answer：Okay, so I have this problem where a graduate student is analyzing two therapeutic approaches using differential equations. I need to find the equilibrium points and analyze their stability for both models. Then, compare the long-term behavior of the improvement scores. Hmm, let me take it step by step.Starting with the first part, Freudian psychoanalysis. The differential equation given is:[frac{dP}{dt} = k_1 P (1 - P) - alpha P^2]Where ( k_1 ) and ( alpha ) are positive constants. I need to find the equilibrium points. Equilibrium points occur where ( frac{dP}{dt} = 0 ). So, I'll set the equation equal to zero and solve for ( P ).Let me rewrite the equation:[k_1 P (1 - P) - alpha P^2 = 0]Factor out ( P ):[P [k_1 (1 - P) - alpha P] = 0]So, either ( P = 0 ) or the term in the brackets is zero. Let's solve for ( P ) when the bracket is zero:[k_1 (1 - P) - alpha P = 0][k_1 - k_1 P - alpha P = 0][k_1 = P (k_1 + alpha)][P = frac{k_1}{k_1 + alpha}]So, the equilibrium points are ( P = 0 ) and ( P = frac{k_1}{k_1 + alpha} ). Got that.Now, I need to analyze their stability. To do this, I'll compute the derivative of ( frac{dP}{dt} ) with respect to ( P ) and evaluate it at each equilibrium point. The derivative is the Jacobian matrix in this case since it's a single equation.Let me denote the right-hand side of the differential equation as ( f(P) ):[f(P) = k_1 P (1 - P) - alpha P^2]Compute ( f'(P) ):[f'(P) = k_1 (1 - P) + k_1 P (-1) - 2 alpha P]Simplify:[f'(P) = k_1 (1 - P - P) - 2 alpha P][f'(P) = k_1 (1 - 2P) - 2 alpha P]Alternatively, I can compute it step by step:First term: ( d/dP [k_1 P (1 - P)] = k_1 (1 - P) + k_1 P (-1) = k_1 (1 - 2P) )Second term: ( d/dP [ - alpha P^2 ] = -2 alpha P )So, combining both:[f'(P) = k_1 (1 - 2P) - 2 alpha P]Yes, that's correct.Now, evaluate ( f'(P) ) at each equilibrium point.First, at ( P = 0 ):[f'(0) = k_1 (1 - 0) - 0 = k_1]Since ( k_1 ) is a positive constant, ( f'(0) > 0 ). Therefore, the equilibrium at ( P = 0 ) is unstable.Next, at ( P = frac{k_1}{k_1 + alpha} ):Let me compute ( f'(P) ) at this point.First, compute ( 1 - 2P ):[1 - 2 left( frac{k_1}{k_1 + alpha} right ) = 1 - frac{2 k_1}{k_1 + alpha}][= frac{(k_1 + alpha) - 2 k_1}{k_1 + alpha}][= frac{alpha - k_1}{k_1 + alpha}]Then, compute ( -2 alpha P ):[-2 alpha left( frac{k_1}{k_1 + alpha} right ) = - frac{2 alpha k_1}{k_1 + alpha}]So, putting it together:[f'left( frac{k_1}{k_1 + alpha} right ) = k_1 left( frac{alpha - k_1}{k_1 + alpha} right ) - frac{2 alpha k_1}{k_1 + alpha}][= frac{k_1 (alpha - k_1) - 2 alpha k_1}{k_1 + alpha}][= frac{k_1 alpha - k_1^2 - 2 alpha k_1}{k_1 + alpha}][= frac{ - k_1^2 - alpha k_1 }{k_1 + alpha}][= frac{ -k_1 (k_1 + alpha) }{k_1 + alpha}][= -k_1]So, ( f'left( frac{k_1}{k_1 + alpha} right ) = -k_1 ). Since ( k_1 ) is positive, this derivative is negative. Therefore, the equilibrium at ( P = frac{k_1}{k_1 + alpha} ) is stable.Alright, so for the first model, we have two equilibrium points: one at zero, which is unstable, and another at ( frac{k_1}{k_1 + alpha} ), which is stable. So, regardless of the initial condition (as long as it's positive, I guess), the improvement score ( P(t) ) will approach ( frac{k_1}{k_1 + alpha} ) as time goes to infinity.Moving on to the second part, existential psychotherapy. The differential equation is:[frac{dE}{dt} = k_2 E left(1 - frac{E}{beta}right) - gamma E]Where ( k_2 ), ( beta ), and ( gamma ) are positive constants. Again, I need to find the equilibrium points and analyze their stability.First, set ( frac{dE}{dt} = 0 ):[k_2 E left(1 - frac{E}{beta}right) - gamma E = 0]Factor out ( E ):[E left[ k_2 left(1 - frac{E}{beta}right) - gamma right ] = 0]So, either ( E = 0 ) or the term in the brackets is zero.Solving for ( E ) when the bracket is zero:[k_2 left(1 - frac{E}{beta}right) - gamma = 0][k_2 - frac{k_2 E}{beta} - gamma = 0][- frac{k_2 E}{beta} = gamma - k_2][E = frac{ (gamma - k_2 ) (- beta) }{ k_2 }][E = frac{ (k_2 - gamma ) beta }{ k_2 }]Wait, let me double-check that algebra:Starting from:[k_2 left(1 - frac{E}{beta}right) - gamma = 0][k_2 - frac{k_2 E}{beta} - gamma = 0][- frac{k_2 E}{beta} = gamma - k_2]Multiply both sides by ( - beta / k_2 ):[E = frac{ (k_2 - gamma ) beta }{ k_2 }]Yes, that's correct.So, the equilibrium points are ( E = 0 ) and ( E = frac{ (k_2 - gamma ) beta }{ k_2 } ).But wait, since ( k_2 ), ( beta ), and ( gamma ) are positive constants, the second equilibrium point is positive only if ( k_2 - gamma > 0 ), that is, if ( k_2 > gamma ). If ( k_2 = gamma ), then the second equilibrium is zero, which coincides with the first one. If ( k_2 < gamma ), then ( E ) would be negative, which doesn't make sense in the context of an improvement score, so we can disregard that.Therefore, the equilibrium points are:1. ( E = 0 )2. ( E = frac{ (k_2 - gamma ) beta }{ k_2 } ) if ( k_2 > gamma )Now, let's analyze the stability of these equilibrium points.First, compute the derivative of ( f(E) = k_2 E (1 - E/beta ) - gamma E ).Compute ( f'(E) ):First term: ( d/dE [k_2 E (1 - E/beta ) ] = k_2 (1 - E/beta ) + k_2 E (-1/beta ) = k_2 (1 - 2 E / beta ) )Second term: ( d/dE [ - gamma E ] = - gamma )So,[f'(E) = k_2 left(1 - frac{2 E}{beta}right) - gamma]Now, evaluate ( f'(E) ) at each equilibrium point.First, at ( E = 0 ):[f'(0) = k_2 (1 - 0 ) - gamma = k_2 - gamma]So, the sign of ( f'(0) ) depends on whether ( k_2 > gamma ) or not.If ( k_2 > gamma ), then ( f'(0) > 0 ), so the equilibrium at ( E = 0 ) is unstable.If ( k_2 = gamma ), then ( f'(0) = 0 ), which is a borderline case, but since we already saw that when ( k_2 = gamma ), the other equilibrium point is zero, so it's a repeated root or something.If ( k_2 < gamma ), then ( f'(0) < 0 ), so the equilibrium at ( E = 0 ) is stable.But wait, if ( k_2 < gamma ), the second equilibrium point is negative, which isn't feasible, so the only equilibrium is ( E = 0 ), which is stable.Now, for the second equilibrium point ( E = frac{ (k_2 - gamma ) beta }{ k_2 } ), let's compute ( f'(E) ) at this point.First, compute ( 1 - 2 E / beta ):[1 - 2 left( frac{ (k_2 - gamma ) beta }{ k_2 } right ) / beta = 1 - frac{2 (k_2 - gamma ) }{ k_2 }][= 1 - 2 + frac{2 gamma }{ k_2 }][= -1 + frac{2 gamma }{ k_2 }]So,[f'left( frac{ (k_2 - gamma ) beta }{ k_2 } right ) = k_2 left( -1 + frac{2 gamma }{ k_2 } right ) - gamma][= -k_2 + 2 gamma - gamma][= -k_2 + gamma][= gamma - k_2]So, ( f'(E) ) at the second equilibrium is ( gamma - k_2 ).Therefore, the stability is determined by the sign of ( gamma - k_2 ):- If ( gamma - k_2 < 0 ), which is ( k_2 > gamma ), then ( f'(E) < 0 ), so the equilibrium is stable.- If ( gamma - k_2 > 0 ), which is ( k_2 < gamma ), then ( f'(E) > 0 ), but in this case, the equilibrium point is negative, so it's not relevant.Wait, but when ( k_2 > gamma ), the second equilibrium is positive, and it's stable because ( f'(E) = gamma - k_2 < 0 ). When ( k_2 < gamma ), the second equilibrium is negative, so we only have ( E = 0 ) as an equilibrium, which is stable.So, summarizing:- If ( k_2 > gamma ), we have two equilibria: ( E = 0 ) (unstable) and ( E = frac{ (k_2 - gamma ) beta }{ k_2 } ) (stable).- If ( k_2 = gamma ), we have a single equilibrium at ( E = 0 ), which is borderline (derivative zero), but since the other equilibrium is zero, it might be a case of transcritical bifurcation or something.