🧠:<think>Alright, so I have this problem about a film festival schedule. Let me try to break it down step by step. First, the film enthusiast has 50 films, none directed by Vincent Gallo. They want to create a 7-day festival, showing 3 films each day. Each day's selection must include at least one film from three different genres: Drama, Comedy, and Documentary. The genres are split as 20 Drama, 15 Comedy, and 15 Documentary.Sub-problem 1 is about figuring out how many distinct ways the enthusiast can select the films for the entire festival while meeting the genre constraints each day.Okay, so for each day, we need to choose 3 films, each from a different genre. Since there are three genres, that means one Drama, one Comedy, and one Documentary each day. So, each day's selection is one film from each genre.Therefore, for each day, the number of ways to choose the films would be the product of the number of choices in each genre. So, on day one, it would be 20 Drama * 15 Comedy * 15 Documentary. But wait, but as the festival progresses, the number of films in each genre decreases because we can't repeat films, right? So, this is a problem of selecting without replacement over 7 days.So, this becomes a problem of counting the number of ways to assign 7 films from each genre to each day, since each day requires one from each genre. But wait, hold on, each day requires one from each genre, so over 7 days, we need 7 Drama, 7 Comedy, and 7 Documentary films. But the total number of films in each genre is 20, 15, and 15 respectively. Wait, 7 days, 3 films each day, so total films shown are 21. Since each day has one from each genre, the total number of films needed per genre is 7. So, we need to choose 7 Drama, 7 Comedy, and 7 Documentary films from the available 20, 15, and 15 respectively.But hold on, 20 Drama is more than enough, but Comedy and Documentary only have 15 each, which is more than 7, so that's fine.So, the first step is to choose 7 Drama films out of 20, 7 Comedy out of 15, and 7 Documentary out of 15. Then, for each day, assign one from each genre. But actually, once we've chosen the 7 films in each genre, we need to assign them to the 7 days. So, for each genre, the number of ways to assign 7 films to 7 days is 7! (since each day gets one film). So, putting it all together, the total number of ways is:C(20,7) * C(15,7) * C(15,7) * (7!)^3Where C(n,k) is the combination of n things taken k at a time, and then for each genre, we multiply by 7! to arrange the selected films over the 7 days.Wait, let me verify that. So, first, choose 7 Drama films from 20: C(20,7). Similarly, choose 7 Comedy from 15: C(15,7), and 7 Documentary from 15: C(15,7). Then, for each genre, we have 7 films, and we need to assign each film to a specific day. Since each day needs one film from each genre, the number of ways to assign the Drama films is 7! (permuting the 7 films over the 7 days), same for Comedy and Documentary. So, the total number of ways is indeed C(20,7)*C(15,7)*C(15,7)*(7!)^3.But wait, is that correct? Because once you choose the 7 films in each genre, the assignment to days is independent across genres. So, yes, for each genre, it's 7! ways, so multiplied together, it's (7!)^3.Alternatively, another way to think about it is: for each day, you choose one Drama, one Comedy, and one Documentary. So, over 7 days, you're selecting 7 Dramas, 7 Comedies, and 7 Documentaries, each assigned to a specific day.So, the total number of ways is:C(20,7) * C(15,7) * C(15,7) * 7! * 7! * 7!Which is the same as above.Alternatively, if we think about it as a multinomial problem, but I think the above approach is correct.So, to compute this, we can write it as:Number of ways = [C(20,7) * C(15,7) * C(15,7)] * (7!)^3So, that's the formula.Sub-problem 2 is about calculating the expected total audience rating for the entire festival schedule, given that each film has a different expected audience rating, uniformly distributed between 1 and 10.So, each film's rating is a uniform random variable on [1,10]. Since there are 21 films shown (7 days * 3 films), the total audience rating is the sum of these 21 independent uniform random variables.The expected value of a single uniform random variable on [1,10] is (1 + 10)/2 = 5.5. Therefore, the expected total audience rating is 21 * 5.5.Calculating that: 21 * 5.5 = 115.5.So, the expected total audience rating is 115.5.Wait, but let me think again. Each film is shown once, and each has a different expected rating. But does the selection process affect the expectation? Since each film is equally likely to be selected, and the expectation is linear, regardless of dependencies, the expected total is just the sum of the expectations of each selected film.But actually, since we are selecting specific films, but the ratings are given as continuous random variables, each film's rating is independent of others, and uniformly distributed. So, the expected total is just the number of films times the expected rating per film.But hold on, the films are selected without replacement, but expectation is linear regardless of dependence. So, even if the films are selected without replacement, the expected total is still the sum of expectations, which is 21 * 5.5 = 115.5.So, yes, that should be correct.So, summarizing:Sub-problem 1: The number of distinct ways is C(20,7)*C(15,7)*C(15,7)*(7!)^3.Sub-problem 2: The expected total audience rating is 115.5.Final AnswerSub-problem 1: The number of distinct ways is boxed{dbinom{20}{7} dbinom{15}{7} dbinom{15}{7} times (7!)^3}.Sub-problem 2: The expected total audience rating is boxed{115.5}.</think>

