🧠:<think>Alright, so I have this problem about a community organizer planning workshops for students who are really committed to their families. The organizer wants to maximize the number of students attending by scheduling workshops optimally. There are two parts to this problem. Let me tackle them one by one.Starting with the first part: Each student can spend a total of 5 hours per week on workshops, split between weekdays (W) and weekends (E). For each student, let ( x_i ) be the hours on weekdays and ( y_i ) on weekends, so ( x_i + y_i = 5 ). The effectiveness of their participation is given by ( f(x_i, y_i) = 3x_i^2 + 4y_i^2 ). We need to find the values of ( x_i ) and ( y_i ) that maximize this function under the given constraint.Hmm, okay. So, this is an optimization problem with a constraint. I remember from my calculus class that we can use the method of Lagrange multipliers for this. Alternatively, since the constraint is linear, we can substitute one variable in terms of the other and then maximize the function.Let me try substitution first because it might be simpler. Since ( x_i + y_i = 5 ), I can express ( y_i ) as ( y_i = 5 - x_i ). Then, substitute this into the effectiveness function:( f(x_i, y_i) = 3x_i^2 + 4(5 - x_i)^2 )Let me expand this:First, expand ( (5 - x_i)^2 ):( (5 - x_i)^2 = 25 - 10x_i + x_i^2 )So, substituting back into f:( f(x_i) = 3x_i^2 + 4(25 - 10x_i + x_i^2) )( f(x_i) = 3x_i^2 + 100 - 40x_i + 4x_i^2 )Combine like terms:( f(x_i) = (3x_i^2 + 4x_i^2) + (-40x_i) + 100 )( f(x_i) = 7x_i^2 - 40x_i + 100 )Now, this is a quadratic function in terms of ( x_i ). Since the coefficient of ( x_i^2 ) is positive (7), the parabola opens upwards, which means the vertex is the minimum point. But we are looking for the maximum. Wait, hold on. If the function is a parabola opening upwards, it doesn't have a maximum; it goes to infinity as ( x_i ) increases or decreases. But in our case, ( x_i ) is constrained between 0 and 5 because it's the number of hours a student can spend on weekdays, and the total is 5 hours.So, the maximum must occur at one of the endpoints of the interval [0,5]. Let me check the value of f at x_i = 0 and x_i = 5.First, at x_i = 0:( f(0) = 7(0)^2 - 40(0) + 100 = 100 )At x_i = 5:( f(5) = 7(25) - 40(5) + 100 = 175 - 200 + 100 = 75 )Wait, so f(0) is 100 and f(5) is 75. So, the maximum is at x_i = 0, which would mean y_i = 5.But that seems counterintuitive. If the student spends all their time on weekends, the effectiveness is higher? Let me double-check my calculations.Wait, let me recast the problem. Maybe I made a mistake in the substitution or the expansion.Original function: ( f(x_i, y_i) = 3x_i^2 + 4y_i^2 )Constraint: ( x_i + y_i = 5 )Express y_i as 5 - x_i:( f(x_i) = 3x_i^2 + 4(5 - x_i)^2 )Expand ( (5 - x_i)^2 ):25 - 10x_i + x_i^2Multiply by 4:100 - 40x_i + 4x_i^2Add 3x_i^2:Total is 7x_i^2 - 40x_i + 100Yes, that seems correct. So, f(x_i) is 7x_i^2 - 40x_i + 100. Since it's a quadratic, and the coefficient is positive, it's convex, so the minimum is at the vertex, but the maximum would be at the endpoints.Wait, but in this case, the function is convex, so it curves upwards, meaning the minimum is in the middle, and the maximums are at the ends. So, the maximum effectiveness is at either x_i=0 or x_i=5.But when x_i=0, y_i=5, f=100.When x_i=5, y_i=0, f=3*(25) + 4*(0) = 75.So, 100 is higher than 75, so the maximum is at x_i=0, y_i=5.Wait, but that seems odd because the coefficient for y_i is higher (4) than for x_i (3). So, perhaps allocating more time to y_i, which has a higher coefficient, gives a higher effectiveness. So, putting all hours into y_i gives a higher f.But is that the case? Let me think. The function is 3x² + 4y². Since 4 is bigger than 3, each hour allocated to y_i gives a higher return in terms of effectiveness.Therefore, to maximize f, we should allocate as much as possible to y_i, which is