🧠:<think>Alright, so I have this problem about Viking FK's recent 10-match performance. The fan missed these matches but got the scores and wants to analyze them mathematically. There are two parts to this problem. Let me tackle them one by one.Starting with the first part: I need to find the sum of the squares of these scores. The set of scores is S = {s₁, s₂, s₃, ..., s₁₀}. It's given that the arithmetic mean is 2.5, and the variance is 1.5. Hmm, okay, I remember that variance is related to the mean of the squares minus the square of the mean. Let me write that down.Variance (σ²) = (Σs_i² / N) - (μ)²Where:- σ² is the variance, which is 1.5- Σs_i² is the sum of the squares of the scores, which is what I need to find- N is the number of matches, which is 10- μ is the arithmetic mean, which is 2.5So plugging in the values:1.5 = (Σs_i² / 10) - (2.5)²Let me compute (2.5)² first. That's 6.25. So:1.5 = (Σs_i² / 10) - 6.25Now, I can solve for Σs_i²:Add 6.25 to both sides:1.5 + 6.25 = Σs_i² / 10That gives:7.75 = Σs_i² / 10Multiply both sides by 10:Σs_i² = 77.5So, the sum of the squares of these scores is 77.5. That seems straightforward. Let me just double-check my steps.1. Variance formula: correct.2. Plugging in the numbers: correct.3. Calculations: 2.5 squared is 6.25, 1.5 + 6.25 is 7.75, multiplied by 10 is 77.5. Yep, that looks right.Moving on to the second part: finding the probability that the next match will have a score greater than 3, given that the scores follow a normal distribution. Okay, so we have a normal distribution with mean μ = 2.5 and variance σ² = 1.5. Therefore, the standard deviation σ is the square root of 1.5.Let me compute that:σ = √1.5 ≈ 1.2247So, the distribution is N(2.5, 1.2247²). Now, I need to find P(X > 3), where X is the score of the next match.To find this probability, I can standardize the score and use the Z-table or a calculator. The formula for Z-score is:Z = (X - μ) / σPlugging in the values:Z = (3 - 2.5) / 1.2247 ≈ 0.5 / 1.2247 ≈ 0.4082So, Z ≈ 0.4082. Now, I need to find the probability that Z is greater than 0.4082. Since standard normal tables give the probability that Z is less than a certain value, I can find P(Z < 0.4082) and subtract it from 1.Looking up Z = 0.4082 in the standard normal table. Hmm, 0.4082 is approximately 0.41. Let me recall the Z-table values. For Z = 0.40, the cumulative probability is 0.6554, and for Z = 0.41, it's 0.6591. Since 0.4082 is closer to 0.41, maybe I can interpolate or just take the average.Alternatively, since I might not have the exact value, I can use a calculator or a more precise method. But for the sake of this problem, let me see.Alternatively, I can use the fact that the Z-score is approximately 0.4082, which is roughly 0.41. So, the cumulative probability up to Z=0.41 is about 0.6591. Therefore, P(Z > 0.4082) ≈ 1 - 0.6591 = 0.3409.So, approximately 34.09% chance that the next match score will exceed 3.Wait, let me verify this calculation. Maybe I should use a more precise method or a calculator to get the exact value.Alternatively, using the error function (erf) to compute the integral. But I think for the purposes of this problem, using the Z-table approximation is acceptable.Alternatively, I can use the formula for the standard normal distribution:P(Z > z) = 1 - Φ(z)Where Φ(z) is the cumulative distribution function. So, Φ(0.4082) is approximately 0.6591 as I thought, so 1 - 0.6591 = 0.3409.Therefore, the probability is approximately 34.09%.Wait, but let me check if I did the Z-score correctly. X is 3, μ is 2.5, so difference is 0.5. Divided by σ, which is sqrt(1.5). Let me compute that again.sqrt(1.5) is approximately 1.22474487. So, 0.5 divided by 1.22474487 is approximately 0.40824829. So, Z ≈ 0.4082.Looking up Z=0.4082 in a standard normal table. Let me recall that for Z=0.40, it's 0.6554, and for Z=0.41, it's 0.6591. The difference between 0.40 and 0.41 is 0.01 in Z, which corresponds to an increase of about 0.0037 in the cumulative probability (from 0.6554 to 0.6591). So, for Z=0.4082, which is 0.40 + 0.0082, the cumulative probability would be approximately 0.6554 + (0.0082 / 0.01) * 0.0037 ≈ 0.6554 + 0.003034 ≈ 0.6584.Therefore, P(Z > 0.4082) ≈ 1 - 0.6584 = 0.3416, which is approximately 34.16%.So, rounding it off, about 34.16%. Alternatively, if I use a calculator or a more precise method, it might be slightly different, but for the purposes of this problem, 34.1% is a reasonable approximation.Alternatively, using a calculator, let me compute the exact value. The integral from 0.4082 to infinity of the standard normal distribution. Using a calculator, I find that the exact value is approximately 0.3413, which is about 34.13%.So, either way, it's approximately 34.1%.Therefore, the probability that the next match will have a score greater than 3 is approximately 34.1%.Wait, let me just make sure I didn't make any mistakes in the calculations.1. Variance is 1.5, so standard deviation is sqrt(1.5) ≈ 1.2247. Correct.2. Z = (3 - 2.5)/1.2247 ≈ 0.4082. Correct.3. Looking up Z=0.4082, cumulative probability is approximately 0.6584, so 1 - 0.6584 ≈ 0.3416. Correct.Yes, that seems right.So, summarizing:1. Sum of squares of scores is 77.5.2. Probability that the next score exceeds 3 is approximately 34.1%.I think that's it. Let me just write down the final answers clearly.</think>

🧠:<think>Alright, so I have these two problems to solve, both related to optimization but in different contexts. Let me take them one by one.Starting with Emily's problem. She needs to distribute 10 independent tasks across 3 processors. Each task has a specific time ( t_i ), and the goal is to minimize the maximum workload on any processor. Hmm, okay. So this sounds like a scheduling problem, specifically the makespan minimization problem. Makespan is the time taken to complete all tasks, which in this case is the maximum workload across all processors.To formulate this as a mathematical optimization problem, I need to define variables and set up the objective function and constraints. Let me think. Let's denote ( x_{ij} ) as a binary variable where ( x_{ij} = 1 ) if task ( i ) is assigned to processor ( j ), and 0 otherwise. Since each task is independent, each task can only be assigned to one processor. So for each task ( i ), the sum over all processors ( j ) of ( x_{ij} ) should be 1. That gives me the constraint:[sum_{j=1}^{m} x_{ij} = 1 quad text{for all } i = 1, 2, ldots, n]Next, the workload on each processor ( j ) is the sum of the times of all tasks assigned to it. So, the workload ( W_j ) is:[W_j = sum_{i=1}^{n} t_i x_{ij} quad text{for all } j = 1, 2, ldots, m]The objective is to minimize the maximum workload across all processors. So, we can write the objective function as:[min max_{j=1}^{m} W_j]Putting it all together, the mathematical optimization problem is:Minimize ( max_{j=1}^{m} W_j )Subject to:1. ( sum_{j=1}^{m} x_{ij} = 1 ) for all ( i )2. ( W_j = sum_{i=1}^{n} t_i x_{ij} ) for all ( j )3. ( x_{ij} in {0, 1} ) for all ( i, j )Now, regarding the computational complexity. This problem is known as the makespan minimization problem on unrelated machines. Wait, actually, in this case, since each task can be assigned to any processor, and the processing times are fixed, it's similar to the problem of partitioning tasks into bins with the goal of minimizing the maximum bin load. I recall that this problem is NP-hard. Specifically, the problem of scheduling on identical machines to minimize makespan is NP-hard, and since here the tasks are independent and the machines are identical (assuming they have the same speed), it falls into that category. For small instances like ( n = 10 ) and ( m = 3 ), exact algorithms or even brute-force methods might be feasible, but in general, it's computationally intensive.Moving on to Mia's music practice problem. Emily wants to maximize Mia's learning efficiency while ensuring that her efficiency doesn't drop below a certain threshold ( E_0 ) for each piece. Mia has ( p ) pieces to practice, each requiring ( h_i ) hours at different tempos. The efficiency model is given by ( E(h) = e^{-lambda h} ).Emily