🧠:<think>Okay, so I have this problem where a crime journalist is making a documentary on unsolved cases. There are two parts to this problem. Let me try to tackle them one by one.Starting with the first part: The producer wants to maximize the overall effectiveness of the documentary series. Each episode's effectiveness is given by the formula (E_i = frac{k cdot C_i}{T_i + log_2(N_i + 1)}). The goal is to allocate the total time (T_{text{total}} = 600) minutes across the 10 episodes to maximize the total effectiveness.Hmm, so I need to maximize the sum of (E_i) for all episodes. Since (k) is a constant factor, it can be factored out, so the problem reduces to maximizing (sum_{i=1}^{10} frac{C_i}{T_i + log_2(N_i + 1)}).This looks like an optimization problem where we need to distribute 600 minutes among 10 variables (T_i), each associated with their own (C_i) and (N_i). I think this is a constrained optimization problem, so maybe I can use Lagrange multipliers.Let me set up the function to maximize:( text{Maximize } sum_{i=1}^{10} frac{C_i}{T_i + log_2(N_i + 1)} )Subject to:( sum_{i=1}^{10} T_i = 600 )And (T_i geq 0) for all (i).To apply Lagrange multipliers, I can form the Lagrangian:( mathcal{L} = sum_{i=1}^{10} frac{C_i}{T_i + log_2(N_i + 1)} - lambda left( sum_{i=1}^{10} T_i - 600 right) )Taking partial derivatives with respect to each (T_i) and setting them equal to zero.The partial derivative of (mathcal{L}) with respect to (T_j) is:( frac{partial mathcal{L}}{partial T_j} = -frac{C_j}{(T_j + log_2(N_j + 1))^2} - lambda = 0 )So,( -frac{C_j}{(T_j + log_2(N_j + 1))^2} - lambda = 0 )Which simplifies to:( frac{C_j}{(T_j + log_2(N_j + 1))^2} = -lambda )Since (lambda) is a constant for all (j), this implies that:( frac{C_j}{(T_j + log_2(N_j + 1))^2} = frac{C_k}{(T_k + log_2(N_k + 1))^2} ) for all (j, k)This suggests that the ratio of (C_j) to the square of (T_j + log_2(N_j + 1)) is constant across all episodes. Let's denote this constant as (c), so:( frac{C_j}{(T_j + log_2(N_j + 1))^2} = c )Which can be rearranged to:( T_j + log_2(N_j + 1) = sqrt{frac{C_j}{c}} )Let me denote ( sqrt{frac{C_j}{c}} ) as ( D_j ), so:( T_j = D_j - log_2(N_j + 1) )But since (D_j) is related to the constant (c), which is the same across all episodes, we can write:( T_j = sqrt{frac{C_j}{c}} - log_2(N_j + 1) )But we also know that the sum of all (T_j) is 600. So,( sum_{j=1}^{10} left( sqrt{frac{C_j}{c}} - log_2(N_j + 1) right) = 600 )This simplifies to:( sum_{j=1}^{10} sqrt{frac{C_j}{c}} - sum_{j=1}^{10} log_2(N_j + 1) = 600 )Let me denote ( S = sum_{j=1}^{10} log_2(N_j + 1) ), which is a constant given the data. Then,( sum_{j=1}^{10} sqrt{frac{C_j}{c}} = 600 + S )Let me denote ( sqrt{frac{1}{c}} = d ), so:( sum_{j=1}^{10} sqrt{C_j} cdot d = 600 + S )Which means,( d = frac{600 + S}{sum_{j=1}^{10} sqrt{C_j}} )Therefore,( sqrt{frac{C_j}{c}} = d cdot sqrt{C_j} = frac{(600 + S) sqrt{C_j}}{sum_{j=1}^{10} sqrt{C_j}} )Thus,( T_j = frac{(600 + S) sqrt{C_j}}{sum_{j=1}^{10} sqrt{C_j}} - log_2(N_j + 1) )But wait, this seems a bit convoluted. Maybe I made a mistake in the substitution.Let me go back. From the partial derivatives, we have:( frac{C_j}{(T_j + log_2(N_j + 1))^2} = lambda ) for all (j)So, for each (j), (T_j + log_2(N_j + 1)) is proportional to (sqrt{C_j}), specifically:( T_j + log_2(N_j + 1) = sqrt{frac{C_j}{lambda}} )Therefore, (T_j = sqrt{frac{C_j}{lambda}} - log_2(N_j + 1))Summing over all (j):( sum_{j=1}^{10} T_j = sum_{j=1}^{10} left( sqrt{frac{C_j}{lambda}} - log_2(N_j + 1) right) = 600 )Which is:( sum_{j=1}^{10} sqrt{frac{C_j}{lambda}} - sum_{j=1}^{10} log_2(N_j + 1) = 600 )Let me denote ( sum_{j=1}^{10} log_2(N_j + 1) = S ), so:( sum_{j=1}^{10} sqrt{frac{C_j}{lambda}} = 600 + S )Let me factor out ( frac{1}{sqrt{lambda}} ):( frac{1}{sqrt{lambda}} sum_{j=1}^{10} sqrt{C_j} = 600 + S )Therefore,( sqrt{lambda} = frac{sum_{j=1}^{10} sqrt{C_j}}{600 + S} )So,( lambda = left( frac{sum_{j=1}^{10} sqrt{C_j}}{600 + S} right)^2 )Substituting back into (T_j):( T_j = sqrt{frac{C_j}{lambda}} - log_2(N_j + 1) = sqrt{frac{C_j (600 + S)^2}{(sum sqrt{C_j})^2}} - log_2(N_j + 1) )Simplifying,( T_j = frac{(600 + S) sqrt{C_j}}{sum_{j=1}^{10} sqrt{C_j}} - log_2(N_j + 1) )So, this gives the optimal (T_j) for each episode. It depends on the complexity (C_j), the number of pieces of evidence (N_j), and the total sum of (sqrt{C_j}) and (S).Therefore, the allocation of (T_i) that maximizes the total effectiveness is:( T_i = frac{(600 + S) sqrt{C_i}}{sum_{j=1}^{10} sqrt{C_j}} - log_2(N_i + 1) )Where (S = sum_{j=1}^{10} log_2(N_j + 1)).Now, moving on to the second part: For each witness, the probability (P_i) is given by (P_i = frac{1}{1 + e^{-(alpha_i + beta_i t)}}). The producer can spend up to 50 minutes interviewing each witness. We need to find the optimal (t) that maximizes (P_i).Wait, but (P_i) is a function of (t). Since the producer can spend up to 50 minutes, we need to find the (t) in [0,50] that maximizes (P_i).Looking at the function (P_i(t) = frac{1}{1 + e^{-(alpha_i + beta_i t)}}), this is a logistic function. It increases with (t) because as (t) increases, the exponent (-(alpha_i + beta_i t)) decreases, making (e^{-(alpha_i + beta_i t)}) decrease, so (P_i(t)) increases.Therefore, (P_i(t)) is an increasing function of (t). Hence, to maximize (P_i), the producer should spend the maximum allowed time, which is 50 minutes, on each witness.Wait, but is that always the case? Let me double-check.The derivative of (P_i(t)) with respect to (t) is:( P_i'(t) = frac{d}{dt} left( frac{1}{1 + e^{-(alpha_i + beta_i t)}} right) = frac{beta_i e^{-(alpha_i + beta_i t)}}{(1 + e^{-(alpha_i + beta_i t)})^2} )Since (beta_i) is a constant specific to witness (W_i). If (beta_i > 0), then (P_i'(t) > 0), meaning (P_i(t)) is increasing. If (beta_i < 0), then (P_i(t)) is decreasing. But in the context, (beta_i) is likely positive because more time spent interviewing would increase the probability of getting crucial information.Assuming (beta_i > 0), then (P_i(t)) is increasing, so the optimal (t) is 50 minutes.But wait, what if (beta_i) is negative? That would mean that more time spent decreases the