the weekend workshops.So, the optimal allocation is x_i=0, y_i=5.Wait, but is that the case? Let me think about the marginal effectiveness.The derivative of f with respect to x_i is 6x_i, and the derivative with respect to y_i is 8y_i.But since we have a constraint, we can set up the Lagrangian.Let me try that approach.The Lagrangian function is:( mathcal{L}(x_i, y_i, lambda) = 3x_i^2 + 4y_i^2 - lambda(x_i + y_i - 5) )Taking partial derivatives:1. ( frac{partial mathcal{L}}{partial x_i} = 6x_i - lambda = 0 ) => ( 6x_i = lambda )2. ( frac{partial mathcal{L}}{partial y_i} = 8y_i - lambda = 0 ) => ( 8y_i = lambda )3. ( frac{partial mathcal{L}}{partial lambda} = -(x_i + y_i - 5) = 0 ) => ( x_i + y_i = 5 )From the first equation, ( lambda = 6x_i )From the second equation, ( lambda = 8y_i )Therefore, ( 6x_i = 8y_i )Simplify: ( 3x_i = 4y_i ) => ( y_i = (3/4)x_i )Now, substitute into the constraint:( x_i + y_i = 5 )( x_i + (3/4)x_i = 5 )( (7/4)x_i = 5 )Multiply both sides by 4/7:( x_i = (5)*(4/7) = 20/7 ≈ 2.857 )Then, ( y_i = 5 - x_i = 5 - 20/7 = 15/7 ≈ 2.143 )Wait, so according to this, the optimal allocation is x_i ≈ 2.857 hours on weekdays and y_i ≈ 2.143 hours on weekends.But earlier, when I substituted, I found that f(0)=100 and f(5)=75, so the maximum was at x_i=0. But according to the Lagrangian method, the maximum is at x_i≈2.857.This is conflicting. Which one is correct?Wait, perhaps I made a mistake in interpreting the function. Let me check.The function f(x_i, y_i) = 3x_i² + 4y_i² is a quadratic function, and since both coefficients are positive, it's a convex function. Therefore, the critical point found by the Lagrangian is a minimum, not a maximum.Ah! That's the key. Since the function is convex, the critical point is a minimum, not a maximum. Therefore, the maximum must occur at the boundaries of the feasible region, which are x_i=0 or x_i=5.So, that explains the discrepancy. The Lagrangian method gave me the minimum, but we are looking for the maximum, which is at the endpoints.Therefore, the maximum effectiveness is achieved when x_i=0, y_i=5, giving f=100, which is higher than f=75 when x_i=5.So, the optimal allocation is x_i=0, y_i=5.Wait, but let me think again. If the function is convex, then the minimum is at the critical point, and the maximum is at the endpoints. So, yes, that makes sense.But just to be thorough, let me compute f(x_i) at x_i=20/7 and x_i=0, x_i=5.Compute f(20/7):x_i = 20/7 ≈ 2.857y_i = 15/7 ≈ 2.143f = 3*(20/7)^2 + 4*(15/7)^2= 3*(400/49) + 4*(225/49)= (1200/49) + (900/49)= 2100/49 ≈ 42.857Compare this with f(0)=100 and f(5)=75. So, 42.857 is less than both, which confirms that it's a minimum.Therefore, the maximum effectiveness is indeed at the endpoints. Since f(0)=100 > f(5)=75, the maximum is at x_i=0, y_i=5.So, the optimal allocation is x_i=0 hours on weekdays and y_i=5 hours on weekends.Wait, but is this practical? If all students attend only weekend workshops, does that mean the organizer should schedule all workshops on weekends? But the problem is about scheduling optimally to maximize participation. If students can attend more on weekends, perhaps that's better.But let me think again. The function f(x_i, y_i) is the effectiveness. So, higher f means more effective participation. Therefore, to maximize effectiveness, each student should allocate all their time to weekend workshops.But wait, the problem says "maximize the number of students who can attend these workshops". Hmm, does that mean maximize the number of students, or maximize the effectiveness per student?Wait, the problem says: "the organizer wants to maximize the number of students who can attend these workshops by scheduling them optimally."So, perhaps the goal is to maximize the number of students attending, not the effectiveness per student. But then, how does the effectiveness function relate to the number of students?Wait, maybe I misread the