🧠:<think>Alright, so I have this problem about a Great Stellated Dodecahedron sculpture. It's a complex polyhedron, and I need to figure out two things: the total surface area and the cost of installing lights at each vertex. Let me take this step by step.Starting with the first part: calculating the total surface area. The sculpture has 12 intersecting pentagrammic faces. Each face is a regular pentagram, and each outer edge is 2 meters. The internal angle at each vertex of the pentagon is 108 degrees. Hmm, okay. So, I need to find the area of one pentagram and then multiply it by 12, but I have to account for overlapping regions because the faces intersect each other.Wait, so each face is a pentagram, which is a star-shaped figure formed by connecting every other vertex of a regular pentagon. But how does that affect the surface area? Is each face contributing its own area, or do the overlapping parts mean I have to subtract some areas?I think in a stellated polyhedron, the faces are extended until they intersect, forming a star. So, each pentagram is a face, but when they intersect, parts of the pentagrams overlap. Therefore, the total surface area isn't just 12 times the area of a single pentagram because some areas are covered multiple times.But how do I calculate the actual surface area without double-counting the overlapping regions? Maybe I need to find the area of one pentagram and then figure out how many times each region is overlapped.Alternatively, perhaps the Great Stellated Dodecahedron can be thought of as a compound of 12 pentagrams, but each face is a pentagram, so maybe the surface area is just 12 times the area of a single pentagram. But I'm not sure if overlapping affects the total surface area in this context.Wait, in terms of surface area, even if the faces intersect, each face is still a separate entity contributing to the overall surface. So, maybe the total surface area is indeed 12 times the area of one pentagram. But I should verify this.Let me recall that the Great Stellated Dodecahedron is one of the Kepler-Poinsot polyhedra. It has 12 faces, each a regular pentagram. It's non-convex and has intersecting faces. So, each face is a pentagram, and they intersect each other, but each face is still a distinct face contributing to the surface area.Therefore, perhaps the total surface area is simply 12 times the area of one pentagram. So, I need to calculate the area of a regular pentagram with an outer edge length of 2 meters.But wait, the formula for the area of a regular pentagram... I remember that a pentagram can be considered as a regular pentagon with five isosceles triangles removed from it. Alternatively, it can be seen as a union of a pentagon and five triangles.Wait, no. Actually, a regular pentagram is a star polygon with Schläfli symbol {5/2}. Its area can be calculated if we know the edge length.I think the area of a regular pentagram can be calculated using the formula:Area = (5/2) * s^2 / (tan(π/5))where s is the edge length.But let me verify this. Alternatively, the area can be calculated as the area of the pentagon plus the area of the five triangles that make up the points of the star.Wait, actually, a regular pentagram can be divided into a regular pentagon and five congruent isosceles triangles. Each triangle has a base equal to the edge length of the pentagram and two equal sides.But I need to figure out the area of the pentagram. Maybe I can use the formula for the area of a regular star polygon, which is:Area = (5/2) * s^2 * (1 / tan(π/5))Yes, that seems familiar. Let me plug in s = 2 meters.So, Area = (5/2) * (2)^2 / tan(π/5)First, calculate tan(π/5). π is approximately 3.1416, so π/5 is about 0.6283 radians. The tangent of that is approximately tan(0.6283) ≈ 0.7265.So, Area ≈ (5/2) * 4 / 0.7265 ≈ (10) / 0.7265 ≈ 13.7638 square meters.Wait, that seems a bit high. Let me double-check.Alternatively, another formula for the area of a regular pentagram is:Area = (5/2) * s^2 * (sqrt(5 + 2*sqrt(5)))/2Wait, that might be the formula for the area of a regular pentagon. Let me recall.The area of a regular pentagon with side length s is:Area = (5/2) * s^2 / (tan(π/5)) ≈ 1.7205 * s^2Wait, so if s = 2, then the area is approximately 1.7205 * 4 ≈ 6.882 square meters.But a pentagram is a star, so its area should be larger than the pentagon. Wait, no. Actually, the pentagram is formed by connecting the vertices of a pentagon, so the area of the pentagram is actually the area of the pentagon plus the area of the five triangles that make up the points.But I think the formula I used earlier is correct for the pentagram. Let me check.Wait, actually, the area of a regular pentagram can be calculated as:Area = (5/2) * s^2 * (1 / tan(π/5)) ≈ (5/2) * 4 / 0.7265 ≈ 10 / 0.7265 ≈ 13.7638 m².But let me think about this differently. A regular pentagram can be considered as a union of a regular pentagon and five congruent isosceles triangles. Each triangle has a base equal to the edge length of the pentagram and two equal sides.Wait, but actually, the triangles are the points of the star, so each triangle has a base equal to the edge length of the pentagram, but the height of each triangle is the distance from the base to the tip of the star.Alternatively, perhaps it's better to use the formula for the area of a regular star polygon, which is:Area = (n * s^2) / (4 * tan(π/n)) * (1 + 2 * cos(2π/n))where n is the number of points. For a pentagram, n = 5.Wait, let me check that formula. I think it might be:Area = (5/2) * s^2 * (1 / tan(π/5)) * (1 + 2 * cos(2π/5))But I'm getting confused. Maybe I should look up the formula for the area of a regular pentagram.Wait, I can't look things up, so I need to derive it.A regular pentagram can be divided into a regular pentagon and five congruent isosceles triangles. Each triangle has a base equal to the edge length of the pentagram, which is 2 meters, and two equal sides which are the edges of the triangles.But to find the area, I need to find the area of the pentagon and the area of the five triangles.First, the area of the regular pentagon with side length s is:Area_pentagon = (5/2) * s^2 / (tan(π/5)) ≈ 1.7205 * s^2So, for s = 2, Area_pentagon ≈ 1.7205 * 4 ≈ 6.882 m².Now, each triangle is an isosceles triangle with base s = 2 meters. The two equal sides are the edges of the star, which are longer than the base.Wait, actually, in a regular pentagram, the triangles are golden triangles. The ratio of the equal sides to the base is the golden ratio, φ = (1 + sqrt(5))/2 ≈ 1.618.So, the equal sides of each triangle are φ * s ≈ 1.618 * 2 ≈ 3.236 meters.But I need the height of each triangle to calculate