🧠:<think>Okay, so I have this problem about two companies merging and trying to optimize their operational costs. Let me try to understand what it's asking and figure out how to solve it step by step.First, the problem gives me two cost functions: one for Company A and one for Company B. Company A's cost function is (C_A(x) = 50x + 200), and Company B's is (C_B(x) = 30x + 500). Here, (x) represents the number of units produced. The CFO wants to merge these two companies and create a new cost function (C_M(x) = alpha x + beta), where (alpha) and (beta) are constants we need to determine.The first part of the problem says that the CFO aims to reduce the annual operating costs by 20% for any given (x) compared to the combined costs of Company A and Company B. So, I think this means that the merged company's cost should be 80% of the combined costs of A and B. Let me write that down.The combined cost of A and B would be (C_A(x) + C_B(x)). Let me compute that:(C_A(x) + C_B(x) = (50x + 200) + (30x + 500)).Combining like terms, that's (50x + 30x = 80x) and (200 + 500 = 700). So, the combined cost is (80x + 700).Now, the merged company wants to reduce this by 20%, which means they want to pay 80% of the combined cost. So, the new cost function (C_M(x)) should be 0.8 times the combined cost.Let me write that equation:(C_M(x) = 0.8 times (80x + 700)).Let me compute that:First, multiply 0.8 by 80x: (0.8 times 80x = 64x).Then, multiply 0.8 by 700: (0.8 times 700 = 560).So, putting it together, (C_M(x) = 64x + 560).Therefore, (alpha = 64) and (beta = 560). That should answer the first part.Wait, let me double-check my calculations to make sure I didn't make a mistake. So, combined cost is (80x + 700). 20% reduction means 80% of that. 80% of 80x is indeed 64x because 80 times 0.8 is 64. Similarly, 700 times 0.8 is 560. Yep, that seems right.So, part 1 is done. Now, moving on to part 2.The second part asks: If the merged company plans to produce 1000 units annually, calculate the total cost savings achieved through the merger and cost-cutting measures.Alright, so cost savings would be the difference between the combined cost of A and B and the merged company's cost at x=1000.First, let me compute the combined cost at x=1000.Combined cost: (80x + 700). Plugging in x=1000:(80 times 1000 + 700 = 80,000 + 700 = 80,700).Now, the merged company's cost at x=1000 is (C_M(1000) = 64x + 560). Plugging in x=1000:(64 times 1000 + 560 = 64,000 + 560 = 64,560).So, the cost savings would be the difference between the combined cost and the merged cost:(80,700 - 64,560 = 16,140).Therefore, the total cost savings achieved is 16,140.Wait, let me verify that again. So, original combined cost is 80,700, merged cost is 64,560. Subtracting, 80,700 minus 64,560 is indeed 16,140. That seems correct.Alternatively, since the merged cost is 80% of the combined cost, the savings should be 20% of the combined cost. Let me compute 20% of 80,700:20% of 80,700 is 0.2 * 80,700 = 16,140. Yep, same result. So that checks out.Therefore, the total cost savings is 16,140.So, summarizing:1. (alpha = 64) and (beta = 560).2. Cost savings at 1000 units is 16,140.I think that's all. Let me just recap to make sure I didn't skip any steps or make any calculation errors.For part 1, combined cost is 80x + 700. 20% reduction means 80% of that, which is 64x + 560. That gives (alpha = 64) and (beta = 560).For part 2, plug in x=1000 into both the combined cost and the merged cost, subtract to find savings. Alternatively, since it's a 20% reduction, just compute 20% of the combined cost at x=1000, which is 16,140.Everything seems consistent. I think I'm confident with these answers.Final Answer1. The values of (alpha) and (beta) are (boxed{64}) and (boxed{560}) respectively.2. The total cost savings achieved is (boxed{16140}) dollars.</think>

🧠:<think>Okay, so I have this problem where an engineer is trying to allocate resources for a bridge project. The goal is to minimize cost and environmental impact while maximizing social benefit. They've given me some functions for cost, environmental impact, and social benefit, and I need to set up a Lagrangian for this optimization problem. Then, if the environmental impact is too high, I have to add a penalty function and adjust the Lagrangian accordingly.First, let me try to understand the problem. We have three variables: x, y, z, which represent resources allocated to construction, environmental protection, and social programs, respectively. The total resources are fixed at T, so x + y + z = T.The cost function is C(x, y, z) = a x² + b y² + c z². So, this is a quadratic function, meaning that as we allocate more resources to any of these areas, the cost increases quadratically. That makes sense because, for example, building more of the bridge would likely cost more, and maybe the cost doesn't increase linearly because of economies of scale or something.The environmental impact is E(x, y, z) = d x y + e y z + f z x. This is interesting because it's a combination of products of the resource allocations. So, the environmental impact isn't just a function of each variable individually but how they interact. For example, more construction (x) and more environmental protection (y) together might have a certain impact, as well as y and z, and z and x. I wonder if this is because each pair of resources affects the environment in a different way.The social benefit is S(x, y, z) = g ln(x) + h sqrt(y) + i exp(z). Okay, so this is a mix of different functions. The natural log of x, which grows slowly, the square root of y, which also grows but maybe a bit faster, and the exponential of