answer：First, determine the total number of hours Alex works each day. He works 8 hours per day.Next, calculate the amount of time Alex spends meditating each day. He meditates for 15 minutes every hour, so for 8 hours, that's 8 multiplied by 15 minutes, which equals 120 minutes per day.Convert the daily meditation time from minutes to hours. Since there are 60 minutes in an hour, 120 minutes is equivalent to 2 hours per day.Finally, find the total meditation time for the entire week. Alex works 5 days a week, so multiply the daily meditation time by 5. That is 2 hours multiplied by 5, resulting in 10 hours of meditation per week.

answer：First, I need to determine how many primary source documents each student group will receive. Dr. Smith has 5 documents and wants to distribute them evenly among 4 groups. Dividing 5 by 4 gives 1.25, which means each group will receive 1 document, and one group will receive an extra document.Next, I'll calculate the number of research tips and strategies each group will get. There are 20 tips to be shared equally among the 4 groups. Dividing 20 by 4 results in 5 tips per group.Finally, to find the total number of resources each group will have, I'll add the number of documents and tips. Each group will have 1 document and 5 tips, totaling 6 resources per group. One group will have an additional document, making their total 7 resources.

answer：First, I need to determine the number of informational sessions held each day, which is 5.Next, I'll calculate the total number of sessions over the 7-day week by multiplying the daily sessions by the number of days: 5 sessions/day × 7 days = 35 sessions.Each session can accommodate 25 people, so to find the total number of people reached, I'll multiply the total number of sessions by the capacity per session: 35 sessions × 25 people/session = 875 people.Therefore, the vaccination campaign can reach a total of 875 people during the entire week.

answer：First, I need to determine the total number of gears and structural frames required for 8 robots. Each robot requires 5 gears and 3 structural frames, so for 8 robots, that's 40 gears and 24 structural frames.Next, I'll calculate how many gears TechPrint Co. can produce in the available 5 hours. Since they can produce 20 gears per hour, in 5 hours they can make 100 gears. This exceeds the required 40 gears, so there's no need to produce additional gears beyond the available time.Now, I'll determine how many structural frames can be produced in the remaining time after producing the necessary gears. Since producing 40 gears takes 2 hours, there are 3 hours left. In those 3 hours, they can produce 30 structural frames. However, they need 24 structural frames, so they have enough capacity within the available time.Finally, since the required number of gears and structural frames can be produced within the 5 hours, TechPrint Co. does not need any additional time beyond the 5 hours they have already allocated.

answer：Okay, so I have this problem where an experienced humanitarian aid worker is setting up three mobile medical clinics in a remote area. The area is a rectangle that's 100 km by 150 km. The population density is given by the function p(x, y) = 500e^{-0.01x - 0.02y} people per square kilometer. The first part is about optimally placing the three clinics so that the total distance from any point in the region to the nearest clinic is minimized. They mention using the Fermat-Weber point, which I think is also called the geometric median. Hmm, okay, so I need to figure out where to place these three clinics.The second part is about scheduling medical supplies. Each clinic needs supplies weekly, and the amount is proportional to the population density around it. The total needed is 3000 units. I need to model this and find out how much each clinic should get.Starting with the first part: optimal placement. I remember that the geometric median minimizes the sum of distances to a set of points. But in this case, it's about covering the entire region, not just a set of points. So maybe it's more about partitioning the region into three areas, each served by a clinic, such that the maximum distance from any point to the nearest clinic is minimized. That sounds like a covering problem.But the question specifically mentions the Fermat-Weber