probability, which doesn't make much sense in this context. So, I think it's safe to assume (beta_i > 0), so (t = 50) minutes is optimal.Therefore, the optimal interview time for each witness is 50 minutes.But let me think again. If (beta_i) is positive, the function increases with (t), so maximum at 50. If (beta_i) is negative, it would decrease with (t), so maximum at 0. But since the producer wants to maximize (P_i), and given that (beta_i) is specific to each witness, perhaps some witnesses have positive (beta_i) and some negative.But in the problem statement, it's not specified whether (beta_i) can be negative. It just says constants specific to each witness. So, maybe we need to consider both cases.However, in the context of an interview, it's more plausible that (beta_i) is positive because more time spent would lead to higher probability of getting crucial information. If (beta_i) were negative, it would imply that longer interviews decrease the probability, which might not make sense unless the witness becomes less cooperative over time, but that's speculative.Given the lack of information, I think the safe assumption is that (beta_i > 0), so the optimal (t) is 50 minutes.Alternatively, if we don't make assumptions about (beta_i), we can find the maximum of (P_i(t)) in the interval [0,50]. The function (P_i(t)) is either increasing or decreasing depending on (beta_i). If (beta_i > 0), it's increasing; if (beta_i < 0), it's decreasing. Therefore, the maximum occurs at the upper bound if (beta_i > 0) and at the lower bound if (beta_i < 0).But since the producer wants to maximize (P_i), and without knowing the sign of (beta_i), perhaps the optimal (t) is 50 minutes if (beta_i > 0) and 0 if (beta_i < 0). However, since the producer can choose to spend time, they would likely only interview witnesses where (beta_i > 0), otherwise, they wouldn't spend any time.But the problem says "for each unsolved case, the producer has identified a set of witnesses", so perhaps each witness has their own (alpha_i) and (beta_i). Therefore, for each witness, if (beta_i > 0), spend 50 minutes; if (beta_i < 0), spend 0 minutes.But the problem says "the producer can spend a maximum of 50 minutes interviewing each witness", so it's per witness, not per case. So, for each witness, decide whether to spend 50 minutes or 0 minutes based on (beta_i).But without knowing (beta_i), we can't determine that. However, the problem asks to determine the optimal (t) for each witness, given that the maximum is 50 minutes. So, assuming (beta_i > 0), which is reasonable, the optimal (t) is 50 minutes.Alternatively, if we consider the mathematical maximum, regardless of context, the function (P_i(t)) is increasing if (beta_i > 0), so maximum at 50; decreasing if (beta_i < 0), so maximum at 0.Therefore, the optimal (t) is:- 50 minutes if (beta_i > 0)- 0 minutes if (beta_i < 0)But since the problem doesn't specify the sign of (beta_i), perhaps the answer is 50 minutes, assuming (beta_i > 0).Alternatively, if we consider that the producer wants to maximize the probability, and without knowing (beta_i), they might have to decide based on that. But since the problem is asking to determine the optimal (t), given that the maximum is 50, and without more info, I think the answer is 50 minutes.Wait, but let's think about the function. The logistic function approaches 1 as (t) increases if (beta_i > 0). So, the closer to 1, the higher the probability. Therefore, to maximize (P_i), set (t) as high as possible, which is 50 minutes.Yes, that makes sense.So, summarizing:1. The optimal allocation of (T_i) is given by ( T_i = frac{(600 + S) sqrt{C_i}}{sum_{j=1}^{10} sqrt{C_j}} - log_2(N_i + 1) ), where (S = sum_{j=1}^{10} log_2(N_j + 1)).2. The optimal interview time (t) for each witness is 50 minutes, assuming (beta_i > 0).But wait, in the first part, the formula for (T_i) might result in negative values if (frac{(600 + S) sqrt{C_i}}{sum sqrt{C_j}} < log_2(N_i + 1)). Since time can't be negative, we need to ensure that (T_i geq 0). Therefore, the actual allocation should be the maximum between the calculated (T_i) and 0.So, ( T_i = maxleft( frac{(600 + S) sqrt{C_i}}{sum_{j=1}^{10} sqrt{C_j}} - log_2(N_i + 1), 0 right) )But since the problem states that the total available time is 600 minutes, and we have 10 episodes, it's likely that the calculated (T_i) will be positive, but it's good to include the max function to ensure non-negative time.Therefore, the final answers are:1. Allocate (T_i) as ( T_i = frac{(600 + S) sqrt{C_i}}{sum_{j=1}^{10} sqrt{C_j}} - log_2(N_i + 1) ), ensuring (T_i geq 0).2. The optimal interview time is 50 minutes for each witness, assuming (beta_i > 0).</think>

🧠:<think>Alright, so I've got this problem about a cosmetics company trying to launch a new line of sustainable beauty products. They have two main components to consider: sourcing fair trade ingredients and ensuring cruelty-free testing. The problem is split into two parts, and I need to tackle both.Starting with part 1: They need to maximize the total weight of ingredients A, B, and C under a budget constraint of 150,000. Let me jot down the details.Ingredient A is sourced from a cooperative that charges x per kilogram with a minimum purchase of 1000 kg. So, the cost for A is straightforward: if they buy 'a' kg, the cost is x*a, but a must be at least 1000.Ingredient B costs y per kilogram but has a 10% discount if more than 2000 kg are purchased. So, if they buy 'b' kg, the cost is y*b if b <= 2000, and 0.9*y*b if b > 2000.Ingredient C costs z per kilogram with no discount, but for every 500 kg purchased, they get 50 kg free. So, for every 500 kg, they get 50 free, meaning for every 550 kg, they pay for 500. So, the effective cost per kg is (z*500)/550. But maybe it's easier to model it as for every 500 kg purchased, they get 50 free. So, if they buy 'c' kg, the