- If ( k_2 < gamma ), only ( E = 0 ) is an equilibrium, and it's stable.Now, comparing the long-term behavior of ( P(t) ) and ( E(t) ).From the first model, ( P(t) ) approaches ( frac{k_1}{k_1 + alpha} ) regardless of initial conditions (assuming positive). For the second model, ( E(t) ) approaches either ( 0 ) or ( frac{ (k_2 - gamma ) beta }{ k_2 } ), depending on whether ( k_2 < gamma ) or ( k_2 > gamma ).So, in terms of long-term behavior:- Freudian psychoanalysis leads to a stable positive improvement score ( frac{k_1}{k_1 + alpha} ).- Existential psychotherapy can lead to either a stable positive improvement score ( frac{ (k_2 - gamma ) beta }{ k_2 } ) if ( k_2 > gamma ), or it can lead to the improvement score tending to zero if ( k_2 < gamma ).Therefore, depending on the parameters, existential psychotherapy might result in either sustained improvement or no improvement, whereas Freudian psychoanalysis always leads to a stable improvement score.Wait, but in the first model, the improvement score approaches a fixed positive value, whereas in the second model, it can either go to zero or a positive value. So, depending on the parameters, one therapy might be better in the long run.But the problem doesn't specify which parameters are larger, so we can't definitively say which is better, but we can describe their behaviors.So, in summary:1. For Freudian psychoanalysis, the improvement score ( P(t) ) tends to ( frac{k_1}{k_1 + alpha} ), a stable equilibrium.2. For existential psychotherapy, if ( k_2 > gamma ), ( E(t) ) tends to ( frac{ (k_2 - gamma ) beta }{ k_2 } ), which is a stable equilibrium. If ( k_2 < gamma ), ( E(t) ) tends to zero, meaning no improvement.Therefore, the long-term behavior depends on the relative values of ( k_2 ) and ( gamma ) in the second model, whereas in the first model, it always stabilizes to a positive value.I think that's about it. Let me just recap:Equilibrium points for P(t): 0 (unstable) and ( frac{k_1}{k_1 + alpha} ) (stable).Equilibrium points for E(t): 0 (stable if ( k_2 < gamma ), unstable if ( k_2 > gamma )) and ( frac{ (k_2 - gamma ) beta }{ k_2 } ) (stable if ( k_2 > gamma )).Long-term behavior: Freudian always leads to a positive stable improvement, while existential can lead to either positive or zero improvement depending on parameters.Yeah, that seems correct.**Final Answer**1. The equilibrium points for ( P(t) ) are ( boxed{0} ) (unstable) and ( boxed{frac{k_1}{k_1 + alpha}} ) (stable).2. The equilibrium points for ( E(t) ) are ( boxed{0} ) (stable if ( k_2 < gamma ), unstable if ( k_2 > gamma )) and ( boxed{frac{(k_2 - gamma)beta}{k_2}} ) (stable if ( k_2 > gamma )). The long-term behavior shows ( P(t) ) approaches a positive equilibrium, while ( E(t) ) may approach zero or a positive equilibrium depending on parameter values.

answer：Alright, so I have two problems here to solve, both related to Coach Alex's methods of tracking athlete performance. Let me tackle them one by one.Starting with the first problem: Coach Alex models the improvement of an athlete's performance over time using the function ( P(t) = 10e^{kt} + c ). Here, ( k ) is a positive constant representing the rate of improvement, and ( c ) is the initial baseline performance. After 5 weeks, the athlete’s performance has doubled. I need to find the value of ( k ) in terms of ( c ).Okay, so let's break this down. The performance function is given by ( P(t) = 10e^{kt} + c ). At time ( t = 0 ), the performance is ( P(0) = 10e^{0} + c = 10 + c ). After 5 weeks, the performance is ( P(5) = 10e^{5k} + c ). According to the problem, this performance has doubled. So, ( P(5) = 2 times P(0) ).Let me write that equation out:( 10e^{5k} + c = 2(10 + c) )Simplify the right side:( 10e^{5k} + c = 20 + 2c )Now, subtract ( c ) from both sides:( 10e^{5k} = 20 + c )Wait, hold on. That doesn't seem right. Let me double-check. If ( P(5) = 2P(0) ), then:( 10e^{5k} + c = 2(10 + c) )Expanding the right side:( 10e^{5k} + c = 20 + 2c )Subtract ( c ) from both sides:( 10e^{5k} = 20 + c )Hmm, okay, so then:( e^{5k} = frac{20 + c}{10} )Simplify that:( e^{5k} = 2 + frac{c}{10} )To solve for ( k ), take the natural logarithm of both sides:( 5k = lnleft(2 + frac{c}{10}right) )Therefore,( k = frac{1}{5} lnleft(2 + frac{c}{10}right) )Wait, but the problem says to express ( k ) in terms of ( c ). So, that's exactly what I have here. Let me just make sure I didn't make any mistakes.Starting from ( P(5) = 2P(0) ):( 10e^{5k} + c = 2(10 + c) )Yes, that's correct. Then subtract ( c ):( 10e^{5k} = 20 + c )Divide both sides by 10:( e^{5k} = 2 + frac{c}{10} )Take natural log:( 5k = lnleft(2 + frac{c}{10}right) )So,( k = frac{1}{5} lnleft(2 + frac{c}{10}right) )Looks good. So that's the value of ( k ) in terms of ( c ).Moving on to the second problem. Coach Alex uses a matrix to analyze the combined performance of his team. The matrix ( A ) is given as:( A = begin{bmatrix} 1 & 2 & 3 0 & 1 & 4 5 & 6 & 0 end{bmatrix} )And the current performance levels of the athletes are represented by the vector ( mathbf{v} = begin{bmatrix} 1 2 3 end{bmatrix} ). I need to compute the resulting performance vector after applying the matrix transformation to ( mathbf{v} ) and interpret its significance in the context of team dynamics.Alright, so this is a matrix multiplication problem. The resulting vector is ( Amathbf{v} ). Let me compute that.First, let me write down the matrix ( A ) and vector ( mathbf{v} ):( A = begin{bmatrix} 1 & 2 & 3 0 & 1 & 4 5 & 6 & 0 end{bmatrix} ), ( mathbf{v} = begin{bmatrix} 1 2 3 end{bmatrix} )So, ( Amathbf{v} ) will be a 3x1 vector. Let's compute each component step by step.First component: Row 1 of A multiplied by vector v.( 1 times 1 + 2 times 2 + 3 times 3 )Compute that:( 1 + 4 + 9 = 14 )Second component: Row 2 of A multiplied by vector v.( 0 times 1 + 1 times 2 + 4 times 3 )Compute that:( 0 + 2 + 12 = 14 )Third component: Row 3 of A multiplied by vector v.( 5 times 1 + 6 times 2 + 0 times 3 )Compute that:( 5 + 12 + 0 = 17 )So, putting it all together, the resulting vector is:( Amathbf{v} = begin{bmatrix} 14 14 17 end{bmatrix} )Now, interpreting this in the context of team dynamics. Hmm. So, the matrix transformation is being applied to the current performance vector. Each component of the resulting vector might represent some combined measure of performance influenced by the interactions or dependencies between the athletes.Looking at the matrix ( A ), it's a 3x3 matrix. Each row corresponds to an athlete, perhaps, and each column corresponds to the influence of each athlete on the others. So, when we multiply by the performance vector, we're essentially calculating a weighted sum of the performances, where the weights are given by the matrix ( A ).In this case, the resulting vector ( begin{bmatrix} 14 14 17 end{bmatrix} ) shows that the first and second athletes have their performances transformed to 14, while the third athlete's performance is transformed to 17. This might indicate that the first athlete's performance is influenced more by the third athlete, as seen by the higher weight in the first row (3 in the third column). Similarly, the second athlete is influenced by the third athlete with a weight of 4. The third athlete is influenced by the first and second athletes with weights 5 and 6, respectively.So, the resulting vector could represent how each athlete's performance is being affected by the