point, which is about minimizing the sum of weighted distances. So perhaps each clinic should be placed at the geometric median of its respective area. But since we have three clinics, we need to divide the region into three parts, each with its own geometric median.Wait, but the region is a rectangle, and the population density isn't uniform—it decreases exponentially with x and y. So the population is denser near the origin (0,0) and less dense as x and y increase. That means more people are concentrated towards the southwest corner, assuming x increases to the east and y increases to the north.So, to place the clinics optimally, we need to consider where the population is higher. Since the density decreases with x and y, the first clinic should probably be near (0,0). But how do we split the region into three parts?Maybe we can divide the region into three regions based on population density. The highest density is near (0,0), then it decreases as we move east or north. So perhaps the three regions are: one near the southwest, one near the southeast, and one near the northwest? Or maybe one in the center and two on the sides? Hmm.Alternatively, maybe it's better to think in terms of Voronoi diagrams. If we place three points, the region will be divided into three Voronoi cells, each containing the area closer to one clinic. The goal is to have each cell's centroid (or geometric median) as the clinic location.But since the population density isn't uniform, the geometric median won't just be the centroid in terms of area, but weighted by population. So each clinic should be placed at the geometric median of its Voronoi cell.But how do we determine the Voronoi cells? This seems like a chicken and egg problem because the cells depend on the clinic locations, and the clinic locations depend on the cells.Maybe an iterative approach is needed. Start with an initial guess for the clinic locations, compute the Voronoi cells, then compute the geometric median for each cell, and update the clinic locations accordingly. Repeat until convergence.But since this is a theoretical problem, maybe there's a smarter way. Let's think about the population distribution. The density function is p(x, y) = 500e^{-0.01x - 0.02y}. So the density decreases as x and y increase. The decay is faster in the y-direction (0.02) than in the x-direction (0.01). So the population is more concentrated along the x-axis than the y-axis.Therefore, the highest population is near (0,0), and it tapers off more quickly as we go north than as we go east.So, if we have three clinics, the first one should be near (0,0). The other two might be placed along the x-axis and y-axis, but considering the decay rates.Wait, but the region is 100 km by 150 km. So x goes from 0 to 100, y from 0 to 150.Given the decay rates, the population density at x=100, y=0 is 500e^{-1} ≈ 500*0.3679 ≈ 183.95 people per km².At x=0, y=150, it's 500e^{-3} ≈ 500*0.0498 ≈ 24.9 people per km².At x=100, y=150, it's 500e^{-1 -3} = 500e^{-4} ≈ 500*0.0183 ≈ 9.15 people per km².So the population is highest near (0,0), and decreases rapidly as we move north and east.Therefore, the first clinic should definitely be near (0,0). The other two clinics should probably be placed along the x-axis and y-axis, but where exactly?Alternatively, maybe the three clinics should be placed in such a way that each serves a region where the population density is roughly similar. So the first clinic serves the high-density area near (0,0), the second serves a medium-density area, and the third serves a low-density area.But how to formalize this?Alternatively, perhaps the optimal placement is to have the three clinics placed such that each is responsible for a third of the total population. So we can calculate the total population, divide by three, and find the regions where each clinic serves a third.But that might not necessarily minimize the total distance. Hmm.Wait, the problem says "the total distance from any point in the region to the nearest clinic is minimized." So it's about minimizing the maximum distance, or the sum of distances? Wait, the wording is a bit unclear. It says "the total distance from any point in the region to the nearest clinic is minimized." Hmm, that could be interpreted as minimizing the maximum distance, i.e., the farthest any point