🧠:<think>Alright, so I have this problem about a Lithuanian historian analyzing the logistical efficiency of their medieval army. It's split into two parts. Let me tackle them one by one.Starting with the first part: The army travels through a series of towns arranged in a linear fashion, each town being a node on a straight line. The distances between consecutive towns are given by the sequence (d_n = 5 + 3n), where (n) is the position of the town starting from 1. I need to calculate the total distance traveled if the army visited every town from the first (starting point) to the 20th town and back to the starting point.Hmm, okay. So, the army starts at the first town, goes to the second, then third, all the way to the 20th town, and then comes back to the first town. So, the total distance will be the sum of the distances from town 1 to 20 and then back from 20 to 1.But wait, since the distance between consecutive towns is given by (d_n = 5 + 3n), each segment from town (n) to (n+1) is (d_n). So, when going from town 1 to 20, they traverse segments (d_1) to (d_{19}), right? Because the distance from town 1 to 2 is (d_1), from 2 to 3 is (d_2), and so on until from 19 to 20 is (d_{19}). Then, when they come back from 20 to 1, they have to traverse the same segments again, but in reverse. So, the distance from 20 to 19 is (d_{19}), 19 to 18 is (d_{18}), etc., all the way back to (d_1) from 2 to 1.Therefore, the total distance is twice the sum of (d_1) through (d_{19}). Because going to town 20 and coming back covers each segment twice.So, the total distance (D) is (2 times sum_{n=1}^{19} d_n).Given that (d_n = 5 + 3n), let's write that out:(D = 2 times sum_{n=1}^{19} (5 + 3n))I can split this sum into two separate sums:(D = 2 times left( sum_{n=1}^{19} 5 + sum_{n=1}^{19} 3n right))Calculating each sum separately:First sum: (sum_{n=1}^{19} 5). Since 5 is constant, this is just 5 multiplied by the number of terms, which is 19.So, (5 times 19 = 95).Second sum: (sum_{n=1}^{19} 3n). I can factor out the 3:(3 times sum_{n=1}^{19} n).The sum of the first (k) natural numbers is given by (frac{k(k+1)}{2}). So, for (k=19):(sum_{n=1}^{19} n = frac{19 times 20}{2} = 190).Therefore, the second sum is (3 times 190 = 570).Adding both sums together:(95 + 570 = 665).Then, multiply by 2 to get the total distance:(2 times 665 = 1330).So, the total distance traveled is 1330 units. Wait, the problem doesn't specify the units, just says "distance traveled." So, I guess it's 1330 whatever units (d_n) is in.Wait, let me double-check my calculations. Maybe I made a mistake in the number of terms.Hold on, the army goes from town 1 to 20, which is 19 segments, right? So, (d_1) to (d_{19}). Then back, another 19 segments. So, total of 38 segments? No, wait, no. Because each segment is traversed twice, so it's 2 times 19 segments, but each segment is a distance of (d_n). So, the total distance is 2*(sum from n=1 to 19 of d_n). Which is what I did.Wait, but let me check the sum again.Sum of 5 from n=1 to 19 is 5*19=95. Sum of 3n from n=1 to 19 is 3*(19*20)/2 = 3*190=570. So, 95+570=665. Then 2*665=1330. Yeah, that seems correct.Wait, but hold on, is the distance from town 1 to town 20 equal to the sum of d1 to d19? Yes, because each d_n is the distance between town n and n+1. So, from 1 to 20, it's 19 distances. Then back is another 19, so total 38 distances, but each is d1 to d19, so 2*(sum from 1 to19). So, yeah, 1330.Okay, so the first part is 1330.Moving on to the second part: The army's rate of resource consumption is given by (R(t) = 200e^{0.05t}), where (R(t)) is the rate in units per day, and (t) is the time in days since