others, perhaps indicating areas where athletes are excelling or needing support. For example, the third athlete's performance is significantly influenced by the first two, resulting in a higher value, while the first and second athletes have similar transformed performances, maybe indicating a balanced influence or some form of synergy between them.Alternatively, this could represent a combined performance metric where each athlete's contribution is weighted differently based on their interactions or roles within the team. The higher value for the third athlete might suggest they have a more significant impact on the team's overall performance due to the weights assigned in the matrix.In any case, the resulting vector gives a transformed view of the team's performance, considering the interdependencies modeled by matrix ( A ). This could help Coach Alex identify which athletes are contributing more or less, or how changes in one athlete's performance affect the others.Wait, just to make sure I did the matrix multiplication correctly. Let me recompute each component.First component:1*1 + 2*2 + 3*3 = 1 + 4 + 9 = 14. Correct.Second component:0*1 + 1*2 + 4*3 = 0 + 2 + 12 = 14. Correct.Third component:5*1 + 6*2 + 0*3 = 5 + 12 + 0 = 17. Correct.Yes, that seems right. So, the resulting vector is indeed [14, 14, 17]^T.So, summarizing my thoughts:1. For the first problem, I found that ( k = frac{1}{5} lnleft(2 + frac{c}{10}right) ).2. For the second problem, I computed the matrix-vector product and got the resulting vector as [14, 14, 17]^T, and interpreted it in terms of team dynamics, considering how each athlete's performance is influenced by others based on the matrix weights.I think that's all. I don't see any mistakes in my calculations, so I feel confident about these answers.**Final Answer**1. The value of ( k ) is boxed{dfrac{1}{5} lnleft(2 + dfrac{c}{10}right)}.2. The resulting performance vector is boxed{begin{bmatrix} 14 14 17 end{bmatrix}}.

answer：Okay, so I have these two problems about cryptography and probability that a PHP developer is dealing with. Let me try to figure them out step by step.Starting with the first problem about symmetric encryption using AES with a 256-bit key. The question is asking for the probability that a randomly chosen 256-bit key will be identical to a predetermined secure key. Hmm, probability in this context. So, I think this is a basic probability question where we have a certain number of possible outcomes and only one successful outcome.AES with a 256-bit key means the key space is all possible 256-bit keys. Each bit can be either 0 or 1, so for each bit, there are 2 possibilities. Since the key is 256 bits long, the total number of possible keys should be 2 multiplied by itself 256 times, which is 2^256. That makes sense because each additional bit doubles the number of possible keys.Now, the probability of randomly selecting the correct key would be the number of successful outcomes divided by the total number of possible outcomes. Here, the successful outcome is just 1 (the predetermined key), and the total number of possible keys is 2^256. So, the probability should be 1 divided by 2^256. That seems straightforward.Let me write that down:Probability = 1 / (2^256)I think that's it for the first part. It's a very small probability because 2^256 is an astronomically large number. So, the chance of randomly guessing the correct key is practically zero, which is why AES is considered secure.Moving on to the second problem about hash function verification using SHA-256. The developer processes 1 million form submissions per day and wants to calculate the expected number of hash collisions per day. They mention using the birthday paradox approximation.Okay, so the birthday paradox is about the probability that in a set of n randomly chosen elements, at least two will be the same. In this case, the elements are the hash values of the form submissions. SHA-256 produces a 256-bit hash, so the number of possible different hash values is 2^256.The birthday paradox tells us that the probability of a collision (two different inputs producing the same hash) becomes significant when the number of elements is on the order of the square root of the number of possible hash values. The expected number of collisions can be approximated by the formula:Expected collisions ≈ (n^2) / (2 * N)Where n is the number of submissions, and N is the number of possible hash values.Given that n is 1,000,000 (1 million) and N is 2^256, let's plug those numbers in.First, calculate n squared: (1,000,000)^2 = 1,000,000,000,000 (1 trillion).Then, divide that by (2 * N): 2 * 2^256 = 2^257.So, the expected number of collisions is approximately 1,000,000,000,000 / 2^257.But 2^257 is a huge number. Let me see if I can simplify this expression.First, note that 2^257 = 2 * 2^256. So, the denominator is 2 * 2^256, and the numerator is 10^12 (since 1,000,000,000,000 is 10^12). Therefore, the expected number of collisions is:Expected collisions ≈ (10^12) / (2 * 2^256) = (10^12) / (2^257)But 2^10 is approximately 10^3 (1024 ≈ 1000), so 2^20 ≈ 10^6, 2^30 ≈ 10^9, and so on. So, 2^257 is 2^(250 + 7) = (2^10)^25 * 2^7 ≈ (10^3)^25 * 128 = 10^75 * 128.Wait, that might not be the most helpful. Alternatively, since 2^256 is such a massive number, dividing 10^12 by it would result in an extremely small number.Let me compute the exponent difference. 2^257 is 2^256 * 2, so 2^256 is about 1.16 x 10^77 (since log10(2^256) = 256 * log10(2) ≈ 256 * 0.3010 ≈ 77.056). So, 2^256 ≈ 1.16 x 10^77, and 2^257 ≈ 2.32 x 10^77.So, 10^12 / 2.32 x 10^77 ≈ (1 / 2.32) x 10^(12 - 77) ≈ 0.431 x 10^-65 ≈ 4.31 x 10^-66.That's an incredibly small number, practically zero. So, the expected number of collisions per day is about 4.31 x 10^-66, which is essentially zero. That makes sense because SHA-256 has a huge number of possible hash values, so even with a million submissions, the chance of a collision is negligible.Wait, but let me double-check the formula. The expected number of collisions is often approximated as n(n - 1)/(2N), which for large N and small n is roughly n^2/(2N). So, yes, that's what I used.Alternatively, the probability of at least one collision is approximately 1 - e^(-n^2/(2N)). But since n^2/(2N) is so small, the probability is approximately equal to n^2/(2N), which is the expected number of collisions. So, either way, it's a very small number.Therefore, the expected number of collisions per day is negligible, around 4.31 x 10^-66. That's way beyond any practical concern, so the developer doesn't have to worry about hash collisions with SHA-256 for 1 million submissions per day.So, summarizing:1. The probability of randomly choosing the correct 256-bit key is 1 / 2^256.2. The expected number of hash collisions per day is approximately 4.31 x 10^-66, which is effectively zero.I think that's it. I don't see any mistakes in my reasoning, but let me just verify the calculations.For the first part, yes, 256 bits mean 2^256 possibilities, so 1 / 2^256 is correct.For the second part, n = 1,000,000, N = 2^256. So, n^2 = 10^12, and 2N = 2^257. So, 10^12 / 2^257 is indeed approximately 4.31 x 10^-66. That seems right.Yeah, I think I've got it.**Final Answer**1. The probability is boxed{dfrac{1}{2^{256}}}.2. The expected number of hash collisions per day is boxed{4.31 times 10^{-66}}.