z, which can grow very rapidly. So, social benefit is a combination of these, each contributing differently based on the allocation.The engineer wants to minimize cost and environmental impact, while maximizing social benefit. So, it's a multi-objective optimization problem with three objectives: minimize C, minimize E, maximize S.But in optimization, especially with multiple objectives, we often combine them into a single function using weights or some other method. But since the problem mentions formulating the Lagrangian, I think we need to set up a Lagrangian that incorporates all these objectives with constraints.Wait, but the main constraint is x + y + z = T. So, we have equality constraints. So, maybe we can use Lagrange multipliers for this.But how do we handle multiple objectives? I remember that in multi-objective optimization, sometimes we can combine the objectives into a single function with weights. So, perhaps we can write a Lagrangian that includes terms for each objective with their respective weights, plus the constraint with a Lagrange multiplier.But the problem says "formulate the Lagrangian for this optimization problem," so maybe it's expecting a Lagrangian that includes all three objectives as separate terms, each multiplied by their own Lagrange multipliers, plus the constraint term.Alternatively, perhaps we can use a weighted sum approach where we combine the objectives into a single function. But since the problem mentions minimizing cost and environmental impact while maximizing social benefit, which are conflicting objectives, we need to balance them somehow.Wait, maybe the problem is expecting us to set up a Lagrangian where we have the cost and environmental impact as functions to be minimized, and the social benefit as a function to be maximized, all subject to the resource constraint.So, in terms of Lagrangian, we can write it as:L = C(x, y, z) + λ (E(x, y, z)) - μ (S(x, y, z)) + ν (x + y + z - T)But wait, no, because in Lagrangian multipliers, we usually have the objective function plus the multiplier times the constraint. But here, we have multiple objectives.Alternatively, perhaps we can consider each objective as a separate constraint, but that might not be the standard approach.Wait, maybe I need to think of it as a constrained optimization problem where we have multiple objectives, but we can use a Lagrangian that includes all of them with their own multipliers.Alternatively, perhaps we can use a scalarization method, where we combine the objectives into a single function. For example, we can write the Lagrangian as:L = C(x, y, z) + λ E(x, y, z) - μ S(x, y, z) + ν (x + y + z - T)But I'm not sure if that's the correct way to set it up. Maybe the signs need to be adjusted because we are minimizing C and E, and maximizing S.Wait, in the Lagrangian, the terms for the objectives should reflect whether we are minimizing or maximizing. So, since we are minimizing C and E, and maximizing S, perhaps we can write the Lagrangian as:L = C(x, y, z) + λ E(x, y, z) - μ S(x, y, z) + ν (x + y + z - T)But I'm not entirely sure. Alternatively, maybe all the objectives are to be minimized, so we can write:L = C(x, y, z) + λ E(x, y, z) - μ (S(x, y, z) - S_min) + ν (x + y + z - T)But I'm not sure if that's the right approach. Maybe it's better to consider that we are minimizing cost and environmental impact, and maximizing social benefit, so we can write the Lagrangian as:L = C(x, y, z) + λ E(x, y, z) - μ S(x, y, z) + ν (x + y + z - T)But I'm not entirely confident. Maybe I should think about how Lagrangian multipliers work in multi-objective optimization.Wait, in multi-objective optimization, one common approach is to use a weighted sum method, where each objective is multiplied by a weight and then combined into a single objective function. So, perhaps the Lagrangian would be:L = w1 C(x, y, z) + w2 E(x, y, z) - w3 S(x, y, z) + ν (x + y + z - T)Where w1, w2, w3 are weights reflecting the priorities of each objective. But the problem doesn't mention weights, so maybe we can assume equal weights or treat them as multipliers.Alternatively, perhaps the problem expects us to set up the Lagrangian with separate multipliers for each objective, but I'm not sure.Wait, another approach is to use the method of Lagrangian multipliers for multiple constraints, but in this case, the objectives are not constraints but separate functions to be optimized.Hmm, maybe I'm overcomplicating it. Let me try to proceed step by step.First, the problem is to minimize C and E, and maximize S, subject to x + y + z = T.So, in terms of optimization, we can think of it as:Minimize C(x, y, z)Minimize E(x, y, z)Maximize S(x, y, z)Subject to x + y + z = TSo, perhaps we can set up a Lagrangian that combines these objectives with their respective multipliers.But in standard Lagrangian multipliers, we have one objective function and multiple constraints. Here, we have multiple objectives and one constraint.So, perhaps we can use a vector optimization approach, but that might be more advanced.Alternatively, maybe we can use a scalarization method, where we combine the objectives into a single function.For example, we can write:L = C(x, y, z) + λ E(x, y, z) - μ S(x, y, z) + ν (x + y + z - T)But I'm not sure if that's the correct way to set it up. Alternatively, maybe we can write:L = C(x, y, z) + λ (E(x, y, z) - E_max) + ν (x + y + z - T) - μ (S(x, y, z) - S_min)But that might not be the right approach either.Wait, perhaps the problem is expecting us to set up the Lagrangian for the original problem without considering the penalty yet, just the initial optimization.So, let's focus