is from a clinic. Or it could be interpreted as minimizing the integral of the distance over the entire region.But the term "total distance" might refer to the sum, but since it's "from any point," it's a bit ambiguous. However, the mention of the Fermat-Weber point, which is about minimizing the sum of weighted distances, suggests that it's about minimizing the sum, weighted by population density.So, the total weighted distance is the integral over the region of the distance from each point to the nearest clinic, multiplied by the population density at that point. So we need to place three clinics such that this integral is minimized.This is a facility location problem with three facilities, where the cost is the weighted distance. The optimal locations are the geometric medians for the regions each clinic serves.But how do we partition the region into three parts such that each part's geometric median is a clinic location, and the total cost is minimized.This seems complex. Maybe we can approximate it by dividing the region into three roughly equal population areas, each served by a clinic.First, let's compute the total population of the region.Total population P = ∫∫ p(x,y) dx dy over the region.So P = ∫_{0}^{100} ∫_{0}^{150} 500e^{-0.01x -0.02y} dy dx.We can separate the integrals:P = 500 ∫_{0}^{100} e^{-0.01x} dx ∫_{0}^{150} e^{-0.02y} dy.Compute ∫ e^{-0.01x} dx from 0 to 100:Integral of e^{-ax} dx = (-1/a)e^{-ax}.So ∫_{0}^{100} e^{-0.01x} dx = (-100)[e^{-1} - 1] = 100(1 - e^{-1}) ≈ 100*(1 - 0.3679) ≈ 100*0.6321 ≈ 63.21.Similarly, ∫_{0}^{150} e^{-0.02y} dy = (-50)[e^{-3} - 1] = 50(1 - e^{-3}) ≈ 50*(1 - 0.0498) ≈ 50*0.9502 ≈ 47.51.So total population P = 500 * 63.21 * 47.51 ≈ 500 * 3014.7 ≈ 1,507,350 people.Wait, let me compute that again.Wait, 63.21 * 47.51 ≈ 63.21*47.51 ≈ let's compute 63*47.5 = 63*40 + 63*7.5 = 2520 + 472.5 = 2992.5. Then 0.21*47.51 ≈ 9.977. So total ≈ 2992.5 + 9.977 ≈ 3002.477.So P ≈ 500 * 3002.477 ≈ 1,501,238.5 people.So approximately 1.5 million people.If we need three clinics, each serving roughly a third, that's about 500,000 people per clinic.But how to partition the region into three parts with roughly equal population.Given the population density decreases with x and y, the first region would be near (0,0), the second might extend further east, and the third further north, but considering the decay rates.Alternatively, maybe the three regions are along the x-axis, y-axis, and somewhere in the middle.But this is getting complicated. Maybe a better approach is to use the concept of the geometric median for multiple points.But since the region is continuous, it's more about finding three points that minimize the integral of the distance from each point to the nearest clinic, weighted by population density.This is similar to k-means clustering, but with a continuous distribution.In k-means, you partition the space into clusters where each cluster's centroid minimizes the sum of squared distances. Here, it's similar but with weighted distances and possibly different distance metrics.But since it's a weighted sum, and the density is given, we need to find three points (clinics) such that the integral over the region of the minimum distance from each point to the clinics, multiplied by the density, is minimized.This is a complex optimization problem. Maybe we can approximate it by dividing the region into three parts where each part has roughly equal population, and then placing each clinic at the geometric median of its part.So first, let's find the cumulative population as we move along x and y.Given the density function, we can compute the cumulative population in a rectangle from (0,0) to (x,y).But perhaps it's easier to think in terms of dividing the region into three parts with equal population.Given that the density decreases exponentially, the first third of the population is concentrated near (0,0), the second third is a bit further out, and the last third is even further.But how to find the boundaries?Alternatively, maybe we can parameterize the region and find the points where the cumulative population reaches 1/3 and 2/3 of the total.But this is a bit involved.Alternatively, maybe we can use the fact that the density is separable into x and y components. So p(x,y) = 500e^{-0.01x}e^{-0.02y}.Therefore, the cumulative population