answer：Okay, so I have this problem about cultural groups on a remote island. There are two parts. Let me try to tackle them one by one.Starting with part 1: It says that each cultural group has a population modeled by the differential equation dP_i(t)/dt = -r_i/100 * P_i(t). I need to solve this differential equation to find P_i(t) in terms of P_i(0), r_i, and t.Hmm, this looks like a standard exponential decay model. The differential equation is linear and separable. The general form is dP/dt = kP, which has the solution P(t) = P(0) e^{kt}. In this case, the rate is negative because it's a decrease, so k would be -r_i/100.Let me write that out:dP_i/dt = - (r_i / 100) P_i(t)This is a first-order linear differential equation. The solution should be:P_i(t) = P_i(0) * e^{ - (r_i / 100) * t }Wait, let me check. If I separate variables:dP_i / P_i = - (r_i / 100) dtIntegrate both sides:∫ (1/P_i) dP_i = ∫ - (r_i / 100) dtWhich gives:ln(P_i) = - (r_i / 100) t + CExponentiating both sides:P_i(t) = e^C * e^{ - (r_i / 100) t }Since P_i(0) is the initial population, when t=0, P_i(0) = e^C. So e^C = P_i(0). Therefore,P_i(t) = P_i(0) * e^{ - (r_i / 100) t }Yes, that seems right. So that's the expression for the population of each group over time.Moving on to part 2: The cultural diversity index D(t) is proportional to the product of all the populations. So D(t) = k * product from i=1 to n of P_i(t). The anthropologist wants the rate of change of D(t) at t=0, which is dD/dt at t=0.I need to find dD/dt evaluated at t=0. Let me think about how to approach this.First, D(t) is a product of functions, each P_i(t). So to find its derivative, I can use the product rule. But since it's a product of multiple functions, the derivative will involve the sum of each function's derivative times the product of all the others.But since we're evaluating at t=0, maybe we can plug in t=0 into the derivative expression.Let me write D(t) = k * P_1(t) * P_2(t) * ... * P_n(t)So, dD/dt = k * [ dP_1/dt * P_2 * ... * P_n + P_1 * dP_2/dt * ... * P_n + ... + P_1 * P_2 * ... * dP_n/dt ]At t=0, each P_i(t) is P_i(0). So, substituting t=0, we have:dD/dt|_{t=0} = k * [ (dP_1/dt|_{t=0}) * P_2(0) * ... * P_n(0) + P_1(0) * (dP_2/dt|_{t=0}) * ... * P_n(0) + ... + P_1(0) * P_2(0) * ... * (dP_n/dt|_{t=0}) ]But each dP_i/dt|_{t=0} is given by the differential equation: - (r_i / 100) P_i(0). So substituting that in:dD/dt|_{t=0} = k * [ (- r_1 / 100 P_1(0)) * P_2(0) * ... * P_n(0) + P_1(0) * (- r_2 / 100 P_2(0)) * ... * P_n(0) + ... + P_1(0) * P_2(0) * ... * (- r_n / 100 P_n(0)) ]Notice that each term in the sum is - (r_i / 100) times the product of all P_j(0) except for P_i(0), but actually, each term is - (r_i / 100) times the product of all P_j(0) including P_i(0) because each term is (- r_i / 100 P_i(0)) multiplied by the product of the other P_j(0). Wait, no, let me clarify.Wait, actually, each term is (- r_i / 100 P_i(0)) multiplied by the product of the other P_j(0). So each term is (- r_i / 100) times the product of all P_j(0) except P_i(0) is not correct. Wait, no, because in the first term, it's (- r_1 / 100 P_1(0)) multiplied by P_2(0) * ... * P_n(0). So that's (- r_1 / 100) times the product of all P_j(0) except P_1(0) is not correct. Wait, no, it's (- r_1 / 100) times P_1(0) times the product of P_2(0) to P_n(0). So actually, each term is (- r_i / 100) times the product of all P_j(0) for j=1 to n, because P_i(0) is included in the product.Wait, no, hold on. Let me think again. The first term is (dP_1/dt) * P_2 * ... * P_n. But dP_1/dt is - r_1 / 100 P_1(0). So the first term is (- r_1 / 100 P_1(0)) * P_2(0) * ... * P_n(0). So that's (- r_1 / 100) times the product of all P_j(0) for j=1 to n. Similarly, the second term is P_1(0) * (- r_2 / 100 P_2(0)) * ... * P_n(0), which is also (- r_2 / 100) times the product of all P_j(0). So each term in the sum is (- r_i / 100) times the product of all P_j(0). Therefore, the entire sum is the sum over i from 1 to n of (- r_i / 100) times the product of all P_j(0).But wait, the product of all P_j(0) is D(t)/k at t=0, because D(0) = k * product P_j(0). So D(0) = k * product P_j(0). Therefore, product P_j(0) = D(0)/k.But in the expression for dD/dt|_{t=0}, we have:dD/dt|_{t=0} = k * sum_{i=1}^n [ (- r_i / 100) * product_{j=1}^n P_j(0) ]Which is equal to:k * [ (-1/100) sum_{i=1}^n r_i ] * product_{j=1}^n P_j(0)But product P_j(0) is D(0)/k, so substituting:dD/dt|_{t=0} = k * (-1/100) sum r_i * (D(0)/k)Simplify:The k cancels out:dD/dt|_{t=0} = (-1/100) sum r_i * D(0)But D(0) is k * product P_j(0). So we can write:dD/dt|_{t=0} = (-1/100) sum_{i=1}^n r_i * D(0)Alternatively, since D(0) = k * product P_j(0), we can express it as:dD/dt|_{t=0} = (-1/100) sum_{i=1}^n r_i * k * product P_j(0)But the question asks for dD/dt|_{t=0} in terms of k, P_i(0), and r_i. So perhaps it's better to express it as:dD/dt|_{t=0} = k * product P_i(0) * (-1/100) sum r_iWhich can be written as:dD/dt|_{t=0} = - (k / 100) * (sum_{i=1}^n r_i) * product_{i=1}^n P_i(0)Alternatively, since D(0) = k * product P_i(0), we can write:dD/dt|_{t=0} = - (1/100) sum r_i * D(0)But the question specifies to express it in terms of k, P_i(0), and r_i, so probably the first expression is better.Wait, let me make sure. The product of P_i(0) is part of D(t), so it's included in the expression. So the derivative is proportional to the sum of r_i times the product of P_i(0), scaled by k and -1/100.So putting it all together:dD/dt|_{t=0} = - (k / 100) * (sum_{i=1}^n r_i) * (product_{i=1}^n P_i(0))Alternatively, since product P_i(0) is D(0)/k, but since we need to express it in terms of k, P_i(0), and r_i, not D(0), the first expression is better.Let me write that:dD/dt|_{t=0} = - (k / 100) * (sum_{i=1}^n r_i) * (product_{i=1}^n P_i(0))Yes, that seems correct.Wait, let me double-check. If I have D(t) = k * product P_i(t). The derivative is k * sum [dP_i/dt * product_{j≠i} P_j(t)]. At t=0, each dP_i/dt = - r_i / 100 P_i(0). So each term is - r_i / 100 P_i(0) * product_{j≠i} P_j(0). But product_{j≠i} P_j(0) is product P_j(0) / P_i(0). So each term is - r_i / 100 * product P_j(0). Therefore, the sum is sum (- r_i / 100) * product P_j(0). So the derivative is k * sum (- r_i / 100) * product P_j(0). Which is - (k / 100) sum r_i * product P_j(0). Yes, that's consistent.So, to summarize:1. The solution to the differential equation is P_i(t) = P_i(0) e^{- (r_i / 100) t}2. The derivative of D(t) at t=0 is dD/dt|_{t=0} = - (k / 100) * (sum_{i=1}^n r_i) * (product_{i=1}^n P_i(0))I think that's it.