on that first.We have three objectives: minimize C, minimize E, maximize S.But in the Lagrangian, we can combine these into a single function with weights. So, perhaps:L = C(x, y, z) + λ E(x, y, z) - μ S(x, y, z) + ν (x + y + z - T)Where λ and μ are the weights (or Lagrange multipliers) associated with minimizing E and maximizing S, respectively.But I'm not entirely sure. Alternatively, maybe we can treat each objective as a separate constraint, but that doesn't seem right.Wait, another thought: in multi-objective optimization, sometimes we use the concept of Pareto optimality, where we find solutions that are optimal in the sense that you can't improve one objective without worsening another. But I don't think that's directly applicable here because the problem is asking for a Lagrangian formulation.Alternatively, perhaps the problem is expecting us to set up the Lagrangian with the three objectives as separate terms, each with their own multipliers, plus the resource constraint.So, maybe:L = C(x, y, z) + λ E(x, y, z) - μ S(x, y, z) + ν (x + y + z - T)But I'm not sure if that's the standard way. Alternatively, perhaps we can write:L = C(x, y, z) + λ E(x, y, z) + μ (S(x, y, z)) + ν (x + y + z - T)But that would be treating all objectives as to be minimized, which isn't the case because S is to be maximized.So, to maximize S, we can write it as minimizing -S, so:L = C(x, y, z) + λ E(x, y, z) - μ S(x, y, z) + ν (x + y + z - T)Yes, that seems more consistent. So, we have the cost and environmental impact as terms to be minimized, and the social benefit is subtracted because we want to maximize it.So, the Lagrangian would be:L = a x² + b y² + c z² + λ (d x y + e y z + f z x) - μ (g ln x + h sqrt y + i exp z) + ν (x + y + z - T)That seems reasonable. So, that's the Lagrangian for the initial problem.Now, moving on to part 2. If the solution results in E(x, y, z) > E_max, we need to add a penalty function P(x, y, z) = k (E - E_max)^2.So, in this case, we need to incorporate this penalty into the Lagrangian. So, perhaps we can add this penalty term to the Lagrangian as another term.So, the new Lagrangian would be:L = C + λ E - μ S + ν (x + y + z - T) + PBut wait, since P is a penalty for exceeding E_max, we can write it as:P = k (E - E_max)^2 when E > E_max, else 0.But in the Lagrangian, we can include it as a term that is added when E exceeds E_max.Alternatively, perhaps we can include it as a separate term in the Lagrangian, so:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2But that would always add the penalty, even if E <= E_max. So, perhaps we need to include it only when E > E_max.But in optimization, it's often easier to include it as a continuous function, so we can write the penalty as k max(0, E - E_max)^2.But in the Lagrangian, we can represent it as:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2But this would add the penalty even when E <= E_max, which might not be desired. Alternatively, perhaps we can use an indicator function, but that complicates the Lagrangian.Alternatively, perhaps we can treat the penalty as a separate constraint, but that might not be straightforward.Wait, maybe the problem is expecting us to incorporate the penalty into the Lagrangian by adding it as a term, so the new Lagrangian becomes:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2But then, when E <= E_max, the term (E - E_max)^2 is non-negative, so it adds a penalty even when E is below the threshold. That might not be ideal because we only want to penalize when E exceeds E_max.Alternatively, perhaps we can write the penalty as k max(0, E - E_max)^2, but in the Lagrangian, we can represent it as:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2But then, when E <= E_max, the term is zero, so it doesn't affect the optimization. When E > E_max, it adds a penalty.Wait, no, because (E - E_max)^2 is always non-negative, so even when E <= E_max, it's adding a term k (E - E_max)^2, which is positive. So, that's not correct because we only want to penalize when E exceeds E_max.So, perhaps the correct way is to write the penalty as k max(0, E - E_max)^2, but in the Lagrangian, we can't directly write max functions because they are non-differentiable. So, perhaps we can use a smooth approximation or treat it as a separate constraint.Alternatively, perhaps we can include the penalty as a term that is only active when E > E_max, but in the Lagrangian, we can represent it as:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2 * I(E > E_max)Where I is an indicator function that is 1 when E > E_max and 0 otherwise. But in practice, this is difficult to handle in optimization because it's non-differentiable.Alternatively, perhaps we can use a different approach, such as adding a constraint E <= E_max and using a penalty method where we add the penalty term only when E exceeds E_max.But in the Lagrangian, we can include the penalty as a separate term, so:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2But as I thought earlier, this adds a penalty even when E <= E_max, which might not be desired. However, perhaps the problem expects us to include it this way, regardless.Alternatively, perhaps the problem is expecting us to treat the environmental constraint as a hard constraint, so E <= E_max, and use a Lagrangian with that constraint, but that's different from the penalty function.Wait, the problem says: "Given the ethical considerations, the engineer wants to ensure that the environmental impact does not exceed a threshold E_max. If the solution to the above problem results in E(x, y, z) > E_max, the engineer must adjust the allocations. Define a penalty