the campaign started. The army's total initial resource supply was 15,000 units. I need to determine the maximum number of days the army could sustain itself before running out of resources. Assume that the army immediately begins traveling at the start of the campaign.So, this is a calculus problem. The total resource consumed over time is the integral of the rate from t=0 to t=T, which should equal 15,000 units.So, total resources consumed: (int_{0}^{T} R(t) dt = 15,000).Given (R(t) = 200e^{0.05t}), so:(int_{0}^{T} 200e^{0.05t} dt = 15,000).Let me compute this integral.First, the integral of (e^{kt}) is (frac{1}{k}e^{kt}). So, here, k=0.05.So, integral becomes:200 * [ (1/0.05) e^{0.05t} ] evaluated from 0 to T.Compute that:200 * (1/0.05) [e^{0.05T} - e^{0}].Simplify:200 / 0.05 = 4000.So, 4000 [e^{0.05T} - 1] = 15,000.So, 4000(e^{0.05T} - 1) = 15,000.Divide both sides by 4000:e^{0.05T} - 1 = 15,000 / 4000 = 3.75.So, e^{0.05T} = 4.75.Take natural logarithm on both sides:0.05T = ln(4.75).Compute ln(4.75). Let me recall that ln(4) is about 1.386, ln(5)=1.609. 4.75 is closer to 5, so maybe around 1.558.But let me calculate it more accurately.Compute ln(4.75):We know that ln(4) = 1.386294ln(4.75) = ln(4 * 1.1875) = ln(4) + ln(1.1875)Compute ln(1.1875). Let's see, ln(1.1875) ≈ 0.1733 (since e^0.1733 ≈ 1.19).So, ln(4.75) ≈ 1.3863 + 0.1733 ≈ 1.5596.So, 0.05T ≈ 1.5596Therefore, T ≈ 1.5596 / 0.05 ≈ 31.192 days.So, approximately 31.192 days. Since the army can't sustain for a fraction of a day beyond 31 days, we need to check whether on day 31, they have enough resources, or if they run out during day 32.But the question says "the maximum number of days the army could sustain itself before running out of resources." So, if it's 31.192 days, that means they can sustain for 31 full days and part of the 32nd day. But depending on interpretation, sometimes it's considered as 31 days, sometimes rounded up. But since the question says "maximum number of days," which is a whole number, I think we should take the floor of 31.192, which is 31 days.But let me verify the calculation.Compute e^{0.05*31} = e^{1.55} ≈ e^1.55.We know that e^1.5 ≈ 4.4817, e^1.6 ≈ 4.953. So, 1.55 is halfway between 1.5 and 1.6. So, e^1.55 ≈ (4.4817 + 4.953)/2 ≈ (9.4347)/2 ≈ 4.717.Then, 4000*(4.717 - 1) = 4000*3.717 ≈ 14,868 units.Which is less than 15,000.Then, at T=31.192, e^{0.05*31.192}= e^{1.5596}≈4.75, as before.So, 4000*(4.75 -1)=4000*3.75=15,000.So, exactly at T≈31.192 days, the resources deplete.Therefore, the army can sustain for approximately 31.192 days. If we need the maximum whole number of days, it's 31 days, because on the 32nd day, they would run out.But let me think again. If the consumption is continuous, then at T=31.192, they run out. So, depending on whether the army can partially sustain on the 32nd day or not, but since the question says "maximum number of days," it's probably 31 days.Alternatively, sometimes in such contexts, they might consider the decimal as a fraction of a day, so 31.192 days is about 31 days and 4.6 hours. So, if they need full days, it's 31 days.But let me see if the question specifies whether it's full days or can be a fractional day. It says "maximum number of days," so probably expects a whole number.Therefore, I think the answer is 31 days.But let me check the integral again to make sure I didn't make a mistake.Integral of 200e^{0.05t} dt from 0 to T is:200*(1/0.05)(e^{0.05T} - 1) = 4000(e^{0.05T} -1 ) =15,000.So, e^{0.05T}=1 +15,000/4000=1 +3.75=4.75.So, ln(4.75)=1.5596, so T=1.5596/0.05≈31.192.Yes, that's correct.So, the maximum number of days is approximately 31.192, so 31 days.Wait, but maybe