answer：Okay, so I need to figure out how to allocate resources to three regions, A, B, and C, to maximize the effectiveness function E(x, y, z) = 3x² + 2y² + z² + 2xy - yz - xz. The total resources are 100 units, and each region needs at least 10 units. Then, in the second part, region B needs at least 20 units. First, I think this is an optimization problem with constraints. Since we're dealing with maximizing a quadratic function subject to linear constraints, I believe we can use the method of Lagrange multipliers. But I'm not entirely sure, so maybe I should also consider if it's a convex function or not because that affects whether the critical point found is a maximum or minimum.Looking at the function E(x, y, z), it's a quadratic function. To determine if it's convex, I can look at the Hessian matrix. The Hessian is the matrix of second partial derivatives. Let me compute that.The function is E = 3x² + 2y² + z² + 2xy - yz - xz.First partial derivatives:- dE/dx = 6x + 2y - z- dE/dy = 4y + 2x - z- dE/dz = 2z - y - xSecond partial derivatives:- d²E/dx² = 6- d²E/dy² = 4- d²E/dz² = 2- d²E/dxdy = 2- d²E/dxdz = -1- d²E/dydz = -1So the Hessian matrix H is:[6 2 -1][2 4 -1][-1 -1 2]To check if this is positive definite (which would mean the function is convex), I can check the leading principal minors.First minor: 6 > 0.Second minor: determinant of the top-left 2x2 matrix:|6 2||2 4| = 6*4 - 2*2 = 24 - 4 = 20 > 0.Third minor: determinant of the full Hessian. Let me compute that.Using the rule of Sarrus or cofactor expansion. Let's do cofactor expansion along the first row.6 * det([4, -1], [-1, 2]) - 2 * det([2, -1], [-1, 2]) + (-1) * det([2, 4], [-1, -1])Compute each term:First term: 6*(4*2 - (-1)*(-1)) = 6*(8 - 1) = 6*7 = 42Second term: -2*(2*2 - (-1)*(-1)) = -2*(4 - 1) = -2*3 = -6Third term: -1*(2*(-1) - 4*(-1)) = -1*(-2 + 4) = -1*2 = -2Adding them up: 42 - 6 - 2 = 34So the determinant is 34, which is positive. Since all leading principal minors are positive, the Hessian is positive definite, meaning the function is convex. Therefore, any critical point found is a global minimum. Wait, but we are trying to maximize E. Hmm, that complicates things because if the function is convex, it doesn't have a maximum unless we're constrained within a compact set.But our domain is constrained by x + y + z = 100 and x, y, z ≥ 10. So it's a compact set (closed and bounded in R^3). Therefore, by Weierstrass theorem, E attains its maximum on this set. However, since E is convex, the maximum will occur at one of the boundary points.Wait, that contradicts my initial thought of using Lagrange multipliers because usually, for convex functions, the maximum is on the boundary. So perhaps the maximum occurs when one of the variables is at its minimum, and the others take up the remaining resources.But before jumping to conclusions, let me think again. The function E is convex, so it curves upward. Therefore, its maximum on a convex set (which our feasible region is) will be on the boundary. So, perhaps the maximum occurs when one or more variables are at their lower bounds.So, the feasible region is defined by x + y + z = 100, x ≥ 10, y ≥ 10, z ≥ 10. So, the boundaries are when x=10, y=10, or z=10.Therefore, to find the maximum, I might need to check the function E at all possible boundary points where one or more variables are at their minimums.But this could be complicated because there are multiple boundaries. Maybe a better approach is to use substitution because of the equality constraint x + y + z = 100.Let me express z = 100 - x - y. Then, substitute this into E to get E in terms of x and y.So, E(x, y) = 3x² + 2y² + (100 - x - y)² + 2xy - y(100 - x - y) - x(100 - x - y)Let me expand this step by step.First, expand (100 - x - y)²:= 10000 - 200x - 200y + x² + 2xy + y²Then, expand - y(100 - x - y):= -100y + xy + y²Similarly, expand -x(100 - x - y):= -100x + x² + xyNow, let's substitute all back into E(x, y):E = 3x² + 2y² + [10000 - 200x - 200y + x² + 2xy + y²] + 2xy + [-100y + xy + y²] + [-100x + x² + xy]Now, let's combine like terms.First, collect all x² terms:3x² + x² + x² = 5x²y² terms:2y² + y² + y² + y² = 5y²xy terms:2xy + 2xy + xy + xy = 6xyConstant terms:10000x terms:-200x -100x = -300xy terms:-200y -100y = -300ySo, putting it all together:E(x, y) = 5x² + 5y² + 6xy - 300x - 300y + 10000Hmm, that's a quadratic function in two variables. Let me write it as:E(x, y) = 5x² + 6xy + 5y² - 300x - 300y + 10000Now, to find the critical points, we can take partial derivatives with respect to x and y, set them to zero, and solve.Compute partial derivatives:dE/dx = 10x + 6y - 300dE/dy = 6x + 10y - 300Set them equal to zero:10x + 6y = 300 ...(1)6x + 10y = 300 ...(2)Now, solve this system of equations.Let me write them as:10x + 6y = 3006x + 10y = 300Let me multiply equation (1) by 5 and equation (2) by 3 to eliminate y:50x + 30y = 1500 ...(1a)18x + 30y = 900 ...(2a)Now subtract (2a) from (1a):(50x - 18x) + (30y - 30y) = 1500 - 90032x = 600x = 600 / 32 = 18.75Now, substitute x = 18.75 into equation (1):10*(18.75) + 6y = 300187.5 + 6y = 3006y = 300 - 187.5 = 112.5y = 112.5 / 6 = 18.75So, x = y = 18.75Then, z = 100 - x - y = 100 - 18.75 - 18.75 = 62.5So, the critical point is at (18.75, 18.75, 62.5)But wait, we have constraints x ≥ 10, y ≥ 10, z ≥ 10. Here, x and y are 18.75, which is above 10, and z is 62.5, which is also above 10. So, this critical point is within the feasible region.But earlier, I thought since the function is convex, the maximum occurs on the boundary. However, this critical point is a minimum because the function is convex. Wait, but we are trying to maximize E, which is convex, so the maximum should be on the boundary.Hmm, this is confusing. Maybe I made a mistake in interpreting the convexity.Wait, if E is convex, then it has a unique minimum. So, the critical point we found is the minimum. Therefore, the maximum must be on the boundary.So, to find the maximum, we need to check the boundaries where x=10, y=10, or z=10.But since the feasible region is defined by x + y + z = 100 and x, y, z ≥ 10, the boundaries are when one or more variables are at their minimums.So, the boundaries are:1. x = 10, y ≥ 10, z = 90 - y2. y = 10, x ≥ 10, z = 90 - x3. z = 10, x ≥ 10, y = 90 - xAdditionally, the intersections where two variables are at their minimums:4. x = 10, y = 10, z = 805. x = 10, z = 10, y = 806. y = 10, z = 10, x = 80So, we need to evaluate E at all these boundary points and also check if the maximum occurs at these points.But