function P(x, y, z) = k(E - E_max)^2 for exceeding the environmental threshold. Reformulate the optimization problem by incorporating this penalty function into the Lagrangian and determine the conditions under which the penalty is minimized."So, the penalty is only applied when E > E_max. So, perhaps we can write the Lagrangian as:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2 * I(E > E_max)But again, this is non-differentiable. Alternatively, perhaps we can use a different approach, such as adding the penalty term to the Lagrangian regardless, but then the conditions for the penalty being minimized would require that E <= E_max.Wait, perhaps the problem is expecting us to include the penalty term in the Lagrangian, so:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2And then, when we take the derivatives, the conditions would include the penalty term, which would effectively push E to be as small as possible, but with the penalty only becoming significant when E exceeds E_max.But I'm not sure. Alternatively, perhaps the problem is expecting us to set up the Lagrangian with the penalty term added, and then find the conditions where the penalty is minimized, which would be when E <= E_max.So, perhaps the new Lagrangian is:L = C + λ E - μ S + ν (x + y + z - T) + k (E - E_max)^2And then, when we take the partial derivatives with respect to x, y, z, and set them to zero, we can find the conditions for optimality, including the penalty term.So, let's proceed with that.So, the Lagrangian is:L = a x² + b y² + c z² + λ (d x y + e y z + f z x) - μ (g ln x + h sqrt y + i exp z) + ν (x + y + z - T) + k (d x y + e y z + f z x - E_max)^2Wait, no, the penalty function is P = k (E - E_max)^2, so E is d x y + e y z + f z x, so the penalty term is k (d x y + e y z + f z x - E_max)^2.So, the Lagrangian becomes:L = a x² + b y² + c z² + λ (d x y + e y z + f z x) - μ (g ln x + h sqrt y + i exp z) + ν (x + y + z - T) + k (d x y + e y z + f z x - E_max)^2That's the Lagrangian with the penalty function included.Now, to find the conditions under which the penalty is minimized, we need to ensure that the derivative of the Lagrangian with respect to each variable includes the penalty term, and we set the derivatives to zero.But perhaps the conditions under which the penalty is minimized would be when the derivative of the penalty term is zero, which would require that the derivative of (E - E_max)^2 with respect to each variable is zero.But since E is a function of x, y, z, the derivative of (E - E_max)^2 with respect to x is 2 (E - E_max) * dE/dx.So, setting the derivative of the Lagrangian to zero would include terms from the penalty function, which would effectively push E towards E_max or below.But perhaps the conditions for the penalty to be minimized would be when E <= E_max, so that the penalty term is zero or minimized.Alternatively, perhaps the conditions would require that the derivative of the penalty term is zero, which would imply that E = E_max, but that might not necessarily be the case.I think I need to proceed step by step.First, for the initial Lagrangian without the penalty:L = a x² + b y² + c z² + λ (d x y + e y z + f z x) - μ (g ln x + h sqrt y + i exp z) + ν (x + y + z - T)To find the optimal x, y, z, we take partial derivatives with respect to x, y, z, and set them to zero.Similarly, for the Lagrangian with the penalty:L = a x² + b y² + c z² + λ (d x y + e y z + f z x) - μ (g ln x + h sqrt y + i exp z) + ν (x + y + z - T) + k (d x y + e y z + f z x - E_max)^2We take partial derivatives with respect to x, y, z, and set them to zero.So, let's compute the partial derivatives.First, for the initial Lagrangian:Partial derivative with respect to x:dL/dx = 2 a x + λ (d y + f z) - μ (g / x) + ν + 0 (since no x in the penalty term yet) = 0Similarly, partial derivative with respect to y:dL/dy = 2 b y + λ (d x + e z) - μ (h / (2 sqrt y)) + ν + 0 = 0Partial derivative with respect to z:dL/dz = 2 c z + λ (e y + f x) - μ (i exp z) + ν + 0 = 0And the constraint:x + y + z = TSo, these are the four equations to solve for x, y, z, and ν.Now, for the Lagrangian with the penalty:Partial derivative with respect to x:dL/dx = 2 a x + λ (d y + f z) - μ (g / x) + ν + 2 k (E - E_max) (d y + f z) = 0Similarly, partial derivative with respect to y:dL/dy = 2 b y + λ (d x + e z) - μ (h / (2 sqrt y)) + ν + 2 k (E - E_max) (d x + e z) = 0Partial derivative with respect to z:dL/dz = 2 c z + λ (e y + f x) - μ (i exp z) + ν + 2 k (E - E_max) (e y + f x) = 0And the constraint:x + y + z = TSo, these are the new conditions when the penalty is included.Now, the conditions under which the penalty is minimized would be when the derivative of the penalty term is zero, which would require that the derivative of (E - E_max)^2 with respect to each variable is zero. But since E is a function of x, y, z, the derivative would be 2 (E - E_max) * dE/dx, etc.But in the partial derivatives above, we have the terms from the penalty function, which are 2 k (E - E_max) times the derivative of E with respect to each variable.So, for the penalty to be minimized, we need that the additional terms from the penalty are zero, which would require that either k = 0 (which would mean no penalty), or (E - E_max) = 0, or the derivatives of E with respect to each variable are zero.But the derivatives of E with respect to x, y, z are d y + f z, d x + e z, and e y + f x, respectively, which are unlikely to be zero unless x, y, z are zero, which is not feasible.Therefore, the only way for the penalty terms to be zero is if (E - E_max) = 0, meaning E = E_max.But wait, if E = E_max, then the penalty term is zero, so