wants to ensure that for each piece, the efficiency ( E(h_i) ) is at least ( E_0 ). So, for each piece ( i ), we have:[e^{-lambda h_i} geq E_0]To find the minimum practice time ( h_i ) for each piece, we can solve for ( h_i ):Taking natural logarithm on both sides:[-lambda h_i geq ln(E_0)]Multiplying both sides by -1 (which reverses the inequality):[lambda h_i leq -ln(E_0)]So,[h_i leq frac{-ln(E_0)}{lambda}]Wait, but this gives an upper bound on ( h_i ). However, since we want the efficiency not to drop below ( E_0 ), we need to ensure that ( h_i ) is such that ( E(h_i) geq E_0 ). So, actually, the maximum allowable ( h_i ) is ( frac{-ln(E_0)}{lambda} ). But if Mia practices for more than this time, her efficiency would drop below ( E_0 ). Therefore, to ensure efficiency doesn't drop below ( E_0 ), the practice time for each piece should be at most ( frac{-ln(E_0)}{lambda} ).But wait, the problem says Emily wants to maximize Mia's learning efficiency. Hmm, perhaps I need to clarify. If the efficiency is modeled as ( E(h) = e^{-lambda h} ), then as ( h ) increases, ( E(h) ) decreases. So, to maximize efficiency, Mia should practice as little as possible. But she needs to practice enough to master the pieces. Maybe I'm misunderstanding.Wait, perhaps the model is that the efficiency decreases with more practice time, so to maximize the efficiency, Mia should practice just enough so that her efficiency doesn't drop below ( E_0 ). So, for each piece, the minimum practice time ( h_i ) is such that ( E(h_i) = E_0 ). Therefore, solving for ( h_i ):[e^{-lambda h_i} = E_0 implies h_i = frac{-ln(E_0)}{lambda}]So, each piece requires ( h_i = frac{-ln(E_0)}{lambda} ) hours of practice. Since there are ( p ) pieces, the total practice time is ( p times frac{-ln(E_0)}{lambda} ).But let me think again. If each piece needs to be practiced until efficiency is at least ( E_0 ), then for each piece, the required practice time is ( h_i geq frac{-ln(E_0)}{lambda} ). So, the minimum total practice time would be ( p times frac{-ln(E_0)}{lambda} ). However, if Mia can practice multiple pieces simultaneously or in some optimized way, maybe the total time can be reduced? But the problem doesn't specify that; it just says she has ( p ) pieces to practice, each requiring ( h_i ) hours. So, I think the total practice time is the sum of the individual times, assuming she practices one piece at a time. So, total practice time ( H ) is:[H = sum_{i=1}^{p} h_i = p times frac{-ln(E_0)}{lambda}]Alternatively, if she can practice multiple pieces simultaneously, but the problem doesn't indicate that, so I think it's safe to assume she practices one piece at a time, so the total time is additive.Wait, but the problem says "Emily devises a schedule that maximizes Mia's learning efficiency." So, maybe she wants to distribute the practice time across the pieces in a way that the total efficiency is maximized, but each piece must have efficiency at least ( E_0 ). Hmm, perhaps it's a resource allocation problem where the total practice time is fixed, and she wants to allocate it across pieces to maximize the total efficiency. But the problem states that Emily wants to ensure efficiency doesn't drop below ( E_0 ) for each piece. So, perhaps she needs to find the minimum total practice time such that each piece is practiced enough to have ( E(h_i) geq E_0 ).In that case, for each piece, the minimum required practice time is ( h_i = frac{-ln(E_0)}{lambda} ). Therefore, the total practice time is ( p times frac{-ln(E_0)}{lambda} ).Alternatively, if the total time is fixed, and she wants to allocate it to maximize the sum of efficiencies, but the problem says she wants to ensure efficiency doesn't drop below a threshold. So, I think the total practice time needed is the sum of the minimum required times for each piece, which is ( p times frac{-ln(E_0)}{lambda} ).So, to express the total practice time ( H ) in terms of ( lambda ) and ( E_0 ), it's:[H = p times frac{-ln(E_0)}{lambda}]But let me double-check. If each piece needs to have ( E(h_i) geq E_0 ), then ( h_i leq frac{-ln(E_0)}{lambda} ). Wait, no, because ( E(h) = e^{-lambda h} ), so higher ( h ) leads to lower ( E(h) ). So, to have ( E(h_i) geq E_0 ), we need ( h_i leq frac{-ln(E_0)}{lambda} ). But that would mean that Mia can't practice each piece for more than ( frac{-ln(E_0)}{lambda} ) hours. But that contradicts the idea of needing to practice enough to master the piece. Maybe I'm misinterpreting the model.Alternatively, perhaps the efficiency is the rate at which she learns, so higher practice time leads to higher efficiency? But the model is ( E(h) = e^{-lambda h} ), which decreases as ( h ) increases. So, that would mean that the longer she practices, the less efficient she becomes, which seems counterintuitive. Maybe the model is the other way around? Or perhaps it's the forgetting curve, where efficiency decreases over time without practice. But in this context, it's about practice time, so maybe the efficiency in learning decreases as she spends more time on a piece? That seems odd.Alternatively, perhaps it's the efficiency in terms of how much she can learn per hour, which decreases as she gets fatigued. So, the longer she practices, the less efficient each additional hour is. Therefore, to ensure that her efficiency doesn't drop below a certain threshold, she needs to limit the time she spends on each piece. But then, how does she master the piece? Maybe she needs to practice each piece until a certain efficiency is achieved, but not beyond that. So, for each piece, the required practice time is such that ( E(h_i) geq E_0 ), which as before, gives ( h_i leq frac{-ln(E_0)}{lambda} ). But that would mean the maximum time she can spend on each piece is ( frac{-ln(E_0)}{lambda} ). But if she needs to master the piece, perhaps she needs to spend at least that much time? I'm confused.Wait, maybe the model is that the efficiency is the rate of learning, so the amount learned is the integral of efficiency over time. If efficiency decreases exponentially, then the total amount learned is ( int_{0}^{h} e^{-lambda t} dt = frac{1 - e^{-lambda h}}{lambda} ). So, to achieve a certain level of mastery, she needs to accumulate enough learning. But the problem doesn't specify that. It just says her efficiency follows ( E(h) = e^{-lambda h} ), and she wants efficiency not to drop below ( E_0 ).So, perhaps for each piece, she needs to practice until her efficiency is at least ( E_0 ). That would mean ( e^{-lambda h_i} geq E_0 ), which implies ( h_i leq frac{-ln(E_0)}{lambda} ). But that seems contradictory because if she practices more, her efficiency per hour decreases. So, to maintain efficiency above ( E_0 ), she can't practice each piece for more than ( frac{-ln(E_0)}{lambda} ) hours. But then, how does she master the piece? Maybe she needs to practice each piece for at least that time? I'm not sure.Alternatively, perhaps the efficiency is the total amount learned, so ( E(h) = e^{-lambda h} ) is the total efficiency, and she needs ( E(h_i) geq E_0 ). Then, solving for ( h_i ):[e^{-lambda h_i} geq E_0 implies h_i leq frac{-ln(E_0)}{lambda}]But again, this suggests that the maximum time she can spend on each piece is ( frac{-ln(E_0)}{lambda} ). But if she needs to reach a certain level of mastery, perhaps she needs to practice until her efficiency is at least ( E_0 ), which would mean she needs to practice for at least ( h_i = frac{-ln(E_0)}{lambda} ) hours. Wait, that makes more sense. Because if she practices less, her efficiency would be higher, but she might not have mastered the piece. So, to ensure that her efficiency doesn't drop below ( E_0 ), she needs to practice each piece for at least ( h_i = frac{-ln(E_0)}{lambda} ) hours. Therefore, the total practice time is the sum over all pieces, which is ( p times frac{-ln(E_0)}{lambda} ).Yes, that seems to make sense. So, the total practice time needed is ( H = p times frac{-ln(E_0)}{lambda} ).Okay, so summarizing:1. For Emily's problem, it's a makespan minimization problem with 10 tasks and 3 processors. The mathematical formulation is as above, and the problem is NP-hard.2. For Mia's practice, the total time needed is ( H = p times frac{-ln(E_0)}{lambda} ).I think that's it.</think>