I should write it as 31 days, or is 31.192 acceptable? The question says "the maximum number of days," so likely expects an integer, so 31 days.But just to be thorough, let me compute the exact value.Compute ln(4.75):We can use a calculator for more precision. Let me recall that ln(4.75)=1.558130906.So, T=1.558130906 /0.05=31.16261812 days.So, approximately 31.1626 days.So, 31 full days, and about 0.1626 of a day, which is roughly 3.9 hours.So, if the army starts on day 0, they can go through day 31, and on day 32, they would run out during that day. So, the maximum number of full days is 31.Hence, the answer is 31 days.So, summarizing:1. Total distance traveled is 1330 units.2. Maximum number of days is 31 days.Final Answer1. The total distance traveled is boxed{1330} units.2. The maximum number of days the army could sustain itself is boxed{31} days.</think>

🧠:<think>Okay, so I have this problem about modeling population growth and resource allocation. Let me try to break it down step by step.First, part 1 is about solving a differential equation for population growth. The equation given is:[ frac{dP}{dt} = kPleft(1 - frac{P}{M}right) ]This looks familiar—it's the logistic growth model, right? I remember that the logistic equation models population growth where the growth rate decreases as the population approaches the carrying capacity, M. So, the solution should be a sigmoidal curve.Given that ( P(0) = 0.4M ), I need to find ( P(t) ) as a function of time. I think the general solution to the logistic equation is:[ P(t) = frac{M}{1 + left(frac{M - P(0)}{P(0)}right)e^{-kt}} ]Let me verify that. If I plug in ( t = 0 ), I should get ( P(0) = 0.4M ). Plugging in:[ P(0) = frac{M}{1 + left(frac{M - 0.4M}{0.4M}right)} = frac{M}{1 + left(frac{0.6M}{0.4M}right)} = frac{M}{1 + 1.5} = frac{M}{2.5} = 0.4M ]Yes, that checks out. So, substituting ( P(0) = 0.4M ) into the general solution, I get:[ P(t) = frac{M}{1 + left(frac{M - 0.4M}{0.4M}right)e^{-kt}} ]Simplify the fraction inside the exponential:[ frac{M - 0.4M}{0.4M} = frac{0.6M}{0.4M} = 1.5 ]So, the solution becomes:[ P(t) = frac{M}{1 + 1.5e^{-kt}} ]That seems right. I think that's the answer for part 1.Moving on to part 2. The budget ( B(t) ) is a quadratic function of ( P(t) ), specifically:[ B(t) = aP(t)^2 + bP(t) + c ]We are given three data points: ( B(0) = B_0 ), ( B(T) = B_T ), and ( B(2T) = B_{2T} ). We need to determine the coefficients ( a ), ( b ), and ( c ).Since ( B(t) ) is quadratic in ( P(t) ), and we have three unknowns, we can set up a system of three equations using the given data points.First, let's write down the equations:1. At ( t = 0 ):[ B(0) = aP(0)^2 + bP(0) + c = B_0 ]2. At ( t = T ):[ B(T) = aP(T)^2 + bP(T) + c = B_T ]3. At ( t = 2T ):[ B(2T) = aP(2T)^2 + bP(2T) + c = B_{2T} ]So, we have three equations:1. ( aP(0)^2 + bP(0) + c = B_0 )2. ( aP(T)^2 + bP(T) + c = B_T )3. ( aP(2T)^2 + bP(2T) + c = B_{2T} )We can solve this system for ( a ), ( b ), and ( c ). Let's denote ( P(0) = P_0 = 0.4M ), ( P(T) = P_T ), and ( P(2T) = P_{2T} ) for simplicity.So, the equations become:1. ( aP_0^2 + bP_0 + c = B_0 ) -- Equation (1)2. ( aP_T^2 + bP_T + c = B_T ) -- Equation (2)3. ( aP_{2T}^2 + bP_{2T} + c = B_{2T} ) -- Equation (3)To solve for ( a ), ( b ), and ( c ), we can subtract Equation (1) from Equation (2) and Equation (2) from Equation (3) to eliminate ( c ).Subtracting Equation (1) from Equation (2):[ a(P_T^2 - P_0^2) + b(P_T - P_0) = B_T - B_0 ] -- Equation (4)Similarly, subtracting Equation (2) from Equation (3):[ a(P_{2T}^2 - P_T^2) + b(P_{2T} - P_T) = B_{2T} - B_T ] -- Equation (5)Now, Equations (4) and (5) form a system of two