🧠:<think>Primero, para resolver el primer problema, sé que el 15% de los textos en la base de datos son escritos por Mariela Campos y que este número corresponde a 450 textos. Por lo tanto, puedo establecer la ecuación 0.15 × N = 450, donde N es el número total de textos. Al resolver esta ecuación, obtengo N = 450 / 0.15 = 3000.Ahora, para el segundo problema, sé que el 60% de los 3000 textos totales están disponibles en al menos dos idiomas. Calculando esto, 0.60 × 3000 = 1800 textos en total están disponibles en al menos dos idiomas.Además, del 15% de los textos que son de Mariela Campos, el 70% están disponibles en al menos dos idiomas. Primero, calculo el número total de textos de Mariela Campos, que ya sabemos es 450. Luego, aplico el 70% a este número: 0.70 × 450 = 315 textos de Mariela Campos están disponibles en al menos dos idiomas.Finalmente, resumo los resultados obtenidos para ambos problemas.</think>

🧠:<think>Alright, so I have this problem where a film student is working on a scene with a mystical forest, and they need to model two things: time dilation and fog density. Let me try to figure out how to approach each part step by step.Starting with the first part: modeling the time dilation effect. The equation given is ( t_f = sqrt{a cdot t_o^2 + b cdot sin(c cdot t_o)} ). We have two data points: when ( t_o = 2 ), ( t_f = 3 ); and when ( t_o = 4 ), ( t_f = 5 ). So, we need to find the constants ( a ), ( b ), and ( c ).Hmm, but wait, we have three unknowns and only two equations. That means we might need to make an assumption or find another condition. Maybe the problem expects a specific behavior or perhaps another point? Let me check the problem statement again. It says "determine the values of ( a ), ( b ), and ( c )" given those two points. So, perhaps we can assume something about the function, like maybe it's linear or something else?Wait, another thought: maybe the sine function is oscillating, but if we have only two points, we might not be able to determine ( c ) uniquely. Maybe the student expects ( c ) to be such that the sine term doesn't affect the given points? Or perhaps the sine term is negligible for these values, but that might not be a safe assumption.Alternatively, maybe the sine term is zero at those specific times, which would simplify the equation. Let's test that idea. If ( sin(c cdot t_o) = 0 ) at ( t_o = 2 ) and ( t_o = 4 ), then ( c cdot t_o ) must be multiples of ( pi ). So, ( c cdot 2 = kpi ) and ( c cdot 4 = mpi ) for integers ( k ) and ( m ). That would mean ( c = frac{kpi}{2} ) and ( c = frac{mpi}{4} ). For these to be consistent, ( frac{k}{2} = frac{m}{4} ), so ( 2k = m ). So, if ( k = 1 ), then ( m = 2 ), which gives ( c = frac{pi}{2} ). Let me test this.If ( c = frac{pi}{2} ), then at ( t_o = 2 ), ( sin(frac{pi}{2} cdot 2) = sin(pi) = 0 ), and at ( t_o = 4 ), ( sin(frac{pi}{2} cdot 4) = sin(2pi) = 0 ). Perfect! So, the sine term is zero at both given points, which means we can ignore it for these specific times. That simplifies our equation to ( t_f = sqrt{a cdot t_o^2} ) for these points, which is ( t_f = sqrt{a} cdot t_o ).Given that, let's plug in the first point: ( t_f = 3 ) when ( t_o = 2 ). So, ( 3 = sqrt{a} cdot 2 ), which gives ( sqrt{a} = frac{3}{2} ), so ( a = left(frac{3}{2}right)^2 = frac{9}{4} ).Similarly, for the second point: ( t_f = 5 ) when ( t_o = 4 ). Plugging in, ( 5 = sqrt{a} cdot 4 ). But we already found ( sqrt{a} = frac{3}{2} ), so ( 5 = frac{3}{2} cdot 4 = 6 ). Wait, that's a problem because 5 ≠ 6. Hmm, that doesn't make sense. Did I make a mistake?Wait, no, because if ( c = frac{pi}{2} ), then the sine term is zero at both points, so the equation reduces to ( t_f = sqrt{a} cdot t_o ). But with ( a = frac{9}{4} ), ( t_f ) at ( t_o = 4 ) would be ( sqrt{frac{9}{4}} cdot 4 = frac{3}{2} cdot 4 = 6 ), but the given ( t_f ) is 5. So, that's inconsistent.Hmm, so my initial assumption that the sine term is zero at both points might be incorrect. Maybe ( c ) isn't ( frac{pi}{2} ). Alternatively, perhaps the sine term isn't zero but contributes to the equation. But with only two points, it's tricky because we have three variables. Maybe I need to set up a system of equations.Let's write the equations:1. When ( t_o = 2 ), ( t_f = 3 ):[ 3 = sqrt{a cdot (2)^2 + b cdot sin(c cdot 2)} ]Simplify:[ 9 = 4a + b cdot sin(2c) ]2. When ( t_o = 4 ), ( t_f = 5 ):[ 5 = sqrt{a cdot (4)^2 + b cdot sin(c cdot 4)} ]Simplify:[ 25 = 16a + b cdot sin(4c) ]So, we have two equations:1. ( 4a + b cdot sin(2c) = 9 )2. ( 16a + b cdot sin(4c) = 25 )We need a third equation. Maybe we can assume another condition? For example, perhaps the sine term is zero at ( t_o = 0 ), but that's not given. Alternatively, maybe the function is symmetric or has a certain property. Alternatively, perhaps we can assume that the sine term is zero at one of the points, but that might not hold.Wait, another idea: perhaps the sine function is periodic, so maybe ( sin(4c) = 2 sin(2c) cos(2c) ). That's a trigonometric identity. So, ( sin(4c) = 2 sin(2c) cos(2c) ). Maybe we can use that to relate the two equations.Let me denote ( s = sin(2c) ) and ( d = cos(2c) ). Then, ( sin(4c) = 2 s d ).So, our equations become:1. ( 4a + b s = 9 )2. ( 16a + b (2 s d) = 25 )But we still have three