up to x and y is the product of the cumulative in x and cumulative in y.So, let me define F(x) = ∫_{0}^{x} 500e^{-0.01t} dt = 500 * [ (-100)(e^{-0.01x} - 1) ] = 50000(1 - e^{-0.01x}).Similarly, G(y) = ∫_{0}^{y} 500e^{-0.02t} dt = 500 * [ (-50)(e^{-0.02y} - 1) ] = 25000(1 - e^{-0.02y}).But wait, actually, the total population is F(100)*G(150)/500, but I think I need to adjust.Wait, no. The total population is ∫∫ p(x,y) dx dy = 500 ∫ e^{-0.01x} dx ∫ e^{-0.02y} dy = 500 * [100(1 - e^{-1})] * [50(1 - e^{-3})] ≈ 500 * 63.21 * 47.51 ≈ 1,501,238 as before.So, the cumulative population up to x and y is ∫_{0}^{x} ∫_{0}^{y} 500e^{-0.01t -0.02s} ds dt = 500 ∫_{0}^{x} e^{-0.01t} [ ∫_{0}^{y} e^{-0.02s} ds ] dt = 500 ∫_{0}^{x} e^{-0.01t} [ (-50)(e^{-0.02y} - 1) ] dt = 500*(-50) ∫_{0}^{x} e^{-0.01t} (e^{-0.02y} - 1) dt.Wait, this is getting messy. Maybe instead, since the density is separable, the cumulative distribution function can be expressed as the product of the marginal distributions.Wait, actually, for independent variables, the joint CDF is the product of the marginal CDFs. But here, x and y are not independent in the density function, but the density is separable, so maybe the CDF can be expressed as the product.Wait, let me think. If p(x,y) = p_x(x) * p_y(y), then the joint CDF is indeed the product of the marginal CDFs. So in this case, p(x,y) = 500e^{-0.01x -0.02y} = (500e^{-0.01x}) * (e^{-0.02y}).Wait, no, because 500 is a constant. So actually, p(x,y) = (500e^{-0.01x}) * (e^{-0.02y}) / (500). Wait, no, that's not correct.Wait, actually, p(x,y) = 500e^{-0.01x -0.02y} = 500e^{-0.01x} * e^{-0.02y}.So, if we consider p_x(x) = 500e^{-0.01x} and p_y(y) = e^{-0.02y}, but that doesn't make sense because p_x(x) integrated over x should be 1, but 500e^{-0.01x} integrated from 0 to 100 is 500*100(1 - e^{-1}) ≈ 500*63.21 ≈ 31,605, which is not 1. So that approach isn't directly applicable.Alternatively, maybe we can normalize the density functions.Let me define f(x) = 500e^{-0.01x} and g(y) = e^{-0.02y}, but then p(x,y) = f(x)g(y)/C, where C is the normalization constant.Wait, but p(x,y) is already given as 500e^{-0.01x -0.02y}, so maybe f(x) = 500e^{-0.01x} and g(y) = e^{-0.02y}, but then the total integral is ∫∫ f(x)g(y) dx dy = (∫ f(x) dx)(∫ g(y) dy) = (500*100(1 - e^{-1}))(∫_{0}^{150} e^{-0.02y} dy).Wait, but we already computed that as 500*63.21*47.51 ≈ 1,501,238.So, to make p(x,y) a proper density, it should integrate to 1. But in this case, it's given as 500e^{-0.01x -0.02y}, which doesn't integrate to 1. So perhaps it's not a probability density, but a population density.So, the total population is 1,501,238 as computed.So, to find the regions where each clinic serves approximately 500,000 people, we can find the boundaries in x and y where the cumulative population reaches 1/3 and 2/3.But since the density is separable, maybe we can find x1 and y1 such that the cumulative population up to x1 and y1 is 500,000.But this is a bit involved. Let me try.The cumulative population up to x and y is:C(x,y) = ∫_{0}^{x} ∫_{0}^{y} 500e^{-0.01t -0.02s} ds dt.We can separate the integrals:C(x,y) = 500 ∫_{0}^{x} e^{-0.01t} dt ∫_{0}^{y} e^{-0.02s} ds.Compute ∫ e^{-0.01t} dt from 0 to x: 100(1 - e^{-0.01x}).Compute ∫ e^{-0.02s} ds from 0 to y: 50(1 - e^{-0.02y}).So, C(x,y) = 500 * 100(1 - e^{-0.01x}) * 50(1 - e^{-0.02y}) = 500*100*50*(1 - e^{-0.01x})(1 - e^{-0.02y}).Wait, that can't be right because 500*100*50 is 2,500,000, which is much larger than the total population we computed earlier (~1.5 million). So I must have made a mistake.Wait, no, actually, the integral of p(x,y) over the entire region is 1,501,238, so C(x,y) should be less than that. Let's re-express.Wait, C(x,y) = ∫_{0}^{x} ∫_{0}^{y} 500e^{-0.01t -0.02s} ds dt = 500 ∫_{0}^{x} e^{-0.01t} [ ∫_{0}^{y} e^{-0.02s} ds ] dt.Compute inner integral: ∫_{0}^{y} e^{-0.02s} ds = (-50)(e^{-0.02y} - 1) = 50(1 - e^{-0.02y}).Then, C(x,y) = 500 ∫_{0}^{x} e^{-0.01t} * 50(1 - e^{-0.02y}) dt = 500*50*(1 - e^{-0.02y}) ∫_{0}^{x} e^{-0.01t} dt.Compute ∫ e^{-0.01t} dt from 0 to x: (-100)(e^{-0.01x} - 1) = 100(1 - e^{-0.01x}).So, C(x,y) = 500*50*(1 - e^{-0.02y})*100*(1 - e^{-0.01x}) = 500*50*100*(1 - e^{-0.01x})(1 - e^{-0.02y}).Wait, that's 25,000,000*(1 - e^{-0.01x})(1 - e^{-0.02y}), which is way too big because the total population is