equations with two unknowns ( a ) and ( b ). Let me write them again:Equation (4):[ a(P_T^2 - P_0^2) + b(P_T - P_0) = B_T - B_0 ]Equation (5):[ a(P_{2T}^2 - P_T^2) + b(P_{2T} - P_T) = B_{2T} - B_T ]Let me denote:Let’s let ( D_1 = P_T - P_0 ) and ( D_2 = P_{2T} - P_T ). Also, let’s denote ( S_1 = P_T^2 - P_0^2 = (P_T - P_0)(P_T + P_0) = D_1(P_T + P_0) ), and similarly ( S_2 = P_{2T}^2 - P_T^2 = D_2(P_{2T} + P_T) ).But maybe it's better to write Equations (4) and (5) in terms of ( a ) and ( b ):Equation (4):[ aS_1 + bD_1 = Delta B_1 ]where ( Delta B_1 = B_T - B_0 )Equation (5):[ aS_2 + bD_2 = Delta B_2 ]where ( Delta B_2 = B_{2T} - B_T )So, we have:1. ( aS_1 + bD_1 = Delta B_1 )2. ( aS_2 + bD_2 = Delta B_2 )This is a linear system which can be solved using substitution or matrix methods. Let me write it in matrix form:[begin{pmatrix}S_1 & D_1 S_2 & D_2end{pmatrix}begin{pmatrix}a bend{pmatrix}=begin{pmatrix}Delta B_1 Delta B_2end{pmatrix}]To solve for ( a ) and ( b ), we can use Cramer's Rule or find the inverse of the coefficient matrix. Let me compute the determinant of the coefficient matrix:Determinant ( Delta = S_1 D_2 - S_2 D_1 )Assuming ( Delta neq 0 ), we can find:[ a = frac{Delta B_1 D_2 - Delta B_2 D_1}{Delta} ][ b = frac{S_1 Delta B_2 - S_2 Delta B_1}{Delta} ]Once ( a ) and ( b ) are found, we can substitute back into Equation (1) to find ( c ):[ c = B_0 - aP_0^2 - bP_0 ]So, the steps are:1. Compute ( S_1 = P_T^2 - P_0^2 ), ( D_1 = P_T - P_0 )2. Compute ( S_2 = P_{2T}^2 - P_T^2 ), ( D_2 = P_{2T} - P_T )3. Compute ( Delta = S_1 D_2 - S_2 D_1 )4. Compute ( a = frac{Delta B_1 D_2 - Delta B_2 D_1}{Delta} )5. Compute ( b = frac{S_1 Delta B_2 - S_2 Delta B_1}{Delta} )6. Compute ( c = B_0 - aP_0^2 - bP_0 )Alternatively, if the determinant is zero, the system might be dependent or inconsistent, but I think in this context, since we have three distinct points, the determinant should not be zero, so a unique solution exists.Wait, but actually, ( P(t) ) is a function of time, so ( P(T) ) and ( P(2T) ) are specific values depending on the logistic growth model. So, unless the population growth is linear, which it isn't, the points ( P(0) ), ( P(T) ), ( P(2T) ) are not equally spaced in population, so the quadratic should be uniquely determined.Therefore, the coefficients ( a ), ( b ), and ( c ) can be uniquely determined using the above method.So, summarizing the process:1. Use the logistic growth solution from part 1 to find ( P(T) ) and ( P(2T) ).2. Plug these into the quadratic budget function to set up three equations.3. Solve the system of equations for ( a ), ( b ), and ( c ).Wait, but in the problem statement, it's given that we have access to detailed census data and local government budgets, so we can compute ( P(T) ) and ( P(2T) ) using the solution from part 1.So, actually, we can express ( P(T) ) and ( P(2T) ) in terms of ( M ) and ( k ), but since we don't have numerical values, we might have to leave the answer in terms of ( P(0) ), ( P(T) ), ( P(2T) ), ( B_0 ), ( B_T ), and ( B_{2T} ).Alternatively, if we consider that ( P(t) ) is known as a function of time, then ( P(T) ) and ( P(2T) ) can be expressed in terms of ( M ) and ( k ), but since the problem doesn't specify numerical values, we can only express ( a ), ( b ), and ( c ) in terms of the given data points.Therefore, the coefficients ( a ), ( b ), and ( c ) can be determined by solving the linear system as outlined above.So, putting it all together, the final expressions for ( a ), ( b ), and ( c ) are:[ a = frac{(B_T - B_0)(P_{2T} - P_T) - (B_{2T} - B_T)(P_T - P_0)}{(P_T^2 - P_0^2)(P_{2T} - P_T) - (P_{2T}^2 - P_T^2)(P_T - P_0)} ][ b = frac{(P_T^2 - P_0^2)(B_{2T} - B_T) - (P_{2T}^2 - P_T^2)(B_T - B_0)}{(P_T^2 - P_0^2)(P_{2T} - P_T) - (P_{2T}^2 - P_T^2)(P_T - P_0)} ][ c = B_0 - aP_0^2 - bP_0 ]Where ( P_0 = 0.4M ), ( P_T = P(T) ), and ( P_{2T} = P(2T) ).Alternatively, if we substitute ( P(t) ) from part 1, we can express ( P(T) ) and ( P(2T) ) in terms of ( M ) and ( k ), but since the problem doesn't provide specific values for ( k ), ( M ), or ( T ), it's probably acceptable to leave the answer in terms of ( P_0 ), ( P_T ), ( P_{2T} ), ( B_0 ), ( B_T ), and ( B_{2T} ).So, in conclusion, the coefficients can be found by solving the linear system using the three given data points, leading to the expressions for ( a ), ( b ), and ( c ) as above.Final Answer1. The population as a function of time is (boxed{P(t) = dfrac{M}{1 + 1.5e^{-kt}}}).2. The coefficients are determined by solving the system of equations with the given data points, resulting in: [ a = frac{(B_T - B_0)(P_{2T} - P_T) - (B_{2T} - B_T)(P_T - P_0)}{(P_T^2 - P_0^2)(P_{2T} - P_T) - (P_{2T}^2 - P_T^2)(P_T - P_0)} ] [ b = frac{(P_T^2 - P_0^2)(B_{2T} - B_T) - (P_{2T}^2 - P_T^2)(B_T - B_0)}{(P_T^2 - P_0^2)(P_{2T} - P_T) - (P_{2T}^2 - P_T^2)(P_T - P_0)} ] [ c = B_0 - aP_0^2 - bP_0 ] where ( P_0 = 0.4M ), ( P_T = P(T) ), and ( P_{2T} = P(2T) ).</think>

🧠:<think>Alright, so I have this problem where an end-user is modeling the performance improvement of a platform using the function ( P(t) = frac{e^{kt}}{1 + e^{kt}} ). They've given me two tasks: first, to find the constant ( k ) when the performance reaches 75% after 5 units of time, and second, to calculate the total time ( T ) required for the performance to reach 90% once ( k ) is known. Okay, let's start with the first part. I need to find ( k ) such that ( P(5) = 0.75 ). So, plugging ( t = 5 ) into the function:[ P(5) = frac{e^{5k}}{1 + e^{5k}} = 0.75 ]Hmm, this looks like a logarithmic equation. Let me rearrange it to solve for ( k ). First, I can write:[ frac{e^{5k}}{1 + e^{5k}} = 0.75 ]Let me denote ( e^{5k} ) as ( x ) for simplicity. So, substituting:[ frac{x}{1 + x} = 0.75 ]Now, solving for ( x ):Multiply both sides by ( 1 + x ):[ x = 0.75(1 + x) ]Expanding the right side:[ x = 0.75 + 0.75x ]Subtract ( 0.75x ) from both sides:[ x - 0.75x = 0.75 ][ 0.25x = 0.75 ]Divide both sides by 0.25:[ x = 3 ]But ( x = e^{5k} ), so:[ e^{5k} = 3 ]To solve for ( k ), take the natural logarithm of both sides:[ ln(e^{5k}) = ln(3) ][ 5k = ln(3) ][ k = frac{ln(3)}{5} ]Alright, so that gives me the value of ( k ). Let me compute that numerically to check. We know that ( ln(3) ) is approximately 1.0986, so:[ k approx frac{1.0986}{5} approx 0.2197 ]So, ( k ) is approximately 0.2197. Let me keep this in mind for the next part.Moving on to the second part, I need to find the time ( T ) when the performance ( P(T) = 0.90 ). Using the same function:[ P(T) = frac{e^{kT}}{1 + e^{kT}} = 0.90 ]Again, let me set ( e^{kT} = y ) for simplicity:[ frac{y}{1 + y} = 0.90 ]Solving for ( y ):Multiply both sides by ( 1 + y ):[ y = 0.90(1 + y) ]Expanding:[ y = 0.90 + 0.90y ]Subtract ( 0.90y ) from both sides:[ y - 0.90y = 0.90 ][ 0.10y = 0.90 ]Divide both sides by 0.10:[ y = 9 ]Since ( y = e^{kT} ), we have:[ e^{kT} = 9 ]Taking the natural logarithm of both sides:[ ln(e^{kT}) = ln(9) ][ kT = ln(9) ][ T = frac{ln(9)}{k} ]We already found ( k = frac{ln(3)}{5} ), so substituting:[ T = frac{ln(9)}{frac{ln(3)}{5}} = frac{5 ln(9)}{ln(3)} ]Simplify ( ln(9) ). Since ( 9 = 3^2 ), ( ln(9) = 2 ln(3) ). Therefore:[ T = frac{5 times 2 ln(3)}{ln(3)} = frac{10 ln(3)}{ln(3)} = 10 ]So, ( T = 10 ) units of time.Wait, let me double-check that. If ( ln(9) = 2 ln(3) ), then