🧠:<think>Okay, so I have this problem about drone delivery routes and battery consumption. Let me try to break it down step by step. First, the drone needs to deliver packages to three locations: A, B, and C, starting from the base station O. The coordinates are given in 3D space, but all the z-coordinates are zero, so it's essentially a 2D problem. The drone starts at O(0,0,0), then goes to A(3,4,0), B(6,8,0), and C(9,12,0). After delivering all packages, it needs to return to O.The battery consumption is a function of the distance traveled and the load carried. Without any load, the drone consumes 1 unit of battery per unit distance. For every unit of weight, it consumes an additional 0.1 units per unit distance. So, if the drone is carrying a load, the total battery consumption per unit distance is 1 + 0.1*(weight). In this case, each package weighs 2 units, and the drone needs to deliver one package to each location. Since the drone can carry multiple packages simultaneously, I think it can carry all three packages at once. That would mean the total weight is 2 + 2 + 2 = 6 units. So, the battery consumption per unit distance would be 1 + 0.1*6 = 1.6 units per unit distance.Wait, but does the drone have to carry all packages at once? Or can it make multiple trips? The problem says it needs to deliver a package to each location, so maybe it can carry all three at once. That would be more efficient. So, I think the drone will carry all three packages, totaling 6 units, for the entire trip.So, the first step is to figure out the route. Since the drone starts at O, it needs to go to A, B, C, and back to O. But what's the most efficient route? Is there a specific order that minimizes the total distance? Because the order can affect the total distance traveled.Let me plot the points mentally. O is at (0,0,0). A is at (3,4), which is 5 units away from O because sqrt(3² + 4²) = 5. Then B is at (6,8), which is further out. The distance from O to B is sqrt(6² + 8²) = sqrt(36 + 64) = sqrt(100) = 10 units. Similarly, C is at (9,12), so distance from O is sqrt(9² + 12²) = sqrt(81 + 144) = sqrt(225) = 15 units.So, the points are colinear, right? Because each point is a multiple of (3,4). A is 1*(3,4), B is 2*(3,4), and C is 3*(3,4). So, they lie on a straight line from O. That simplifies things because the drone can just go from O to A to B to C and back to O, but that might not be the most efficient.Wait, but if it goes from O to A to B to C and back to O, the total distance would be O to A: 5, A to B: distance between A and B. Let's calculate that. The coordinates of A are (3,4) and B are (6,8). The distance between A and B is sqrt((6-3)² + (8-4)²) = sqrt(9 + 16) = sqrt(25) = 5 units. Similarly, distance from B to C is sqrt((9-6)² + (12-8)²) = sqrt(9 + 16) = 5 units. Then from C back to O is 15 units. So total distance is 5 + 5 + 5 + 15 = 30 units.But is that the shortest possible route? Or is there a more efficient path? Since all points are colinear, the drone can't really take a shortcut. It has to go through each point in order. So, the minimal route is O -> A -> B -> C -> O, which is 30 units.Alternatively, maybe going from O to C to B to A to O would be the same distance, 15 + 5 + 5 + 5 = 30. So, regardless of the order, the total distance is 30 units.Wait, but actually, if the drone can choose the order, maybe it's more efficient to go from O to C directly, then to B, then to A, and back. But since all points are on a straight line, the distance remains the same. So, regardless of the order, the total distance is 30 units.But wait, is there a way to make the return trip shorter? For example, going from C back to O is 15 units, but maybe if the drone doesn't have to carry the packages back, it can take a different route? But the problem says the drone needs to return to the base station, so it has to go back to O regardless.But actually, the drone is delivering packages, so once it drops off the package at C, it doesn't need to carry that package anymore. So, does the load decrease as it delivers each package? Hmm, the problem says "the drone can carry multiple packages simultaneously." So, does that mean it can carry all three packages at once, or it can choose to carry them one by one?Wait, the problem states: "the drone can carry multiple packages simultaneously." So, it can carry all three packages at once. So, the weight is 6 units for the entire trip except when it's returning. Wait, no. When it's returning, it doesn't have any packages, right? Because it delivered all of them.Wait, no, actually, it needs to return to the base station after delivering all packages. So, the drone starts at O with 6 units of weight, goes to A, delivers a package (so now it has 4 units left), then goes to B, delivers another package (2 units left), then goes to C, delivers the last package (0 units left), and then returns to O without any load.So, the battery consumption is different for each segment because the weight carried changes.So, let me break it down:1. O to A: 5 units distance, carrying 6 units. So, battery consumed: 5 * (1 + 0.1*6) = 5 * 1.6 = 8 units.2. A to B: 5 units distance, carrying 4 units (since one package is delivered). So, battery consumed: 5 * (1 + 0.1*4) = 5 * 1.4 = 7 units.3. B to C: 5 units distance, carrying 2 units. Battery consumed: 5 * (1 + 0.1*2) = 5 * 1.2 = 6 units.4. C to O: 15 units distance, carrying 0 units. Battery consumed: 15 * 1 = 15 units.Total battery consumed: 8 + 7 + 6 + 15 = 36 units.Wait, that's different from my initial thought where I considered the drone carrying all packages the entire trip. But actually, since it's delivering packages along the way, the weight decreases, which affects the battery consumption.So, the total battery consumption is 36 units.But wait, is there a more efficient route? Maybe not going in the order O -> A -> B -> C -> O, but a different order? For example, O -> B -> A -> C -> O? Let's see.Calculating the distances:O to B: 10 units, carrying 6 units. Battery: 10 * 1.6 = 16 units.B to A: distance between B and A is 5 units, carrying 4 units. Battery: 5 * 1.4 = 7 units.A to C: distance between A and C is sqrt((9-3)^2 + (12-4)^2) = sqrt(36 + 64) = sqrt(100) = 10 units, carrying 2 units. Battery: 10 * 1.2 = 12 units.C to O: 15 units, carrying 0. Battery: 15 * 1 = 15 units.Total: 16 + 7 + 12 + 15 = 50 units. That's worse than the previous total of 36.Alternatively, O -> C -> B -> A -> O.O to C: 15 units, carrying 6 units. Battery: 15 * 1.6 = 24 units.C to B: 5 units, carrying 4 units. Battery: 5 * 1.4 = 7 units.B to A: 5 units, carrying 2 units. Battery: 5 * 1.2 = 6 units.A to O: 5 units, carrying 0. Battery: 5 * 1 = 5 units.Total: 24 + 7 + 6 + 5 = 42 units. Still worse than 36.So, the initial order O -> A -> B -> C -> O seems to be the most efficient, giving a total battery consumption of 36 units.Wait, but is there a way to minimize the distance? For example, if the drone can choose a different path that's shorter? But since all points are colinear, the distance is fixed. So, regardless of the order, the total distance is 30 units, but the battery consumption varies because the load changes.Wait, no. Actually, the total distance isn't fixed because the order affects the total distance. For example, going from O to C to B to A to O is 15 + 5 + 5 + 5 = 30 units, same as O to A to B to C to O.But the battery consumption is different because the load is different in each segment.So, the key is to minimize the total battery consumption, which depends on the order because the load is decreasing as packages are delivered.So, to minimize battery consumption, we should carry the heaviest load over the shortest distances possible.In the initial order, O -> A -> B -> C -> O:- The heaviest load (6 units) is carried over the shortest distance (5 units from O to A).- Then, 4 units over 5 units (A to B).- Then, 2 units over 5 units (B to C).