only ~1.5 million. So clearly, I'm making a mistake in the integration constants.Wait, let's go back.The density is p(x,y) = 500e^{-0.01x -0.02y}.So, the cumulative population up to x and y is:C(x,y) = ∫_{0}^{x} ∫_{0}^{y} 500e^{-0.01t -0.02s} ds dt.We can separate the integrals:C(x,y) = 500 ∫_{0}^{x} e^{-0.01t} dt ∫_{0}^{y} e^{-0.02s} ds.Compute ∫ e^{-0.01t} dt from 0 to x: [ (-1/0.01)e^{-0.01t} ] from 0 to x = (-100)(e^{-0.01x} - 1) = 100(1 - e^{-0.01x}).Similarly, ∫ e^{-0.02s} ds from 0 to y: [ (-1/0.02)e^{-0.02s} ] from 0 to y = (-50)(e^{-0.02y} - 1) = 50(1 - e^{-0.02y}).So, C(x,y) = 500 * 100(1 - e^{-0.01x}) * 50(1 - e^{-0.02y}) = 500 * 100 * 50 * (1 - e^{-0.01x})(1 - e^{-0.02y}).Wait, 500*100*50 = 2,500,000, which is much larger than the total population. That can't be right. So I must have messed up the units.Wait, no, actually, the density is 500 people per km², so when we integrate over x and y, we get the total population. But the way I separated the integrals, I think I have an extra factor.Wait, let me re-express:C(x,y) = ∫_{0}^{x} ∫_{0}^{y} 500e^{-0.01t -0.02s} ds dt = 500 ∫_{0}^{x} e^{-0.01t} dt ∫_{0}^{y} e^{-0.02s} ds.But ∫_{0}^{x} e^{-0.01t} dt = 100(1 - e^{-0.01x}).Similarly, ∫_{0}^{y} e^{-0.02s} ds = 50(1 - e^{-0.02y}).So, C(x,y) = 500 * 100(1 - e^{-0.01x}) * 50(1 - e^{-0.02y}) = 500 * 100 * 50 * (1 - e^{-0.01x})(1 - e^{-0.02y}).Wait, that's 500*100*50 = 2,500,000, which is way too big. But the total population is only ~1.5 million. So clearly, I'm making a mistake.Wait, no, actually, the total population is ∫_{0}^{100} ∫_{0}^{150} 500e^{-0.01x -0.02y} dy dx = 500 * ∫_{0}^{100} e^{-0.01x} dx ∫_{0}^{150} e^{-0.02y} dy.We computed ∫ e^{-0.01x} dx from 0 to 100 as ~63.21, and ∫ e^{-0.02y} dy from 0 to 150 as ~47.51.So, total population is 500 * 63.21 * 47.51 ≈ 1,501,238.Therefore, C(x,y) = 500 * ∫_{0}^{x} e^{-0.01t} dt * ∫_{0}^{y} e^{-0.02s} ds = 500 * [100(1 - e^{-0.01x})] * [50(1 - e^{-0.02y})] = 500 * 100 * 50 * (1 - e^{-0.01x})(1 - e^{-0.02y}) / (100*50).Wait, no, that doesn't make sense. Wait, I think I need to normalize the integrals.Wait, actually, ∫_{0}^{x} e^{-0.01t} dt = 100(1 - e^{-0.01x}).Similarly, ∫_{0}^{y} e^{-0.02s} ds = 50(1 - e^{-0.02y}).So, C(x,y) = 500 * [100(1 - e^{-0.01x})] * [50(1 - e^{-0.02y})] / (100*50).Wait, that is, because the total integral is 500*100*50*(1 - e^{-1})(1 - e^{-3}) ≈ 500*63.21*47.51 ≈ 1.5 million.So, to get C(x,y), we have:C(x,y) = [500 * 100 * 50 * (1 - e^{-0.01x})(1 - e^{-0.02y})] / (100*50) = 500*(1 - e^{-0.01x})(1 - e^{-0.02y}).Wait, that makes more sense because when x=100 and y=150, C(100,150) = 500*(1 - e^{-1})(1 - e^{-3}) ≈ 500*0.6321*0.9502 ≈ 500*0.600 ≈ 300, which is not matching the total population. Wait, no, 500*(1 - e^{-1})(1 - e^{-3}) ≈ 500*0.6321*0.9502 ≈ 500*0.600 ≈ 300, but the total population is ~1.5 million. So clearly, this approach is flawed.I think I need to abandon trying to compute C(x,y) directly and instead think differently.Since the density is separable, maybe the optimal clinic locations can be found by considering the one-dimensional cases separately.In one dimension, the geometric median minimizes the sum of weighted distances. For a density function, it's the point where the cumulative density is half.But in two dimensions, it's more complex.Alternatively, since the density is separable, maybe the optimal points are along the axes where the cumulative density in x and y reaches certain thresholds.But I'm not sure.Alternatively, maybe the three clinics should be placed at points where the population density is one-third of the maximum.The maximum density is at (0,0): p(0,0) = 500.So one-third is ~166.67.Set p(x,y) = 166.67: 500e^{-0.01x -0.02y} = 166.67.Divide both sides by 500: e^{-0.01x -0.02y} = 166.67/500 ≈ 0.3333.Take natural log: -0.01x -0.02y = ln(0.3333) ≈ -1.0986.So 0.01x + 0.02y ≈ 1.0986.This is a straight line in the x-y plane. So the region where p(x,y) ≥ 166.67 is below this line.Similarly, for p(x,y) = 500e^{-0.01x -0.02y} = 500/3 ≈ 166.67.So, the first clinic should be near (0,0), covering the area where p(x,y) ≥ 166.67.Then, the second and third clinics can be placed in regions where the density is lower.But this is just a rough idea.Alternatively, maybe the optimal clinics are placed at the points where the cumulative population is 1/3 and 2/3.But given the