variables: ( a ), ( b ), and ( s d ). Hmm, not sure if that helps.Alternatively, maybe we can express ( b ) from the first equation and substitute into the second.From equation 1:( b s = 9 - 4a )So, ( b = frac{9 - 4a}{s} )Plug into equation 2:( 16a + left(frac{9 - 4a}{s}right) cdot sin(4c) = 25 )But ( sin(4c) = 2 sin(2c) cos(2c) = 2 s d ), so:( 16a + left(frac{9 - 4a}{s}right) cdot 2 s d = 25 )Simplify:( 16a + 2 (9 - 4a) d = 25 )Which is:( 16a + 18d - 8a d = 25 )Hmm, still complicated. Maybe we need another approach.Alternatively, let's assume that ( c ) is such that ( sin(2c) ) and ( sin(4c) ) are known or can be expressed in terms of each other. For example, if ( c = frac{pi}{4} ), then ( 2c = frac{pi}{2} ), so ( sin(2c) = 1 ), and ( 4c = pi ), so ( sin(4c) = 0 ). Let's test this.If ( c = frac{pi}{4} ), then:Equation 1:( 4a + b cdot 1 = 9 ) => ( 4a + b = 9 )Equation 2:( 16a + b cdot 0 = 25 ) => ( 16a = 25 ) => ( a = frac{25}{16} )Then, from equation 1: ( 4*(25/16) + b = 9 ) => ( 25/4 + b = 9 ) => ( b = 9 - 25/4 = (36 -25)/4 = 11/4 )So, ( a = 25/16 ), ( b = 11/4 ), ( c = pi/4 ). Let me check if this works.At ( t_o = 2 ):( t_f = sqrt( (25/16)*4 + (11/4)*sin(pi/2) ) = sqrt(25/4 + 11/4*1) = sqrt(25/4 + 11/4) = sqrt(36/4) = sqrt(9) = 3 ). Good.At ( t_o = 4 ):( t_f = sqrt( (25/16)*16 + (11/4)*sin(pi) ) = sqrt(25 + 0) = 5 ). Perfect.So, that works! So, ( a = 25/16 ), ( b = 11/4 ), ( c = pi/4 ).Wait, but is this the only solution? Because I assumed ( c = pi/4 ) to make the sine terms take specific values. Maybe there are other values of ( c ) that could also satisfy the equations. But since the problem doesn't specify any other conditions, this seems like a valid solution.Moving on to the second part: modeling the fog density with a Gaussian distribution. The equation is ( f(x) = frac{1}{sigma sqrt{2pi}} e^{-frac{(x - mu)^2}{2sigma^2}} ). We know that at ( x = mu ), the density is 0.5, and at ( x = mu + 10 ), it's 0.1.First, at ( x = mu ), the exponent becomes zero, so ( f(mu) = frac{1}{sigma sqrt{2pi}} ). So, ( frac{1}{sigma sqrt{2pi}} = 0.5 ). Let's solve for ( sigma ):( sigma sqrt{2pi} = 2 ) => ( sigma = frac{2}{sqrt{2pi}} = sqrt{frac{2}{pi}} ). Let me compute that numerically: ( sqrt{2/pi} approx sqrt{0.6366} approx 0.7979 ). But maybe we can leave it in exact form.Next, at ( x = mu + 10 ), the density is 0.1. So:( 0.1 = frac{1}{sigma sqrt{2pi}} e^{-frac{(10)^2}{2sigma^2}} )But we already know ( frac{1}{sigma sqrt{2pi}} = 0.5 ), so:( 0.1 = 0.5 e^{-frac{100}{2sigma^2}} )Divide both sides by 0.5:( 0.2 = e^{-frac{100}{2sigma^2}} )Take natural logarithm:( ln(0.2) = -frac{100}{2sigma^2} )Simplify:( ln(0.2) = -frac{50}{sigma^2} )Multiply both sides by -1:( -ln(0.2) = frac{50}{sigma^2} )Compute ( -ln(0.2) ). Since ( ln(0.2) = ln(1/5) = -ln(5) approx -1.6094 ), so ( -ln(0.2) approx 1.6094 ).Thus:( 1.6094 = frac{50}{sigma^2} )Solve for ( sigma^2 ):( sigma^2 = frac{50}{1.6094} approx 31.06 )So, ( sigma approx sqrt{31.06} approx 5.57 ). Wait, but earlier we had ( sigma = sqrt{2/pi} approx 0.7979 ). That's a contradiction. Hmm, that can't be. Wait, no, because in the first part, we found ( sigma ) based on the maximum density, and in the second part, we're using another condition. Wait, no, actually, the first part gives us ( sigma ) as ( sqrt{2/pi} ), but the second condition gives a different ( sigma ). That suggests that my approach is wrong.Wait, no, actually, the first part gives us ( sigma ) in terms of the maximum density, but the second condition relates to the same ( sigma ). So, perhaps I need to use the first condition to express ( sigma ) in terms of the maximum, and then plug into the second condition.Wait, let me re-express:From ( x = mu ), ( f(mu) = 0.5 = frac{1}{sigma sqrt{2pi}} ). So, ( sigma = frac{1}{0.5 sqrt{2pi}} = frac{2}{sqrt{2pi}} = sqrt{frac{2}{pi}} ). So, ( sigma ) is fixed as ( sqrt{2/pi} ).Then, at ( x = mu + 10 ), ( f(x) = 0.1 ). So:( 0.1 = frac{1}{sqrt{2/pi} cdot sqrt{2pi}} e^{-frac{100}{2 cdot (2/pi)}} )Wait, let's compute ( frac{1}{sigma sqrt{2pi}} ):( sigma = sqrt{2/pi} ), so ( sigma sqrt{2pi} = sqrt{2/pi} cdot sqrt{2pi} = sqrt{(2/pi)(2pi)} = sqrt{4} = 2 ). So, ( frac{1}{sigma sqrt{2pi}} = 1/2 ). So, the equation becomes:( 0.1 = 0.5 e^{-frac{100}{2 cdot (2/pi)}} )Simplify the exponent:( frac{100}{2 cdot (2/pi)} = frac{100}{4/pi} = 25 pi approx 78.54 )So, ( 0.1 = 0.5 e^{-78.54} ). But ( e^{-78.54} ) is an extremely small number, practically zero. So, ( 0.5 e^{-78.54} ) is almost zero, which is nowhere near 0.1. That's a problem.Wait, that suggests that with ( sigma = sqrt{2/pi} ), the density at ( x = mu +10 ) is practically zero, but we need it to be 0.1. So, my initial assumption that ( mu ) is the center is correct, but perhaps ( mu ) isn't zero? Wait, no, ( mu ) is the mean, so it's the center. But the problem is that with the given maximum density, the standard deviation is too small to have a significant density at 10 units away.Wait, maybe I made a mistake in interpreting the problem. Let me read it again: "the fog density at the center of the forest (( x = mu )) is 0.5 units and at a point 10 units away from the center (( x = mu + 10 )) is 0.1 units." So, ( mu ) is the center, and ( x = mu +10 ) is 10 units away. So, the distance is 10 units, but in terms of the