- Finally, 0 units over 15 units (C to O).This seems optimal because the heaviest load is on the shortest segment, which minimizes the total battery consumed.If we reverse the order, carrying the heaviest load over the longest distance, it would be worse.So, the total battery consumption is 36 units.Therefore, the answer to part 1 is 36 units.Now, moving on to part 2. The drone's battery capacity is limited to 50 units. We need to determine if the drone can complete the delivery route calculated in part 1 without recharging. Since 36 < 50, yes, it can complete the route without recharging.But wait, the second part says: "If not, find the maximum number of packages (weighing 2 units each) it can deliver in a single trip and return to the base station O."Since 36 < 50, the drone can complete the route. But maybe the question is implying that if the battery wasn't enough, we have to find the maximum number. But in this case, it's enough. However, perhaps the problem is more general, and we need to find the maximum number of packages the drone can carry, given the battery limit.Wait, let me read the question again."Assume the drone's battery capacity is limited to 50 units. Determine if the drone can complete the delivery route calculated in the previous sub-problem without recharging. If not, find the maximum number of packages (weighing 2 units each) it can deliver in a single trip and return to the base station O."So, since in part 1, the drone needs 36 units, which is less than 50, it can complete the route. So, the answer is yes, it can complete without recharging. But the second part is only if it cannot complete, then find the maximum number. So, in this case, since it can complete, maybe we don't need to do anything else. But perhaps the problem is asking, given the battery limit, what's the maximum number of packages it can deliver in a single trip, considering the same route.Wait, maybe I misinterpreted part 1. Maybe part 1 is just calculating the battery consumption for delivering one package to each location, but part 2 is about delivering multiple packages, each weighing 2 units, and finding the maximum number such that the drone can return without recharging.Wait, the problem says: "the drone can carry multiple packages simultaneously." So, in part 1, it's delivering one package to each location, but in part 2, it's about delivering multiple packages, each weighing 2 units, in a single trip.Wait, let me read the problem again."1. A drone needs to deliver packages to three different locations, A, B, and C, starting from a base station at point O... deliver a package weighing 2 units to each of the locations A, B, and C..."So, part 1 is delivering one package to each location, each weighing 2 units, so total weight is 6 units.Part 2: "Assume the drone's battery capacity is limited to 50 units. Determine if the drone can complete the delivery route calculated in the previous sub-problem without recharging. If not, find the maximum number of packages (weighing 2 units each) it can deliver in a single trip and return to the base station O."So, part 2 is asking, given the same route as part 1, but with the battery limit of 50 units, can it complete the same delivery? Since in part 1, the battery consumption was 36 units, which is less than 50, so yes, it can complete. But perhaps the problem is more about, if the battery was less, say, if the battery was 35 units, then it couldn't complete, and we would have to find the maximum number of packages.But in this case, since 36 < 50, it can complete. However, maybe the problem is asking, given the battery capacity, what's the maximum number of packages (each 2 units) it can deliver in a single trip, following the same route.Wait, the wording is a bit confusing. It says, "find the maximum number of packages (weighing 2 units each) it can deliver in a single trip and return to the base station O."So, perhaps, instead of delivering one package to each location, it can deliver multiple packages to each location, but the total number of packages is what we need to maximize, given the battery limit.But the problem in part 1 was delivering one package to each location. So, maybe in part 2, it's about delivering multiple packages to each location, but the same route.Wait, the problem says: "find the maximum number of packages (weighing 2 units each) it can deliver in a single trip and return to the base station O."So, perhaps, the drone can carry multiple packages, but the number of packages is the total number, not per location. So, for example, delivering n packages, each weighing 2 units, to the three locations, but the total number is n.But the problem is a bit ambiguous. Let me try to parse it."the maximum number of packages (weighing 2 units each) it can deliver in a single trip and return to the base station O."So, it's about the total number of packages, each 2 units, that can be delivered in a single trip, following the same route as part 1, which is O -> A -> B -> C -> O.So, in part 1, the drone delivered 3 packages (one to each location), each 2 units, total weight 6 units, battery consumed 36 units.In part 2, the battery is limited to 50 units. So, we need to find the maximum number of packages (each 2 units) that can be delivered in a single trip, following the same route, without exceeding the battery limit.So, the total weight carried would be 2n units, where n is the number of packages. But wait, the drone can carry multiple packages simultaneously, so it can carry all n packages at once, but the weight is 2n units.Wait, no. Because in part 1, it delivered one package to each location, so it had to carry 6 units total. But if it's delivering more packages, say, delivering multiple packages to each location, then the total weight would be higher.Wait, but the problem doesn't specify that the drone has to deliver one package to each location. It just says "deliver packages to three different locations." So, perhaps, the drone can deliver multiple packages to each location, but the total number of packages is what we need to maximize.But the problem says "the maximum number of packages (weighing 2 units each) it can deliver in a single trip and return to the base station O."So, it's about the total number of packages, regardless of the locations. So, the drone can deliver multiple packages to each location, but the total number is n, each 2 units, so total weight is 2n units.But wait, the route is fixed as O -> A -> B -> C -> O, as in part 1. So, the drone starts at O, goes to A, delivers some packages, then to B, delivers some, then to C, delivers some, then back to O.Each time it delivers, the weight decreases.So, the total battery consumption would be:- O to A: distance 5, carrying 2n units. Battery: 5*(1 + 0.1*(2n)) = 5*(1 + 0.2n)- A to B: distance 5, carrying 2n - 2k1 units, where k1 is the number of packages delivered at A. Wait, but if we're delivering multiple packages, it's more complicated.Wait, perhaps it's better to assume that the drone delivers all packages at each location in sequence. So, it starts with 2n units, delivers some at A, then some at B, then some at C, and returns.But the problem is that the number of packages delivered at each location can vary. To maximize the total number of packages, we need to distribute the deliveries in a way that minimizes the battery consumption.But this is getting complicated. Maybe a better approach is to consider that the drone can carry all packages at once, but the weight affects the battery consumption on each segment.Wait, but when delivering packages, the weight decreases as packages are delivered. So, the battery consumption on each segment depends on the number of packages carried during that segment.So, let's model it as follows:Let n be the total number of packages, each 2 units, so total weight is 2n units.The drone starts at O with 2n units.It goes to A, delivers k1 packages (each 2 units), so the weight after A is 2n - 2k1.Then goes to B, delivers k2 packages, weight becomes 2n - 2k1 - 2k2.Then goes to C, delivers k3 packages, weight becomes 2n - 2k1 - 2k2 - 2k3.Then returns to O, delivering the remaining packages? Wait, no, it has to return to O, so it can't deliver on the way back.Wait, actually, the drone delivers packages only at A, B, and C. So, after delivering at C, it returns to O with no packages.So, the total weight carried on each segment is:- O to A: 2n units.