yes, substituting back, the ( ln(3) ) cancels out, leaving 10. That seems correct.Just to make sure, let me verify with the approximate value of ( k ). Earlier, I found ( k approx 0.2197 ). So, plugging into ( T = frac{ln(9)}{k} ):First, ( ln(9) approx 2.1972 ). Then,[ T approx frac{2.1972}{0.2197} approx 10 ]Yep, that checks out. So, the total time ( T ) required for the performance to reach 90% is 10 units of time.Let me recap the steps to ensure I didn't skip anything:1. For the first part, set ( P(5) = 0.75 ), substituted into the function, solved for ( e^{5k} = 3 ), took natural log to get ( k = ln(3)/5 ).2. For the second part, set ( P(T) = 0.90 ), substituted into the function, solved for ( e^{kT} = 9 ), took natural log to get ( T = ln(9)/k ). Substituted ( k ) and simplified to find ( T = 10 ).Everything seems consistent. I think I've got it right.Final Answer1. The value of ( k ) is ( boxed{dfrac{ln 3}{5}} ).2. The total time ( T ) required for the performance to reach 90% is ( boxed{10} ) units of time.</think>

🧠:<think>Okay, so I've got this problem about Alex, a young musician who needs regular blood transfusions. The problem has two parts, and I need to solve both using calculus. Let me start by understanding each part step by step.Problem 1: Calculate the total volume of blood Alex requires over a 30-day period.Alright, the volume of blood needed is given by the function ( V(t) = 2e^{0.1t} ) liters, where ( t ) is the time in days since the last transfusion. They want the total volume over 30 days, so I think I need to integrate this function from day 0 to day 30.Let me write that down:Total volume ( = int_{0}^{30} 2e^{0.1t} dt )Hmm, integrating an exponential function. I remember that the integral of ( e^{kt} ) is ( frac{1}{k}e^{kt} ). So, applying that here.First, factor out the constant 2:( 2 int_{0}^{30} e^{0.1t} dt )Let me compute the integral:( int e^{0.1t} dt = frac{1}{0.1}e^{0.1t} + C = 10e^{0.1t} + C )So, evaluating from 0 to 30:( 2 [10e^{0.1 times 30} - 10e^{0.1 times 0}] )Simplify the exponents:( 0.1 times 30 = 3 ), so ( e^{3} ), and ( 0.1 times 0 = 0 ), so ( e^{0} = 1 ).So, plugging those in:( 2 [10e^{3} - 10 times 1] = 2 [10(e^{3} - 1)] )Multiply the 2 in:( 20(e^{3} - 1) )Let me compute ( e^{3} ). I know ( e ) is approximately 2.71828, so ( e^3 ) is about 20.0855.So, ( 20(20.0855 - 1) = 20(19.0855) = 381.71 ) liters.Wait, that seems like a lot. Let me double-check my steps.1. The integral of ( e^{0.1t} ) is indeed ( 10e^{0.1t} ).2. Evaluated from 0 to 30: ( 10e^{3} - 10 ).3. Multiply by 2: ( 20(e^{3} - 1) ).4. Calculating ( e^{3} approx 20.0855 ), so ( 20(19.0855) approx 381.71 ).Hmm, 381 liters over 30 days? That seems high because 2 liters per day would be 60 liters, but since it's exponential, it's increasing over time. Let me see, on day 30, the rate is ( 2e^{3} approx 2 times 20.0855 approx 40.17 ) liters per day. So, the average rate over 30 days would be somewhere between 2 and 40 liters per day. The integral is the area under the curve, which would be more than 2*30=60 and less than 40*30=1200. 381 is in that range, so maybe it's correct.But let me think again. The function is ( V(t) = 2e^{0.1t} ). So, the volume needed per day is increasing exponentially. So, over 30 days, the total would indeed be significantly higher than 60 liters. So, 381 liters might be correct. I'll go with that for now.Problem 2: Solve the differential equation ( frac{dB}{dt} = -0.05B(t) ) and find the steady-state amount of blood just before each transfusion, given that Alex receives 1 liter every 7 days.Alright, this is a first-order linear differential equation. It looks like an exponential