its area. The height can be found using Pythagoras' theorem.Each triangle has a base of 2 meters and two equal sides of approximately 3.236 meters. The height h can be calculated as:h = sqrt((3.236)^2 - (1)^2) ≈ sqrt(10.472 - 1) ≈ sqrt(9.472) ≈ 3.078 meters.Wait, but actually, the base is 2 meters, so half the base is 1 meter. So, yes, h ≈ sqrt(3.236² - 1²) ≈ sqrt(10.472 - 1) ≈ sqrt(9.472) ≈ 3.078 meters.So, the area of one triangle is (base * height)/2 ≈ (2 * 3.078)/2 ≈ 3.078 m².Since there are five such triangles, the total area of the triangles is 5 * 3.078 ≈ 15.39 m².Therefore, the total area of the pentagram is the area of the pentagon plus the area of the five triangles:Area_pentagram ≈ 6.882 + 15.39 ≈ 22.272 m².Wait, that seems quite large. Let me check my calculations.Alternatively, maybe I made a mistake in calculating the height of the triangles. Let me recalculate.The equal sides are φ * s ≈ 1.618 * 2 ≈ 3.236 meters.Half the base is 1 meter.So, the height h = sqrt((3.236)^2 - (1)^2) ≈ sqrt(10.472 - 1) ≈ sqrt(9.472) ≈ 3.078 meters.So, the area of one triangle is (2 * 3.078)/2 ≈ 3.078 m².Five triangles: 5 * 3.078 ≈ 15.39 m².Adding to the pentagon area: 6.882 + 15.39 ≈ 22.272 m².Hmm, that seems correct, but I'm not sure because I've heard that the area of a pentagram is about 1.122 times the area of the pentagon. Wait, 22.272 / 6.882 ≈ 3.24, which is more than 3 times. That doesn't seem right.Wait, maybe I'm overcomplicating this. Let me try another approach.The area of a regular pentagram can be calculated using the formula:Area = (5/2) * s^2 * (sqrt(5 + 2*sqrt(5)))/2Wait, let me compute that.First, compute sqrt(5 + 2*sqrt(5)).sqrt(5) ≈ 2.236, so 2*sqrt(5) ≈ 4.472.Then, 5 + 4.472 ≈ 9.472.sqrt(9.472) ≈ 3.078.So, the formula becomes:Area = (5/2) * (2)^2 * (3.078)/2= (5/2) * 4 * 3.078 / 2= (5/2) * 4 * 1.539= (5/2) * 6.156= 5 * 3.078= 15.39 m².Wait, that's the same as the area of the five triangles. So, perhaps the area of the pentagram is just the area of the five triangles, which is 15.39 m², and the pentagon is inside, but not part of the star. Hmm, that might make sense.Wait, no. The pentagram includes the pentagon and the five triangles. So, if the pentagon is part of the star, then the total area should be the sum of both. But according to this formula, it's just 15.39 m², which is the area of the five triangles.I'm confused now. Maybe I need to clarify: is the pentagram considered as just the star without the inner pentagon, or including it?In the context of the sculpture, each face is a pentagram, which is a star. So, perhaps the area we're considering is just the star part, not including the inner pentagon. Therefore, the area would be 15.39 m² per face.But wait, in reality, the pentagram is a single continuous face, so it's the star itself, which includes the inner pentagon. Therefore, the area should be the sum of the pentagon and the five triangles, which is 22.272 m².But I'm not sure. Maybe I should look for another way to calculate the area.Alternatively, I can use the formula for the area of a regular star polygon, which is:Area = (n * s^2) / (4 * tan(π/n)) * (1 + 2 * cos(2π/n))For a pentagram, n = 5.So, plugging in n = 5 and s = 2:Area = (5 * 4) / (4 * tan(π/5)) * (1 + 2 * cos(2π/5))Simplify:= (20) / (4 * tan(π/5)) * (1 + 2 * cos(72°))= (5) / tan(π/5) * (1 + 2 * cos(72°))We know that tan(π/5) ≈ 0.7265, and cos(72°) ≈ 0.3090.So,= 5 / 0.7265 * (1 + 2 * 0.3090)= 5 / 0.7265 * (1 + 0.618)= 5 / 0.7265 * 1.618≈ 6.8819 * 1.618 ≈ 11.135 m².Wait, that's different from before. So, now I have another value: approximately 11.135 m² per pentagram.This is conflicting with my previous calculations. I need to resolve this.Wait, perhaps the formula I used is for the area of the star polygon without the inner pentagon. So, if I use this formula, I get 11.135 m², which is less than the sum of the pentagon and triangles.Alternatively, maybe the formula is correct, and my previous approach was wrong.Let me check the formula again.The formula for the area of a regular star polygon {n/m} is:Area = (n * s^2) / (4 * tan(π/n)) * (1 + 2 * cos(2π*m/n))For a pentagram, it's {5/2}, so m = 2.Thus,Area = (5 * 4) / (4 * tan(π/5)) * (1 + 2 * cos(4π/5))Wait, hold on. 2π*m/n = 2π*2/5 = 4π/5, which is 144 degrees, whose cosine is negative.cos(144°) ≈ -0.8090.So,Area = (20) / (4 * tan(π/5)) * (1 + 2*(-0.8090))= (5) / tan(π/5) * (1 - 1.618)= (5 / 0.7265) * (-0.618)≈ 6.8819 * (-0.618) ≈ -4.26 m².Wait, that can't be right. Area can't be negative.I must have messed up the formula. Maybe the formula is different.Alternatively, perhaps the formula is:Area = (n * s^2) / (4 * tan(π/n)) * (1 + 2 * cos(2π*m/n))But for m = 2, 2π*m/n = 4π/5, which is 144 degrees, and cos(144°) is negative, which would make the term (1 + 2*cos(144°)) negative, leading to a negative area, which doesn't make sense.Therefore, maybe the formula is not applicable here, or I'm using it incorrectly.Alternatively, perhaps the formula is for the area of the star polygon without considering the inner polygon. So, maybe the area is just the area of the star without the inner pentagon.But in that case, the area would be the area of the five triangles, which is 15.39 m², as calculated earlier.But then, why does the formula give a different result?Wait, maybe I should use the formula for the area of a regular star polygon, which is:Area = (5/2) * s^2 * (sqrt(5 + 2*sqrt(5)))/2Wait, let me compute that.sqrt(5 + 2*sqrt(5)) ≈ sqrt(5 + 4.472) ≈ sqrt(9.472) ≈ 3.078.So,Area = (5/2) * 4 * (3.078)/2= (5/2) * 4 * 1.539= (5/2) * 6.156= 5 * 3.078 ≈ 15.39 m².So, that's the same as the area of the five triangles. Therefore, perhaps the area of the pentagram is 15.39 m², which is just the area of the five triangles, not including the inner pentagon.But that seems contradictory because the pentagram includes the inner pentagon. So, maybe the formula is considering only the star part, excluding the inner pentagon.Alternatively, perhaps the formula is correct, and the area of the pentagram is 15.39 m², which is the area of the star without the inner pentagon.But in the context of the sculpture, each face is a pentagram, which is a star. So, perhaps the area we need is just the area of the star, which is 15.39 m² per face.But I'm still confused because the pentagram is a single continuous face, so it should include the inner pentagon.Wait, perhaps the area of the pentagram is the area of the star, which is the union of the five triangles and the inner pentagon. So, the total area would be the sum of the pentagon and the five triangles, which is 6.882 + 15.39 ≈ 22.272 m².But then, why does the formula give 15.39 m²? Maybe I need to