- A to B: 2n - 2k1 units.- B to C: 2n - 2k1 - 2k2 units.- C to O: 0 units.The total battery consumption is the sum of the battery consumed on each segment.So, total battery consumption:= (5 * (1 + 0.1*(2n))) + (5 * (1 + 0.1*(2n - 2k1))) + (5 * (1 + 0.1*(2n - 2k1 - 2k2))) + (15 * 1)Simplify:= 5*(1 + 0.2n) + 5*(1 + 0.2n - 0.2k1) + 5*(1 + 0.2n - 0.2k1 - 0.2k2) + 15= 5 + 1n + 5 + 1n - 1k1 + 5 + 1n - 1k1 - 1k2 + 15Combine like terms:= (5 + 5 + 5 + 15) + (n + n + n) + (-k1 - k1 - k2)= 30 + 3n - 2k1 - k2But we also know that the total number of packages delivered is k1 + k2 + k3 = n. Wait, but in the above, we have k1 and k2, but not k3. Because after delivering at C, the drone returns to O, so k3 is the number of packages delivered at C, which is 2n - 2k1 - 2k2 units, but since each package is 2 units, k3 = (2n - 2k1 - 2k2)/2 = n - k1 - k2.So, the total battery consumption is:30 + 3n - 2k1 - k2But we also have that k3 = n - k1 - k2.But we need to express the battery consumption in terms of n and the k's, but it's a bit tangled.Alternatively, perhaps we can assume that the drone delivers the same number of packages at each location, but that might not be optimal.Wait, maybe to minimize the battery consumption, we should deliver as many packages as possible at the earliest locations, so that the weight decreases earlier, reducing the battery consumption on the longer segments.So, for example, delivering more packages at A (earlier) would reduce the weight for the longer segments (B to C and C to O).Similarly, delivering more at B would reduce the weight for the C to O segment.So, to minimize the total battery consumption, we should deliver as many packages as possible at the earliest possible points.Therefore, to maximize the number of packages n, given the battery limit of 50 units, we need to distribute the deliveries such that the total battery consumption is <= 50.But this is getting complex. Maybe a better approach is to consider that the drone can carry all packages at once, but the weight affects the battery consumption on each segment.Wait, but the problem is that the drone can deliver packages at each location, so the weight decreases as it delivers.So, the total battery consumption is:From O to A: 5*(1 + 0.1*(2n)) = 5 + nFrom A to B: 5*(1 + 0.1*(2n - 2k1)) = 5 + (2n - 2k1)*0.1*5 = 5 + (n - k1)Wait, no, let me recast it.Battery consumption per segment is distance * (1 + 0.1*weight). So, for O to A:5*(1 + 0.1*(2n)) = 5*(1 + 0.2n) = 5 + nFrom A to B:5*(1 + 0.1*(2n - 2k1)) = 5*(1 + 0.2n - 0.2k1) = 5 + n - k1From B to C:5*(1 + 0.1*(2n - 2k1 - 2k2)) = 5*(1 + 0.2n - 0.2k1 - 0.2k2) = 5 + n - k1 - k2From C to O:15*(1 + 0.1*0) = 15Total battery consumption:(5 + n) + (5 + n - k1) + (5 + n - k1 - k2) + 15= 5 + n + 5 + n - k1 + 5 + n - k1 - k2 + 15= (5+5+5+15) + (n + n + n) + (-k1 - k1 - k2)= 30 + 3n - 2k1 - k2But we also know that k1 + k2 + k3 = n, and k3 = n - k1 - k2.So, the total battery consumption is 30 + 3n - 2k1 - k2.We need this to be <= 50.So, 30 + 3n - 2k1 - k2 <= 50Which simplifies to:3n - 2k1 - k2 <= 20But we also have k1 + k2 + k3 = n, and k3 = n - k1 - k2.To maximize n, we need to minimize the left side, which is 3n - 2k1 - k2.But since k1 and k2 are non-negative integers, the minimum value of 3n - 2k1 - k2 occurs when k1 and k2 are as large as possible.Wait, but we need to find the maximum n such that 3n - 2k1 - k2 <= 20.But this is a bit abstract. Maybe we can set k1 and k2 to their maximum possible values given n.Since k1 <= n, k2 <= n - k1, etc.But perhaps a better approach is to consider that to minimize the battery consumption, we should deliver as many packages as possible at the earliest points, so k1 is as large as possible, then k2, etc.So, to minimize 3n - 2k1 - k2, we need to maximize 2k1 + k2.Given that k1 + k2 <= n.So, the maximum of 2k1 + k2 given k1 + k2 <= n is achieved when k1 is as large as possible.So, set k1 = n - k2, but to maximize 2k1 + k2, set k2 = 0, then 2k1 = 2n.So, the maximum of 2k1 + k2 is 2n.Therefore, the minimum value of 3n - 2k1 - k2 is 3n - 2n = n.So, we have n <= 20.Therefore, n <= 20.Wait, but that seems too high because in part 1, n=3, and the battery consumption was 36, which is less than 50.Wait, maybe I made a mistake in the reasoning.Let me try plugging in n=10.If n=10, total weight is 20 units.If we deliver all 10 packages at A, then k1=10, k2=0, k3=0.Then, battery consumption:O to A: 5*(1 + 0.2*10) = 5*(1 + 2) = 15A to B: 5*(1 + 0.2*(10 - 10)) = 5*(1 + 0) = 5B to C: 5*(1 + 0.2*(10 - 10 - 0)) = 5*(1 + 0) = 5C to O: 15*1 = 15Total: 15 + 5 + 5 + 15 = 40 <=50So, n=10 is possible.What about n=15.If n=15, total weight=30.Deliver all 15 at A:O to A: 5*(1 + 0.2*15)=5*(1 +3)=20A to B: 5*(1 +0)=5B to C:5*(1 +0)=5C to O:15Total:20+5+5+15=45<=50n=15 is possible.n=20:Deliver all 20 at A:O to A:5*(1 +0.2*20)=5*(1 +4)=25A to B:5*(1 +0)=5B to C:5*(1 +0)=5C to O:15Total:25+5+5+15=50So, n=20 is possible, consuming exactly 50 units.If n=21:Deliver all 21 at A:O to A:5*(1 +0.2*21)=5*(1 +4.2)=5*5.2=26A to B:5*(1 +0)=5B to C:5*(1 +0)=5C to O:15Total:26+5+5+15=51>50So, n=21 is not possible.But wait, maybe we can deliver some packages at A and some at B, so that the total battery consumption is <=50.For example, n=20:If we deliver 19 at A, and 1 at B.Then:O to A:5*(1 +0.2*20)=25A to B:5*(1 +0.2*(20 -19))=5*(1 +0.2)=5*1.2=6B to C:5*(1 +0.2*(20 -19 -1))=5*(1 +0)=5C to O:15Total:25+6+5+15=51>50Still over.Alternatively, deliver 18 at A, 2 at B.O to A:5*(1 +0.2*20)=25A to B:5*(1 +0.2*(20 -18))=5*(1 +0.4)=5*1.4=7B to C:5*(1 +0.2*(20 -18 -2))=5*(1 +0)=5C to O:15Total:25+7+5+15=52>50Still over.Wait, maybe deliver 17 at A, 3 at B.O to A:25A to B:5*(1 +0.2*(20 -17))=5*(1 +0.6)=5*1.6=8B to C:5*(1 +0.2*(20 -17 -3))=5*(1 +0)=5C to O:15Total:25+8+5+15=53>50Still over.Wait, maybe deliver 16 at A, 4 at B.O to A:25A to B:5*(1 +0.2*(20 -16))=5*(1 +0.8)=5*1.8=9B to C:5*(1 +0.2*(20 -16 -4))=5*(1 +0)=5C to O:15Total:25+9+5+15=54>50Still over.Wait, maybe deliver 15 at A, 5 at B.O to A:5*(1 +0.2*15)=5*(1 +3)=20A to B:5*(1 +0.2*(15 -15))=5*(1 +0)=5B to C:5*(1 +0.2*(15 -15 -5))=5*(1 +0)=5C to O:15Total:20+5+5+15=45<=50But n=15 is possible, but we were trying to get n=20.Wait, but if we deliver 15 at A, 5 at B, that's n=20, but the battery consumption is 45, which is under 50. Wait, no, n=15 is delivering 15 packages, each 2 units, so total weight 30 units.Wait, I'm getting confused.Wait, n is the total number of packages, each 2 units. So, n=20 means total weight 40 units.Wait, no, n=20 packages, each 2 units, total weight 40 units.Wait, but earlier, when n=20, delivering all at A, the battery consumption was 25 +5 +5 +15=50.But if n=20, and we deliver 15 at A, and 5 at B, then the total weight is 40 units.Wait, no, n=20, each package is 2 units, so total weight is 40 units.But if we deliver 15 packages at A, that's 30 units, leaving 10 units (5 packages) to deliver at B and C.Wait, no, n=20, so total packages is 20, each 2 units, so total weight is 40 units.If we deliver 15 packages at A (30 units), then 5 packages remain, which can be delivered at B and C.But the problem is that the battery consumption is calculated based on the weight carried during each segment.So, delivering 15 at A, then 5 at B, then 0 at C.So, the battery consumption would be:O to A:5*(1 +0.2*20)=5*(1 +4)=25A to B:5*(1 +0.2*(20 -15))=5*(1 +1)=10B to C:5*(1 +0.2*(20 -15 -5))=5*(1 +0)=5C to O:15Total:25+10+5+15=55>50Still over.Alternatively, deliver 16 at A, 4 at B.O to A:25A to B:5*(1 +0.2*(20 -16))=5*(1 +0.8)=9B to C:5*(1 +0.2*(20 -16 -4))=5*(1 +0)=5C to O:15Total:25+9+5+15=54>50Still over.Wait, maybe deliver 17 at A, 3 at B.O to A:25A to B:5*(1 +0.2*(20 -17))=5*(1 +0.6)=8B to C:5*(1 +0.2*(20 -17 -3))=5*(1 +0)=5C to O:15Total:25+8+5+15=53>50Still over.Wait, maybe deliver 18 at A, 2 at B.O to A:25A to B:5*(1 +0.2*(20 -18))=5*(1 +0.4)=7B to C:5*(1 +0.2*(20 -18 -2))=5*(1 +0)=5C to O:15Total:25+7+5+15=52>50Still over.Wait, deliver 19 at