this seems like a lot of work, but let's proceed step by step.First, let's consider the case where x = 10.Then, z = 90 - y.So, E(10, y, 90 - y) = 3*(10)^2 + 2y² + (90 - y)^2 + 2*10*y - y*(90 - y) - 10*(90 - y)Compute this:= 300 + 2y² + (8100 - 180y + y²) + 20y - 90y + y² - 900 + 10ySimplify term by term:300 + 2y² + 8100 - 180y + y² + 20y - 90y + y² - 900 + 10yCombine like terms:Constants: 300 + 8100 - 900 = 7500y² terms: 2y² + y² + y² = 4y²y terms: -180y + 20y -90y +10y = (-180 + 20 -90 +10)y = (-240)ySo, E = 4y² - 240y + 7500This is a quadratic in y. To find its maximum, since the coefficient of y² is positive, it opens upwards, so it has a minimum, not a maximum. Therefore, on the boundary x=10, the maximum of E would occur at the endpoints of y.The endpoints are when y=10 and y=80 (since z=90 - y must be ≥10, so y ≤80).So, compute E at y=10 and y=80.At y=10:E = 4*(10)^2 -240*10 +7500 = 400 -2400 +7500 = 5500At y=80:E = 4*(80)^2 -240*80 +7500 = 4*6400 -19200 +7500 = 25600 -19200 +7500 = 13900So, on the boundary x=10, the maximum E is 13900 at y=80, z=10.Wait, but z=90 - y=10 when y=80, so that's correct.Now, let's consider the case where y=10.Then, z=90 -x.So, E(x,10,90 -x) = 3x² + 2*(10)^2 + (90 -x)^2 + 2x*10 -10*(90 -x) -x*(90 -x)Compute this:= 3x² + 200 + (8100 - 180x +x²) + 20x -900 +10x -90x +x²Simplify term by term:3x² +200 +8100 -180x +x² +20x -900 +10x -90x +x²Combine like terms:Constants: 200 +8100 -900 = 7400x² terms: 3x² +x² +x² =5x²x terms: -180x +20x +10x -90x = (-180 +20 +10 -90)x = (-240)xSo, E =5x² -240x +7400Again, quadratic in x with positive coefficient, so minimum at vertex, maximum at endpoints.Endpoints are x=10 and x=80 (since z=90 -x ≥10, so x ≤80).Compute E at x=10:E=5*(10)^2 -240*10 +7400=500 -2400 +7400=5500At x=80:E=5*(80)^2 -240*80 +7400=5*6400 -19200 +7400=32000 -19200 +7400=19200+7400=26600? Wait, 32000 -19200=12800, 12800 +7400=20200.Wait, let me compute again:5*(80)^2 =5*6400=32000-240*80= -19200+7400So total: 32000 -19200=12800; 12800 +7400=20200.So, E=20200 at x=80, y=10, z=10.That's higher than the previous case.Now, consider the case where z=10.Then, x + y =90.So, E(x, y,10)=3x² +2y² +100 +2xy - y*10 -x*10=3x² +2y² +100 +2xy -10y -10xBut since x + y =90, y=90 -x.Substitute y=90 -x:E(x)=3x² +2*(90 -x)^2 +100 +2x*(90 -x) -10*(90 -x) -10xExpand:3x² +2*(8100 -180x +x²) +100 +180x -2x² -900 +10x -10xSimplify term by term:3x² +16200 -360x +2x² +100 +180x -2x² -900 +10x -10xCombine like terms:x² terms:3x² +2x² -2x²=3x²Constants:16200 +100 -900=15400x terms:-360x +180x +10x -10x= (-360 +180 +10 -10)x= (-180)xSo, E=3x² -180x +15400Again, quadratic in x with positive coefficient, so minimum at vertex, maximum at endpoints.Endpoints are x=10 and x=80 (since y=90 -x ≥10, so x ≤80).Compute E at x=10:E=3*(10)^2 -180*10 +15400=300 -1800 +15400=13900At x=80:E=3*(80)^2 -180*80 +15400=3*6400 -14400 +15400=19200 -14400 +15400=19200 -14400=4800; 4800 +15400=20200.So, E=20200 at x=80, y=10, z=10.So, on the boundary z=10, the maximum E is 20200 at x=80, y=10.Now, we also need to check the cases where two variables are at their minimums.Case 4: x=10, y=10, z=80.Compute E(10,10,80)=3*(10)^2 +2*(10)^2 +80^2 +2*10*10 -10*80 -10*80=300 +200 +6400 +200 -800 -800Compute step by step:300 +200=500500 +6400=69006900 +200=71007100 -800=63006300 -800=5500So, E=5500.Case 5: x=10, z=10, y=80.E(10,80,10)=3*100 +2*6400 +100 +2*10*80 -80*10 -10*10=300 +12800 +100 +1600 -800 -100Compute:300 +12800=1310013100 +100=1320013200 +1600=1480014800 -800=1400014000 -100=13900So, E=13900.Case 6: y=10, z=10, x=80.E(80,10,10)=3*6400 +2*100 +100 +2*80*10 -10*10 -80*10=19200 +200 +100 +1600 -100 -800Compute:19200 +200=1940019400 +100=1950019500 +1600=2110021100 -100=2100021000 -800=20200So, E=20200.So, among all these boundary points, the maximum E is 20200, which occurs at two points: (80,10,10) and (10,80,10). Wait, no, in case 5, (10,80,10) gives E=13900, but in case 6, (80,10,10) gives E=20200. Similarly, in the case where z=10, x=80, y=10 gives E=20200.Wait, so the maximum E is 20200, achieved when either x=80, y=10, z=10 or y=80, x=10, z=10? Wait, no, in case 5, when x=10, y=80, z=10, E=13900, which is less than 20200. So, only when x=80, y=10, z=10 and when y=80, x=10, z=10? Wait, no, when y=80, x=10, z=10, E=13900, which is less than 20200.Wait, perhaps I made a mistake. Let me check again.In case 4: x=10, y=10, z=80: E=5500Case 5: x=10, z=10, y=80: E=13900Case 6: y=10, z=10, x=80: E=20200So, only when x=80, y=10, z=10, E=20200.Similarly, when y=80, x=10, z=10, E=13900.So, the maximum is 20200 at (80,10,10).Wait, but earlier, when we considered the boundary y=10, x=80, z=10, E=20200.Similarly, when z=10, x=80, y=10, E=20200.So, it's the same point.Therefore, the maximum occurs at x=80, y=10, z=10.But wait, earlier, when we found the critical point inside the feasible region, it was a minimum because the function is convex. So, the maximum is indeed on the boundary at (80,10,10).But let me confirm this by checking another point. Suppose we take x=70, y=20, z=10. Then, E=3*(70)^2 +2*(20)^2 +10^2 +2*70*20 -20*10 -70*10.Compute:3*4900=147002*400=8001002*70*20=2800-20*10=-200-70*10=-700Total:14700 +800=15500; 15500 +100=15600; 15600 +2800=18400; 18400 -200=18200; 18200 -700=17500.So, E=17500, which is less than 20200.Another point: x=90, y=0, z=10. But y must be at least 10, so y=0 is invalid.Wait, y must be ≥10, so x=90 is invalid because z=10, y=0 is not allowed.Wait, actually, x=90 would require y + z=10, but y and z must be at least 10 each, so x cannot be more than 80.So, x=80 is the maximum x can be when y and z are at their minimums.Therefore, the maximum E is 20200 at (80,10,10).But wait, earlier, when we considered the case where y=10, x=80, z=10, E=20200.Similarly, when z=10, x=80, y=10, E=20200.So, the maximum is achieved at (80,10,10).But let me check another point: x=70, y=10, z=20.E=3*(70)^2 +2*(10)^2 +20^2 +2*70*10 -10*20 -70*20=3*4900=147002*100=2004002*700=1400-200-1400Total:14700 +200=14900; 14900 +400=15300; 15300 +1400=16700; 16700 -200=16500; 16500 -1400=15100.So, E=15100, which is less than 20200.Another point: x=60, y=10, z=30.E=3*3600 +2*100 +900 +2*600 -10*30 -60*30=10800 +200 +900 +1200 -300 -1800Total:10800 +200=11000; 11000 +900=11900; 11900 +1200=13100; 13100 -300=12800; 12800 -1800=11000.E=11000, which is less than 20200.So, it seems that the maximum is indeed at (80,10,10).Wait, but earlier, when we considered the case where y=10, x=80, z=10, E=20200.Similarly, when z=10, x=80, y=10, E=20200.So, the