clarify.Wait, let me think about the structure of the pentagram. It's a five-pointed star, formed by connecting every other vertex of a regular pentagon. The area of the pentagram is the area of the star, which includes the inner pentagon. So, the area should be the area of the star, which is the area of the five triangles plus the inner pentagon.But according to the formula, it's 15.39 m², which is just the area of the five triangles. So, perhaps the formula is considering only the star part, excluding the inner pentagon.Alternatively, maybe the formula is correct, and the area of the pentagram is 15.39 m², which is the area of the star without the inner pentagon. Therefore, the total area of the sculpture would be 12 * 15.39 ≈ 184.68 m².But I'm not sure. Let me try to find another way.Alternatively, I can calculate the area of the pentagram using the formula for the area of a regular star polygon, which is:Area = (5/2) * s^2 * (sqrt(5 + 2*sqrt(5)))/2As before, this gives 15.39 m².But if I consider the pentagram as a union of a pentagon and five triangles, the total area would be 6.882 + 15.39 ≈ 22.272 m².But I think the formula is correct, and the area of the pentagram is 15.39 m², which is the area of the star without the inner pentagon. Therefore, the total surface area of the sculpture would be 12 * 15.39 ≈ 184.68 m².But wait, the problem says "each face of the sculpture must be precisely cut and assembled". So, each face is a pentagram, which is a star. Therefore, the area of each face is the area of the star, which is 15.39 m².Therefore, the total surface area is 12 * 15.39 ≈ 184.68 m².But I'm still not entirely confident. Let me check another source in my mind.Wait, I recall that the area of a regular pentagram is approximately 1.122 times the area of the regular pentagon with the same edge length. So, if the pentagon area is 6.882 m², then the pentagram area would be approximately 6.882 * 1.122 ≈ 7.72 m². But that contradicts the previous calculations.Wait, no, that can't be right because the pentagram is larger than the pentagon. So, the area should be larger, not smaller.Wait, perhaps the factor is different. Maybe it's 1.122 times the area of the pentagon, but that would still make it smaller. Hmm, that doesn't make sense.Alternatively, maybe the factor is 1.122 times the area of the pentagon, but that would mean the pentagram is slightly larger, which still doesn't make sense because the pentagram is a star that extends beyond the pentagon.Wait, perhaps the factor is actually higher. Let me think.If the pentagram is formed by extending the sides of the pentagon, the area increases. So, the area of the pentagram should be larger than the pentagon.Given that, and considering the earlier calculation of 15.39 m², which is about twice the area of the pentagon, that seems plausible.Therefore, I think the area of one pentagram is approximately 15.39 m², and the total surface area is 12 * 15.39 ≈ 184.68 m².But let me try to calculate it using another method to confirm.Another approach: The area of a regular pentagram can be calculated using the formula:Area = (5/2) * s^2 * (sqrt(5 + 2*sqrt(5)))/2As before, plugging in s = 2:Area = (5/2) * 4 * (sqrt(5 + 2*sqrt(5)))/2= (5/2) * 4 * (sqrt(9.472))/2= (5/2) * 4 * 3.078 / 2= (5/2) * 6.156 / 2= (5/2) * 3.078= 5 * 1.539= 7.695 m².Wait, that's different. Now I'm getting 7.695 m². That's even more confusing.Wait, no, I think I made a mistake in the calculation.Let me recalculate:sqrt(5 + 2*sqrt(5)) ≈ sqrt(5 + 4.472) ≈ sqrt(9.472) ≈ 3.078.So,Area = (5/2) * (2)^2 * (3.078)/2= (5/2) * 4 * 3.078 / 2= (5/2) * (12.312)/2= (5/2) * 6.156= 5 * 3.078= 15.39 m².Yes, that's correct. So, the area is 15.39 m².Therefore, the total surface area is 12 * 15.39 ≈ 184.68 m².But wait, the problem mentions that each face is a regular pentagram, and the internal angle at each vertex of the pentagon is 108 degrees. So, maybe the area calculation should consider that.Wait, the internal angle of the pentagon is 108 degrees, which is consistent with a regular pentagon. So, perhaps the formula I used is correct.Therefore, I think the total surface area is approximately 184.68 m².But let me check if there's any overlapping regions that I need to account for. The problem says to account for any overlapping regions due to the intersection of the faces.So, each face is a pentagram, and they intersect each other. Therefore, when calculating the total surface area, I might be double-counting the overlapping regions.But how much do they overlap? Each face intersects with others, but the overlapping regions are internal and not part of the exterior surface. Therefore, the total surface area is just the sum of the areas of all the faces, without subtracting any overlapping regions because those overlaps are internal and not part of the exterior.Wait, no. In a polyhedron, the surface area is the sum of the areas of all the faces, regardless of their intersections, because each face is a separate entity contributing to the surface. So, even if they intersect, each face's area is counted fully.Therefore, the total surface area is indeed 12 * 15.39 ≈ 184.68 m².But let me confirm this with another approach.Alternatively, I can calculate the area of one pentagram and then multiply by 12.Using the formula:Area = (5/2) * s^2 * (sqrt(5 + 2*sqrt(5)))/2= (5/2) * 4 * (sqrt(5 + 2*sqrt(5)))/2= (5/2) * 4 * (3.078)/2= (5/2) * 6.156= 5 * 3.078= 15.39 m².So, 12 * 15.39 ≈ 184.68 m².Therefore, the total surface area is approximately 184.68 square meters.Now, moving on to the second part: determining the total cost for the lighting system.The artist wants to install a light at each vertex of the Great Stellated Dodecahedron. The cost per light is 50. So, I need to find the number of distinct vertices and then multiply by 50.But the Great Stellated Dodecahedron is a non-convex polyhedron. I need to recall how many vertices it has.I remember that the Great Stellated Dodecahedron has the same vertices as the regular dodecahedron, but arranged differently. A regular dodecahedron has 20 vertices. Wait, no, a regular dodecahedron has 20 vertices, 12 faces, and 30 edges.But the Great Stellated Dodecahedron is a Kepler-Poinsot polyhedron, and it has 12 faces, each a pentagram, and it's dual to the Great Icosahedron.Wait, let me recall the properties.The Great Stellated Dodecahedron has:- Faces: 12 (each a regular pentagram)- Edges: 30- Vertices: 20Wait, is that correct? Let me think.In the regular dodecahedron, there are 12 faces, 20 vertices, and 30 edges. The Great Stellated Dodecahedron is a stellation of the dodecahedron, so it should have the same number of vertices as the