decay model because the derivative is proportional to the negative of the function.The equation is ( frac{dB}{dt} = -0.05B(t) ). To solve this, I can separate variables.Let me write it as:( frac{dB}{B} = -0.05 dt )Integrating both sides:( int frac{1}{B} dB = int -0.05 dt )Left side integral is ( ln|B| + C ), right side is ( -0.05t + C ).So,( ln B = -0.05t + C )Exponentiating both sides:( B(t) = e^{-0.05t + C} = e^{C}e^{-0.05t} )Let me denote ( e^{C} ) as ( B_0 ), the initial amount of blood at time t=0.So, the solution is ( B(t) = B_0 e^{-0.05t} ).But wait, Alex receives a transfusion every 7 days. So, the process is periodic. Each time, Alex gets 1 liter, and then the blood is metabolized over the next 7 days.So, we need to model this as a periodic function where every 7 days, 1 liter is added, and in between, the blood decays exponentially.This sounds like a problem that can be modeled using the concept of steady-state in periodic systems. Maybe using the integrating factor method or considering the system over each interval.Alternatively, since the metabolism is a linear process, the steady-state can be found by considering the balance between the input and the output.Let me think about it.Each 7-day period, Alex receives 1 liter. Between transfusions, the blood decays according to ( B(t) = B_0 e^{-0.05t} ). So, just before the next transfusion, the amount of blood is ( B(7) = B_0 e^{-0.05 times 7} ).But since the transfusion is periodic, the amount just before each transfusion should reach a steady state, meaning that the amount just before the transfusion is the same each time.Let me denote ( B_s ) as the steady-state amount just before each transfusion.So, after receiving 1 liter, the amount becomes ( B_s + 1 ). Then, over the next 7 days, this amount decays to ( B_s ) again.So, mathematically:( B_s = (B_s + 1) e^{-0.05 times 7} )Let me solve for ( B_s ).First, compute ( e^{-0.05 times 7} ).( 0.05 times 7 = 0.35 ), so ( e^{-0.35} approx e^{-0.35} approx 0.7047 ).So, the equation becomes:( B_s = (B_s + 1) times 0.7047 )Let me write that:( B_s = 0.7047 B_s + 0.7047 )Bring the ( 0.7047 B_s ) term to the left:( B_s - 0.7047 B_s = 0.7047 )Factor out ( B_s ):( B_s (1 - 0.7047) = 0.7047 )Compute ( 1 - 0.7047 = 0.2953 )So,( B_s = frac{0.7047}{0.2953} )Calculate that:( 0.7047 / 0.2953 ≈ 2.386 )So, approximately 2.386 liters.Wait, let me check my calculations again.1. The decay factor is ( e^{-0.35} ≈ 0.7047 ).2. Equation: ( B_s = (B_s + 1) times 0.7047 )3. Distribute: ( B_s = 0.7047 B_s + 0.7047 )4. Subtract ( 0.7047 B_s ): ( B_s - 0.7047 B_s = 0.7047 )5. Factor: ( 0.2953 B_s = 0.7047 )6. So, ( B_s = 0.7047 / 0.2953 ≈ 2.386 )Yes, that seems correct.Alternatively, to express this more precisely, without approximating ( e^{-0.35} ), let's keep it symbolic.Let me denote ( k = 0.05 ), so the decay rate is ( k ).The time between transfusions is ( T = 7 ) days.So, the equation is:( B_s = (B_s + 1) e^{-kT} )Solving for ( B_s ):( B_s = (B_s + 1) e^{-kT} )( B_s = B_s e^{-kT} + e^{-kT} )Bring ( B_s e^{-kT} ) to the left:( B_s - B_s e^{-kT} = e^{-kT} )Factor ( B_s ):( B_s (1 - e^{-kT}) = e^{-kT} )Thus,( B_s = frac{e^{-kT}}{1 - e^{-kT}} )Simplify numerator and denominator:Multiply numerator and denominator by ( e^{kT} ):( B_s = frac{1}{e^{kT} - 1} )So, plugging in ( k = 0.05 ) and ( T = 7 ):( B_s = frac{1}{e^{0.05 times 7} - 1} = frac{1}{e^{0.35} - 1} )Compute ( e^{0.35} ):( e^{0.35} ≈ 1.41906754 )So,( B_s ≈ frac{1}{1.41906754 - 1} = frac{1}{0.41906754} ≈ 2.386 ) liters.Yes, that matches my earlier calculation. So, the steady-state amount just before each transfusion is approximately 2.386 liters.But let me think