a}^{b} f(x) dx ≈ frac{h}{2} [f(a) + 2(f(a+h) + f(a+2h) + ... + f(b-h)) + f(b)] )Here, h=3, a=0, b=12.So,( int_{0}^{12} ln R_A(s) ds ≈ frac{3}{2} [f(0) + 2(f(3) + f(6) + f(9)) + f(12)] )Plugging in the values:= (3/2) [ (-0.10536) + 2*(-0.051293 + (-0.10536) + (-0.1625)) + (-0.10536) ]First, compute the terms inside:= (3/2) [ -0.10536 + 2*(-0.051293 -0.10536 -0.1625) -0.10536 ]Compute the sum inside the brackets:First, compute the sum inside the 2*():-0.051293 -0.10536 -0.1625 = (-0.051293 -0.10536) = -0.156653 -0.1625 = -0.319153Multiply by 2: -0.638306Now, add the other terms:-0.10536 + (-0.638306) + (-0.10536) = (-0.10536 -0.10536) = -0.21072 -0.638306 = -0.849026Now, multiply by (3/2):= (3/2)*(-0.849026) ≈ 1.5*(-0.849026) ≈ -1.27354So, the approximate integral of ( ln R_A(s) ) from 0 to12 is -1.27354.Similarly, for Company B:( R_B(s) = 0.85 + 0.1 cos(pi s /6) )Compute ( ln R_B(s) ) at s=0,3,6,9,12.At s=0:( R_B(0) = 0.85 + 0.1 cos(0) = 0.85 + 0.1*1 = 0.95 )So, ( ln(0.95) ≈ -0.051293 )At s=3:( R_B(3) = 0.85 + 0.1 cos(pi *3 /6) = 0.85 + 0.1 cos(pi/2) = 0.85 + 0 = 0.85 )So, ( ln(0.85) ≈ -0.1625 )At s=6:( R_B(6) = 0.85 + 0.1 cos(pi *6 /6) = 0.85 + 0.1 cos(pi) = 0.85 + 0.1*(-1) = 0.75 )So, ( ln(0.75) ≈ -0.28768 )At s=9:( R_B(9) = 0.85 + 0.1 cos(pi *9 /6) = 0.85 + 0.1 cos(3pi/2) = 0.85 + 0 = 0.85 )So, ( ln(0.85) ≈ -0.1625 )At s=12:( R_B(12) = 0.85 + 0.1 cos(pi *12 /6) = 0.85 + 0.1 cos(2pi) = 0.85 + 0.1*1 = 0.95 )So, ( ln(0.95) ≈ -0.051293 )Now, applying the trapezoidal rule for Company B:( int_{0}^{12} ln R_B(s) ds ≈ frac{3}{2} [f(0) + 2(f(3) + f(6) + f(9)) + f(12)] )Plugging in the values:= (3/2) [ (-0.051293) + 2*(-0.1625 + (-0.28768) + (-0.1625)) + (-0.051293) ]Compute the terms inside:First, compute the sum inside the 2*():-0.1625 -0.28768 -0.1625 = (-0.1625 -0.1625) = -0.325 -0.28768 = -0.61268Multiply by 2: -1.22536Now, add the other terms:-0.051293 + (-1.22536) + (-0.051293) = (-0.051293 -0.051293) = -0.102586 -1.22536 = -1.327946Multiply by (3/2):= (3/2)*(-1.327946) ≈ 1.5*(-1.327946) ≈ -1.99192So, the approximate integral of ( ln R_B(s) ) from 0 to12 is -1.99192.Now, going back to the total retained customers.For Company A:( text{Total retained}_A = int_{0}^{12} A(t) expleft( int_{t}^{12} ln R_A(s) ds right) dt )We can rewrite the inner integral as:( int_{t}^{12} ln R_A(s) ds = int_{0}^{12} ln R_A(s) ds - int_{0}^{t} ln R_A(s) ds )We already approximated ( int_{0}^{12} ln R_A(s) ds ≈ -1.27354 )So, ( int_{t}^{12} ln R_A(s) ds ≈ -1.27354 - int_{0}^{t} ln R_A(s) ds )Similarly, for Company B:( int_{t}^{12} ln R_B(s) ds ≈ -1.99192 - int_{0}^{t} ln R_B(s) ds )But this still leaves us with the problem of computing ( int_{0}^{t} ln R_A(s) ds ) and ( int_{0}^{t} ln R_B(s) ds ) for each t, which is not straightforward without numerical methods.Alternatively, perhaps I can approximate the survival probability as ( exp(-kt) ), but I'm not sure.Wait, another approach: since the retention rate is periodic, maybe the integral over a full period can be used to find an average decay rate.But I'm not sure. Alternatively, perhaps I can approximate the survival function as a constant decay rate.Wait, let me think differently. The survival function S(t) is the probability that a customer acquired at time t is still retained at time 12. So, for each t, S(t) = product of R(s) from s=t to 12.But