original dodecahedron, which is 20.But wait, no. When you stellate a polyhedron, you extend its faces until they meet again, forming new vertices. So, the number of vertices might change.Wait, actually, the Great Stellated Dodecahedron has 12 faces, each a pentagram, and it has 20 vertices. Each vertex is where five pentagrams meet.Wait, let me check.In the Great Stellated Dodecahedron, each vertex is formed by the intersection of five pentagrams. So, the number of vertices is 20.Yes, that's correct. So, there are 20 distinct vertices.Therefore, the number of lights needed is 20, each costing 50. So, the total cost is 20 * 50 = 1000.But let me confirm the number of vertices.The Great Stellated Dodecahedron has 20 vertices, 30 edges, and 12 faces. It's dual is the Great Icosahedron, which has 12 vertices, 30 edges, and 20 faces.Wait, no, actually, the dual of the Great Stellated Dodecahedron is the Great Icosahedron, which has 12 vertices, 30 edges, and 20 faces. Therefore, the Great Stellated Dodecahedron has 20 vertices, 30 edges, and 12 faces.Yes, that's correct. So, 20 vertices.Therefore, the total cost is 20 * 50 = 1000.So, summarizing:1. Total surface area ≈ 184.68 m².2. Total cost for lighting system = 1000.But let me write the exact value for the surface area instead of the approximate.Earlier, I calculated the area of one pentagram as 15.39 m², which is approximately 15.39. But let me express it more precisely.The exact formula is:Area = (5/2) * s^2 * (sqrt(5 + 2*sqrt(5)))/2= (5/4) * s^2 * sqrt(5 + 2*sqrt(5))For s = 2,Area = (5/4) * 4 * sqrt(5 + 2*sqrt(5))= 5 * sqrt(5 + 2*sqrt(5))So, the exact area per face is 5 * sqrt(5 + 2*sqrt(5)) m².Therefore, the total surface area is 12 * 5 * sqrt(5 + 2*sqrt(5)) = 60 * sqrt(5 + 2*sqrt(5)) m².But if I want to write it in decimal form, it's approximately 60 * 3.078 ≈ 184.68 m².So, the exact value is 60 * sqrt(5 + 2*sqrt(5)) m², and the approximate is 184.68 m².Therefore, the answers are:1. Total surface area: 60 * sqrt(5 + 2*sqrt(5)) m² ≈ 184.68 m².2. Total cost: 1000.But let me write the exact value for the surface area.So, 5 * sqrt(5 + 2*sqrt(5)) per face, times 12 faces: 60 * sqrt(5 + 2*sqrt(5)).Yes, that's correct.Therefore, the final answers are:1. Total surface area: 60 * sqrt(5 + 2*sqrt(5)) square meters.2. Total cost: 1000.But let me check if the number of vertices is indeed 20.Yes, the Great Stellated Dodecahedron has 20 vertices, each where five pentagrams meet. So, 20 lights at 50 each is 1000.Therefore, I think I've got it.</think>

🧠:<think>Alright, so I have this problem about Dr. Smith analyzing media coverage intensity for two newspapers, The Daily Chronicle and The Evening Gazette. She's looking at how their coverage intensity has changed over the last century, starting from 1923. The functions given are f(t) for The Daily Chronicle and g(t) for The Evening Gazette. First, let me parse the problem. The first part asks me to find the year t since 1923 when the combined intensity of both newspapers is maximized. The functions are given as:f(t) = 5e^{0.03t} + 2sin(0.5t)g(t) = 3e^{0.04t} + 4cos(0.4t)So, the combined intensity would be f(t) + g(t). I need to find the t that maximizes this sum. Then, I have to check if this t falls within a relevant historical period, like the last century, so t should be between 0 and 100, since 1923 + 100 is 2023.Alright, so step one is to write the combined intensity function:C(t) = f(t) + g(t) = 5e^{0.03t} + 2sin(0.5t) + 3e^{0.04t} + 4cos(0.4t)Now, to find the maximum of C(t), I need to take its derivative with respect to t, set it equal to zero, and solve for t. Then, verify if it's a maximum by checking the second derivative or using some other method.So, let me compute C'(t):C'(t) = d/dt [5e^{0.03t}] + d/dt [2sin(0.5t)] + d/dt [3e^{0.04t}] + d/dt [4cos(0.4t)]Calculating each term:d/dt [5e^{0.03t}] = 5 * 0.03 e^{0.03t} = 0.15 e^{0.03t}d/dt [2sin(0.5t)] = 2 * 0.5 cos(0.5t) = cos(0.5t)d/dt [3e^{0.04t}] = 3 * 0.04 e^{0.04t} = 0.12 e^{0.04t}d/dt [4cos(0.4t)] = 4 * (-0.4) sin(0.4t) = -1.6 sin(0.4t)So, putting it all together:C'(t) = 0.15 e^{0.03t} + cos(0.5t) + 0.12 e^{0.04t} - 1.6 sin(0.4t)Now, to find the critical points, set C'(t) = 0:0.15 e^{0.03t} + cos(0.5t) + 0.12 e^{0.04t} - 1.6 sin(0.4t) = 0This is a transcendental equation, meaning it can't be solved algebraically. I'll need to use numerical methods to approximate the solution. Given that t is between 0 and 100, I can use methods like Newton-Raphson or the bisection method. But since I don't have a calculator here, maybe I can estimate or see where the function crosses zero.Alternatively, I can analyze the behavior of C'(t) over the interval [0, 100] to see where it might cross zero.First, let's evaluate C'(t) at some points to get an idea.At t=0:C'(0) = 0.15 e^{0} + cos(0) + 0.12 e^{0} - 1.6 sin(0) = 0.15 + 1 + 0.12 - 0 = 1.27So, positive at t=0.At t=100:Compute each term:0.15 e^{0.03*100} = 0.15 e^{3} ≈ 0.15 * 20.0855 ≈ 3.0128cos(0.5*100) = cos(50) ≈ cos(50 radians). Wait, 50 radians is a lot. Let me compute 50 radians in terms of multiples of 2π. 2π ≈ 6.283, so 50 / 6.283 ≈ 7.96. So, 50 radians is about 7 full circles plus 0.96*6.283 ≈ 6.03 radians. So, cos(6.03) ≈ cos(6.03 - 2π*0) ≈ cos(6.03) ≈ -0.996Similarly, 0.12 e^{0.04*100} = 0.12 e^{4} ≈ 0.12 * 54.598 ≈ 6.5518-1.6 sin(0.4*100) = -1.6 sin(40 radians). 40 radians is about 6.366*2π, so 40 - 6*2π ≈ 40 - 37.699 ≈ 2.301 radians. sin(2.301) ≈ 0.7457. So, -1.6*0.7457 ≈ -1.193So, adding up:3.0128 + (-0.996) + 6.5518 + (-1.193) ≈ 3.0128 - 0.996 + 6.5518 - 1.193 ≈ (3.0128 - 0.996) + (6.5518 - 1.193) ≈ 2.0168 + 5.3588 ≈ 7.3756So, positive at t=100.Wait, so at both t=0 and t=100, C'(t) is positive. So, the derivative is positive at both ends. Hmm, but maybe it dips below zero somewhere in between.Let me check at t=50.Compute each term:0.15 e^{0.03*50} = 0.15 e^{1.5} ≈ 0.15 * 4.4817 ≈ 0.6723cos(0.5*50) = cos(25 radians). 25 radians is about 3.97*2π, so 25 - 3*2π ≈ 25 - 18.849 ≈ 6.151 radians. cos(6.151) ≈ 0.9996 (since cos(6π) is 1, 6.151 is just a bit more, so maybe around 0.9996)0.12 e^{0.04*50} = 0.12 e^{2} ≈ 0.12 * 7.389 ≈ 0.8867-1.6 sin(0.4*50) = -1.6 sin(20 radians). 20 radians is about 3.183*2π, so 20 - 3*2π ≈ 20 - 18.849 ≈ 1.151 radians. sin(1.151) ≈ 0.913. So, -1.6*0.913 ≈ -1.4608Adding up:0.6723 + 0.9996 + 0.8867 - 1.4608 ≈ (0.6723 + 0.9996) + (0.8867 - 1.4608) ≈ 1.6719 - 0.5741 ≈ 1.0978Still positive at t=50.Hmm, maybe try t=20.Compute each term:0.15 e^{0.03*20} = 0.15 e^{0.6} ≈ 0.15 * 1.8221 ≈ 0.2733cos(0.5*20) = cos(10 radians). 10 radians is about 1.5915*2π, so 10 - 1*2π ≈ 10 - 6.283 ≈ 3.717 radians. cos(3.717) ≈ -0.75390.12 e^{0.04*20} = 0.12 e^{0.8} ≈ 0.12 * 2.2255 ≈ 0.2671-1.6 sin(0.4*20) = -1.6 sin(8 radians). 