complexity, maybe it's better to approximate the clinic locations.Given the density is highest near (0,0), the first clinic is definitely near there. Let's say (x1, y1) ≈ (0,0).Then, the second and third clinics should be placed in regions where the population is next highest.Given the decay rates, moving east (increasing x) decreases density slower than moving north (increasing y). So, perhaps the second clinic is placed further east, and the third further north.But how far?Alternatively, maybe the three clinics are placed along the line where the density is equal, i.e., where 0.01x + 0.02y is constant.But I'm not sure.Alternatively, maybe the optimal points are the centroids of the regions where the density is above certain thresholds.But this is getting too vague.Alternatively, perhaps the problem expects us to use the concept that the geometric median for a region with density p(x,y) is the point (x,y) that minimizes the integral of distance from (x,y) to all other points, weighted by p.But with three clinics, it's a multi-facility location problem.Given the complexity, maybe the answer expects us to place the clinics at specific points based on the density function.Alternatively, maybe the optimal points are where the partial derivatives of the total distance function are zero.But without a clear method, I think I need to make an educated guess.Given the density decreases with x and y, and the decay is faster in y, the population is more spread out along the x-axis.So, perhaps the three clinics are placed along the x-axis at certain intervals.But considering the decay, the first clinic is near (0,0), the second somewhere in the middle, and the third near (100,0). But that might not account for the y-direction.Alternatively, maybe the clinics are placed in a triangular formation, but given the rectangle, maybe along the main diagonals.But without a clear method, I think I need to simplify.Let me assume that the optimal clinics are placed at points where the density is one-third of the maximum, so p(x,y) = 500/3 ≈ 166.67.As before, 500e^{-0.01x -0.02y} = 166.67.So, e^{-0.01x -0.02y} = 1/3.Taking natural log: -0.01x -0.02y = -ln(3) ≈ -1.0986.So, 0.01x + 0.02y ≈ 1.0986.This is a straight line. So, the region where p(x,y) ≥ 166.67 is below this line.So, the first clinic is in the region below this line, which is the highest density area.Then, the second and third clinics can be placed in regions where p(x,y) is next highest.But this is still vague.Alternatively, maybe the optimal points are where the cumulative population is 1/3 and 2/3.Given the total population is ~1.5 million, each clinic should serve ~500,000.So, find x and y such that the cumulative population up to x and y is 500,000.But given the separable density, maybe we can find x and y such that:C(x,y) = 500,000 = 500 ∫_{0}^{x} e^{-0.01t} dt ∫_{0}^{y} e^{-0.02s} ds.But as we saw earlier, this leads to a complex equation.Alternatively, maybe we can find x such that the cumulative population along x is 500,000, ignoring y.But that's not accurate because y also contributes.Alternatively, maybe we can find x and y such that the product of the cumulative in x and cumulative in y equals 500,000 / 500 = 1000.Wait, no, because C(x,y) = 500 * ∫_{0}^{x} e^{-0.01t} dt * ∫_{0}^{y} e^{-0.02s} ds.So, 500 * A(x) * B(y) = 500,000.Thus, A(x) * B(y) = 1000.Where A(x) = ∫_{0}^{x} e^{-0.01t} dt = 100(1 - e^{-0.01x}).B(y) = ∫_{0}^{y} e^{-0.02s} ds = 50(1 - e^{-0.02y}).So, 100(1 - e^{-0.01x}) * 50(1 - e^{-0.02y}) = 1000.Simplify: 5000(1 - e^{-0.01x})(1 - e^{-0.02y}) = 1000.Divide both sides by 5000: (1 - e^{-0.01x})(1 - e^{-0.02y}) = 0.2.So, we have (1 - e^{-0.01x})(1 - e^{-0.02y}) = 0.2.This equation defines a curve in the x-y plane. To find specific points, we can make assumptions.For example, assume that x and y are such that (1 - e^{-0.01x}) = (1 - e^{-0.02y}) = sqrt(0.2) ≈ 0.4472.But sqrt(0.2) is ~0.447, so:1 - e^{-0.01x} = 0.447 => e^{-0.01x} = 0.553 => -0.01x = ln(0.553) ≈ -0.592 => x ≈ 59.2 km.Similarly, 1 - e^{-0.02y} = 0.447 => e^{-0.02y} = 0.553 => -0.02y = ln(0.553) ≈ -0.592 => y ≈ 29.6 km.So, one possible point is (59.2, 29.6). But this is just one point on the curve.Alternatively, we can set x=0 and solve for y:(1 - e^{0})(1 - e^{-0.02y}) = 0.2 => 0*(1 - e^{-0.02y}) = 0.2, which is impossible. So x can't be 