since R(s) is a function, and we're dealing with continuous time, the product becomes an exponential of the integral of ln R(s).So, S(t) = exp(∫_{t}^{12} ln R(s) ds )But we can write this as exp( C - ∫_{0}^{t} ln R(s) ds ), where C = ∫_{0}^{12} ln R(s) ds.So, S(t) = exp(C) * exp(-∫_{0}^{t} ln R(s) ds )But exp(C) is a constant for each company.So, for Company A:S_A(t) = exp(-1.27354) * exp(-∫_{0}^{t} ln R_A(s) ds )Similarly, for Company B:S_B(t) = exp(-1.99192) * exp(-∫_{0}^{t} ln R_B(s) ds )But this still requires knowing ∫_{0}^{t} ln R(s) ds, which is not easy without numerical integration.Alternatively, perhaps I can approximate the integral ∫_{0}^{t} ln R(s) ds for each t by using the average value of ln R(s) over the period.Wait, for Company A, the average value of ln R_A(s) over 0 to12 is (1/12)*(-1.27354) ≈ -0.10613 per month.Similarly, for Company B, the average is (1/12)*(-1.99192) ≈ -0.16599 per month.So, if I approximate ∫_{0}^{t} ln R(s) ds ≈ average * t = (-0.10613)t for Company A and (-0.16599)t for Company B.Then, S_A(t) ≈ exp(-1.27354) * exp(0.10613 t )Similarly, S_B(t) ≈ exp(-1.99192) * exp(0.16599 t )Wait, but this seems counterintuitive because the survival probability should decrease as t increases, but with this approximation, S(t) increases as t increases, which is incorrect.Wait, no, because the integral ∫_{t}^{12} ln R(s) ds = C - ∫_{0}^{t} ln R(s) ds ≈ C - (-0.10613 t ) = C + 0.10613 tBut C is negative, so S(t) = exp(C + 0.10613 t )But C is -1.27354, so S(t) = exp(-1.27354 + 0.10613 t )Similarly for Company B: S(t) = exp(-1.99192 + 0.16599 t )Wait, but this would mean that as t increases, the exponent becomes less negative, so S(t) increases, which is correct because a customer acquired later has less time to be lost, so higher survival probability.But this is an approximation. Let me check if this makes sense.At t=0, S(0) = exp(C) = exp(-1.27354) ≈ 0.28 for Company A, and exp(-1.99192) ≈ 0.135 for Company B.At t=12, S(12) = exp(C + 0.10613*12 ) = exp(-1.27354 + 1.27356 ) ≈ exp(0.00002 ) ≈ 1 for Company A.Similarly, for Company B: exp(-1.99192 + 0.16599*12 ) = exp(-1.99192 + 1.99188 ) ≈ exp(-0.00004 ) ≈ 1.So, this approximation seems to hold at the endpoints.Therefore, perhaps this is a reasonable approximation.So, for Company A:S_A(t) ≈ exp(-1.27354 + 0.10613 t )Similarly, for Company B:S_B(t) ≈ exp(-1.99192 + 0.16599 t )Now, the total retained customers for Company A is:( text{Total retained}_A = int_{0}^{12} A(t) times S_A(t) dt = int_{0}^{12} 100e^{0.05t} times exp(-1.27354 + 0.10613 t ) dt )Simplify the exponent:= 100 * exp(-1.27354) * ∫_{0}^{12} e^{0.05t + 0.10613 t} dt= 100 * exp(-1.27354) * ∫_{0}^{12} e^{(0.05 + 0.10613)t} dt= 100 * exp(-1.27354) * ∫_{0}^{12} e^{0.15613 t} dtSimilarly, for Company B:( text{Total retained}_B = int_{0}^{12} B(t) times S_B(t) dt = int_{0}^{12} 120e^{0.04t} times exp(-1.99192 + 0.16599 t ) dt )Simplify:= 120 * exp(-1.99192) * ∫_{0}^{12} e^{0.04t + 0.16599 t} dt= 120 * exp(-1.99192) * ∫_{0}^{12} e^{0.20599 t} dtNow, let's compute these integrals.For Company A:Integral of e^{0.15613 t} from 0 to12:= [1/0.15613 e^{0.15613 t}] from 0 to12= (1/0.15613)(e^{1.87356} - 1 )Compute e^{1.87356}: approximately e^1.87356 ≈ 6.51 (since e^1.8 ≈ 6.05, e^1.9 ≈ 6.73, so 1.87356 is closer to 1.87, which is about 6.51)So, ≈ (1/0.15613)(6.51 -1 ) ≈ (6.425)(5.51) ≈ 6.425*5.51 ≈ let's compute:6 *5.51 = 33.060.425*5.51 ≈ 2.34Total ≈ 33.06 + 2.34 ≈ 35.4So, the integral ≈35.4Therefore, Total retained_A ≈ 100 * exp(-1.27354) *35.4Compute exp(-1.27354) ≈ 0.28So, ≈100 *0.28 *35.4 ≈ 28 *35.4 ≈ 991.2Similarly, for Company B:Integral of e^{0.20599 t} from 0 to12:= [1/0.20599 e^{0.20599 t}] from 0 to12= (1/0.20599)(e^{2.47188} -1 )Compute e^{2.47188}: e^2 ≈7.389, e^2.47188 ≈11.7 (since e^2.3026≈10, e^2.47188≈11.7)So, ≈ (1/0.20599)(11.7 -1 ) ≈ (4.857)(10.7) ≈ 4.857*10 +4.857*0.7 ≈48.57 +3.4≈51.97So, the integral ≈51.97Therefore, Total retained_B ≈120 * exp(-1.99192) *51.97Compute exp(-1.99192) ≈0.135So, ≈120 *0.135 *51.97 ≈16.2 *51.97 ≈843.5Wait, that can't be right because Company A's total retained is 991 and Company B's is 843, but Company B had a higher acquisition. Hmm, but this is just an approximation.Wait, but let me check my calculations again.For Company A:Integral of e^{0.15613 t} from 0 to12:= (1/0.15613)(e^{0.15613*12} -1 ) = (1/0.15613)(e^{1.87356} -1 )e^{1.87356} ≈6.51, so 6.51 -1=5.515.51 /0.15613 ≈35.3Yes, that's correct.Then, 100 * exp(-1.27354) ≈100 *0.28≈2828 *35.3≈988.4≈991.2For Company B:Integral of e^{0.20599 t} from 0 to12:= (1/0.20599)(e^{0.20599*12} -1 ) = (1/0.20599)(e^{2.47188} -1 )e^{2.47188}≈11.7, so 11.7 -1=10.710.7 /0.20599≈51.97Then, 120 * exp(-1.99192)≈120 *0.135≈16.216.2 *51.97≈843.5So, according to this approximation, Company A retains approximately 991 customers, and Company B retains approximately 843 customers. Therefore, Company A has a more effective retention strategy.But wait, this is an approximation. The actual values might differ because we used the average value of ln R(t) to approximate the integral, which might not capture the true behavior, especially since the retention rates vary sinusoidally.Alternatively, perhaps I can use a better approximation by considering more points in the trapezoidal rule.But given the time constraints, let me proceed with this approximation.So, summarizing:- Company A: Total acquired ≈1644, Total retained≈991- Company B: Total acquired≈1848, Total retained≈843Therefore, Company B is better at acquisition, but Company A is better at retention.But wait, the problem asks to compare the total retained customer base at the end of 12 months. So, based on this approximation, Company A retains more customers despite having a lower acquisition rate.But let me think again. The retention rate for Company A is higher on average? Let's check the average retention rate.For Company A, R_A(t) =0.9 +0.05 sin(πt/6). The average of sin over a period is zero, so average R_A =0.9.For Company B, R_B(t)=0.85 +0.1 cos(πt/6). Similarly, average R_B=0.85.So, Company A has a higher average retention rate, which would lead to better retention over time, even if they acquire fewer customers.Therefore, my conclusion is that Company A has a more effective retention strategy.But let me check if my approximation makes sense. The survival probability for Company A is higher because their average retention rate is higher, so even though they acquire fewer customers, they retain a higher proportion.Yes, that makes sense.Therefore, the answers are:1. Company B has a more effective acquisition strategy over the first 12 months.2. Company A has a more effective retention strategy.</think>