the penalty is minimized.Therefore, the conditions under which the penalty is minimized are when E = E_max, and the other optimality conditions are satisfied.So, in summary, when we include the penalty function, the optimal solution will satisfy E = E_max, and the other conditions from the partial derivatives.Therefore, the conditions under which the penalty is minimized are when the environmental impact E equals the threshold E_max, and the other optimality conditions (from the partial derivatives) are satisfied.So, putting it all together, the Lagrangian with the penalty function is as above, and the conditions are that E = E_max and the partial derivatives equal zero.I think that's the approach. So, to answer the questions:1. The Lagrangian for the initial problem is:L = a x² + b y² + c z² + λ (d x y + e y z + f z x) - μ (g ln x + h sqrt y + i exp z) + ν (x + y + z - T)2. The Lagrangian with the penalty function is:L = a x² + b y² + c z² + λ (d x y + e y z + f z x) - μ (g ln x + h sqrt y + i exp z) + ν (x + y + z - T) + k (d x y + e y z + f z x - E_max)^2And the conditions under which the penalty is minimized are when E = E_max and the partial derivatives of the Lagrangian with respect to x, y, z are zero.I think that's the solution.</think>

🧠:<think>Okay, so I have this problem about an expat who wants to invest in a stock to send money back home. The stock price is modeled by a differential equation, which is given as:[ frac{dP}{dt} = kP(t)left(1 - frac{P(t)}{M}right) ]Hmm, this looks familiar. I think it's the logistic growth model. Yeah, that's right. The logistic equation models population growth with a carrying capacity, but here it's applied to stock prices. Interesting.Alright, part 1 asks me to find the price of the stock ( P(t) ) as a function of time ( t ), given the initial price ( P(0) = 50 ), carrying capacity ( M = 200 ), and growth constant ( k = 0.1 ).I remember that the general solution to the logistic differential equation is:[ P(t) = frac{M}{1 + left(frac{M - P(0)}{P(0)}right)e^{-kMt}} ]Wait, let me double-check that. Maybe I should derive it to be sure.Starting with the differential equation:[ frac{dP}{dt} = kPleft(1 - frac{P}{M}right) ]This is a separable equation, so I can rewrite it as:[ frac{dP}{Pleft(1 - frac{P}{M}right)} = k dt ]To integrate the left side, I think I need to use partial fractions. Let me set:[ frac{1}{Pleft(1 - frac{P}{M}right)} = frac{A}{P} + frac{B}{1 - frac{P}{M}} ]Multiplying both sides by ( Pleft(1 - frac{P}{M}right) ):[ 1 = Aleft(1 - frac{P}{M}right) + BP ]Expanding:[ 1 = A - frac{A}{M}P + BP ]Grouping terms:[ 1 = A + left(B - frac{A}{M}right)P ]For this to hold for all ( P ), the coefficients must be equal on both sides. So:1. Constant term: ( A = 1 )2. Coefficient of ( P ): ( B - frac{A}{M} = 0 ) => ( B = frac{A}{M} = frac{1}{M} )So, the partial fractions decomposition is:[ frac{1}{Pleft(1 - frac{P}{M}right)} = frac{1}{P} + frac{1}{Mleft(1 - frac{P}{M}right)} ]Therefore, integrating both sides:[ int left( frac{1}{P} + frac{1}{Mleft(1 - frac{P}{M}right)} right) dP = int k dt ]Let me compute the integrals:Left side:[ int frac{1}{P} dP + int frac{1}{Mleft(1 - frac{P}{M}right)} dP ]First integral is ( ln|P| ).Second integral: Let me substitute ( u = 1 - frac{P}{M} ), so ( du = -frac{1}{M} dP ). Therefore, ( -M du = dP ). So,[ int frac{1}{M u} (-M du) = -int frac{1}{u} du = -ln|u| + C = -lnleft|1 - frac{P}{M}right| + C ]So, combining both integrals:[ ln|P| - lnleft|1 - frac{P}{M}right| = kt + C ]Simplify the left side using logarithm properties:[ lnleft|frac{P}{1 - frac{P}{M}}right| = kt + C ]Exponentiate both sides:[ frac{P}{1 - frac{P}{M}} = e^{kt + C} = e^{C} e^{kt} ]Let me denote ( e^{C} ) as another constant, say ( C' ). So,[ frac{P}{1 - frac{P}{M}} = C' e^{kt} ]Solving for ( P ):Multiply both sides by denominator:[ P = C' e^{kt} left(1 - frac{P}{M}right) ][ P = C' e^{kt} - frac{C'}{M} e^{kt} P ]Bring the term with ( P ) to the left:[ P + frac{C'}{M} e^{kt} P = C' e^{kt} ]Factor out ( P ):[ P left(1 + frac{C'}{M} e^{kt}right) = C' e^{kt} ]Solve for ( P ):[ P = frac{C' e^{kt}}{1 + frac{C'}{M} e^{kt}} ]Let me simplify this expression. Let's factor out ( e^{kt} ) in the denominator:[ P = frac{C' e^{kt}}{1 + frac{C'}{M} e^{kt}} = frac{C'}{e^{-kt} + frac{C'}{M}} ]Alternatively, to make it look neater, let me write it as:[ P(t) = frac{M}{1 + left( frac{M - P(0)}{P(0)} right) e^{-k t}} ]Wait, how did I get that? Let me see. At ( t = 0 ), ( P(0) = 50 ). Let's plug ( t = 0 ) into my expression:[ P(0) = frac{C'}{1 + frac{C'}{M}} ]So,[ 50 = frac{C'}{1 + frac{C'}{200}} ]Multiply both sides by denominator:[ 50 left(1 + frac{C'}{200}right) = C' ][ 50 + frac{50 C'}{200} = C' ]Simplify:[ 50 + frac{C'}{4} = C' ]Subtract ( frac{C'}{4} ) from both sides:[ 50 = C' - frac{C'}{4} = frac{3C'}{4} ]So,[ C' = frac{50 times 4}{3} = frac{200}{3} approx 66.6667 ]Therefore, plugging back into the expression for ( P(t) ):[ P(t) = frac{frac{200}{3}}{1 + frac{frac{200}{3}}{200} e^{-0.1 t}} ]Simplify the denominator:[ frac{frac{200}{3}}{200} = frac{1}{3} ]So,[ P(t) = frac{frac{200}{3}}{1 + frac{1}{3} e^{-0.1 t}} ]Multiply numerator and denominator by 3 to simplify:[ P(t) = frac{200}{3 + e^{-0.1 t}} ]Alternatively, factor out the negative exponent:[ P(t) = frac{200}{3 + e^{-0.1 t}} ]Wait, but I think the standard form is:[ P(t) = frac{M}{1 + left( frac{M - P(0)}{P(0)} right) e^{-kt}} ]Let me check that. Plugging in ( M = 200 ), ( P(0) = 50 ), ( k = 0.1 ):[ P(t) = frac{200}{1 + left( frac{200 - 50}{50} right) e^{-0.1 t}} ]Simplify ( frac{150}{50} = 3 ):[ P(t) = frac{200}{1 + 3 e^{-0.1 t}} ]Yes, that's the same as what I derived earlier. So, that's