total weight they get is c + (c // 500)*50. Hmm, but since we're trying to maximize the total weight, maybe we can express c in terms of the amount they pay for. Let me think.Wait, actually, the company is purchasing c kg, but for every 500 kg, they get an extra 50 kg. So, the total weight they receive is c + floor(c / 500)*50. But since we're dealing with optimization, we might need to model this without the floor function. Alternatively, we can express the total weight as c + (c / 500)*50, but that would be c + 0.1c = 1.1c. However, this is only accurate if c is a multiple of 500. Otherwise, it's a bit more complicated. Maybe to simplify, we can consider that for every 500 kg purchased, they get 50 free, so the total weight is c + (c / 500)*50, but since it's per 500, it's actually c + 50*(c / 500) = c + 0.1c = 1.1c. But this assumes that c is a multiple of 500. If c isn't, then the free kg would be less. Hmm, this complicates things.Alternatively, maybe we can model the total weight as c + 50*(c // 500). But in optimization, dealing with integer divisions can be tricky. Perhaps, for the sake of simplicity, we can approximate it as 1.1c, knowing that it's an upper bound, or maybe use a piecewise function. But since the problem is about maximizing total weight, perhaps we can express the total weight as c + 50*(c / 500) = 1.1c, and then note that this is an approximation. Alternatively, maybe we can set up the problem with variables for the number of 500 kg increments.But perhaps it's better to model it as: for every 500 kg purchased, you get 50 kg free. So, if you buy c kg, the total weight is c + floor(c / 500)*50. To avoid dealing with floor functions, maybe we can let c be a multiple of 500. But that might restrict the solution space. Alternatively, we can use a variable t, which is the number of 500 kg increments. So, c = 500*t + r, where r is the remainder, 0 <= r < 500. Then, the total weight is c + 50*t = 500*t + r + 50*t = 550*t + r. But this complicates the model with integer variables, which might not be ideal for a linear programming problem.Given that, maybe it's acceptable to approximate the total weight as 1.1c, understanding that it's an upper bound, and then adjust later if necessary. Alternatively, since the problem is about maximizing total weight, perhaps the optimal solution will have c as a multiple of 500, making the approximation exact. I'll proceed with that assumption for now.So, total weight from C is 1.1c.Now, the total cost is x*a + y*b + z*c, subject to x*a + y*b + z*c <= 150,000.But wait, for ingredient B, if b > 2000, the cost is 0.9*y*b. So, the cost function for B is piecewise linear. Similarly, for C, the cost is z*c, but the total weight is 1.1c.So, the total weight to maximize is a + b + 1.1c.Subject to:x*a + (if b <= 2000 then y*b else 0.9*y*b) + z*c <= 150,000a >= 1000b >= 0c >= 0But since this is an optimization problem, we need to handle the piecewise cost for B. One way to handle this is to split it into two cases: b <= 2000 and b > 2000, and solve both cases, then choose the better solution.Alternatively, we can model it with a binary variable, but that would make it a mixed-integer linear program. Since the problem doesn't specify, I'll assume we can handle it as two separate cases.So, first, let's consider case 1: b <= 2000.Then, the cost for B is y*b.Total cost: x*a + y*b + z*c <= 150,000Total weight: a + b + 1.1cWe need to maximize this.But we also have a >= 1000.So, variables: a >= 1000, b >= 0, c >= 0.Similarly, case 2: b > 2000.Then, the cost for B is 0.9*y*b.Total cost: x*a + 0.9*y*b + z*c <= 150,000Total weight: a + b + 1.1cAgain, maximize this.So, we can solve both cases and see which gives a higher total weight.But before that, let's think about the objective function. We need to maximize a + b + 1.1c, subject to x*a + y*b + z*c <= 150,000 (for case 1) or x*a + 0.9*y*b + z*c <= 150,000 (for case 2), and a >= 1000, b >=0, c >=0.This is a linear programming problem in both cases.But since we don't have specific values for x, y, z, we need to express the solution in terms of these variables.Wait, actually, the problem says "formulate and solve an optimization problem". So, perhaps we need to set up the problem and express the solution in terms of x, y, z.But without specific values, we can't compute numerical quantities. Hmm, maybe I'm misunderstanding. Perhaps the problem expects us to set up the model, not necessarily solve it numerically.Wait, the problem says "formulate and solve an optimization problem". So, perhaps we need to set up the model, but since it's a general case, we can't solve it numerically without more information. Alternatively, maybe the problem expects us to express the optimal quantities in terms of x, y, z.Alternatively, perhaps the problem is expecting us to recognize that to maximize total weight, we should allocate as much as possible to the ingredient with the lowest cost per kg, considering the discounts and freebies.Wait, let's think about the cost per kg for each ingredient, considering their discounts and freebies.For ingredient A: cost per kg is x.For ingredient B: if b <= 2000, cost per kg is y. If b > 2000, cost per kg is 0.9y.For ingredient C: the effective cost per kg is z / 1.1, because for every kg you get, you paid for 1/1.1 kg. Wait, no. Wait, for every 500 kg purchased, you get 50 free. So, for 500 kg paid, you get 550 kg total. So, the cost per kg is (z*500)/550 = (10/11)z ≈ 0.909z.So, the effective cost per kg for C is (10/11)z.So, comparing the cost per kg:A: xB: y (if <=2000) or 0.9y (if >2000)C: (10/11)z ≈ 0.909zSo, to maximize total weight, we should prioritize buying the ingredient with the lowest cost per kg.So, the order would be:1. If 0.9y < (10/11)z and 0.9y < x, then buy as much B as possible beyond 2000 kg.2. If (10/11)z is the lowest, then buy as much C as possible.3. Then A, but A has a