Gaussian, the exponent is ( (x - mu)^2 / (2sigma^2) ). So, the distance in terms of standard deviations is ( 10 / sigma ).Given that, the ratio of densities is ( f(mu +10)/f(mu) = 0.1 / 0.5 = 0.2 ). So, ( e^{-100/(2sigma^2)} = 0.2 ). So, taking natural log:( -100/(2sigma^2) = ln(0.2) )Which is:( -50/sigma^2 = -1.6094 )So, ( 50/sigma^2 = 1.6094 ) => ( sigma^2 = 50 / 1.6094 approx 31.06 ) => ( sigma approx 5.57 ).But earlier, from ( f(mu) = 0.5 ), we have ( 1/(sigma sqrt{2pi}) = 0.5 ) => ( sigma = 1/(0.5 sqrt{2pi}) = 2 / sqrt{2pi} = sqrt{2/pi} approx 0.7979 ). But this contradicts the previous result. So, something's wrong.Wait, perhaps I misapplied the maximum density. The maximum density of a Gaussian is ( 1/(sigma sqrt{2pi}) ). So, if ( f(mu) = 0.5 ), then ( 1/(sigma sqrt{2pi}) = 0.5 ). So, ( sigma = 1/(0.5 sqrt{2pi}) = 2 / sqrt{2pi} = sqrt{2/pi} approx 0.7979 ). But then, the density at ( x = mu +10 ) is ( 0.5 e^{-100/(2sigma^2)} ). Plugging ( sigma^2 = 2/pi approx 0.6366 ), so ( 100/(2*0.6366) approx 100/1.2732 approx 78.54 ). So, ( e^{-78.54} ) is practically zero, which can't be 0.1. So, this is impossible.Wait, that suggests that with the given maximum density of 0.5 at the center, the density at 10 units away can't be 0.1 because the Gaussian decays too quickly. So, maybe the student made a mistake in the problem setup, or perhaps I'm misunderstanding the units.Alternatively, perhaps the distance isn't 10 units in the x-axis, but 10 standard deviations? But the problem says 10 units away. Hmm.Wait, another approach: maybe the student wants the density at 10 units to be 0.1, regardless of the maximum. So, perhaps we need to solve for ( sigma ) such that ( f(mu +10) = 0.1 ), given ( f(mu) = 0.5 ).So, let's denote ( f(mu) = 0.5 = frac{1}{sigma sqrt{2pi}} ), so ( sigma = frac{1}{0.5 sqrt{2pi}} = sqrt{frac{2}{pi}} ).Then, ( f(mu +10) = 0.5 e^{-100/(2sigma^2)} = 0.1 ).So, ( e^{-100/(2sigma^2)} = 0.2 ).Take natural log:( -100/(2sigma^2) = ln(0.2) approx -1.6094 ).So, ( 100/(2sigma^2) = 1.6094 ).Thus, ( 2sigma^2 = 100 / 1.6094 approx 62.12 ).So, ( sigma^2 approx 31.06 ), so ( sigma approx 5.57 ).But wait, earlier we had ( sigma = sqrt{2/pi} approx 0.7979 ). So, this is a contradiction. Therefore, it's impossible to have both ( f(mu) = 0.5 ) and ( f(mu +10) = 0.1 ) with the same ( sigma ). Therefore, perhaps the student made a mistake in the problem, or perhaps I'm misunderstanding something.Wait, perhaps the density is not normalized? Wait, no, the Gaussian is normalized by ( 1/(sigma sqrt{2pi}) ), so the maximum is ( 1/(sigma sqrt{2pi}) ). So, if the maximum is 0.5, then ( sigma ) is fixed as ( sqrt{2/pi} ), and the density at 10 units away is practically zero, which can't be 0.1. Therefore, perhaps the student wants the density to be 0.1 at 10 units, but without fixing the maximum. Or maybe the maximum is not at the center? Wait, no, the center is the mean, so the maximum is at the center.Alternatively, perhaps the student wants the density to be 0.5 at the center and 0.1 at 10 units, but without the Gaussian being normalized. Wait, but the equation given is a normalized Gaussian. So, perhaps the student wants to adjust the amplitude. Wait, but the equation is given as ( f(x) = frac{1}{sigma sqrt{2pi}} e^{-frac{(x - mu)^2}{2sigma^2}} ), which is the standard Gaussian PDF. So, the maximum is ( 1/(sigma sqrt{2pi}) ).Therefore, to have ( f(mu) = 0.5 ), ( sigma ) must be ( sqrt{2/pi} approx 0.7979 ). But then, ( f(mu +10) ) is practically zero, which can't be 0.1. Therefore, perhaps the student wants a different scaling. Maybe the Gaussian is scaled by a factor ( A ), so ( f(x) = A e^{-frac{(x - mu)^2}{2sigma^2}} ). Then, ( A ) would be the maximum density. So, if ( f(mu) = 0.5 ), then ( A = 0.5 ). Then, ( f(mu +10) = 0.5 e^{-100/(2sigma^2)} = 0.1 ). So, solving for ( sigma ):( e^{-100/(2sigma^2)} = 0.2 )Take ln:( -100/(2sigma^2) = ln(0.2) approx -1.6094 )So, ( 100/(2sigma^2) = 1.6094 ) => ( sigma^2 = 100/(2*1.6094) approx 31.06 ) => ( sigma approx 5.57 ).But in this case, the Gaussian isn't normalized, because the integral would be ( A sigma sqrt{2pi} ). So, if the student wants a normalized Gaussian, it's impossible to have both ( f(mu) = 0.5 ) and ( f(mu +10) = 0.1 ). Therefore, perhaps the student made a mistake in the problem setup, or perhaps they don't require the Gaussian to be normalized.Assuming that the Gaussian doesn't need to be normalized, then we can proceed. So, let's redefine the problem: ( f(x) = A e^{-frac{(x - mu)^2}{2sigma^2}} ), with ( f(mu) = 0.5 ) and ( f(mu +10) = 0.1 ).Then, ( A = 0.5 ).At ( x = mu +10 ):( 0.1 = 0.5 e^{-100/(2sigma^2)} )So, ( e^{-100/(2sigma^2)} = 0.2 )Take ln:( -100/(2sigma^2) = ln(0.2) approx -1.6094 )Thus, ( 100/(2sigma^2) = 1.6094 ) => ( sigma^2 = 100/(2*1.6094) approx 31.06 ) => ( sigma approx 5.57 ).So, in this case, ( mu ) is the center, and ( sigma approx 5.57 ). But the problem didn't specify whether the Gaussian is normalized or not. Since the given equation is the standard PDF, which is normalized, but the maximum density is ( 1/(sigma sqrt{2pi}) ). So, if the student wants the maximum density to be 0.5, then ( sigma ) is fixed, and the density at 10 units away is practically zero. Therefore, perhaps the student wants a non-normalized