0.Similarly, set y=0:(1 - e^{-0.01x})(1 - e^{0}) = 0.2 => (1 - e^{-0.01x})*0 = 0.2, which is also impossible.So, the curve doesn't intersect the axes.Alternatively, set x=50:1 - e^{-0.5} ≈ 1 - 0.6065 ≈ 0.3935.Then, 0.3935*(1 - e^{-0.02y}) = 0.2 => (1 - e^{-0.02y}) ≈ 0.2/0.3935 ≈ 0.508.So, e^{-0.02y} ≈ 1 - 0.508 ≈ 0.492 => -0.02y ≈ ln(0.492) ≈ -0.707 => y ≈ 35.35 km.So, another point is (50, 35.35).Similarly, set y=50:1 - e^{-0.02*50} = 1 - e^{-1} ≈ 0.6321.Then, (1 - e^{-0.01x})*0.6321 = 0.2 => (1 - e^{-0.01x}) ≈ 0.2/0.6321 ≈ 0.316.So, e^{-0.01x} ≈ 1 - 0.316 ≈ 0.684 => -0.01x ≈ ln(0.684) ≈ -0.380 => x ≈ 38 km.So, another point is (38, 50).So, these are points on the curve where the cumulative population is 500,000.But how does this help us place the clinics?Well, if we consider that each clinic should serve a region with cumulative population 500,000, then the first clinic is in the region where p(x,y) ≥ 166.67, which is below the line 0.01x + 0.02y ≈ 1.0986.Then, the second and third clinics can be placed in regions where the cumulative population is 500,000 each.But this is getting too vague.Alternatively, maybe the optimal clinics are placed at the points where the cumulative population is 1/3 and 2/3 along the line of decreasing density.But I'm not sure.Given the time I've spent and the complexity, I think I need to make an educated guess.Given the density is highest near (0,0), and decreases with x and y, the first clinic is near (0,0). The second and third clinics should be placed in regions where the population is next highest, which would be along the x-axis and y-axis.But considering the decay rates, moving east (x) is less steep than moving north (y), so the population decreases more slowly along x.Therefore, the second clinic should be placed further east, and the third further north.But how far?Alternatively, maybe the optimal points are where the density is one-third of the maximum, which we found earlier as the line 0.01x + 0.02y ≈ 1.0986.So, points on this line could be potential locations.For example, if we set y=0, then 0.01x ≈ 1.0986 => x ≈ 109.86 km, which is beyond the region (max x=100). So, not possible.If we set x=100, then 0.01*100 + 0.02y ≈ 1.0986 => 1 + 0.02y ≈ 1.0986 => 0.02y ≈ 0.0986 => y ≈ 4.93 km.So, one point is (100, 4.93).Similarly, if we set y=150, 0.01x + 0.02*150 ≈ 1.0986 => 0.01x + 3 ≈ 1.0986 => 0.01x ≈ -1.9014, which is negative, so not possible.Alternatively, set x=50: 0.5 + 0.02y ≈ 1.0986 => 0.02y ≈ 0.5986 => y ≈ 29.93 km.So, another point is (50, 29.93).Similarly, set y=50: 0.01x + 1 ≈ 1.0986 => 0.01x ≈ 0.0986 => x ≈ 9.86 km.So, another point is (9.86, 50).So, these are points on the line where p(x,y) = 166.67.But how does this help us place the clinics?Perhaps the three clinics are placed at these points: (0,0), (50, 29.93), and (9.86, 50). But I'm not sure.Alternatively, maybe the clinics are placed at the centroids of the regions where p(x,y) is above certain thresholds.But without a clear method, I think I need to conclude that the optimal clinics are placed near (0,0), (50, 30), and (10, 50), but I'm not certain.Given the time I've spent, I think I'll proceed to the second part, as it might be more straightforward.The second part is about scheduling medical supplies. Each clinic needs supplies proportional to the population density in its vicinity. Total weekly supply is 3000 units.So, we need to model the supply distribution.Assuming that the supply needed at each clinic is proportional to the integral of the population density in its region.So, if we have three regions R1, R2, R3, each served by a clinic, then the supply for clinic i is proportional to ∫∫_{Ri} p(x,y) dx dy.Given that the total supply is 3000, we can compute the supplies as fractions of the total population.But since we don't have the exact regions, we can assume that each clinic serves a region with equal population, so each gets 1/3 of the total supply.But wait, the problem says "the amount of supplies needed at each clinic is proportional to the population density in its vicinity."So, it's proportional to the population in its vicinity, not necessarily equal regions.But without knowing the regions, we can't compute the exact supplies.But perhaps, given that the clinics are placed at the geometric medians, which are points that minimize the sum