maximum is achieved at (80,10,10).But wait, in the case where x=80, y=10, z=10, E=20200.Similarly, if we set y=80, x=10, z=10, E=13900, which is less.So, the maximum is achieved when x is as large as possible, given the constraints, and y and z are at their minimums.Therefore, the optimal allocation is x=80, y=10, z=10.But wait, let me check if there's a higher value when two variables are increased beyond their minimums.Wait, for example, if x=70, y=20, z=10, E=17500 as computed earlier.Which is less than 20200.Similarly, x=60, y=30, z=10: E=?E=3*3600 +2*900 +100 +2*60*30 -30*10 -60*10=10800 +1800 +100 +3600 -300 -600Total:10800 +1800=12600; 12600 +100=12700; 12700 +3600=16300; 16300 -300=16000; 16000 -600=15400.E=15400, still less than 20200.Alternatively, x=50, y=40, z=10.E=3*2500 +2*1600 +100 +2*50*40 -40*10 -50*10=7500 +3200 +100 +4000 -400 -500Total:7500 +3200=10700; 10700 +100=10800; 10800 +4000=14800; 14800 -400=14400; 14400 -500=13900.E=13900, still less.Alternatively, x=40, y=50, z=10.E=3*1600 +2*2500 +100 +2*40*50 -50*10 -40*10=4800 +5000 +100 +4000 -500 -400Total:4800 +5000=9800; 9800 +100=9900; 9900 +4000=13900; 13900 -500=13400; 13400 -400=13000.E=13000.So, it's clear that as we increase x beyond 80, we can't because y and z have to be at least 10. So, x=80 is the maximum x can be.Therefore, the maximum E is achieved at x=80, y=10, z=10.But wait, earlier, when we considered the case where x=80, y=10, z=10, E=20200.Similarly, when y=80, x=10, z=10, E=13900, which is less.So, the maximum is indeed at x=80, y=10, z=10.But let me check another point: x=75, y=15, z=10.E=3*(75)^2 +2*(15)^2 +10^2 +2*75*15 -15*10 -75*10=3*5625=168752*225=4501002*1125=2250-150-750Total:16875 +450=17325; 17325 +100=17425; 17425 +2250=19675; 19675 -150=19525; 19525 -750=18775.E=18775, which is less than 20200.So, yes, 20200 is higher.Therefore, the optimal allocation is x=80, y=10, z=10.But wait, let me check if there's a higher value when z is increased beyond 10, but keeping x and y above 10.Wait, for example, x=70, y=15, z=15.E=3*4900 +2*225 +225 +2*70*15 -15*15 -70*15=14700 +450 +225 +2100 -225 -1050Total:14700 +450=15150; 15150 +225=15375; 15375 +2100=17475; 17475 -225=17250; 17250 -1050=16200.E=16200, which is less than 20200.Another point: x=60, y=20, z=20.E=3*3600 +2*400 +400 +2*60*20 -20*20 -60*20=10800 +800 +400 +2400 -400 -1200Total:10800 +800=11600; 11600 +400=12000; 12000 +2400=14400; 14400 -400=14000; 14000 -1200=12800.E=12800.Still less.Alternatively, x=50, y=25, z=25.E=3*2500 +2*625 +625 +2*50*25 -25*25 -50*25=7500 +1250 +625 +2500 -625 -1250Total:7500 +1250=8750; 8750 +625=9375; 9375 +2500=11875; 11875 -625=11250; 11250 -1250=10000.E=10000.Less.So, it seems that the maximum is indeed at x=80, y=10, z=10.Therefore, the answer to part 1 is x=80, y=10, z=10.Now, for part 2, the constraint changes to y ≥20.So, the new constraints are x + y + z=100, x ≥10, y ≥20, z ≥10.So, we need to find the new optimal allocation.Again, since E is convex, the maximum will be on the boundary.But now, the boundaries are different.The boundaries are:1. x=10, y ≥20, z=90 - y2. y=20, x ≥10, z=80 -x3. z=10, x ≥10, y=90 -xAdditionally, the intersections where two variables are at their minimums:4. x=10, y=20, z=705. x=10, z=10, y=806. y=20, z=10, x=70But let's proceed step by step.First, consider the case where x=10.Then, z=90 - y, with y ≥20, z ≥10 ⇒ y ≤80.So, E(10, y, 90 - y)=3*100 +2y² + (90 - y)^2 +2*10*y - y*(90 - y) -10*(90 - y)Compute:=300 +2y² +8100 -180y +y² +20y -90y +y² -900 +10ySimplify:300 +8100 -900=75002y² +y² +y²=4y²-180y +20y -90y +10y= -240ySo, E=4y² -240y +7500Quadratic in y, positive coefficient, so minimum at vertex, maximum at endpoints.Endpoints: y=20 and y=80.Compute E at y=20:E=4*(20)^2 -240*20 +7500=4*400 -4800 +7500=1600 -4800 +7500=4300At y=80:E=4*(80)^2 -240*80 +7500=4*6400 -19200 +7500=25600 -19200 +7500=13900So, on x=10 boundary, maximum E=13900 at y=80, z=10.Now, consider y=20.Then, z=80 -x, with x ≥10, z ≥10 ⇒x ≤70.So, E(x,20,80 -x)=3x² +2*(20)^2 + (80 -x)^2 +2x*20 -20*(80 -x) -x*(80 -x)Compute:=3x² +800 +6400 -160x +x² +40x -1600 +20x -80x +x²Simplify term by term:3x² +x² +x²=5x²Constants:800 +6400 -1600=5600x terms:-160x +40x +20x -80x= (-160 +40 +20 -80)x= (-180)xSo, E=5x² -180x +5600Quadratic in x, positive coefficient, so minimum at vertex, maximum at endpoints.Endpoints: x=10 and x=70.Compute E at x=10:E=5*100 -1800 +5600=500 -1800 +5600=4300At x=70:E=5*(70)^2 -180*70 +5600=5*4900 -12600 +5600=24500 -12600 +5600=17500So, on y=20 boundary, maximum E=17500 at x=70, z=10.Now, consider z=10.Then, x + y=90, with x ≥10, y ≥20.So, y=90 -x, with x ≤70 (since y ≥20 ⇒x ≤70).So, E(x,90 -x,10)=3x² +2*(90 -x)^2 +100 +2x*(90 -x) - (90 -x)*10 -x*10Compute:=3x² +2*(8100 -180x +x²) +100 +180x -2x² -900 +10x -10xSimplify:3x² +16200 -360x +2x² +100 +180x -2x² -900 +10x -10xCombine like terms:3x² +2x² -2x²=3x²Constants:16200 +100 -900=15400x terms:-360x +180x +10x -10x= (-360 +180 +10 -10)x= (-180)xSo, E=3x² -180x +15400Quadratic in x, positive coefficient, so minimum at vertex, maximum at endpoints.Endpoints: x=10 and x=70.Compute E at x=10:E=3*100 -1800 +15400=300 -1800 +15400=13900At x=70:E=3*(70)^2 -180*70 +15400=3*4900 -12600 +15400=14700 -12600 +15400=17500So, on z=10 boundary, maximum E=17500 at x=70, y=20.Now, check the intersection points where two variables are at their minimums.Case 4: x=10, y=20, z=70.Compute E(10,20,70)=3*100 +2*400 +4900 +2*10*20 -20*70 -10*70=300 +800 +4900 +400 -1400 -700Total:300 +800=1100; 1100 +4900=6000; 6000 +400=6400; 6400 -1400=5000; 5000 -700=4300.E=4300.Case 5: x=10, z=10, y=80.E(10,80,10)=3*100 +2*6400 +100 +2*10*80 -80*10 -10*10=300 +12800 +100 +1600 -800 -100Total:300 +12800=13100; 13100 +100=13200; 13200 +1600=14800; 14800 -800=14000; 14000 -100=13900.E=13900.Case 6: y=20, z=10, x=70.E(70,20,10)=3*4900 +2*400 +100 +2*70*20 -20*10 -70*10=14700 +800 +100 +2800 -200 -700Total:14700 +800=15500; 15500 +100=15600; 15600 +2800=18400; 18400 -200=18200; 18200 -700=17500.E=17500.So, the maximum E on the boundaries is 17500, achieved at (70,20,10) and (10,80,10). Wait, but when x=10, y=80, z=10, E=13900, which is less than 17500. So, the maximum is at (70,20,10).But wait, in case 6, y=20, z=10, x=70, E=17500.Similarly, when z=10, x=70, y=20, E=17500.So, the maximum is achieved at (70,20,10).But let me check another point: x=60, y=30, z=10.E=3*3600 +2*900 +100 +2*60*30 -30*10 -60*10=10800 +1800 +100 +3600 -300 -600Total:10800 +1800=12600; 12600 +100=12700; 12700 +3600=16300; 16300 -300=16000; 16000 -600=15400.E=15400, which is less than 17500.Another point: x=75, y=15, z=10. But y must be ≥20, so y=15 is invalid.Alternatively, x=65, y=25, z=10.E=3*4225 +2*625 +100 +2*65*25 -25*10 -65*10=12675 +1250 +100 +3250 -250 -650Total:12675 +1250=13925; 13925 +100=14025; 14025 +3250=17275; 17275 -250=17025; 17025 -650=16375.E=16375, which is less than 17500.Another point: x=70, y=20, z=10, E=17500.Another point: x=70, y=20, z=10.Wait, that's the same as the maximum point.So, the maximum E is 17500 at (70,20,10).But let me check if there's a higher value when two variables are increased beyond their minimums.Wait, for example, x=60, y=30, z=10: E=15400 as before.Another point: x=50, y=40, z=10.E=3*2500 +2*1600 +100 +2*50*40 -40*10 -50*10=7500 +3200 +100 +4000 -400 -500Total:7500 +3200=10700; 10700 +100=10800; 10800 +4000=14800; 14800 -400=14400; 14400 -500=13900.E=13900.Less than 17500.Alternatively, x=80, y=10, z=10 is invalid because y must be ≥20.So, the maximum is indeed at x=70, y=20, z=10.But wait, let me check another point: x=70, y=20, z=10.E=3*4900 +2*400 +100 +2*70*20 -20*10 -70*10=14700 +800 +100 +2800 -200 -700Total:14700 +800=15500; 15500 +100=15600; 15600 +2800=18400; 18400 -200=18200; 18200 -700=17500.Yes, E=17500.Another point: x=65, y=25, z=10: E=16375 as before.So, the maximum is indeed at x=70, y=20, z=10.Therefore, the optimal allocation under the new constraint y ≥20 is x=70, y=20, z=10.But wait, let me check if there's a higher value when z is increased beyond 10, but keeping x and y above their minimums.For example, x=60, y=20, z=20.E=3*3600 +2*400 +400 +2*60*20 -20*20 -60*20=10800 +800 +400 +2400 -400 -1200Total:10800 +800=11600; 11600 +400=12000; 12000 +2400=14400; 14400 -400=14000; 14000 -1200=12800.E=12800, which is less than 17500.Another point: x=50, y=30, z=20.E=3*2500 +2*900 +400 +2*50*30 -30*20 -50*20=7500 +1800 +400 +3000 -600 -1000Total:7500 +1800=9300; 9300 +400=9700; 9700 +3000=12700; 12700 -600=12100; 12100 -1000=11100.E=11100.Less.Alternatively, x=70, y=20, z=10: E=17500.Another point: x=75, y=15, z=10: invalid because y must be ≥20.So, yes, the maximum is at x=70, y=20, z=10.Therefore, the answer to part 2 is x=70, y=20, z=10.

answer：Okay, so I have two problems to solve here related to filmmaking and lighting. Let me take them one at a time.Starting with the first problem: We have 12 different types of lights, each with unique power consumption rates. The goal is to maximize the total power consumption without exceeding 10,000 watts. Hmm, this sounds like an optimization problem, specifically a knapsack problem where we want to maximize the total value (which here is power consumption) without exceeding the capacity (10,000 watts). But wait, the problem says "using any number of each type," so it's actually an unbounded knapsack problem because we can use multiple units of each light type.But the user mentioned to use integer programming techniques. So, I should model this as an integer linear program. Let me recall the structure of integer programming. The variables will be the number of each type of light, say x_i for each light type i, where i ranges from 1 to 12. The objective is to maximize the total power, which is the sum over all i of P_i * x_i. The constraint is that this total must be less than or equal to 10,000 watts. Also, each x_i must be a non-negative integer.So, in mathematical terms, the problem is:Maximize: Σ (P_i * x_i) for i=1 to 12Subject to: Σ (P_i * x_i) ≤ 10,000And x_i ≥ 0, integers.To prove the optimality, I need to show that the solution found by the integer programming model is indeed the maximum possible without exceeding the power limit. Since it's an integer linear program, if we solve it using a solver, it will give us the optimal solution. But since I don't have the specific P_i values, I can't compute the exact numbers. However, I can explain the method.First, we can relax the integer constraint to make it a linear program, solve it, and then use techniques like branch and bound to find the integer solution. The optimal solution will be the one where the total power is as close as possible to 10,000 without exceeding it, considering the integer quantities of each light type.Now, moving on to the second problem. The historian wants the light intensity at a specific point P to be uniformly distributed with minimal variance. The intensity from each light is given by I_i = L_i / d_i². The total intensity is the sum of all I_i, which equals I_total. We need to find the distances d_i such that the variance of the intensities is minimized.Variance is a measure of how spread out the numbers are. To minimize the variance, we want all I_i to be as equal as possible. So, if all I_i are equal, the variance would be zero, which is the minimum possible.Let me denote each I_i as equal to some constant k. Then, since the total intensity is I_total, we have 12k = I_total, so k = I_total / 12.Therefore, each I_i should be equal to I_total / 12. So, for each light, we have:L_i / d_i² = I_total / 12Solving for d_i, we get:d_i = sqrt(12 * L_i / I_total)So, each distance d_i is the square root of (12 * L_i divided by I_total). This ensures that each intensity I_i is equal, thus making the variance zero, which is the minimum possible variance.But wait, is this feasible? We need to make sure that all d_i are positive real numbers, which they are as long as L_i and I_total are positive, which they are in the context of lighting.So, the system of equations would be:For each i from 1 to 12:I_i = L_i / d_i²And the sum of all I_i equals I_total:Σ (L_i / d_i²) = I_totalBut to minimize the variance, we set each I_i equal, leading to each d_i being sqrt(12 * L_i / I_total). Therefore, the system of equations is:d_i² = 12 * L_i / I_total for each iWhich gives us the distances needed.I think that's the approach. Let me double-check. If all I_i are equal, variance is zero, which is minimal. So yes, that should be the solution.**Final Answer**1. The optimal combination is found by solving the integer linear program, ensuring the total power is maximized without exceeding 10,000 watts. The solution is proven optimal through integer programming techniques. boxed{text{Optimal solution found via integer programming}}2. The distances are determined by setting each intensity equal, resulting in ( d_i = sqrt{frac{12 L_i}{I_{text{total}}}} ). The system of equations ensures uniform intensity distribution with minimal variance. boxed{d_i = sqrt{dfrac{12 L_i}{I_{text{total}}}}}