Gaussian, which would allow both conditions to be satisfied.Alternatively, perhaps the student made a mistake in the problem, and the density at 10 units is 0.1 times the maximum, not 0.1 units. That would make more sense. So, if ( f(mu +10) = 0.1 f(mu) ), then:( 0.1 = e^{-100/(2sigma^2)} )Which would give ( sigma approx 5.57 ), and ( f(mu) = 0.5 ) would require ( A = 0.5 ), but then the integral wouldn't be 1. So, perhaps the student wants the density to be 0.5 at the center and 0.1 at 10 units, regardless of normalization.Given that, I think the answer is ( mu ) is the center (so it's a parameter, but since it's the mean, it's just the center point, so perhaps ( mu ) is given as the center, so it's just a point, not a value to solve for. Wait, the problem says "find the values of ( mu ) and ( sigma )". But ( mu ) is the mean, which is the center of the forest. So, unless the forest's center is at a specific x-coordinate, but the problem doesn't specify. It just says the fog density at the center is 0.5 and at 10 units away is 0.1. So, ( mu ) is the center, but we don't have its specific value. Wait, but the problem says "find the values of ( mu ) and ( sigma )", so perhaps ( mu ) is zero? Or is it arbitrary?Wait, no, the problem doesn't specify the location of the center, just that the density at the center is 0.5 and at 10 units away is 0.1. So, ( mu ) is just the center, but we don't need to find its numerical value because it's a parameter that can be set to any value, but in the equation, it's just the mean. So, perhaps the problem expects ( mu ) to be zero, but that's an assumption. Alternatively, since the problem doesn't specify, maybe ( mu ) can be any value, but we need to express ( sigma ) in terms of the distance from the center.Wait, but the problem says "find the values of ( mu ) and ( sigma )", implying that both are to be determined. But without more information, ( mu ) can't be determined numerically. So, perhaps the problem expects ( mu ) to be zero, or perhaps it's a typo, and only ( sigma ) is to be found. Alternatively, maybe the problem expects ( mu ) to be the center, so it's a known value, but since it's not given, perhaps it's arbitrary, and we can set ( mu = 0 ) for simplicity.Assuming ( mu = 0 ), then the density at ( x = 10 ) is 0.1. So, let's proceed with that.So, with ( mu = 0 ), we have:1. ( f(0) = 0.5 = frac{1}{sigma sqrt{2pi}} )2. ( f(10) = 0.1 = frac{1}{sigma sqrt{2pi}} e^{-100/(2sigma^2)} )From equation 1, ( sigma = sqrt{2/pi} approx 0.7979 ). But then, equation 2 gives ( f(10) approx 0.5 e^{-78.54} approx 0 ), which is not 0.1. So, contradiction.Therefore, unless the Gaussian is not normalized, which would allow us to set ( A = 0.5 ) and solve for ( sigma approx 5.57 ), but then the integral wouldn't be 1.Alternatively, perhaps the student wants the density at 10 units to be 0.1 times the maximum, so ( f(10) = 0.1 f(0) ). Then, ( 0.1 = e^{-100/(2sigma^2)} ), leading to ( sigma approx 5.57 ), and ( f(0) = 0.5 ) would require ( A = 0.5 ), but then the integral isn't 1. So, perhaps the student doesn't care about normalization.Given that, I think the answer is ( mu ) is the center (so it's a parameter, but since it's not given, perhaps it's zero or arbitrary), and ( sigma approx 5.57 ). But since the problem asks for exact values, let's compute it precisely.From ( e^{-100/(2sigma^2)} = 0.2 ), so:( -100/(2sigma^2) = ln(0.2) )( 100/(2sigma^2) = -ln(0.2) = ln(5) approx 1.6094 )Thus, ( sigma^2 = 100/(2 ln(5)) = 50 / ln(5) )So, ( sigma = sqrt{50 / ln(5)} )Compute ( ln(5) approx 1.6094 ), so ( 50 / 1.6094 approx 31.06 ), so ( sigma approx sqrt{31.06} approx 5.57 ).But since the problem didn't specify whether the Gaussian is normalized, and given that the equation provided is the standard PDF, which is normalized, but the maximum density is ( 1/(sigma sqrt{2pi}) ), which would be 0.5, leading to a contradiction with the second condition, I think the problem expects us to ignore normalization and solve for ( sigma ) such that ( f(mu +10) = 0.1 ) with ( f(mu) = 0.5 ). Therefore, ( mu ) is the center, which is a known point, but since it's not given numerically, perhaps it's just the mean, so we can set ( mu = 0 ) for simplicity, and ( sigma = sqrt{50 / ln(5)} ).But wait, if we set ( mu = 0 ), then the density at ( x = 10 ) is 0.1, but the maximum density is 0.5, which requires ( sigma approx 5.57 ). So, the answer is ( mu = 0 ) and ( sigma = sqrt{50 / ln(5)} ).Alternatively, if the problem expects ( mu ) to be a specific value, but since it's not given, perhaps it's just the center, so we can leave it as ( mu ) (the center), and ( sigma = sqrt{50 / ln(5)} ).But the problem says "find the values of ( mu ) and ( sigma )", so perhaps ( mu ) is zero, and ( sigma ) is as above.Alternatively, perhaps the problem expects ( mu ) to be the center, so it's a known value, but since it's not given, we can't determine it numerically. Therefore, perhaps the problem expects ( mu ) to be zero, and ( sigma ) as above.In conclusion, for the first part, ( a = 25/16 ), ( b = 11/4 ), ( c = pi/4 ). For the second part, ( mu ) is the center (so perhaps zero), and ( sigma = sqrt{50 / ln(5)} approx 5.57 ).But wait, let me double-check the first part. If ( c = pi/4 ), then at ( t_o = 2 ), ( sin(2c) = sin(pi/2) = 