A, 1 at B.O to A:25A to B:5*(1 +0.2*(20 -19))=5*(1 +0.2)=6B to C:5*(1 +0.2*(20 -19 -1))=5*(1 +0)=5C to O:15Total:25+6+5+15=51>50Still over.So, the only way to get exactly 50 is to deliver all 20 packages at A, which gives 25+5+5+15=50.But wait, delivering all 20 at A, the drone would have to carry 40 units of weight from O to A, which is 5 units distance, consuming 25 units. Then, from A to B, it's carrying 0 units, so 5 units distance, consuming 5 units. Then, B to C, 5 units, 5 units. Then, C to O, 15 units, 15 units. Total 25+5+5+15=50.But wait, if the drone delivers all 20 packages at A, then it doesn't need to go to B and C, right? Because it's only delivering to A.Wait, but the problem says "deliver packages to three different locations, A, B, and C." So, in part 1, it's delivering one package to each location. In part 2, it's about delivering multiple packages, but still to the three locations.So, the drone must visit A, B, and C, delivering at least one package to each, but can deliver more.Therefore, in part 2, the drone must still visit A, B, and C, delivering at least one package to each, but can deliver more.So, the total number of packages n must be at least 3, but can be more.So, in that case, the drone cannot just deliver all packages at A; it must also deliver at least one at B and C.Therefore, the minimum number of packages is 3, but we can have more.So, to maximize n, we need to deliver as many as possible, but still have to deliver at least one to B and C.So, let's denote:k1: number of packages delivered at Ak2: number delivered at Bk3: number delivered at CWith k1 + k2 + k3 = nAnd k1 >=1, k2 >=1, k3 >=1We need to maximize n such that the total battery consumption is <=50.The total battery consumption is:O to A:5*(1 +0.2n)A to B:5*(1 +0.2(n -k1))B to C:5*(1 +0.2(n -k1 -k2))C to O:15Total:5*(1 +0.2n) + 5*(1 +0.2(n -k1)) + 5*(1 +0.2(n -k1 -k2)) +15=5 + n +5 + n -0.2*5k1 +5 + n -0.2*5k1 -0.2*5k2 +15Wait, let me compute each term:O to A:5*(1 +0.2n)=5 +nA to B:5*(1 +0.2(n -k1))=5 +n -k1B to C:5*(1 +0.2(n -k1 -k2))=5 +n -k1 -k2C to O:15Total:(5 +n) + (5 +n -k1) + (5 +n -k1 -k2) +15=5 +n +5 +n -k1 +5 +n -k1 -k2 +15= (5+5+5+15) + (n +n +n) + (-k1 -k1 -k2)=30 +3n -2k1 -k2We need 30 +3n -2k1 -k2 <=50So, 3n -2k1 -k2 <=20But since k1 >=1, k2 >=1, and k3 >=1, we have k1 +k2 +k3 =n >=3.To maximize n, we need to minimize 3n -2k1 -k2.Given that k1 and k2 are at least 1, the minimum value of 3n -2k1 -k2 is achieved when k1 and k2 are as large as possible.But since k1 +k2 <=n -1 (because k3 >=1), the maximum k1 and k2 can be is n -1 (if k3=1).But to minimize 3n -2k1 -k2, we need to maximize 2k1 +k2.Given that k1 +k2 <=n -1, the maximum of 2k1 +k2 is achieved when k1 is as large as possible.So, set k1 =n -1 -k2, but to maximize 2k1 +k2, set k2=0, but k2 >=1.So, set k2=1, then k1 can be as large as n -2.Thus, 2k1 +k2 =2(n -2) +1=2n -4 +1=2n -3.Therefore, 3n -2k1 -k2=3n - (2k1 +k2)=3n - (2n -3)=n +3.So, n +3 <=20 =>n <=17.Wait, that seems possible.Wait, let me verify.If n=17, then 3n -2k1 -k2=17 +3=20, which is equal to 20.So, n=17 is possible.Let me check with n=17.k1=17 -1 -k2=16 -k2.But to maximize 2k1 +k2, set k2=1, then k1=16.So, k1=16, k2=1, k3=0? Wait, no, k3 must be at least 1.Wait, k3= n -k1 -k2=17 -16 -1=0, which is less than 1. So, that's not allowed.So, we need k3 >=1, so k1 +k2 <=16.Thus, to maximize 2k1 +k2, set k2=1, k1=15.Then, k3=17 -15 -1=1.So, 2k1 +k2=2*15 +1=31.Thus, 3n -2k1 -k2=3*17 -31=51 -31=20.So, 20<=20, which is acceptable.So, n=17 is possible.Let me calculate the total battery consumption.k1=15, k2=1, k3=1.O to A:5*(1 +0.2*17)=5*(1 +3.4)=5*4.4=22A to B:5*(1 +0.2*(17 -15))=5*(1 +0.4)=5*1.4=7B to C:5*(1 +0.2*(17 -15 -1))=5*(1 +0.2)=5*1.2=6C to O:15Total:22+7+6+15=50.Perfect, exactly 50 units.So, n=17 is possible.Can we go higher?n=18.Then, 3n -2k1 -k2=3*18 - (2k1 +k2)=54 - (2k1 +k2)=?We need 54 - (2k1 +k2) <=20 =>2k1 +k2 >=34.But k1 +k2 <=17 (since k3 >=1).The maximum of 2k1 +k2 given k1 +k2 <=17 is achieved when k1 is as large as possible.Set k2=1, then k1=16.2k1 +k2=32 +1=33 <34.Not enough.Alternatively, set k1=17, k2=0, but k2 must be >=1.So, k1=16, k2=1: 2*16 +1=33 <34.Not enough.Thus, n=18 is not possible.Therefore, the maximum n is 17.So, the drone can deliver 17 packages (each 2 units) in a single trip, visiting A, B, and C, delivering 15 at A, 1 at B, and 1 at C, consuming exactly 50 units of battery.Therefore, the answer to part 2 is 17 packages.</think>

🧠:<think>Okay, so I have this problem about a photographer trying to capture a hotel lobby. There are two parts to the problem, and I need to figure out both. Let me start with the first one.Problem 1: The camera is positioned at a height of 1.5 meters, and the lens has a horizontal field of view of 90 degrees. I need to find the minimum distance from the camera to the nearest wall so that the entire width of the lobby is captured in the shot. The lobby is 30 meters wide, right? Because it's 20 meters by 30 meters, so the width is 30 meters.Hmm, so the camera is at 1.5 meters height, and the field of view is 90 degrees. I think this is a problem involving trigonometry, specifically the tangent function because we're dealing with angles and opposite and adjacent sides.Let me visualize this. The camera is somewhere in the lobby, and it's looking towards the nearest wall. The field of view is 90 degrees, which means the camera can capture a 90-degree angle horizontally. Since the lobby is 30 meters wide, the camera needs to capture 15 meters on either side of its position to cover the entire width.Wait, no. Actually, if the camera is positioned somewhere in the lobby, the distance from the camera to the nearest wall will determine how much of the lobby is captured. If the camera is too close to one wall, the other wall might not be in the frame. So, the idea is that the camera's field of view should cover the entire 30-meter width.But the field of view is 90 degrees. So, the angle between the two walls as seen from the camera is 90 degrees. So, if I imagine a triangle where the camera is at one vertex, and the two walls are at the other two vertices, forming a right angle at the camera.Wait, no. Actually, the field of view is the angle that the camera can see. So, the horizontal field of view is 90 degrees, meaning that from the camera's position, the angle between the leftmost and rightmost points of the lobby should be 90 degrees.So, if the camera is at a certain distance from the nearest wall, say 'd' meters, then the distance to the farthest wall would be 30 - d meters. The angle between these two points, as seen from the camera, should be 90 degrees.So, I can model this as a triangle where the camera is at one point, and the two walls are at the other two points, with the angle at the camera being 90 degrees. The sides adjacent to the angle are 'd' and '30 - d', and the opposite side would be the height, but wait, the height is 1.5 meters, but I don't think that's directly relevant here because the field of view is horizontal. So maybe I can ignore the height for this part?Wait, no, actually, the height might affect the angle because the camera is elevated. Hmm, but the field of view is given as horizontal, so maybe it's just the horizontal angle. So, perhaps the height doesn't matter for the horizontal field of view.Wait, I'm getting confused. Let me think again. The horizontal field of view is the angle that the camera can see from side to side, so it's purely horizontal. So, the vertical position (height) of the camera doesn't affect the horizontal field of view. So, maybe I can treat this as a 2D problem on the ground.So, if the camera is at a point, and it can see 90 degrees horizontally, then the distance from the camera to each wall should be such that the angle between them is 90 degrees.So, if I denote the distance from the camera to the nearest wall as 'd', then the distance to the farthest wall is '30 - d'. The angle between these two lines is 90 degrees. So, in trigonometry, if I have two sides and the included angle, I can use the tangent function or maybe the law of cosines.Wait, actually, since it's a triangle