the solution.So, part 1 is done. The function is ( P(t) = frac{200}{1 + 3 e^{-0.1 t}} ).Moving on to part 2. The expat plans to invest 10,000 when the stock price reaches 100 and sell when it reaches 150. I need to find the time duration between when the stock is at 100 and when it's at 150.So, essentially, I need to find the time ( t_1 ) when ( P(t_1) = 100 ) and the time ( t_2 ) when ( P(t_2) = 150 ). Then, the duration is ( t_2 - t_1 ).Alternatively, since the expat is investing when the stock reaches 100, maybe he is waiting until the stock hits 100, then invests, and then waits until it hits 150 to sell. So, the time he holds the stock is from ( t_1 ) to ( t_2 ), which is ( t_2 - t_1 ).So, first, let's find ( t_1 ) such that ( P(t_1) = 100 ).Using the solution from part 1:[ 100 = frac{200}{1 + 3 e^{-0.1 t_1}} ]Solve for ( t_1 ).Multiply both sides by denominator:[ 100 (1 + 3 e^{-0.1 t_1}) = 200 ]Divide both sides by 100:[ 1 + 3 e^{-0.1 t_1} = 2 ]Subtract 1:[ 3 e^{-0.1 t_1} = 1 ]Divide by 3:[ e^{-0.1 t_1} = frac{1}{3} ]Take natural logarithm:[ -0.1 t_1 = lnleft( frac{1}{3} right) ]Simplify:[ -0.1 t_1 = -ln(3) ]Multiply both sides by -1:[ 0.1 t_1 = ln(3) ]Therefore,[ t_1 = frac{ln(3)}{0.1} = 10 ln(3) ]Calculate ( ln(3) approx 1.0986 ), so:[ t_1 approx 10 times 1.0986 = 10.986 ] years.Similarly, find ( t_2 ) when ( P(t_2) = 150 ):[ 150 = frac{200}{1 + 3 e^{-0.1 t_2}} ]Multiply both sides by denominator:[ 150 (1 + 3 e^{-0.1 t_2}) = 200 ]Divide both sides by 150:[ 1 + 3 e^{-0.1 t_2} = frac{200}{150} = frac{4}{3} ]Subtract 1:[ 3 e^{-0.1 t_2} = frac{4}{3} - 1 = frac{1}{3} ]Divide by 3:[ e^{-0.1 t_2} = frac{1}{9} ]Take natural logarithm:[ -0.1 t_2 = lnleft( frac{1}{9} right) = -ln(9) ]Multiply both sides by -1:[ 0.1 t_2 = ln(9) ]So,[ t_2 = frac{ln(9)}{0.1} = 10 ln(9) ]But ( ln(9) = ln(3^2) = 2 ln(3) ), so:[ t_2 = 10 times 2 ln(3) = 20 ln(3) approx 20 times 1.0986 = 21.972 ] years.Therefore, the time duration is ( t_2 - t_1 = 20 ln(3) - 10 ln(3) = 10 ln(3) approx 10.986 ) years.Wait, that seems a bit long. Let me verify.Alternatively, perhaps I can compute ( t_2 - t_1 ) directly without calculating each separately.From the equation:When ( P(t) = 100 ):[ 100 = frac{200}{1 + 3 e^{-0.1 t}} implies 1 + 3 e^{-0.1 t} = 2 implies 3 e^{-0.1 t} = 1 implies e^{-0.1 t} = 1/3 implies -0.1 t = ln(1/3) implies t = -10 ln(1/3) = 10 ln(3) ]Similarly, when ( P(t) = 150 ):[ 150 = frac{200}{1 + 3 e^{-0.1 t}} implies 1 + 3 e^{-0.1 t} = 200 / 150 = 4/3 implies 3 e^{-0.1 t} = 1/3 implies e^{-0.1 t} = 1/9 implies -0.1 t = ln(1/9) implies t = -10 ln(1/9) = 10 ln(9) ]So, ( t_2 - t_1 = 10 ln(9) - 10 ln(3) = 10 (ln(9) - ln(3)) = 10 ln(9/3) = 10 ln(3) approx 10.986 ) years.Hmm, so approximately 10.986 years. That's about 11 years. That seems quite a long time for a stock investment. Maybe the growth constant ( k = 0.1 ) is too low? Let me check the calculations again.Wait, ( k = 0.1 ) per year. So, the growth rate is 10% per year, but it's logistic, so it's limited by the carrying capacity.Alternatively, maybe I made a mistake in the algebra.Wait, let's re-express the solution:[ P(t) = frac{200}{1 + 3 e^{-0.1 t}} ]So, when ( P(t) = 100 ):[ 100 = frac{200}{1 + 3 e^{-0.1 t}} implies 1 + 3 e^{-0.1 t} = 2 implies 3 e^{-0.1 t} = 1 implies e^{-0.1 t} = 1/3 implies -0.1 t = ln(1/3) implies t = -10 ln(1/3) = 10 ln(3) ]Similarly, for ( P(t) = 150 ):[ 150 = frac{200}{1 + 3 e^{-0.1 t}} implies 1 + 3 e^{-0.1 t} = 200 / 150 = 4/3 implies 3 e^{-0.1 t} = 1/3 implies e^{-0.1 t} = 1/9 implies -0.1 t = ln(1/9) implies t = -10 ln(1/9) = 10 ln(9) ]So, ( t_2 - t_1 = 10 (ln(9) - ln(3)) = 10 ln(3) ) as before.So, it's correct. The time is approximately 10.986 years. So, about 11 years.Alternatively, maybe the expat doesn't need to wait for the stock to reach 100 from 50, but he is investing when it's at 100, regardless of the initial condition. Wait, the initial condition is 50, so the stock starts at 50 and grows to 100, then to 150.So, the time from 100 to 150 is 10 ln(3) years, which is about 10.986 years.Alternatively, maybe the expat is just investing when the stock is at 100, regardless of when that happens, and then sells when it's at 150. So, the time between those two points is 10 ln(3) years.Alternatively, perhaps I can compute the time it takes for the stock to go from 100 to 150 using the logistic equation.But since we have the solution, it's straightforward to compute the times when the stock is at those prices and subtract.So, I think my calculations are correct. So, the duration is ( 10 ln(3) ) years, which is approximately 10.986 years.But let me check if the model makes sense. The stock starts at 50, grows to 100 in about 10.986 years, and then to 150 in another 10.986 years? Wait, no, from 50 to 100 is 10 ln(3) years, and from 100 to 150 is another 10 ln(3) years? That seems symmetric, but in reality, logistic growth is faster in the beginning and slows down as it approaches the carrying capacity.Wait, actually, looking at the solution ( P(t) = frac{200}{1 + 3 e^{-0.1 t}} ), as ( t ) increases, the denominator decreases, so ( P(t) ) increases. The growth rate is highest when ( P(t) ) is around half of ( M ), which is 100. So, the growth is fastest around 100, which is why the time to go from 50 to 100 is the same as from 100 to 150? Wait, that seems counterintuitive.Wait, actually, the logistic curve is symmetric around the inflection point, which is at ( P = M/2 ). So, the time to go from ( M/2 - x ) to ( M/2 + x ) is the same. So, in this case, ( M = 200 ), so ( M/2 = 100 ). So, going from 50 to 100 is the same time as going from 100 to 150. That's why both intervals take the same