8 radians is about 1.273*2π, so 8 - 1*2π ≈ 8 - 6.283 ≈ 1.717 radians. sin(1.717) ≈ 0.987. So, -1.6*0.987 ≈ -1.579Adding up:0.2733 + (-0.7539) + 0.2671 + (-1.579) ≈ (0.2733 - 0.7539) + (0.2671 - 1.579) ≈ (-0.4806) + (-1.3119) ≈ -1.7925So, negative at t=20.Wait, so at t=20, C'(t) ≈ -1.7925, which is negative, while at t=0, it's positive. So, somewhere between t=0 and t=20, the derivative goes from positive to negative, implying a maximum in that interval.Similarly, at t=50, it's positive again. So, between t=20 and t=50, the derivative goes from negative to positive, implying a minimum. Then, between t=50 and t=100, it's positive, so maybe another maximum? Or perhaps just increasing.Wait, but at t=100, it's still positive. So, perhaps the function C(t) has a maximum somewhere between t=0 and t=20, then a minimum between t=20 and t=50, and then continues to increase after t=50.So, the first critical point is a maximum between t=0 and t=20, and then a minimum between t=20 and t=50, and then it keeps increasing.But since we're looking for the maximum combined intensity, we need to check both the maximum between t=0 and t=20 and see if there's another maximum after t=50. Wait, but at t=100, the derivative is still positive, so the function is increasing at t=100, meaning it might not have a maximum beyond t=50, unless it starts decreasing again, but since the derivative is positive at t=100, it's still increasing.So, the maximum could be either the first peak between t=0 and t=20 or the last value at t=100. But since the function is increasing at t=100, the maximum might actually be at t=100, but we need to check.Wait, but let's think about the functions. Both f(t) and g(t) have exponential terms with positive exponents, so they are increasing functions overall. The sine and cosine terms are oscillating, but their amplitudes are fixed. So, as t increases, the exponential terms dominate, making the overall function C(t) increase. However, the oscillating terms can cause local maxima and minima.So, perhaps the overall maximum is at t=100, but there might be a local maximum somewhere before that. But since the derivative at t=100 is positive, the function is still increasing, so the maximum would be at t=100. But wait, let's check C(t) at t=100 and see if it's higher than, say, t=50.Compute C(50):f(50) = 5e^{0.03*50} + 2sin(0.5*50) = 5e^{1.5} + 2sin(25)5e^{1.5} ≈ 5*4.4817 ≈ 22.4085sin(25 radians). 25 radians is about 3.97*2π, so 25 - 3*2π ≈ 25 - 18.849 ≈ 6.151 radians. sin(6.151) ≈ sin(6.151 - 2π) ≈ sin(6.151 - 6.283) ≈ sin(-0.132) ≈ -0.1317So, 2sin(25) ≈ 2*(-0.1317) ≈ -0.2634g(50) = 3e^{0.04*50} + 4cos(0.4*50) = 3e^{2} + 4cos(20)3e^{2} ≈ 3*7.389 ≈ 22.167cos(20 radians). 20 radians is about 3.183*2π, so 20 - 3*2π ≈ 20 - 18.849 ≈ 1.151 radians. cos(1.151) ≈ 0.4067So, 4cos(20) ≈ 4*0.4067 ≈ 1.6268So, C(50) = f(50) + g(50) ≈ 22.4085 - 0.2634 + 22.167 + 1.6268 ≈ (22.4085 + 22.167) + (-0.2634 + 1.6268) ≈ 44.5755 + 1.3634 ≈ 45.9389Now, compute C(100):f(100) = 5e^{0.03*100} + 2sin(0.5*100) = 5e^{3} + 2sin(50)5e^{3} ≈ 5*20.0855 ≈ 100.4275sin(50 radians). 50 radians is about 7.96*2π, so 50 - 7*2π ≈ 50 - 43.982 ≈ 6.018 radians. sin(6.018) ≈ sin(6.018 - 2π) ≈ sin(6.018 - 6.283) ≈ sin(-0.265) ≈ -0.261So, 2sin(50) ≈ 2*(-0.261) ≈ -0.522g(100) = 3e^{0.04*100} + 4cos(0.4*100) = 3e^{4} + 4cos(40)3e^{4} ≈ 3*54.598 ≈ 163.794cos(40 radians). 40 radians is about 6.366*2π, so 40 - 6*2π ≈ 40 - 37.699 ≈ 2.301 radians. cos(2.301) ≈ -0.6691So, 4cos(40) ≈ 4*(-0.6691) ≈ -2.6764Thus, C(100) = f(100) + g(100) ≈ 100.4275 - 0.522 + 163.794 - 2.6764 ≈ (100.4275 + 163.794) + (-0.522 - 2.6764) ≈ 264.2215 - 3.1984 ≈ 261.0231So, C(100) ≈ 261.02, which is much higher than C(50) ≈ 45.94. So, the function is indeed increasing overall, but with oscillations.But wait, the derivative at t=100 is positive, so it's still increasing. So, the maximum might be beyond t=100, but since we're limited to the last century, t=100 is the end. So, the maximum within the interval [0,100] would be at t=100.But wait, earlier, we saw that the derivative was negative at t=20, implying a local maximum somewhere between t=0 and t=20. So, there might be a local maximum in that interval, but the overall maximum at t=100 is much higher.So, the question is, does the combined intensity have a maximum at t=100, or is there a higher peak somewhere else? Since the exponential terms are increasing, and their growth rates are positive, the overall trend is upwards. The oscillating terms cause fluctuations, but the exponential growth dominates.Therefore, the maximum combined intensity would be at t=100, which is the year 2023. But wait, the problem says "over the last century," which would be from 1923 to 2023, so t=100 is included. So, the maximum is at t=100.But wait, let me double-check. Maybe the function has a higher peak before t=100 due to the oscillations. For example, if the sine and cosine terms align to add constructively with the exponential growth, maybe there's a higher peak somewhere else.But given that the exponential terms are growing, and the oscillating terms have fixed amplitudes, the overall function is increasing. So, the maximum would be at the end of the interval, t=100.However, let's check another point, say t=80.Compute C'(80):0.15 e^{0.03*80} + cos(0.5*80) + 0.12 e^{0.04*80} - 1.6 sin(0.4*80)Compute each term:0.15 e^{2.4} ≈ 0.15 * 11.023 ≈ 1.6535cos(40 radians) ≈ -0.6691 (as before)0.12 e^{3.2} ≈ 0.12 * 24.538 ≈ 2.9446-1.6 sin(32 radians). 32 radians is about 5.106*2π, so 32 - 5*2π ≈ 32 - 31.416 ≈ 0.584 radians. sin(0.584) ≈ 0.551. So, -1.6*0.551 ≈ -0.8816Adding up:1.6535 + (-0.6691) + 2.9446 + (-0.8816) ≈ (1.6535 - 0.6691) + (2.9446 - 0.8816) ≈ 0.9844 + 2.063 ≈ 3.0474Positive at t=80.So, the derivative is positive at t=80, meaning the function is still increasing there.So, it seems that after t=20, the function starts increasing again, reaches a minimum somewhere, then continues to increase. So, the maximum is indeed at t=100.But wait, let's check t=70.C'(70):0.15 e^{0.03*70} ≈ 0.15 e^{2.1} ≈ 0.15 * 8.166 ≈ 1.2249cos(35 radians). 35 radians is about 5.566*2π, so 35 - 5*2π ≈ 35 - 31.416 ≈ 3.584 radians. cos(3.584) ≈ -0.9360.12 e^{0.04*70} ≈ 0.12 e^{2.8} ≈ 0.12 * 16.4446 ≈ 1.9734-1.6 sin(28 radians). 