if this makes sense. Each time, Alex gets 1 liter, and over 7 days, the blood decays. The decay factor is about 0.7047, so 1 liter would decay to about 0.7047 liters in 7 days. But since the steady-state is 2.386 liters, which is higher than 1 liter, that seems a bit confusing.Wait, no. The steady-state is the amount just before the transfusion. So, after receiving 1 liter, the amount becomes ( B_s + 1 ), which then decays over 7 days to ( B_s ). So, the decay from ( B_s + 1 ) to ( B_s ) over 7 days.So, ( B_s = (B_s + 1) e^{-0.35} ). So, solving for ( B_s ), we get approximately 2.386 liters.Yes, that seems correct because each transfusion adds 1 liter, but the decay over 7 days reduces it by a factor of ~0.7047, so the steady-state is higher than 1 liter.Alternatively, if we think about it in terms of the balance: the amount lost per transfusion interval is equal to the amount added.The amount added is 1 liter every 7 days.The amount lost is the difference between the amount just after transfusion and just before the next transfusion.So, the amount lost is ( (B_s + 1) - B_s = 1 ) liter? Wait, no, that's not considering the decay.Wait, actually, the amount lost is the decay over 7 days, which is ( (B_s + 1) - B_s = 1 ) liter. But that can't be because the decay is exponential, not linear.Wait, perhaps another way: the rate of loss is proportional to the amount present. So, the total loss over 7 days is the integral of the loss rate over that period.But since we have a steady-state, the total loss over 7 days should equal the amount added, which is 1 liter.So, the total loss is ( int_{0}^{7} 0.05 B(t) dt ), where ( B(t) = (B_s + 1) e^{-0.05t} ).So, total loss:( int_{0}^{7} 0.05 (B_s + 1) e^{-0.05t} dt )Factor out constants:( 0.05 (B_s + 1) int_{0}^{7} e^{-0.05t} dt )Compute the integral:( int e^{-0.05t} dt = frac{1}{-0.05} e^{-0.05t} + C = -20 e^{-0.05t} + C )Evaluate from 0 to 7:( -20 [e^{-0.35} - e^{0}] = -20 [e^{-0.35} - 1] )So, total loss:( 0.05 (B_s + 1) times (-20 [e^{-0.35} - 1]) )Simplify:( 0.05 times (-20) = -1 ), so:( -1 (B_s + 1) [e^{-0.35} - 1] )But since loss is positive, we can take absolute value:Total loss ( = (B_s + 1) [1 - e^{-0.35}] )This total loss should equal the amount added, which is 1 liter.So,( (B_s + 1) [1 - e^{-0.35}] = 1 )Solve for ( B_s ):( B_s + 1 = frac{1}{1 - e^{-0.35}} )So,( B_s = frac{1}{1 - e^{-0.35}} - 1 )Simplify:( B_s = frac{1 - (1 - e^{-0.35})}{1 - e^{-0.35}} = frac{e^{-0.35}}{1 - e^{-0.35}} )Which is the same as before:( B_s = frac{1}{e^{0.35} - 1} approx 2.386 ) liters.Yes, that confirms the earlier result. So, the steady-state amount just before each transfusion is approximately 2.386 liters.But let me compute it more precisely.Given ( e^{0.35} approx 1.41906754 ), so ( e^{0.35} - 1 ≈ 0.41906754 ).Thus, ( B_s ≈ 1 / 0.41906754 ≈ 2.386 ) liters.So, rounding to three decimal places, it's approximately 2.386 liters.Alternatively, if we want to express it exactly, it's ( frac{1}{e^{0.35} - 1} ), but since the question doesn't specify, probably the approximate value is fine.Summary of Thoughts:1. For the first part, integrating the exponential function over 30 days gives a total volume of approximately 381.71 liters. That seems high, but considering the exponential growth, it might be correct.2. For the second part, solving the differential equation gives an exponential decay model. Considering the periodic transfusions, the steady-state amount just before each transfusion is approximately 2.386 liters.I think I've covered all the steps and double-checked my calculations. I don't see any mistakes in my reasoning, so I'll proceed to write the final answers.</think>