time, 10 ln(3) years.So, that makes sense. So, the duration is indeed 10 ln(3) years.Therefore, the answer is 10 ln(3) years, which is approximately 10.986 years.But since the problem says to use the solution from part 1, I think expressing it in terms of ln(3) is acceptable, unless they want a numerical value.But let me see the question: "calculate the time duration for which he should hold the stock to achieve this price increase." It doesn't specify the form, so maybe both exact and approximate are fine.But in exams, sometimes they prefer exact forms. So, 10 ln(3) years is exact, and approximately 10.986 years.But let me compute 10 ln(3):ln(3) ≈ 1.098612289So, 10 * 1.098612289 ≈ 10.98612289 years.So, approximately 10.99 years.But maybe we can write it as 10 ln(3) years or approximately 11 years.But since the question is about calculation, probably expects a numerical answer.So, 10.986 years, which is roughly 11 years.But let me check if I can write it in a more precise fractional form.Alternatively, since ln(3) is approximately 1.0986, 10 * 1.0986 = 10.986, which is about 10 years and 11.5 months (since 0.986 years * 12 ≈ 11.83 months). So, approximately 10 years and 12 months, which is almost 11 years.But maybe the answer expects the exact expression, 10 ln(3). So, perhaps better to write both.But the question says "calculate the time duration", so maybe they expect a numerical value.Alternatively, let me see if I can express it in a different way.Wait, another approach: since the logistic function is symmetric around the inflection point, the time to go from 50 to 100 is the same as from 100 to 150, which is 10 ln(3). So, that's consistent.Therefore, the time duration is 10 ln(3) years, approximately 10.986 years.So, I think that's the answer.Final Answer1. The price of the stock as a function of time is (boxed{dfrac{200}{1 + 3e^{-0.1t}}}).2. The time duration for which the expat should hold the stock is (boxed{10 ln 3}) years, approximately (boxed{10.99}) years.</think>

🧠:<think>Okay, so I have this problem where an elected representative needs to allocate a 10 million budget among three initiatives: healthcare (H), education (E), and sustainable development (S). The goal is to maximize the impact score, which is based on how much each dollar contributes to each initiative. First, let me break down the information given. Each dollar in healthcare gives 3 points, education gives 5 points, and sustainable development gives 4 points. So, clearly, education has the highest impact per dollar, followed by sustainable development, and then healthcare. But there are some constraints here.The constraints are:1. At least 25% of the total budget must be allocated to healthcare. Since the total budget is 10 million, 25% of that is 2.5 million. So, H has to be at least 2.5 million.2. No more than 40% of the total budget can be allocated to sustainable development. 40% of 10 million is 4 million, so S has to be less than or equal to 4 million.Also, the total budget is 10 million, so H + E + S = 10 million.I need to formulate this as a linear programming model. Let me recall that linear programming involves maximizing or minimizing a linear objective function subject to linear constraints.So, the variables are H, E, and S, which are the amounts allocated to healthcare, education, and sustainable development respectively.The objective function is the impact score, which is 3H + 5E + 4S. We want to maximize this.The constraints are:1. H >= 2.5 million2. S <= 4 million3. H + E + S = 10 million4. H, E, S >= 0 (since you can't allocate negative money)Wait, actually, the last constraint is already covered by the first and the fact that H, E, S can't be negative. But just to be thorough, I should include them.So, putting it all together, the linear programming model is:Maximize Z = 3H + 5E + 4SSubject to:H >= 2.5S <= 4H + E + S = 10H, E, S >= 0But since H + E + S = 10, we can express one variable in terms of the others. Maybe express E as E = 10 - H - S. Then substitute into the objective function.So, substituting E, the objective becomes Z = 3H + 5(10 - H - S) + 4S.Let me compute that:Z = 3H + 50 - 5H - 5S + 4SZ = (3H - 5H) + (-5S + 4S) + 50Z = (-2H) + (-S) + 50So, Z = -2H - S + 50Wait, that's interesting. So, to maximize Z, which is equal to -2H - S + 50, we need to minimize 2H + S because Z is 50 minus that. So, the problem reduces to minimizing 2H + S, given the constraints.But let me think if that's correct. Because the original impact score is 3H + 5E + 4S, which is equal to -2H - S + 50. So, yes, maximizing Z is equivalent to minimizing 2H + S.But maybe it's easier to stick with the original variables and constraints.Alternatively, since we have H >= 2.5 and S <= 4, and H + E + S = 10, perhaps we can express E in terms of H and S, and then substitute into the objective.Wait, but in the objective function, E has a coefficient of 5, which is higher than both H and S. So, to maximize the impact score, we should allocate as much as possible to E, right? But we have constraints on H and S.So, let me think about the constraints.We have H >= 2.5, S <= 4, and H + E + S = 10.So, E = 10 - H - S.Given that E has the highest impact per dollar, we want to maximize E. To maximize E, we need to minimize H and S as much as possible, subject to their constraints.So, the minimal H is 2.5 million, and the minimal S is 0 (since S can be as low as 0, but it's not required to be anything else except not exceeding 4 million). So, if we set H = 2.5 and S = 0, then E would be 10 - 2.5 - 0 = 7.5 million.But wait, is S allowed to be 0? The problem only states that S cannot exceed 4 million, but