with sides 'd', '30 - d', and the angle between them is 90 degrees, maybe I can use the Pythagorean theorem? Because if the angle is 90 degrees, then the triangle is right-angled.But wait, the triangle isn't necessarily right-angled. The angle at the camera is 90 degrees, but the sides are 'd' and '30 - d', and the opposite side would be the distance across the lobby, but I don't know that.Wait, maybe I can use the law of cosines here. The law of cosines relates the sides of a triangle to the cosine of one of its angles. The formula is:c² = a² + b² - 2ab cos(C)Where C is the angle opposite side c.In this case, the angle at the camera is 90 degrees, so C = 90°, and sides a and b are 'd' and '30 - d'. The side opposite the angle C would be the distance between the two walls, which is 30 meters. Wait, no, the distance between the two walls is 30 meters, but in this case, the two walls are 30 meters apart, so the side opposite the angle is 30 meters.Wait, that doesn't make sense because the two walls are 30 meters apart, but the side opposite the angle is actually the distance between the two walls as seen from the camera, which is not necessarily 30 meters because the camera is at some distance from the walls.Wait, I'm getting confused. Let me try to draw this mentally.Imagine the camera is somewhere in the lobby, not necessarily in the center. Let's say it's 'd' meters away from the nearest wall, so it's '30 - d' meters away from the farthest wall. The angle between these two lines of sight is 90 degrees. So, the triangle formed by the camera and the two walls has sides 'd', '30 - d', and the angle between them is 90 degrees.So, using the law of cosines, the side opposite the 90-degree angle would be the distance between the two walls as seen from the camera, but actually, the two walls are 30 meters apart, but the side opposite the angle is not 30 meters because the camera is at a distance from both walls.Wait, maybe I'm overcomplicating this. Let's think about the horizontal field of view. The horizontal field of view is 90 degrees, so the camera can see 45 degrees to the left and 45 degrees to the right from its central axis.So, if the camera is 'd' meters away from the nearest wall, then the angle between the camera's central axis and the line of sight to the wall is 45 degrees. Similarly, the angle to the farthest wall would also be 45 degrees on the other side.Wait, no. If the total field of view is 90 degrees, then the angle from the center to each wall is 45 degrees. So, the distance from the camera to each wall can be found using the tangent of 45 degrees.But wait, the tangent of 45 degrees is 1, so tan(45) = opposite/adjacent. In this case, the opposite side would be the distance from the camera to the wall, and the adjacent side would be the distance along the ground from the camera to the point directly in front of it.Wait, no, actually, if the camera is at a height of 1.5 meters, but the field of view is horizontal, so maybe the vertical height doesn't come into play here. So, perhaps it's just a matter of the horizontal distances.Wait, I'm getting confused again. Let me try to clarify.The horizontal field of view is the angle that the camera can see from side to side. So, if it's 90 degrees, that means from the camera's position, the angle between the leftmost and rightmost points it can capture is 90 degrees.So, if the camera is 'd' meters away from the nearest wall, then the angle between the camera's central axis and the line of sight to the wall is 45 degrees (since 90 degrees total, split equally on both sides). So, tan(45) = (distance to wall) / (distance along the ground).Wait, but the distance along the ground is the same as the distance from the camera to the wall? No, wait, no. If the camera is at a height, then the distance along the ground is different from the distance to the wall.Wait, but the field of view is horizontal, so maybe it's purely about the horizontal distances, not considering the height.Wait, I think I need to model this as a right triangle where the camera is at a point, and the line of sight to the wall makes a 45-degree angle with the central axis.But since the field of view is horizontal, the height of the camera doesn't affect the horizontal field of view. So, maybe I can ignore the height for this part.So, if the camera has a horizontal field of view of 90 degrees, then the angle from the camera to each wall is 45 degrees on either side.So, if the camera is 'd' meters away from the nearest wall, then the distance from the camera to the farthest wall is 30 - d meters.But the angle between these two lines of sight is 90 degrees, so using the law of cosines:(30)^2 = d^2 + (30 - d)^2 - 2 * d * (30 - d) * cos(90°)Wait, is that correct? Let me think.Wait, no. The side opposite the 90-degree angle is the distance between the two walls, which is 30 meters. So, using the law of cosines:30² = d² + (30 - d)² - 2 * d * (30 - d) * cos(90°)But cos(90°) is 0, so the equation simplifies to:900 = d² + (30 - d)²Let me expand (30 - d)²:(30 - d)² = 900 - 60d + d²So, plugging back into the equation:900 = d² + 900 - 60d + d²Simplify:900 = 2d² - 60d + 900Subtract 900 from both sides:0 = 2d² - 60dFactor out 2d:0 = 2d(d - 30)So, the solutions are d = 0 or d = 30. But d = 0 would mean the camera is right at the wall, which doesn't make sense because then the other wall would be 30 meters away, but the field of view is 90 degrees, so it's possible? Wait, no, if the camera is at the wall, the angle to the farthest wall would be larger than 90 degrees, right?Wait, maybe I made a mistake in setting up the equation. Let me think again.If the camera is 'd' meters from the nearest wall, then the distance to the farthest wall is 30 - d meters. The angle between these two lines is 90 degrees. So, using the law of cosines:(30)^2 = d^2 + (30 - d)^2 - 2 * d * (30 - d) * cos(90°)But cos(90°) is 0, so:900 = d² + (30 - d)²Which leads to 900 = 2d² - 60d + 900, which simplifies to 0 = 2d² - 60d, so d = 0 or d = 30. But that can't be right because if d = 0, the camera is at the wall, and the angle to the farthest wall would be 90 degrees, but the distance is 30 meters, so the angle would actually be larger.Wait, maybe I'm misunderstanding the setup. Maybe the angle is not between the two walls, but the total field of view is 90 degrees, meaning that the camera can see 45 degrees to the left and 45 degrees to the right.So, if the camera is 'd' meters away from the nearest wall, then the distance to the farthest wall is 30 - d meters. The angle from the camera to the nearest wall is 45 degrees, and the angle to the farthest wall is also 45 degrees on the other side.Wait, no, that doesn't make sense because the total field of view is 90 degrees, so each side is 45 degrees from the center.Wait, maybe I should model this as two right triangles, each with an angle of 45 degrees, and the opposite side being 'd' and '30 - d'.But since the field of view is horizontal, the height of the camera doesn't affect the horizontal distances. So, maybe I can use the tangent of 45 degrees, which is 1, to find the distances.So, tan(45) = opposite / adjacent = 1 = (distance to wall) / (distance along the ground)But wait, the distance along the ground is the same as the distance from the camera to the wall because the field of view is horizontal. So, if tan(45) = 1, then the distance to the wall is equal to the distance along the ground.Wait, that doesn't make sense because the distance along the ground is the same as the distance to the wall. So, if the camera is 'd' meters from the wall, then tan(45) = d / d = 1, which is true, but that doesn't help me find 'd'.Wait, maybe I'm overcomplicating it. If the horizontal field of view is 90 degrees, then the camera can see 45 degrees to the left and 45 degrees to the right. So, the distance from the camera to each wall should be such that the angle from the camera to each wall is 45 degrees.So, if the camera is 'd' meters from the nearest wall, then the angle to that wall is 45 degrees. Similarly, the angle to the farthest wall is also 45 degrees on the other side.But wait, the