of weighted distances, the supplies would be proportional to the population in each Voronoi cell.But since we don't have the exact regions, maybe we can assume that each clinic serves a region with population proportional to the density at the clinic location.But that might not be accurate.Alternatively, maybe the supplies are proportional to the density at the clinic locations.But the density function is p(x,y) = 500e^{-0.01x -0.02y}.So, if we have clinics at (x1,y1), (x2,y2), (x3,y3), then the supplies would be proportional to p(x1,y1), p(x2,y2), p(x3,y3).But the total supply is 3000, so we can compute the supplies as:S1 = 3000 * p(x1,y1) / (p(x1,y1) + p(x2,y2) + p(x3,y3)),and similarly for S2 and S3.But without knowing the clinic locations, we can't compute this.Alternatively, if the clinics are placed such that each serves a region with equal population, then each would get 1000 units.But the problem says "proportional to the population density in its vicinity," which might mean proportional to the integral of p(x,y) over the region served by the clinic.But without knowing the regions, it's hard to compute.Alternatively, maybe the supplies are proportional to the density at the clinic locations.But again, without knowing the locations, we can't compute.Alternatively, maybe the problem expects us to assume that the supplies are divided equally, so each clinic gets 1000 units.But that might not be correct.Alternatively, maybe the supplies are proportional to the density at the centroid of the region.But the centroid is (50,75), but the density there is p(50,75) = 500e^{-0.5 -1.5} = 500e^{-2} ≈ 500*0.1353 ≈ 67.65.But if we have three clinics, maybe each is placed at a point where the density is one-third of the maximum.But as before, p(x,y) = 500/3 ≈ 166.67.So, the density at each clinic is 166.67, so the supplies would be proportional to 166.67 each, so total is 3*166.67 ≈ 500, but the total supply is 3000, so each gets 3000*(166.67/500) ≈ 3000*0.333 ≈ 1000 each.So, each clinic gets 1000 units.But this is a rough approximation.Alternatively, maybe the supplies are divided based on the regions each clinic serves.But without knowing the regions, it's hard.Given the time I've spent, I think I'll conclude that each clinic gets 1000 units, as the total is 3000 and three clinics.But I'm not entirely sure.So, to summarize:1. Optimal Placement: Clinics are placed near (0,0), (50,30), and (10,50). But I'm not certain.2. Supplies: Each clinic gets 1000 units.But I think the supplies should be proportional to the population in their regions, which might not be equal.Given that the density is highest near (0,0), the first clinic should get more supplies, and the others less.But without knowing the regions, it's hard.Alternatively, maybe the supplies are proportional to the density at the clinic locations.If the clinics are placed at (0,0), (50,30), and (10,50), then:p(0,0) = 500,p(50,30) = 500e^{-0.5 -0.6} = 500e^{-1.1} ≈ 500*0.3329 ≈ 166.45,p(10,50) = 500e^{-0.1 -1} = 500e^{-1.1} ≈ same as above ≈ 166.45.So, total density is 500 + 166.45 + 166.45 ≈ 832.9.So, the supplies would be:S1 = 3000*(500/832.9) ≈ 3000*0.600 ≈ 1800,S2 = 3000*(166.45/832.9) ≈ 3000*0.200 ≈ 600,S3 = same as S2 ≈ 600.So, approximately 1800, 600, 600.But this is just an assumption.Alternatively, if the regions are equal population, each gets 1000.But given the density is highest near (0,0), it's more likely that the first clinic serves a smaller area but with higher density, so it needs more supplies.Therefore, the supplies should be distributed as approximately 1800, 600, 600.But I'm not entirely sure.Given the time I've spent, I think I'll go with this.

answer：First, I need to determine how many cookies each person should bake. There are four people in total: Jamie, Taylor, Alex, and Bailey. They want to make 60 cookies in total.To find out how many cookies each person should bake, I'll divide the total number of cookies by the number of people. So, 60 cookies divided by 4 people equals 15 cookies per person.Next, after baking, they want to divide the cookies equally among the four of them to take home. Since they've already baked 60 cookies in total, I'll divide this number by 4 again. This means each person will take home 15 cookies.Therefore, each person bakes 15 cookies and takes home 15 cookies.