minimum purchase of 1000 kg.But we also have to consider the budget.Wait, but without knowing the relationship between x, y, z, it's hard to say. So, perhaps the optimal solution is to buy as much as possible of the cheapest ingredient, then the next, etc., while respecting the minimum purchase for A.But since A has a minimum purchase of 1000 kg, we have to buy at least 1000 kg of A.So, let's structure the problem.Let’s denote:a = kg of A purchasedb = kg of B purchasedc = kg of C purchasedTotal cost:If b <= 2000: x*a + y*b + z*c <= 150,000If b > 2000: x*a + 0.9y*b + z*c <= 150,000Total weight: a + b + 1.1cWe need to maximize this.Constraints:a >= 1000b >= 0c >= 0So, to set up the linear program, we can consider both cases.Case 1: b <= 2000Maximize: a + b + 1.1cSubject to:x*a + y*b + z*c <= 150,000a >= 1000b <= 2000b >= 0c >= 0Case 2: b > 2000Maximize: a + b + 1.1cSubject to:x*a + 0.9y*b + z*c <= 150,000a >= 1000b >= 2000c >= 0We can solve both cases and choose the one with the higher total weight.But without specific values for x, y, z, we can't compute the exact quantities. However, we can express the solution in terms of these variables.Alternatively, perhaps the problem expects us to set up the model, not necessarily solve it numerically. So, perhaps the answer is to set up the linear program as above, considering both cases for B.But the problem says "formulate and solve", so maybe we need to proceed further.Wait, perhaps the problem expects us to consider that the optimal solution will have the company buying as much as possible of the cheapest ingredient, considering the discounts and freebies, while respecting the minimum purchase for A.So, let's assume that the company will buy the minimum required for A (1000 kg), and then allocate the remaining budget to the other ingredients.But let's see:Total budget: 150,000Cost for A: x*1000Remaining budget: 150,000 - 1000xNow, with the remaining budget, we can buy B and C.But we need to decide whether to buy B at y or 0.9y, depending on quantity.Alternatively, perhaps the optimal strategy is to buy as much as possible of the ingredient with the lowest effective cost per kg.So, let's compute the effective cost per kg for each ingredient:A: xB: y (if <=2000) or 0.9y (if >2000)C: (10/11)zSo, if 0.9y < (10/11)z and 0.9y < x, then buy as much B as possible beyond 2000 kg.If (10/11)z is the lowest, buy as much C as possible.If x is the lowest, buy more A, but A has a minimum, so we already bought 1000 kg.But since we don't know the relationships, perhaps we can express the solution in terms of these variables.Alternatively, perhaps the problem expects us to set up the model with variables and constraints, and then express the optimal solution in terms of x, y, z.But since the problem is about formulating and solving, perhaps we need to proceed with setting up the linear program.So, for case 1 (b <= 2000):Maximize: a + b + 1.1cSubject to:x*a + y*b + z*c <= 150,000a >= 1000b <= 2000a, b, c >= 0For case 2 (b > 2000):Maximize: a + b + 1.1cSubject to:x*a + 0.9y*b + z*c <= 150,000a >= 1000b >= 2000a, b, c >= 0Now, to solve these, we can use the simplex method or other LP techniques, but without specific values, we can't compute the exact quantities. However, we can express the optimal solution in terms of x, y, z.Alternatively, perhaps the problem expects us to recognize that the optimal solution will have the company buying the minimum required for A, and then allocating the remaining budget to the cheapest ingredient, considering the discounts.So, let's proceed step by step.First, buy the minimum of A: 1000 kg, costing 1000x.Remaining budget: 150,000 - 1000x.Now, with this remaining budget, we can buy B and C.We need to decide whether to buy B at y or 0.9y, depending on quantity.But since buying more than 2000 kg of B gives a discount, we might want to consider whether buying beyond 2000 kg is beneficial.So, let's compare the effective cost per kg for B beyond 2000 kg: 0.9y vs. the effective cost for C: (10/11)z.If 0.9y < (10/11)z, then it's cheaper to buy B beyond 2000 kg.Otherwise, buy C.Similarly, compare with A's cost x.But since we've already bought the minimum A, unless x is cheaper than both, we wouldn't buy more A.But let's assume x is higher than both 0.9y and (10/11)z, which is likely since A has a minimum purchase.So, the remaining budget will be allocated to B and C, prioritizing the cheaper one.So, let's assume that 0.9y < (10/11)z, so we buy as much B as possible beyond 2000 kg.So, total cost for B beyond 2000 kg: 0.9y*bBut we need to buy at least 2000 kg to get the discount, so let's first buy 2000 kg of B at y per kg, costing 2000y.Wait, no. Wait, if we buy more than 2000 kg, the entire amount gets the 10% discount. So, if we buy b kg, where b > 2000, the cost is 0.9y*b.So, perhaps it's better to buy as much as possible of B beyond 2000 kg if 0.9y is cheaper than C's effective cost.So, let's structure it:After buying 1000 kg of A, remaining budget: 150,000 - 1000x.Now, we can buy B and C.If 0.9y < (10/11)z, then buy as much B as possible beyond 2000 kg.So, the cost for B would be 0.9y*b.But we need to buy at least 2000 kg to get the discount. Wait, no. The discount applies if more than 2000 kg are purchased. So, if we buy exactly 2000 kg, we don't get the discount. Only if we buy more than 2000 kg.So, to get the discount, we need to buy b > 2000 kg.So, the cost for B is 0.9y*b for b > 2000.So, perhaps we can model it as:If we decide to buy B beyond 2000 kg, the cost is 0.9y*b, and we can buy as much as possible.Alternatively, if buying B beyond 2000 kg is cheaper than C, we should buy as much B as possible, then use remaining budget for C.Similarly, if C is cheaper, buy as much C as possible.But let's formalize this.Let’s denote:After buying 1000 kg of A, remaining budget: B = 150,000 - 1000x.Now, we can buy B and C.Let’s compare the effective