28 radians is about 4.448*2π, so 28 - 4*2π ≈ 28 - 25.133 ≈ 2.867 radians. sin(2.867) ≈ 0.334. So, -1.6*0.334 ≈ -0.5344Adding up:1.2249 + (-0.936) + 1.9734 + (-0.5344) ≈ (1.2249 - 0.936) + (1.9734 - 0.5344) ≈ 0.2889 + 1.439 ≈ 1.7279Still positive.So, it seems that after t=20, the derivative remains positive, except for a dip to negative between t=20 and t=50, but then it goes back to positive and stays positive. So, the function has a local maximum around t=0-20, then a local minimum around t=20-50, and then increases again beyond t=50.Therefore, the overall maximum in the interval [0,100] is at t=100, since the function is increasing there.But wait, let me check t=10:C'(10):0.15 e^{0.3} ≈ 0.15 * 1.3499 ≈ 0.2025cos(5 radians). 5 radians is about 0.7958*2π, so cos(5) ≈ 0.28370.12 e^{0.4} ≈ 0.12 * 1.4918 ≈ 0.179-1.6 sin(4 radians). 4 radians is about 0.6366*2π, so sin(4) ≈ -0.7568. So, -1.6*(-0.7568) ≈ 1.2109Adding up:0.2025 + 0.2837 + 0.179 + 1.2109 ≈ (0.2025 + 0.2837) + (0.179 + 1.2109) ≈ 0.4862 + 1.3899 ≈ 1.8761Positive at t=10.Wait, but earlier at t=20, it was negative. So, between t=10 and t=20, the derivative goes from positive to negative, implying a local maximum somewhere in that interval.So, the function increases from t=0 to t around, say, 15, then decreases from t=15 to t=20, reaches a local minimum, then starts increasing again.But since the function is increasing at t=100, the overall maximum is at t=100.However, the problem is asking for the year when the combined intensity is maximized. So, if the maximum is at t=100, that's 2023, which is within the last century (1923-2023). But perhaps the local maximum around t=15 is higher than the value at t=100? Unlikely, because the exponential terms are growing.Wait, let's compute C(t) at t=15 and t=100 to compare.Compute C(15):f(15) = 5e^{0.03*15} + 2sin(0.5*15) = 5e^{0.45} + 2sin(7.5)5e^{0.45} ≈ 5*1.5683 ≈ 7.8415sin(7.5 radians). 7.5 radians is about 1.1937*2π, so 7.5 - 1*2π ≈ 7.5 - 6.283 ≈ 1.217 radians. sin(1.217) ≈ 0.939. So, 2sin(7.5) ≈ 2*0.939 ≈ 1.878g(15) = 3e^{0.04*15} + 4cos(0.4*15) = 3e^{0.6} + 4cos(6)3e^{0.6} ≈ 3*1.8221 ≈ 5.4663cos(6 radians). 6 radians is about 0.9549*2π, so cos(6) ≈ 0.96024cos(6) ≈ 4*0.9602 ≈ 3.8408So, C(15) ≈ 7.8415 + 1.878 + 5.4663 + 3.8408 ≈ (7.8415 + 5.4663) + (1.878 + 3.8408) ≈ 13.3078 + 5.7188 ≈ 19.0266Compare to C(100) ≈ 261.02, which is much higher. So, the local maximum at t=15 is much lower than the value at t=100.Therefore, the combined intensity is maximized at t=100, which is the year 2023.But wait, the problem says "over the last century," which would be from 1923 to 2023, so t=100 is included. So, the maximum is at t=100.However, I need to confirm if the function actually reaches its maximum at t=100 or if it's still increasing beyond that. Since t=100 is the end of the interval, and the derivative is positive there, the function is increasing at t=100, meaning that if we extended beyond t=100, the function would continue to increase. But within the interval [0,100], the maximum is at t=100.Therefore, the year is 1923 + 100 = 2023.But wait, let me check if the function could have a higher value somewhere else. For example, maybe at t=90.Compute C(90):f(90) = 5e^{0.03*90} + 2sin(0.5*90) = 5e^{2.7} + 2sin(45)5e^{2.7} ≈ 5*14.880 ≈ 74.4sin(45 radians). 45 radians is about 7.161*2π, so 45 - 7*2π ≈ 45 - 43.982 ≈ 1.018 radians. sin(1.018) ≈ 0.846. So, 2sin(45) ≈ 2*0.846 ≈ 1.692g(90) = 3e^{0.04*90} + 4cos(0.4*90) = 3e^{3.6} + 4cos(36)3e^{3.6} ≈ 3*36.603 ≈ 109.809cos(36 radians). 36 radians is about 5.73*2π, so 36 - 5*2π ≈ 36 - 31.416 ≈ 4.584 radians. cos(4.584) ≈ -0.21084cos(36) ≈ 4*(-0.2108) ≈ -0.8432So, C(90) ≈ 74.4 + 1.692 + 109.809 - 0.8432 ≈ (74.4 + 109.809) + (1.692 - 0.8432) ≈ 184.209 + 0.8488 ≈ 185.0578Compare to C(100) ≈ 261.02, which is higher. So, yes, the function is increasing from t=90 to t=100.Therefore, the maximum combined intensity occurs at t=100, which is the year 2023.Now, for the second part of the problem, Dr. Smith wants to model the impact of a significant public relations event in 1963. The spike is modeled by adding a Gaussian function h(t) to both f(t) and g(t). The parameters are:h(t) = A / (sqrt(2πσ²)) e^{-(t-μ)^2 / (2σ²)}, where μ = 1963 - 1923 = 40, A = 100, σ = 2.So, the new functions become:f_new(t) = f(t) + h(t)g_new(t) = g(t) + h(t)Therefore, the combined intensity is C_new(t) = f_new(t) + g_new(t) = f(t) + g(t) + 2h(t) = C(t) + 2h(t)Wait, no, because h(t) is added to both f(t) and g(t), so C_new(t) = f(t) + h(t) + g(t) + h(t) = C(t) + 2h(t)So, the new combined intensity function is C(t) + 2h(t).Now, the question is to calculate the new combined intensity function and determine the total intensity at the peak of the Gaussian function.First, let's write the new function:C_new(t) = 5e^{0.03t} + 2sin(0.5t) + 3e^{0.04t} + 4cos(0.4t) + 2*(100 / sqrt(2π*2²)) e^{-(t-40)^2 / (2*2²)}Simplify the Gaussian term:2*(100 / sqrt(8π)) e^{-(t-40)^2 / 8}Because σ=2, so σ²=4, and 2σ²=8.So, 2*(100 / sqrt(8π)) = 200 / sqrt(8π) ≈ 200 / (2.828*1.772) ≈ 200 / 5.013 ≈ 39.91But let's keep it exact for now.So, C_new(t) = C(t) + (200 / sqrt(8π)) e^{-(t-40)^2 / 8}Now, the peak of the Gaussian function occurs at t=μ=40, since the Gaussian is centered there. So, the peak intensity is C_new(40).Compute C_new(40):C_new(40) = C(40) + 2h(40)But h(40) is the peak of the Gaussian, which is A / sqrt(2πσ²). So, h(40) = 100 / sqrt(8π)Therefore, C_new(40) = C(40) + 2*(100 / sqrt(8π)) = C(40) + 200 / sqrt(8π)But let's compute C(40) first.Compute C(40):f(40) = 5e^{0.03*40} + 2sin(0.5*40) = 5e^{1.2} + 2sin(20)5e^{1.2} ≈ 5*3.3201 ≈ 16.6005sin(20 radians). As before, 20 radians ≈ 1.151 radians after subtracting 3*2π. sin(1.151) ≈ 0.913. So, 2sin(20) ≈ 2*0.913 ≈ 1.826g(40) = 3e^{0.04*40} + 4cos(0.4*40) = 3e^{1.6} + 4cos(16)3e^{1.6} ≈ 3*4.953 ≈ 14.859cos(16 radians). 16 radians is about 2.546*2π, so 16 - 2*2π ≈ 16 - 12.566 ≈ 3.434 radians. cos(3.434) ≈ -0.9364cos(16) ≈ 4*(-0.936) ≈ -3.744So, C(40) = f(40) + g(40) ≈ 16.6005 + 1.826 + 14.859 - 3.744 ≈ (16.6005 + 14.859) + (1.826 - 3.744) ≈ 31.4595 - 1.918 ≈ 29.5415Now, compute 200 / sqrt(8π):sqrt(8π) ≈ sqrt(25.1327) ≈ 5.013So, 200 / 5.013 ≈ 39.91Therefore, C_new(40) ≈ 29.5415 + 39.91 ≈ 69.4515So, the total intensity at the peak of the Gaussian function is approximately 69.45.But let's compute it more accurately.First, compute C(40):f(40):5e^{1.2} ≈ 5*3.3201169228 ≈ 16.6005846142sin(20):sin(20) ≈ sin(20 - 3*2π) ≈ sin(20 - 18.8495559215) ≈ sin(1.1504440785) ≈ 0.9135454576So, 2sin(20) ≈ 1.8270909152g(40):3e^{1.6} ≈ 3*4.953032146 ≈ 14.8590964384cos(16):cos(16) ≈ cos(16 - 2*2π) ≈ cos(16 - 12.566370614) ≈ cos(3.433629386) ≈ -0.9364804745So, 4cos(16) ≈ -3.745921898Thus, C(40) ≈ 16.600584614 + 1.8270909152 + 14.859096438 - 3.745921898 ≈16.600584614 + 1.8270909152 = 18.42767552914.859096438 - 3.745921898 = 11.1131745418.427675529 + 11.11317454 ≈ 29.54085007Now, compute 200 / sqrt(8π):sqrt(8π) ≈ sqrt(25.1327412287) ≈ 5.013274182200 / 5.013274182 ≈ 39.90381057So, C_new(40) ≈ 29.54085007 + 39.90381057 ≈ 69.44466064Rounding to four decimal places, approximately 69.4447.But since the problem asks for the total intensity at the peak, we can present it as approximately 69.44.Alternatively, if we want to keep it exact, we can write it as C(40) + 200 / sqrt(8π), but since the problem asks for the value, we can compute it numerically.So, the total intensity at the peak is approximately 69.44.But let me double-check the calculation of h(40):h(t) = (100 / sqrt(2π*2²)) e^{-(t-40)^2 / (2*2²)} = (100 / sqrt(8π)) e^{-(t-40)^2 /8}At t=40, the exponent is zero, so h(40) = 100 / sqrt(8π)Therefore, 2h(40) = 200 / sqrt(8π) ≈ 39.9038Adding to C(40) ≈ 29.5409 gives ≈ 69.4447Yes, that seems correct.So, summarizing:1. The combined intensity is maximized at t=100, which is the year 2023, within the last century.2. The new combined intensity function is C(t) + 2h(t), and the total intensity at the peak (t=40) is approximately 69.44.</think>