1 ), and at ( t_o =4 ), ( sin(4c) = sin(pi) = 0 ). So, the equations become:1. ( 4a + b = 9 )2. ( 16a = 25 )From equation 2, ( a = 25/16 ). Then, equation 1: ( 4*(25/16) + b = 9 ) => ( 25/4 + b = 9 ) => ( b = 9 - 25/4 = (36 -25)/4 = 11/4 ). So, yes, that's correct.For the second part, since the problem didn't specify whether the Gaussian is normalized, and given that the maximum density is 0.5, but the density at 10 units is 0.1, which is impossible with a normalized Gaussian, I think the problem expects us to ignore normalization and solve for ( sigma ) such that ( f(mu +10) = 0.1 ) with ( f(mu) = 0.5 ). Therefore, ( sigma = sqrt{50 / ln(5)} ).But let me compute ( ln(5) ) precisely: ( ln(5) approx 1.60943791 ). So, ( 50 / 1.60943791 approx 31.06 ), so ( sigma approx sqrt{31.06} approx 5.573 ).But since the problem might expect an exact form, ( sigma = sqrt{50 / ln(5)} ).So, summarizing:1. ( a = 25/16 ), ( b = 11/4 ), ( c = pi/4 )2. ( mu ) is the center (could be zero or any specific point, but since not given, perhaps it's just the mean, so we can denote it as ( mu )), and ( sigma = sqrt{50 / ln(5)} )But the problem asks to "find the values of ( mu ) and ( sigma )", so perhaps ( mu ) is zero, and ( sigma ) is as above.Alternatively, if the problem expects ( mu ) to be a specific value, but since it's not given, perhaps it's just the center, so we can't determine it numerically. Therefore, the answer is ( mu ) is the center, and ( sigma = sqrt{50 / ln(5)} ).But since the problem didn't specify units or a specific location for the center, perhaps ( mu ) is zero, and ( sigma ) is as above.So, final answers:1. ( a = frac{25}{16} ), ( b = frac{11}{4} ), ( c = frac{pi}{4} )2. ( mu ) is the center (could be zero), ( sigma = sqrt{frac{50}{ln(5)}} )</think>

🧠:<think>Okay, so I have this problem where I need to help a tech magnate optimize their schedule for the upcoming week. They have two main tasks: distributing 60 hours of work over 5 days to maximize productivity, and calculating the expected total travel time and its variance for three international trips. Let me try to break this down step by step.Starting with the first part: maximizing productivity. The productivity function given is ( P(t) = frac{100}{1 + e^{-0.1(t-50)}} ). Hmm, this looks like a logistic function. It starts at a lower productivity when t is low and increases as t increases, approaching 100 as t becomes very large. But since the magnate can only work 60 hours in total over 5 days, we need to distribute these hours in a way that the sum of daily productivity is maximized.I think I need to set up an optimization problem where the variables are the hours worked each day, ( t_1, t_2, t_3, t_4, t_5 ), with the constraint that ( t_1 + t_2 + t_3 + t_4 + t_5 = 60 ). The objective is to maximize ( sum_{i=1}^5 P(t_i) ).Since the productivity function is concave or convex? Let me check the second derivative to see. If it's concave, then by Jensen's inequality, the maximum would be achieved when all ( t_i ) are equal. If it's convex, then we might need to concentrate the hours on a single day.Calculating the first derivative of P(t):( P'(t) = frac{100 times 0.1 e^{-0.1(t-50)}}{(1 + e^{-0.1(t-50)})^2} )Simplify that:( P'(t) = frac{10 e^{-0.1(t-50)}}{(1 + e^{-0.1(t-50)})^2} )Now, the second derivative:Let me denote ( u = e^{-0.1(t-50)} ), then ( P(t) = frac{100}{1 + u} ), so ( P'(t) = frac{-100 u'}{(1 + u)^2} ), and ( P''(t) = frac{-100 [u''(1 + u)^2 - u'(2(1 + u)u')]}{(1 + u)^4} ). Hmm, this is getting complicated. Maybe instead of computing it directly, I can analyze the shape.Looking at the function ( P(t) ), as t increases, P(t) increases but at a decreasing rate because the derivative ( P'(t) ) is positive but decreasing. So the function is concave. If the function is concave, then the maximum of the sum is achieved when the variables are equal, due to Jensen's inequality.Therefore, to maximize the total productivity, each day should have the same number of hours. Since the total is 60 hours over 5 days, each day should have 12 hours.Wait, but let me think again. If the function is concave, then the sum is maximized when the inputs are equal. So yes, distributing equally would give the maximum total productivity.So, the optimal distribution is 12 hours each day.Now, moving on to the second part: calculating the expected total travel time and its variance for three international trips A, B, and C.Each trip has a normally distributed travel time with given means and standard deviations:- Trip A: ( mu_A = 5 ) hours, ( sigma_A = 1 )- Trip B: ( mu_B = 7 ) hours, ( sigma_B = 1.5 )- Trip C: ( mu_C = 4 ) hours, ( sigma_C = 0.5 )Since the travel times are independent, the expected total travel time is just the sum of the expected times for each trip.So, ( E[T_{total}] = E[T_A] + E[T_B] + E[T_C] = 5 + 7 + 4 = 16 ) hours.For the variance, since the trips are independent, the variance of the total travel time is the sum of the variances of each trip.Variance of A: ( sigma_A^2 = 1^2 = 1 )Variance of B: ( sigma_B^2 = 1.5^2 = 2.25 )Variance of C: ( sigma_C^2 = 0.5^2 = 0.25 )So, ( Var(T_{total}) = 1 + 2.25 + 0.25 = 3.5 )Therefore, the expected total travel time is 16 hours with a variance of 3.5.Wait, but hold on. The problem says the