cost per kg:- B beyond 2000 kg: 0.9y- C: (10/11)zSo, if 0.9y < (10/11)z, buy as much B as possible beyond 2000 kg.Otherwise, buy as much C as possible.But we also have to consider that buying B beyond 2000 kg requires that we buy at least 2000 kg to get the discount. Wait, no, the discount applies if more than 2000 kg are purchased. So, if we buy 2001 kg, we get the discount on all 2001 kg. So, it's not that we have to buy exactly 2000 kg to get the discount; rather, any amount over 2000 kg gets the discount.So, the cost for B is:If b <= 2000: y*bIf b > 2000: 0.9y*bSo, to decide whether to buy B beyond 2000 kg, we need to see if 0.9y is cheaper than C's effective cost.If yes, then we should buy as much B as possible beyond 2000 kg, using the remaining budget.If not, buy as much C as possible.But we also have to consider that buying B beyond 2000 kg might allow us to get more total weight, but we have to pay the higher cost for the first 2000 kg.Wait, no. The discount applies to the entire purchase if it's over 2000 kg. So, if we buy b kg where b > 2000, the entire b kg is at 0.9y per kg.So, it's better to buy as much B as possible beyond 2000 kg if 0.9y is cheaper than C's effective cost.So, let's proceed.Case 1: 0.9y < (10/11)zThen, we should buy as much B as possible beyond 2000 kg.So, the cost for B is 0.9y*b.We have remaining budget B = 150,000 - 1000x.So, the maximum b we can buy is b = (B) / (0.9y).But we need to ensure that b > 2000.So, if (B) / (0.9y) > 2000, then we can buy b = (B) / (0.9y).Otherwise, we can't buy enough to get the discount, so we have to buy b <= 2000 kg at y per kg.Wait, but if B is the remaining budget after buying A, which is 150,000 - 1000x, and if 0.9y is cheaper than C's effective cost, we should try to buy as much B as possible beyond 2000 kg.So, let's compute the maximum b we can buy:b_max = (150,000 - 1000x) / (0.9y)If b_max > 2000, then we buy b = b_max.Otherwise, we can't buy enough to get the discount, so we buy b = (150,000 - 1000x)/y, but since b_max < 2000, we can't get the discount, so we have to buy b at y per kg.But wait, if b_max < 2000, then buying b at y per kg is the only option, but since 0.9y is cheaper than y, but we can't buy enough to get the discount, so we have to buy at y per kg.But in this case, since 0.9y < (10/11)z, but we can't buy enough B to get the discount, then we have to buy B at y per kg, but since y > 0.9y, it's more expensive than if we could buy more. But since we can't, we have to buy as much as possible at y per kg, and then use remaining budget for C.Wait, this is getting complicated. Maybe it's better to structure it as:If 0.9y < (10/11)z, then:- Try to buy as much B as possible beyond 2000 kg.- If the remaining budget allows buying more than 2000 kg of B at 0.9y, then buy b = (150,000 - 1000x)/0.9y.- If not, buy b = (150,000 - 1000x)/y, which will be <=2000 kg, and then buy C with the remaining budget.But since we don't know the values, we can't compute exact quantities.Alternatively, perhaps the problem expects us to set up the model without considering the piecewise cost for B, but that seems unlikely.Alternatively, maybe the problem expects us to assume that the discount applies only if more than 2000 kg are purchased, so we can model it as a binary variable, but that complicates things.Alternatively, perhaps the problem expects us to recognize that to maximize total weight, we should buy as much as possible of the cheapest ingredient, considering discounts and freebies, and then allocate the remaining budget to the next cheapest.So, let's proceed with that approach.First, buy the minimum of A: 1000 kg, costing 1000x.Remaining budget: 150,000 - 1000x.Now, compare the effective cost per kg for B and C.If 0.9y < (10/11)z, then buy as much B as possible beyond 2000 kg.So, the cost for B is 0.9y*b.The maximum b we can buy is b = (150,000 - 1000x)/0.9y.If this b > 2000, then we buy b kg of B.Otherwise, we can't get the discount, so we buy b = (150,000 - 1000x)/y kg of B, which will be <=2000 kg, and then buy C with the remaining budget.Similarly, if (10/11)z < 0.9y, then buy as much C as possible.So, the total weight will be 1000 + b + 1.1c.But without specific values, we can't compute the exact quantities.Alternatively, perhaps the problem expects us to set up the model with variables and constraints, and then express the optimal solution in terms of x, y, z.So, perhaps the answer is to set up the linear program as follows:Case 1: b <= 2000Maximize: a + b + 1.1cSubject to:x*a + y*b + z*c <= 150,000a >= 1000b <= 2000a, b, c >= 0Case 2: b > 2000Maximize: a + b + 1.1cSubject to:x*a + 0.9y*b + z*c <= 150,000a >= 1000b >= 2000a, b, c >= 0Then, solve both cases and choose the one with the higher total weight.But since we can't solve it numerically, we can express the optimal solution in terms of x, y, z.Alternatively, perhaps the problem expects us to recognize that the optimal solution will have the company buying the minimum required for A, and then allocating the remaining budget to the cheapest ingredient, considering the discounts and freebies.So, in summary, the optimal quantities depend on the relative costs of the ingredients, considering discounts and freebies. The company should buy the minimum required for A, then allocate the remaining budget to the ingredient with the lowest effective cost per kg, which could be B beyond 2000 kg or C, depending on the values of y and z.Now, moving on to part 2: The cruelty-free testing process requires that the total weight of ingredients (A, B, and C combined) must not exceed 5000 kg. Additionally, the weight of ingredient A must be at least 30% of the total weight, and ingredient B must be no more than 40% of the total weight.So, we have additional constraints:Total weight: a + b + 1.1c <= 5000A >= 0.3*(a + b + 1.1c)B <= 0.4*(a + b + 1.1c)And we still have the budget constraint: x*a + y*b + z*c <= 150,000 (with the discount for B if applicable)And the minimum purchase for A: a >= 1000So, now we have to maximize the total weight (a + b + 1.1c) subject to these constraints.But again, without specific values for x, y, z, we can't compute exact quantities. However, we can set up the model.Let’s denote:Total weight: W = a + b + 1.1cConstraints:1. W <= 50002. a >= 0.3W3. b <= 0.4W4. x*a + y*b + z*c <= 150,000 (with the discount for B if applicable)5. a >= 10006. a, b, c >= 0But since W is a function of a, b, c, we can substitute W in the constraints.So, constraint 2: a >= 0.3(a + b + 1.1c)Which simplifies to:a >= 0.3a + 0.3b + 0.33c0.7a >= 0.3b + 0.33cSimilarly, constraint 3: b <= 0.4(a + b + 1.1c)Which simplifies to:b <= 0.4a + 0.4b + 0.44c0 <= 0.4a - 0.6b + 0.44cOr:0.4a + 0.44c >= 0.6bWhich can be rewritten as:0.4a + 0.44c - 0.6b >= 0So, now we have:Maximize W = a + b + 1.1cSubject to:1. a + b + 1.1c <= 50002. 0.7a - 0.3b - 0.33c >= 03. 0.4a - 0.6b + 0.44c >= 04. x*a + y*b + z*c <= 150,000 (with discount for B if applicable)5. a >= 10006. a, b, c >= 0This is a more complex linear program, but again, without specific values for x, y, z, we can't solve it numerically.However, we can note that the optimal solution must satisfy all these constraints, and the total weight W will be as large as possible, up to 5000 kg, but also subject to the budget constraint.So, the company needs to balance between buying enough A to meet the 30% requirement, not buying too much B to exceed the 40% limit, and staying within the budget.Given that, perhaps the optimal solution will have W = 5000 kg, with a = 0.3*5000 = 1500 kg, and b <= 0.4*5000 = 2000 kg.But we also have to consider the budget.So, let's assume that the company wants to maximize W to 5000 kg.Then, a = 1500 kg (minimum required for A is 1000, but to meet 30%, it's 1500).Then, b <= 2000 kg.Then, c = (5000 - a - b)/1.1But since W = a + b + 1.1c = 5000, we have c = (5000 - a - b)/1.1But we also have the budget constraint: x*a + y*b + z*c <= 150,000Substituting a = 1500, b = 2000 (maximum allowed), then c = (5000 - 1500 - 2000)/1.1 = 1500/1.1 ≈ 1363.64 kgSo, total cost: x*1500 + y*2000 + z*1363.64 <= 150,000But we don't know x, y, z, so we can't verify if this is within budget.Alternatively, perhaps the company can't reach 5000 kg due to budget constraints, so W will be less than 5000.But again, without specific values, we can't compute exact quantities.In conclusion, the optimal quantities depend on the relative costs of the ingredients and the budget. The company should prioritize buying the cheapest ingredients while respecting the constraints on total weight, A's minimum, and B's maximum percentage.But since the problem asks to "determine the optimal quantities", perhaps the answer is to set up the model as above, considering both the budget and testing constraints.So, summarizing:For part 1, the company should buy the minimum required for A (1000 kg), then allocate the remaining budget to the ingredient with the lowest effective cost per kg, which could be B beyond 2000 kg or C, depending on the values of y and z.For part 2, the company must also ensure that the total weight does not exceed 5000 kg, A is at least 30%, and B is no more than 40%. This adds additional constraints to the model, potentially reducing the total weight from part 1 if the budget is too tight.But without specific values for x, y, z, we can't compute exact quantities. Therefore, the answer is to set up the linear programs as described, considering the constraints and optimizing for total weight.</think>

🧠:<think>Okay, so I have this problem where a teenage vlogger is analyzing the popularity of Aubree Riley's latest album by looking at streaming data and concert attendance. There are two parts to this problem: one about streaming data and another about concert attendance. Let me try to tackle each part step by step.Starting with the first part: Streaming Data Analysis. The function given is ( S(t) = 5000e^{0.05t} ), where ( S(t) ) is the number of streams on day ( t ). The task is to calculate the total number of streams in the first 30 days after the album’s release.Hmm, so this is a continuous growth model, right? It's an exponential function, which makes sense for streams because often things like music streams can grow exponentially, especially if the album is gaining popularity. So, to find the total streams over 30 days, I think I need to integrate this function from day 0 to day 30. That should give me the area under the curve, which in this context would be the total streams.Let me recall how to integrate an exponential function. The integral of ( e^{kt} ) with respect to ( t ) is ( frac{1}{k}e^{kt} + C ), where ( C ) is the constant of integration. So, applying that to ( S(t) ), the integral should be straightforward.So, let's set up the integral:Total streams ( = int_{0}^{30} 5000e^{0.05t} dt )Let me factor out the 5000:( = 5000 int_{0}^{30} e^{0.05t} dt )Now, let me compute the integral. Let ( u = 0.05t ), so ( du = 0.05 dt ), which means ( dt = frac{du}{0.05} ). But maybe it's easier to just apply the formula directly.The integral of ( e^{0.05t} ) is ( frac{1}{0.05}e^{0.05t} ), so plugging in the limits:( 5000 times left[ frac{1}{0.05}e^{0.05t} right]_0^{30} )Simplify ( frac{1}{0.05} ) which is 20, so:( 5000 times 20 times left[ e^{0.05 times 30} - e^{0} right] )Calculating ( 0.05 times 30 = 1.5 ), so:( 5000 times 20 times (e^{1.5} - 1) )Now, compute ( e^{1.5} ). I remember that ( e ) is approximately 2.71828, so ( e^{1} ) is about 2.71828, ( e^{0.5} ) is about 1.64872. So, ( e^{1.5} ) is ( e^{1} times e^{0.5} ) which is approximately 2.71828 * 1.64872. Let me calculate that:2.71828 * 1.64872 ≈ Let's see, 2 * 1.64872 = 3.29744, 0.71828 * 1.64872 ≈ approximately 1.183. So adding them together, 3.29744 + 1.183 ≈ 4.48044. So, ( e^{1.5} ) is