🧠:<think>Alright, so I need to help this school administrator figure out how to improve parent-teacher communication. They're looking at three methods: emails, phone calls, and in-person meetings. Each has its own effectiveness score based on parent surveys. The administrator wants to analyze both the frequency and effectiveness of these methods and then optimize their usage for the next month.Starting with part 1: calculating the weighted average effectiveness score for each communication method. Hmm, okay. So, for each method, we have vectors of communication counts and effectiveness scores. For example, emails have ( mathbf{C_e} = [c_{e1}, c_{e2}, ldots, c_{en}] ) and ( mathbf{E_e} = [e_{e1}, e_{e2}, ldots, e_{en}] ). I think the weighted average would be the sum of each communication's effectiveness multiplied by its count, divided by the total number of communications for that method.Wait, let me think. The weighted average effectiveness score is usually the sum of (each effectiveness score multiplied by its weight) divided by the total weight. In this case, the weight is the number of communications. So for emails, it would be:( text{Weighted Average for Emails} = frac{sum_{i=1}^{n} c_{ei} times e_{ei}}{sum_{i=1}^{n} c_{ei}} )Similarly, for phone calls:( text{Weighted Average for Phone Calls} = frac{sum_{i=1}^{n} c_{pi} times e_{pi}}{sum_{i=1}^{n} c_{pi}} )And for in-person meetings:( text{Weighted Average for In-Person Meetings} = frac{sum_{i=1}^{n} c_{mi} times e_{mi}}{sum_{i=1}^{n} c_{mi}} )So, each method's effectiveness is a weighted average where the weights are the number of communications. That makes sense because more frequent communications would have a bigger impact on the overall effectiveness score.Moving on to part 2: setting up an optimization problem. The administrator wants to maximize overall communication effectiveness next month with two constraints. First, the total number of communications can't exceed 500. Second, each method must be used at least 50 times.Alright, so the variables here are the number of emails ( c_e ), phone calls ( c_p ), and in-person meetings ( c_m ). The objective function should be the overall effectiveness, which I think is the sum of the weighted average effectiveness scores for each method multiplied by their respective counts? Wait, no. Wait, actually, the weighted average effectiveness scores are already calculated in part 1. So, maybe the overall effectiveness is the sum of each method's effectiveness multiplied by the number of communications? Or perhaps it's a combined effectiveness.Wait, let me clarify. The effectiveness scores are on a scale of 1 to 10. So, if we have a weighted average effectiveness for each method, say ( E_e ) for emails, ( E_p ) for phone calls, and ( E_m ) for in-person meetings, then the overall effectiveness would be the sum of each method's effectiveness multiplied by the number of times it's used, right?So, the total effectiveness ( E_{total} ) would be:( E_{total} = c_e times E_e + c_p times E_p + c_m times E_m )And the administrator wants to maximize this ( E_{total} ).But wait, actually, the weighted average effectiveness scores are already aggregating the effectiveness per method. So, if we have the weighted average for each method, then the total effectiveness is just the sum of each method's effectiveness multiplied by the number of communications. So, yes, that seems right.So, the objective function is to maximize ( E_{total} = c_e times E_e + c_p times E_p + c_m times E_m ).Now, the constraints. The first constraint is that the total number of communications cannot exceed 500:( c_e + c_p + c_m leq 500 )The second constraint is that each method must be used at least 50 times:( c_e geq 50 )( c_p geq 50 )( c_m geq 50 )Also, we should consider that the number of communications can't be negative, but since each method must be used at least 50 times, the non-negativity constraints are already covered by the second set of constraints.So, putting it all together, the optimization problem is:Maximize ( E_{total} = c_e times E_e + c_p times E_p + c_m times E_m )Subject to:1. ( c_e + c_p + c_m leq 500 )2. ( c_e geq 50 )3. ( c_p geq 50 )4. ( c_m geq 50 )Where ( E_e ), ( E_p ), and ( E_m ) are the weighted average effectiveness scores calculated in part 1.Wait, but in the problem statement, it says to use the weighted average effectiveness scores derived in part 1 as the objective function. So, does that mean the objective function is the sum of the weighted averages? Or is it the sum of the products of counts and effectiveness?I think it's the latter because the weighted average is a score per method, so to get the total effectiveness, you need to multiply by the number of communications. Otherwise, if you just sum the weighted averages, you'd get a number between 3 and 30, which doesn't take into account how many communications you're doing.So, yeah, I think the objective function is correctly formulated as ( E_{total} = c_e times E_e + c_p times E_p + c_m times E_m ).Therefore, the optimization problem is set up as a linear programming problem where we maximize the total effectiveness given the constraints on total communications and minimum usage per method.I should also note that this is a linear optimization problem because the objective function is linear in terms of ( c_e ), ( c_p ), and ( c_m ), and the constraints are also linear.So, to recap:1. For each method, calculate the weighted average effectiveness score by summing the product of each communication's count and effectiveness, then dividing by the total count for that method.2. Set up a linear program where we maximize the total effectiveness, which is the sum of each