magnate must travel to three international locations over the week. Does this mean that each trip is made once, so we have three trips in total? Or is it that each trip is made multiple times? The problem states "three international locations (A, B, and C)", so I think each trip is made once. So, the total travel time is the sum of T_A, T_B, and T_C, each being a single trip.Therefore, the calculations above are correct.So, summarizing:1. The optimal distribution is 12 hours each day.2. The expected total travel time is 16 hours, and the variance is 3.5.I think that's it. Let me just double-check if there's any other consideration for the first part. For example, is there a maximum number of hours per day? The problem doesn't specify any constraints on individual days, just the total over 5 days. So, distributing equally is fine.Also, since the productivity function is concave, equal distribution maximizes the sum. If it were convex, we might have a different distribution, but in this case, equal hours per day is optimal.For the second part, since the travel times are independent, their variances add up, so 1 + 2.25 + 0.25 is indeed 3.5. The expected value is straightforward, just the sum of the means.Yeah, I think I covered everything.Final Answer1. The optimal distribution of work hours is boxed{12} hours per day.2. The expected total travel time is boxed{16} hours and the variance is boxed{3.5}.</think>

🧠:<think>Alright, so I have this problem about a Pakistani entrepreneur who's using a logistic growth model for her startup's revenue. The differential equation given is:[ frac{dR}{dt} = rRleft(1 - frac{R}{K}right) ]where ( R(t) ) is the revenue in millions of Rupees, ( r = 0.2 ) is the growth rate, and ( K = 100 ) is the carrying capacity. The initial revenue is ( R(0) = 5 ) million Rupees. First, I need to solve this differential equation to find ( R(t) ). I remember that the logistic equation is a common model for growth with a carrying capacity. The standard solution involves integrating both sides, probably using partial fractions. Let me recall the steps.The logistic equation is separable, so I can write:[ frac{dR}{Rleft(1 - frac{R}{K}right)} = r dt ]I need to integrate both sides. The left side can be integrated using partial fractions. Let me set up the partial fractions decomposition for the integrand.Let me denote:[ frac{1}{Rleft(1 - frac{R}{K}right)} = frac{A}{R} + frac{B}{1 - frac{R}{K}} ]Multiplying both sides by ( Rleft(1 - frac{R}{K}right) ) gives:[ 1 = Aleft(1 - frac{R}{K}right) + B R ]Expanding this:[ 1 = A - frac{A R}{K} + B R ]Grouping like terms:[ 1 = A + Rleft(-frac{A}{K} + Bright) ]Since this must hold for all R, the coefficients of like powers of R must be equal on both sides. So, the constant term:[ A = 1 ]And the coefficient of R:[ -frac{A}{K} + B = 0 ]Plugging in ( A = 1 ):[ -frac{1}{K} + B = 0 implies B = frac{1}{K} ]So, the partial fractions decomposition is:[ frac{1}{Rleft(1 - frac{R}{K}right)} = frac{1}{R} + frac{1}{Kleft(1 - frac{R}{K}right)} ]Therefore, the integral becomes:[ int left( frac{1}{R} + frac{1}{Kleft(1 - frac{R}{K}right)} right) dR = int r dt ]Let me compute the left integral term by term.First term:[ int frac{1}{R} dR = ln |R| + C_1 ]Second term:Let me make a substitution. Let ( u = 1 - frac{R}{K} ), then ( du = -frac{1}{K} dR ), so ( -K du = dR ).So,[ int frac{1}{K u} (-K du) = -int frac{1}{u} du = -ln |u| + C_2 = -ln left|1 - frac{R}{K}right| + C_2 ]Putting it all together, the left integral is:[ ln |R| - ln left|1 - frac{R}{K}right| + C ]Which can be written as:[ ln left| frac{R}{1 - frac{R}{K}} right| + C ]The right integral is straightforward:[ int r dt = r t + C' ]So, combining both sides:[ ln left( frac{R}{1 - frac{R}{K}} right) = r t + C ]Exponentiating both sides to eliminate the logarithm:[ frac{R}{1 - frac{R}{K}} = e^{r t + C} = e^{C} e^{r t} ]Let me denote ( e^{C} ) as another constant, say ( C'' ). So,[ frac{R}{1 - frac{R}{K}} = C'' e^{r t} ]Let me solve for R. Multiply both sides by ( 1 - frac{R}{K} ):[ R = C'' e^{r t} left(1 - frac{R}{K}right) ]Expanding the right side:[ R = C'' e^{r t} - frac{C'' e^{r t} R}{K} ]Bring the term with R to the left:[ R + frac{C'' e^{r t} R}{K} = C'' e^{r t} ]Factor out R:[ R left(1 + frac{C'' e^{r t}}{K}right) = C'' e^{r t} ]Solve for R:[ R = frac{C'' e^{r t}}{1 + frac{C'' e^{r t}}{K}} ]Multiply numerator and denominator by K to simplify:[ R = frac{C'' K e^{r t}}{K + C'' e^{r t}} ]Now, let's apply the initial condition ( R(0) = 5 ). So, when ( t = 0 ):[ 5 = frac{C'' K e^{0}}{K + C'' e^{0}} = frac{C'' K}{K + C''} ]Solving for ( C'' ):Multiply both sides by ( K + C'' ):[ 5(K + C'') = C'' K ]Expand:[ 5K + 5 C'' = C'' K ]Bring all terms with ( C'' ) to one side:[ 5K = C'' K - 5 C'' ]Factor out ( C'' ):[ 5K = C'' (K - 5) ]Solve for ( C'' ):[ C'' = frac{5K}{K - 5} ]Plugging in ( K = 100 ):[ C'' = frac{5 times 100}{100 - 5} = frac{500}{95} = frac{100}{19} ]So, plugging back into the expression for R(t):[ R(t) = frac{left( frac{100}{19} times 100 right) e^{0.2 t}}{100 + frac{100}{19} e^{0.2 t}} ]Simplify numerator and denominator:Numerator: ( frac{10000}{19} e^{0.2 t} )Denominator: ( 100 + frac{100}{19} e^{0.2 t} )Factor out 100 in the denominator:[ 100 left(1 + frac{1}{19} e^{0.2 t}right) ]So, R(t) becomes:[ R(t) = frac{frac{10000}{19} e^{0.2 t}}{100 left(1 + frac{1}{19} e^{0.2 t}right)} ]Simplify numerator and denominator by dividing numerator and denominator by 100:[ R(t) = frac{frac{100}{19} e^{0.2 t}}{1 + frac{1}{19} e^{0.2 t}} ]We can write this as:[ R(t) = frac{100 e^{0.2 t}}{19 + e^{0.2 t}} ]Alternatively, factor out ( e^{0.2 t} ) in the denominator:[ R(t) = frac{100}{19 e^{-0.2 t} + 1} ]But the first form is probably more straightforward.So, that's the solution for part 1.Moving on to part 2: the entrepreneur wants to reinvest a portion of the revenue back into the startup, with the reinvestment rate given by ( I(t) = alpha R(t)^2 ), where ( alpha = 0.01 ). We need to find the total amount reinvested over the first 10 years.So, the total reinvestment is the integral of ( I(t) ) from t=0 to t=10:[ text{Total Reinvestment} = int_{0}^{10} I(t) dt = int_{0}^{10} 0.01 R(t)^2 dt ]We already have ( R(t) ) from part 1, so we can plug that in.So, first, let's write ( R(t) = frac{100 e^{0.2 t}}{19 + e^{0.2 t}} ). Therefore, ( R(t)^2 ) is:[ R(t)^2 = left( frac{100 e^{0.2 t}}{19 + e^{0.2 t}} right)^2 = frac{10000 e^{0.4 t}}{(19 + e^{0.2 t})^2} ]So, the integral becomes:[ int_{0}^{10} 0.01 times frac{10000 e^{0.4 t}}{(19 + e^{0.2 t})^2} dt ]Simplify constants:0.01 * 10000 = 100, so:[ 100 int_{0}^{10} frac{e^{0.4 t}}{(19 + e^{0.2 t})^2} dt ]Let me make a substitution to solve this integral. Let me set:Let ( u = 19 + e^{0.2 t} )Then, ( du/dt = 0.2 e^{0.2 t} implies du = 0.2 e^{0.2 t} dt )Hmm, but in the integral, I have ( e^{0.4 t} ). Let me see:Note that ( e^{0.4 t} = (e^{0.2 t})^2 ). Since ( u = 19 + e^{0.2 t} ), then ( e^{0.2 t} = u - 19 ), so ( e^{0.4 t} = (u - 19)^2 ).So, substituting:The integral becomes:[ 100 int frac{(u - 19)^2}{u^2} times frac{du}{0.2 (u - 19)} ]Wait, let's see:From ( du = 0.2 e^{0.2 t} dt ), so ( dt = frac{du}{0.2 e^{0.2 t}} = frac{du}{0.2 (u - 19)} )So, substituting into the integral:[ 100 times int frac{(u - 19)^2}{u^2} times frac{du}{0.2 (u - 19)} ]Simplify:The ( (u - 19)^2 ) in the numerator and ( (u - 19) ) in the denominator gives ( (u - 19) ). So:[ 100 times frac{1}{0.2} int frac{(u - 19)}{u^2} du ]Simplify constants:100 / 0.2 = 500, so:[ 500 int frac{u - 19}{u^2} du ]Break the fraction into two terms:[ 500 int left( frac{u}{u^2} - frac{19}{u^2} right) du = 500 int left( frac{1}{u} - frac{19}{u^2} right) du ]Integrate term by term:Integral of ( 1/u ) is ( ln |u| ), integral of ( 19/u^2 ) is ( -19/u ). So:[ 500 left( ln |u| + frac{19}{u} right) + C ]Substitute back ( u = 19 + e^{0.2 t} ):[ 500 left( ln (19 + e^{0.2 t}) + frac{19}{19 + e^{0.2 t}} right) + C ]Now, evaluate this from t=0 to t=10.So, the definite integral is:[ 500 left[ ln (19 + e^{0.2 times 10}) + frac{19}{19 + e^{0.2 times 10}} right] - 500 left[ ln (19 + e^{0}) + frac{19}{19 + e^{0}} right] ]Compute each term step by step.First, compute at t=10:Compute ( e^{0.2 times 10} = e^{2} approx 7.389056 )So,First part: ( ln (19 + 7.389056) = ln (26.389056) approx 3.272 )Second part: ( frac{19}{19 + 7.389056} = frac{19}{26.389056} approx 0.720 )So, total at t=10: ( 3.272 + 0.720 = 3.992 )Multiply by 500: ( 500 times 3.992 = 1996 )Now, compute at t=0:( e^{0} = 1 )First part: ( ln (19 + 1) = ln (20) approx 2.9957 )Second part: ( frac{19}{19 + 1} = frac{19}{20} = 0.95 )Total at t=0: ( 2.9957 + 0.95 = 3.9457 )Multiply by 500: ( 500 times 3.9457 = 1972.85 )So, the definite integral is:1996 - 1972.85 = 23.15So, approximately 23.15 million Rupees.But let me check the calculations again because the numbers seem a bit off.Wait, let me recalculate the definite integral:At t=10:( u = 19 + e^{2} approx 19 + 7.389056 = 26.389056 )So,( ln(26.389056) approx 3.272 )( 19 / 26.389056 approx 0.720 )So, 3.272 + 0.720 = 3.992Multiply by 500: 3.992 * 500 = 1996At t=0:( u = 19 + 1 = 20 )( ln(20) approx 2.9957 )( 19 / 20 = 0.95 )So, 2.9957 + 0.95 = 3.9457Multiply by 500: 3.9457 * 500 = 1972.85Subtracting: 1996 - 1972.85 = 23.15So, approximately 23.15 million Rupees.But let me check if I did the substitution correctly.Wait, when I did the substitution, I had:Integral becomes:500 [ ln(u) + 19/u ] evaluated from u(0) to u(10)So, u(10) = 19 + e^{2} ≈ 26.389u(0) = 20So, 500 [ (ln(26.389) + 19/26.389) - (ln(20) + 19/20) ]Compute each term:ln(26.389) ≈ 3.27219 / 26.389 ≈ 0.720ln(20) ≈ 2.995719 / 20 = 0.95So, 3.272 + 0.720 = 3.9922.9957 + 0.95 = 3.9457Difference: 3.992 - 3.9457 = 0.0463Multiply by 500: 0.0463 * 500 ≈ 23.15Yes, that seems correct.So, the total reinvestment over the first 10 years is approximately 23.15 million Rupees.But let me check if I did the substitution correctly.Wait, in the substitution step:We had:Integral of ( frac{e^{0.4 t}}{(19 + e^{0.2 t})^2} dt )Let ( u = 19 + e^{0.2 t} ), so ( du = 0.2 e^{0.2 t} dt implies dt = du / (0.2 e^{0.2 t}) )But ( e^{0.2 t} = u - 19 ), so ( dt = du / (0.2 (u - 19)) )So, substituting into the integral:( int frac{(u - 19)^2}{u^2} times frac{du}{0.2 (u - 19)} )Simplify:( int frac{(u - 19)}{u^2} times frac{du}{0.2} )Which is:( frac{1}{0.2} int frac{u - 19}{u^2} du = 5 int left( frac{1}{u} - frac{19}{u^2} right) du )Wait, hold on, earlier I had 500, but let's see:Wait, the original integral was:100 * integral(...)After substitution, the integral became:100 * [ integral(...) ] which was transformed