distance to the farthest wall is 30 - d meters, and the angle to that wall is also 45 degrees. So, using the tangent function:tan(45) = (distance to wall) / (distance along the ground)But since the field of view is horizontal, the distance along the ground is the same as the distance to the wall. So, tan(45) = 1 = (distance to wall) / (distance to wall), which is just 1, which is always true. So, this approach isn't helping.Wait, maybe I need to consider that the total width of the lobby is 30 meters, and the camera's field of view is 90 degrees, so the width captured is 30 meters, which is the opposite side of a triangle with angle 90 degrees.Wait, maybe I should think of it as the camera's field of view forming a triangle where the base is 30 meters, and the angle at the camera is 90 degrees. So, using the formula for the area of a triangle, but I'm not sure.Alternatively, maybe I can use the formula for the width captured by a camera:Width = 2 * distance * tan(field of view / 2)But wait, that formula is usually for the width at a certain distance, but in this case, the width is fixed at 30 meters, and the field of view is 90 degrees. So, maybe:30 = 2 * distance * tan(90° / 2)Which is:30 = 2 * distance * tan(45°)Since tan(45°) is 1, this simplifies to:30 = 2 * distance * 1So, distance = 15 meters.Wait, that seems too straightforward. So, the minimum distance from the camera to the nearest wall would be 15 meters? But that would mean the camera is in the center of the lobby, 15 meters from each wall, and the field of view is 90 degrees, capturing the entire 30 meters.But wait, the camera is at a height of 1.5 meters. Does that affect the calculation? Because if the camera is elevated, the actual distance to the wall might be different due to the angle of elevation.Wait, but the field of view is given as horizontal, so maybe the height doesn't matter for the horizontal field of view. So, the calculation would still be 15 meters.But let me double-check. If the camera is 1.5 meters high, and it's 15 meters from the wall, then the line of sight to the wall would be at an angle. But since the field of view is horizontal, the horizontal distance is still 15 meters, regardless of the height.Wait, but actually, the horizontal field of view is determined by the angle between the left and right sides of the frame, so the height doesn't affect the horizontal field of view. So, the calculation of 15 meters is correct.But wait, the problem says "the minimum distance from the camera to the nearest wall". So, if the camera is 15 meters from the nearest wall, it's also 15 meters from the farthest wall, which is the center. But if the camera is closer to one wall, say 'd' meters, then the distance to the farthest wall is 30 - d meters. The field of view needs to cover the entire 30 meters, so the angle between the two walls as seen from the camera must be 90 degrees.Wait, so maybe the minimum distance is when the camera is as close as possible to one wall while still capturing the entire width. So, if the camera is closer to one wall, the angle to the farthest wall would be larger, but we need it to be exactly 90 degrees.Wait, so perhaps the minimum distance is when the angle to the farthest wall is 90 degrees, but that doesn't make sense because the angle can't be more than 90 degrees if the field of view is 90 degrees.Wait, I'm getting confused again. Let me try to set up the equation properly.Let me denote 'd' as the distance from the camera to the nearest wall. The distance to the farthest wall is then 30 - d meters. The angle between these two lines of sight is 90 degrees.Using the law of cosines:(30)^2 = d^2 + (30 - d)^2 - 2 * d * (30 - d) * cos(90°)But cos(90°) is 0, so:900 = d^2 + (30 - d)^2Expanding (30 - d)^2:900 = d^2 + 900 - 60d + d^2Simplify:900 = 2d^2 - 60d + 900Subtract 900 from both sides:0 = 2d^2 - 60dFactor:0 = 2d(d - 30)So, d = 0 or d = 30. But d = 0 would mean the camera is at the wall, which doesn't make sense because then the angle to the farthest wall would be larger than 90 degrees. Similarly, d = 30 would mean the camera is at the farthest wall, which is the same issue.Wait, this suggests that the only solutions are when the camera is at one of the walls, but that can't be right because the angle would be more than 90 degrees. So, maybe my initial approach is wrong.Alternatively, perhaps I should consider that the camera's field of view is 90 degrees, so the angle from the camera to each wall is 45 degrees. So, using the tangent function:tan(45) = (distance to wall) / (distance along the ground)But since the field of view is horizontal, the distance along the ground is the same as the distance to the wall. So, tan(45) = 1 = (distance to wall) / (distance to wall), which is always true, so this doesn't help.Wait, maybe I need to consider the height of the camera. Since the camera is 1.5 meters high, the line of sight to the wall is at an angle, so the horizontal distance is different from the actual distance to the wall.Wait, but the field of view is horizontal, so the horizontal distance is what matters. So, maybe the height doesn't affect the horizontal field of view.Wait, I'm stuck. Let me try a different approach.If the camera has a horizontal field of view of 90 degrees, then the width of the lobby that can be captured is determined by the distance from the camera to the walls. The formula for the width captured is:Width = 2 * distance * tan(field of view / 2)But in this case, the width is 30 meters, and the field of view is 90 degrees, so:30 = 2 * distance * tan(45°)Since tan(45°) is 1, this simplifies to:30 = 2 * distanceSo, distance = 15 meters.Therefore, the camera needs to be 15 meters away from the nearest wall to capture the entire width of the lobby with a 90-degree horizontal field of view.But wait, the problem says "the minimum distance from the camera to the nearest wall". So, if the camera is 15 meters from the nearest wall, it's also 15 meters from the farthest wall, meaning it's in the center. But if the camera is closer to one wall, say 'd' meters, then the distance to the farthest wall is 30 - d meters. The field of view needs to cover the entire 30 meters, so the angle between the two walls as seen from the camera must be 90 degrees.Wait, but earlier, when I tried using the law of cosines, I ended up with d = 0 or d = 30, which doesn't make sense. So, maybe the only way to have a 90-degree angle between the walls is to be equidistant from both walls, i.e., 15 meters from each.Therefore, the minimum distance from the camera to the nearest wall is 15 meters.Wait, but the problem says "minimum distance", so if the camera is closer than 15 meters to one wall, the angle to the farthest wall would be more than 90 degrees, which exceeds the camera's field of view. Therefore, the minimum distance is 15 meters.Okay, I think that makes sense. So, the answer to part 1 is 15 meters.Problem 2: The photographer wants to place the camera at a point where the angle of elevation to the chandelier is exactly 30 degrees. The chandelier is 10 meters high, and the camera is at 1.5 meters. So, the vertical distance from the camera to the chandelier is 10 - 1.5 = 8.5 meters.We need to find the horizontal distance from the point directly beneath the chandelier to the camera's position.This is a straightforward trigonometry problem. The angle of elevation is 30 degrees, the opposite side is 8.5 meters, and we need to find the adjacent side (horizontal distance).Using the tangent function:tan(30°) = opposite / adjacent = 8.5 / adjacentSo, adjacent = 8.5 / tan(30°)We know that tan(30°) = 1/√3 ≈ 0.57735So, adjacent = 8.5 / (1/√3) = 8.5 * √3Calculating that:√3 ≈ 1.732So, 8.5 * 1.732 ≈ 14.722 metersBut the problem asks for the exact value, so we can leave it in terms of √3.Therefore, the exact horizontal distance is 8.5√3 meters.But let me write that as a fraction to be precise. 8.5 is 17/2, so:(17/2) * √3 = (17√3)/2 meters.So, the exact horizontal distance is (17√3)/2 meters.Final Answer1. The minimum distance is boxed{15} meters.2. The exact horizontal distance is boxed{dfrac{17sqrt{3}}{2}} meters.</think>