method's weighted average effectiveness multiplied by their respective counts, subject to the constraints on total communications and minimum usage per method.I think that's it. I don't see any immediate mistakes in this reasoning. It makes sense that the administrator would want to use more of the more effective methods, but also has to balance the total number of communications and ensure each method is used a minimum number of times.Final Answer1. The weighted average effectiveness scores are calculated as: - For emails: ( boxed{frac{sum_{i=1}^{n} c_{ei} e_{ei}}{sum_{i=1}^{n} c_{ei}}} ) - For phone calls: ( boxed{frac{sum_{i=1}^{n} c_{pi} e_{pi}}{sum_{i=1}^{n} c_{pi}}} ) - For in-person meetings: ( boxed{frac{sum_{i=1}^{n} c_{mi} e_{mi}}{sum_{i=1}^{n} c_{mi}}} )2. The optimization problem is formulated as: - Maximize ( E_{total} = c_e times E_e + c_p times E_p + c_m times E_m ) - Subject to: - ( c_e + c_p + c_m leq 500 ) - ( c_e geq 50 ) - ( c_p geq 50 ) - ( c_m geq 50 ) The problem is set up as a linear program with the objective function and constraints as stated above.The final answer for part 2 is the optimization problem setup, which is presented in the boxed equations above.boxed{begin{aligned}& text{Maximize} & & E_{total} = c_e times E_e + c_p times E_p + c_m times E_m & text{Subject to} & & c_e + c_p + c_m leq 500 & & & c_e geq 50 & & & c_p geq 50 & & & c_m geq 50 end{aligned}}</think>

🧠:<think>Okay, so I have two problems here related to a health initiative for seniors. Let me try to tackle them one by one.Starting with the first problem. It says that the improvement in physical fitness, I(t), over time t (in weeks) is modeled by the differential equation dI/dt = k(100 - I(t)). We're given that initially, I(0) = 20, and after 12 weeks, I(12) = 85. We need to find the value of k.Hmm, this looks like a differential equation problem. The equation is dI/dt = k(100 - I). That seems like a linear differential equation, maybe an exponential growth model? Let me recall. The standard form for such an equation is dI/dt = k(M - I), which is similar to a logistic growth model but without the carrying capacity term. Wait, actually, it's more like an exponential decay towards a carrying capacity. So, the solution should be of the form I(t) = M + (I0 - M)e^(-kt), where M is the carrying capacity, I0 is the initial value.In this case, M is 100, I0 is 20. So, plugging that in, I(t) = 100 + (20 - 100)e^(-kt) = 100 - 80e^(-kt). That seems right.We can check the initial condition: at t=0, I(0) = 100 - 80e^0 = 100 - 80 = 20. Perfect, that matches.Now, we need to find k such that I(12) = 85. So, plug t=12 into the equation:85 = 100 - 80e^(-12k)Let me solve for k.First, subtract 100 from both sides:85 - 100 = -80e^(-12k)-15 = -80e^(-12k)Divide both sides by -80:(-15)/(-80) = e^(-12k)15/80 = e^(-12k)Simplify 15/80: that's 3/16.So, 3/16 = e^(-12k)Take the natural logarithm of both sides:ln(3/16) = -12kSo, k = - (ln(3/16))/12I can compute this value numerically if needed, but maybe it's better to leave it in terms of ln. Alternatively, I can write it as ln(16/3)/12 because ln(3/16) is negative.Wait, yes, ln(3/16) = ln(3) - ln(16) = negative value. So, k = (ln(16/3))/12.Let me compute ln(16/3). 16 is 2^4, 3 is prime. So, ln(16/3) = ln(16) - ln(3) = 4 ln(2) - ln(3). I know ln(2) is approximately 0.6931 and ln(3) is approximately 1.0986.So, 4*0.6931 = 2.7724Subtract ln(3): 2.7724 - 1.0986 = 1.6738So, ln(16/3) ≈ 1.6738Therefore, k ≈ 1.6738 / 12 ≈ 0.1395 per week.Let me double-check my steps.1. Wrote the differential equation as dI/dt = k(100 - I). Correct.2. Recognized it as a linear ODE and solved it using integrating factor or separation of variables. The solution I(t) = 100 - 80e^(-kt). Correct.3. Plugged in t=12, I=85: 85 = 100 - 80e^(-12k). Correct.4. Solved for e^(-12k) = 15/80 = 3/16. Correct.5. Took natural log: ln(3/16) = -12k => k = -ln(3/16)/12 = ln(16/3)/12. Correct.6. Calculated ln(16/3) ≈ 1.6738, so k ≈ 0.1395 per week. That seems reasonable.I think that's solid.Moving on to the second problem. It says that the healthcare professional studies the correlation between time spent in exercise sessions (in hours) and improvement in a cardiovascular fitness metric. The relationship is linear. Given two data points: 3 hours per week results in 10 points improvement, and 5 hours per week results in 20 points improvement. We need to derive the equation of the line and predict the improvement for 4 hours per week.Alright, so this is a linear regression problem. We have two points: (3,10) and (5,20). We can find the equation of the line passing through these two points.First, let's find the slope (m). The formula for slope is (y2 - y1)/(x2 - x1).So, m = (20 - 10)/(5 - 3) = 10/2 = 5.So, the slope is 5. That means for each additional hour, the improvement increases by 5 points.Now, using the point-slope form to find the equation. Let's use point (3,10):y - 10 = 5(x - 3)Simplify:y - 10 = 5x - 15y = 5x - 15 + 10y = 5x - 5So, the equation is y = 5x - 5.Let me verify with the other point (5,20):y = 5*5 -5 = 25 -5 = 20. Correct.So, the equation is correct.Now, predict the improvement for a participant who spends 4 hours per week.Plug x=4 into the equation:y = 5*4 -5 = 20 -5 = 15.So, the predicted improvement is 15 points.Wait, let me think again. Is that the only way? Alternatively, since it's a linear relationship, we can also compute the average rate of change, which we did, and then express it as a linear equation.Alternatively, we can compute the intercept. Let's see:We have two points: (3,10) and (5,20). The slope is 5, as above.So, using point-slope form with (3,10):y = 5x + bPlug in x=3, y=10:10 = 5*3 + b => 10 = 15 + b => b = -5. So, same result.Thus, the equation is y = 5x -5.Therefore, for x=4, y=5*4 -5=20-5=15.That seems correct.Alternatively, if I didn't use the equation, I could compute the improvement per hour. From 3 to 5 hours is an increase of 2 hours, resulting in an increase of 10 points (from 10 to 20). So, 10 points over 2 hours is 5 points per hour. So, the rate is 5 points per hour.Therefore, starting from 3 hours: 10 points. So, for each hour beyond 3, add 5 points. So, 4 hours is 1 hour more than 3, so 10 +5=15. Same result.Therefore, the prediction is 15 points.I think that's solid as well.So, summarizing:1. The value of k is approximately 0.1395 per week, or exactly (ln(16/3))/12.2. The equation is y = 5x -5, and the predicted improvement for 4 hours is 15 points.Final Answer1. The value of ( k ) is (boxed{dfrac{lnleft(dfrac{16}{3}right)}{12}}).2. The predicted improvement for a participant who spends 4 hours per week is (boxed{15}) points.</think>