approximately 4.4817 (I think the exact value is around 4.4816890703).So, ( e^{1.5} - 1 ≈ 4.4817 - 1 = 3.4817 ).Now, plugging back into the equation:5000 * 20 * 3.4817First, 5000 * 20 = 100,000.Then, 100,000 * 3.4817 = 348,170.So, approximately 348,170 streams over the first 30 days.Wait, let me double-check my calculations because 5000 * 20 is 100,000, that's correct. Then 100,000 * 3.4817 is indeed 348,170. So, that seems right.But just to make sure, maybe I can use a calculator for ( e^{1.5} ). Let me recall that ( e^{1.5} ) is approximately 4.4816890703. So, subtracting 1 gives 3.4816890703.So, 5000 * 20 = 100,000, then 100,000 * 3.4816890703 ≈ 348,168.90703. So, approximately 348,169 streams. So, rounding to the nearest whole number, it's 348,169.But since the question didn't specify rounding, maybe I should present it as 348,169 or perhaps keep it in terms of exact exponentials? Wait, the question says "calculate the total number of streams", so it's expecting a numerical value. So, 348,169 is the approximate total.Alternatively, maybe I can write it as ( 100,000(e^{1.5} - 1) ), but since they asked for the total, it's better to compute the numerical value.So, I think 348,169 is the answer for the first part.Moving on to the second part: Concert Attendance Analysis. The function given is ( A(x) = 2000 + 1500 sinleft(frac{pi x}{6}right) ), where ( A(x) ) is the attendance on the ( x )-th concert. We need to calculate the average attendance over the first 12 concerts.Alright, average attendance over the first 12 concerts. So, average is typically the sum divided by the number of terms. So, I can think of two approaches: either compute the sum of ( A(x) ) from x=1 to x=12 and then divide by 12, or use the properties of the sine function to find the average.Let me first recall that the average value of a sinusoidal function over a full period is equal to its vertical shift. Because the sine function oscillates symmetrically around its midline, so the average over a full period is just the midline.Looking at the function ( A(x) = 2000 + 1500 sinleft(frac{pi x}{6}right) ). The midline is 2000, and the amplitude is 1500. The period of the sine function is ( frac{2pi}{pi/6} = 12 ). So, the period is 12 concerts. That means over the first 12 concerts, we complete exactly one full period of the sine wave.Therefore, the average attendance over one full period should just be the midline, which is 2000. So, the average attendance is 2000.But let me verify that by actually computing the sum and dividing by 12, just to make sure.So, the sum ( sum_{x=1}^{12} A(x) = sum_{x=1}^{12} [2000 + 1500 sin(frac{pi x}{6})] )This can be split into two sums:( sum_{x=1}^{12} 2000 + sum_{x=1}^{12} 1500 sinleft(frac{pi x}{6}right) )The first sum is straightforward: 2000 added 12 times, so 2000 * 12 = 24,000.The second sum is 1500 times the sum of ( sinleft(frac{pi x}{6}right) ) from x=1 to 12.Let me compute ( sum_{x=1}^{12} sinleft(frac{pi x}{6}right) ).Let me note that ( frac{pi x}{6} ) for x from 1 to 12 gives angles from ( pi/6 ) to ( 2pi ) in increments of ( pi/6 ). So, it's a full circle, 12 points around the unit circle.Now, the sine function is symmetric, and over a full period, the sum of sine functions at equally spaced intervals is zero. Because for every positive sine value, there is a corresponding negative value that cancels it out.Alternatively, I can compute each term:x=1: ( sin(pi/6) = 0.5 )x=2: ( sin(2pi/6) = sin(pi/3) ≈ 0.8660 )x=3: ( sin(3pi/6) = sin(pi/2) = 1 )x=4: ( sin(4pi/6) = sin(2pi/3) ≈ 0.8660 )x=5: ( sin(5pi/6) = 0.5 )x=6: ( sin(6pi/6) = sin(pi) = 0 )x=7: ( sin(7pi/6) = -0.5 )x=8: ( sin(8pi/6) = sin(4pi/3) ≈ -0.8660 )x=9: ( sin(9pi/6) = sin(3pi/2) = -1 )x=10: ( sin(10pi/6) = sin(5pi/3) ≈ -0.8660 )x=11: ( sin(11pi/6) = -0.5 )x=12: ( sin(12pi/6) = sin(2pi) = 0 )Now, let's list all these sine values:0.5, 0.8660, 1, 0.8660, 0.5, 0, -0.5, -0.8660, -1, -0.8660, -0.5, 0Now, let's add them up:Start adding term by term:0.5 + 0.8660 = 1.36601.3660 + 1 = 2.36602.3660 + 0.8660 = 3.23203.2320 + 0.5 = 3.73203.7320 + 0 = 3.73203.7320 + (-0.5) = 3.23203.2320 + (-0.8660) = 2.36602.3660 + (-1) = 1.36601.3660 + (-0.8660) = 0.50.5 + (-0.5) = 00 + 0 = 0So, the total sum is 0. Therefore, the second sum is 1500 * 0 = 0.Therefore, the total sum of attendances is 24,000 + 0 = 24,000.Thus, the average attendance is 24,000 / 12 = 2000.So, that confirms it. The average attendance over the first 12 concerts is 2000.Wait, just to make sure, let me think about why the sum of sine over a full period is zero. It's because the sine function is symmetric and oscillates equally above and below the x-axis. So, over a full period, the positive and negative areas cancel out, resulting in an average of zero. Therefore, when you add up all the sine terms over a full period, they indeed sum to zero.Therefore, the average attendance is just the constant term, which is 2000.So, both methods confirm that the average attendance is 2000.So, summarizing:1. Total streams in the first 30 days: approximately 348,169.2. Average attendance over the first 12 concerts: 2000.I think that's it. I don't see any mistakes in my calculations, but let me just quickly recap.For the first part, integrating the exponential function over 30 days, I correctly applied the integral formula, computed the exponential term, and multiplied through. The result was approximately 348,169 streams.For the second part, recognizing that the sine function over a full period averages out to zero, so the average attendance is just the midline, which is 2000. I also verified this by computing the sum of each sine term, which indeed summed to zero, confirming that the average is 2000.So, I feel confident with these answers.Final Answer1. The total number of streams in the first 30 days is boxed{348169}.2. The average attendance over the first 12 concerts is boxed{2000}.</think>