there's no lower bound. So, yes, S can be 0.But let me check if that's the optimal solution.If we set H = 2.5, S = 0, then E = 7.5.Impact score would be 3*2.5 + 5*7.5 + 4*0 = 7.5 + 37.5 + 0 = 45.Is this the maximum?Alternatively, if we set S to its maximum, 4 million, then H would be 2.5 million, and E would be 10 - 2.5 - 4 = 3.5 million.Impact score would be 3*2.5 + 5*3.5 + 4*4 = 7.5 + 17.5 + 16 = 41.That's less than 45, so worse.Alternatively, what if we set S to some other value between 0 and 4, and see if that gives a higher impact score.But since E has the highest impact, we should allocate as much as possible to E. So, to maximize E, set H to its minimum and S to its minimum.But S's minimum is 0, so yes, H = 2.5, S = 0, E = 7.5.But wait, is there a constraint that S must be at least something? No, only that it can't exceed 4.So, that should be the optimal.But let me think again. Since the impact per dollar is 5 for E, which is higher than 4 for S and 3 for H, so yes, we should prioritize E.But let me consider the shadow prices or something. Wait, maybe not necessary here.Alternatively, let's set up the equations.We have:Maximize Z = 3H + 5E + 4SSubject to:H >= 2.5S <= 4H + E + S = 10H, E, S >= 0We can write this in standard form for linear programming.But since it's a small problem, maybe we can solve it graphically or by substitution.Let me express E in terms of H and S: E = 10 - H - S.Substitute into Z:Z = 3H + 5(10 - H - S) + 4SZ = 3H + 50 - 5H - 5S + 4SZ = -2H - S + 50So, to maximize Z, we need to minimize 2H + S.Given the constraints:H >= 2.5S <= 4H + S <= 10 (since E = 10 - H - S >= 0, so H + S <= 10)But H >= 2.5, S <= 4, and H + S <= 10.So, the feasible region is defined by these constraints.We need to minimize 2H + S.So, let's consider the feasible region.The variables H and S must satisfy:H >= 2.5S <= 4H + S <= 10Also, since E = 10 - H - S >= 0, so H + S <= 10.But since H >= 2.5, S <= 4, and H + S <= 10, the feasible region is a polygon.Let me plot this mentally.H ranges from 2.5 to 10 - S.But S ranges from 0 to 4.Wait, actually, H can be from 2.5 to 10 - S, but since S can be up to 4, H can be as low as 2.5 and as high as 10 - 0 = 10, but since S can't exceed 4, H can be up to 10 - 0 = 10, but in reality, H + S <= 10, so when S is 4, H can be up to 6.Wait, maybe I should find the corner points of the feasible region.The corner points occur where the constraints intersect.So, let's find the intersection points.1. Intersection of H = 2.5 and S = 4.At H = 2.5, S = 4, then E = 10 - 2.5 - 4 = 3.5.2. Intersection of H = 2.5 and H + S = 10.At H = 2.5, S = 10 - 2.5 = 7.5. But S cannot exceed 4, so this point is not feasible because S = 7.5 > 4.So, the next point is where S = 4 and H + S = 10.So, S = 4, H = 10 - 4 = 6.So, that's another corner point: H = 6, S = 4, E = 0.But wait, E = 0 is allowed? The problem doesn't specify a minimum for E, so yes.3. Intersection of H = 2.5 and S = 0.H = 2.5, S = 0, E = 7.5.4. Intersection of S = 4 and H = 2.5 is already considered.5. Intersection of S = 0 and H + S = 10 is H = 10, S = 0, but H must be >= 2.5, so that's feasible, but E would be 0.Wait, but if H = 10, S = 0, then E = 0. But H cannot exceed 10, but since H + S <= 10, H can be up to 10 if S is 0.But let's check if that's a corner point.So, the corner points are:A. H = 2.5, S = 0, E = 7.5B. H = 2.5, S = 4, E = 3.5C. H = 6, S = 4, E = 0D. H = 10, S = 0, E = 0Wait, but point D is H = 10, S = 0, which would require E = 0. But is that feasible? Yes, because H >= 2.5 is satisfied, S <= 4 is satisfied, and H + S = 10 is satisfied.But let's check if point C is feasible. H = 6, S = 4, which is within the constraints: H >= 2.5, S <= 4, and H + S = 10.Similarly, point A: H = 2.5, S = 0, E = 7.5.Point B: H = 2.5, S = 4, E = 3.5.So, these are the four corner points.Now, let's compute the value of Z at each corner point.But wait, Z is equal to -2H - S + 50, which we need to maximize. Alternatively, since Z = 3H + 5E + 4S, we can compute that as well.Let's compute Z for each point.Point A: H = 2.5, E = 7.5, S = 0.Z = 3*2.5 + 5*7.5 + 4*0 = 7.5 + 37.5 + 0 = 45.Point B: H = 2.5, E = 3.5, S = 4.Z = 3*2.5 + 5*3.5 + 4*4 = 7.5 + 17.5 + 16 = 41.Point C: H = 6, E = 0, S = 4.Z = 3*6 + 5*0 + 4*4 = 18 + 0 + 16 = 34.Point D: H = 10, E = 0, S = 0.Z = 3*10 + 5*0 + 4*0 = 30 + 0 + 0 = 30.So, the maximum Z is at point A, which is 45.Therefore, the optimal allocation is H = 2.5 million, E = 7.5 million, S = 0 million.But wait, is S = 0 allowed? The problem only states that S cannot exceed 4 million, but doesn't specify a minimum. So, yes, S can be 0.Therefore, the optimal solution is to allocate 2.5 million to healthcare, 7.5 million to education, and 0 to sustainable development.But let me double-check if there's any other point that could give a higher Z.Wait, what if we set S to something between 0 and 4, and H to something between 2.5 and 6, would that give a higher Z?But since E has the highest coefficient, we should allocate as much as possible to E, which is achieved by setting H to its minimum and S to its minimum.So, yes, point A is indeed the optimal.Alternatively, let's consider the objective function Z = -2H - S + 50.To maximize Z, we need to minimize 2H + S.Given that, the minimal value of 2H + S would be at the minimal H and minimal S.H is minimal at 2.5, S is minimal at 0.So, 2*2.5 + 0 = 5.Thus, Z = 50 - 5 = 45.Which is the same as point A.Therefore, the optimal solution is H = 2.5, E = 7.5, S = 0.So, the representative should allocate 2.5 million to healthcare, 7.5 million to education, and 0 to sustainable development to achieve the maximum impact score of 45.</think>