into:100 * [ 5 * integral(...) ]Wait, no, let's step back.Original integral:100 * integral [ e^{0.4 t} / (19 + e^{0.2 t})^2 dt ]After substitution, we have:100 * integral [ (u - 19)^2 / u^2 * (du / (0.2 (u - 19))) ]Which is:100 * integral [ (u - 19) / u^2 * du / 0.2 ]Which is:100 / 0.2 * integral [ (u - 19)/u^2 du ] = 500 * integral [ (1/u - 19/u^2) du ]Yes, that's correct.So, the integral is 500 [ ln u + 19/u ] evaluated from u=20 to u≈26.389.So, 500 [ (ln(26.389) + 19/26.389) - (ln(20) + 19/20) ] ≈ 500 [ (3.272 + 0.720) - (2.9957 + 0.95) ] ≈ 500 [3.992 - 3.9457] ≈ 500 * 0.0463 ≈ 23.15So, approximately 23.15 million Rupees.But let me check if I can compute this more accurately.Compute ln(26.389):26.389 is e^3.272, since e^3 ≈ 20.085, e^3.272 ≈ 26.389, so ln(26.389)=3.272Similarly, 19/26.389 ≈ 0.720ln(20)=2.995719/20=0.95So, 3.272 + 0.720 = 3.9922.9957 + 0.95 = 3.9457Difference: 3.992 - 3.9457 = 0.0463Multiply by 500: 0.0463 * 500 = 23.15So, yes, 23.15 million Rupees.But let me see if I can compute this more precisely.Alternatively, perhaps I can compute the integral numerically to check.Alternatively, perhaps using substitution, but I think the analytical approach is correct.So, the total reinvestment is approximately 23.15 million Rupees.But let me see if I can express it in exact terms.Wait, the integral was:500 [ ln(u) + 19/u ] from 20 to 19 + e^{2}So, exact expression is:500 [ ln(19 + e^{2}) + 19/(19 + e^{2}) - ln(20) - 19/20 ]But perhaps we can leave it in terms of exponentials, but since the question asks for the total amount, it's fine to compute numerically.So, approximately 23.15 million Rupees.But let me check if I made any mistake in the substitution.Wait, when I did the substitution, I had:I(t) = 0.01 R(t)^2R(t) = 100 e^{0.2 t} / (19 + e^{0.2 t})So, R(t)^2 = 10000 e^{0.4 t} / (19 + e^{0.2 t})^2Thus, I(t) = 0.01 * 10000 e^{0.4 t} / (19 + e^{0.2 t})^2 = 100 e^{0.4 t} / (19 + e^{0.2 t})^2So, the integral is:∫₀¹⁰ 100 e^{0.4 t} / (19 + e^{0.2 t})² dtWhich is what I had.Then substitution u = 19 + e^{0.2 t}, du = 0.2 e^{0.2 t} dt, so dt = du / (0.2 e^{0.2 t}) = du / (0.2 (u - 19))Thus, integral becomes:∫ [100 e^{0.4 t} / u² ] * [ du / (0.2 (u - 19)) ]But e^{0.4 t} = (e^{0.2 t})² = (u - 19)^2So, integral becomes:∫ [100 (u - 19)^2 / u² ] * [ du / (0.2 (u - 19)) ] = ∫ [100 (u - 19) / u² ] * [ du / 0.2 ]Which is:(100 / 0.2) ∫ (u - 19)/u² du = 500 ∫ (1/u - 19/u²) duWhich is correct.So, the integral is 500 [ ln u + 19/u ] from 20 to 19 + e²So, the calculation is correct.Thus, the total reinvestment is approximately 23.15 million Rupees.But let me compute it more accurately.Compute ln(26.389056):Using calculator, ln(26.389056) ≈ 3.27219 / 26.389056 ≈ 0.720ln(20) ≈ 2.99573219 / 20 = 0.95So,At upper limit: 3.272 + 0.720 = 3.992At lower limit: 2.995732 + 0.95 = 3.945732Difference: 3.992 - 3.945732 = 0.046268Multiply by 500: 0.046268 * 500 = 23.134So, approximately 23.134 million Rupees.Rounding to two decimal places, 23.13 million Rupees.Alternatively, perhaps we can express it as 23.13 million.But let me check if I can compute it more precisely.Alternatively, perhaps using a calculator for the integral.But since we've done the substitution correctly, and the result is approximately 23.13 million Rupees.So, the total reinvestment over the first 10 years is approximately 23.13 million Rupees.But let me see if I can express it in exact terms.Wait, the exact expression is:500 [ ln(19 + e²) + 19/(19 + e²) - ln(20) - 19/20 ]But perhaps we can compute it more accurately.Compute e² ≈ 7.38905609893So, 19 + e² ≈ 26.38905609893Compute ln(26.38905609893):Using calculator: ln(26.38905609893) ≈ 3.272019649Compute 19 / 26.38905609893 ≈ 0.720000000Compute ln(20) ≈ 2.99573227355Compute 19 / 20 = 0.95So,Upper limit: 3.272019649 + 0.72 ≈ 3.992019649Lower limit: 2.99573227355 + 0.95 ≈ 3.94573227355Difference: 3.992019649 - 3.94573227355 ≈ 0.04628737545Multiply by 500: 0.04628737545 * 500 ≈ 23.143687725So, approximately 23.1437 million Rupees.Rounding to two decimal places, 23.14 million Rupees.So, about 23.14 million.But perhaps the exact value is better expressed as 23.14 million.Alternatively, if we want to be more precise, 23.1437 million.But in the context of the problem, probably two decimal places are sufficient.So, the total reinvestment is approximately 23.14 million Rupees.But let me check if I can express it in terms of fractions.Wait, 0.04628737545 * 500 = 23.143687725So, 23.143687725 ≈ 23.1437So, 23.1437 million Rupees.Alternatively, we can write it as 23.14 million.But perhaps the exact value is better left as is.Alternatively, perhaps the problem expects an exact expression, but since it's a definite integral, it's fine to compute numerically.So, in conclusion, the total reinvestment over the first 10 years is approximately 23.14 million Rupees.But let me check if I can compute it more accurately.Alternatively, perhaps using a calculator for the integral.But given that we've done the substitution correctly, and the result is approximately 23.14 million Rupees.So, I think that's the answer.Final Answer1. The revenue function is (boxed{dfrac{100 e^{0.2 t}}{19 + e^{0.2 t}}}) million Rupees.2. The total amount reinvested over the first 10 years is approximately (boxed{23.14}) million Rupees.</think>