problem. Let me check again.The problem says: "the organizer wants to maximize the number of students who can attend these workshops by scheduling them optimally."Then, it says: "the effectiveness of a student's participation is given by the function f(x_i, y_i) = 3x_i^2 + 4y_i^2."Wait, perhaps the effectiveness is a measure of how much each student can contribute or benefit, and the organizer wants to maximize the total effectiveness, which would be the sum over all students of f(x_i, y_i). But the problem says "maximize the number of students", so maybe it's just about how many can attend, not the total effectiveness.Wait, but the problem is a bit ambiguous. Let me read it again."A community organizer is planning a series of workshops for students who are deeply committed to their families. The organizer has noticed that students' participation is influenced by their family responsibilities. The organizer wants to maximize the number of students who can attend these workshops by scheduling them optimally."So, the goal is to maximize the number of students attending. The next part says: "The organizer has identified that each student can allocate a total of 5 hours per week for workshops, but they must split this time between weekday (W) and weekend (E) workshops. Let ( x_i ) be the number of hours a student ( i ) spends on weekday workshops, and ( y_i ) be the number of hours on weekend workshops, such that ( x_i + y_i = 5 ). The effectiveness of a student's participation is given by the function ( f(x_i, y_i) = 3x_i^2 + 4y_i^2 ). Determine the values of ( x_i ) and ( y_i ) that maximize ( f(x_i, y_i) ) under the constraint ( x_i + y_i = 5 )."So, the first part is about each student's effectiveness, and the second part is about the transportation cost.Wait, perhaps the first part is about each student's optimal allocation to maximize their own effectiveness, which would then allow the organizer to schedule workshops in a way that accommodates as many students as possible.But I'm not sure. Maybe the first part is just a separate optimization problem for each student, and the second part is about the total number of students given the transportation budget.So, perhaps the two parts are separate. The first part is about each student's optimal time allocation, and the second part is about how many students can be supported given the transportation cost.So, for the first part, each student wants to maximize their own effectiveness, which is f(x_i, y_i). So, as we found, each student should allocate all their time to weekend workshops, x_i=0, y_i=5.But wait, if all students are attending only weekend workshops, does that mean the organizer can schedule all workshops on weekends, allowing more students to attend because they don't have weekday conflicts? Or is it about the total time each student can contribute?Wait, perhaps the effectiveness function is a measure of how much each student can contribute, and the organizer wants to maximize the total contribution. But the problem says "maximize the number of students", so maybe it's just about how many can attend, regardless of their contribution.But the problem statement is a bit unclear. However, given that the first part specifically asks to maximize f(x_i, y_i), which is the effectiveness, I think the first part is about each student's optimal allocation, and the second part is about the number of students given the transportation budget.So, for the first part, each student should allocate all their time to weekend workshops to maximize their effectiveness, so x_i=0, y_i=5.But wait, let me think again. If the function is f(x_i, y_i) = 3x_i² + 4y_i², and we found that the maximum is at x_i=0, y_i=5, then that's the optimal allocation for each student.But perhaps the organizer wants to schedule workshops in such a way that the total time across all students is maximized, but that might not be the case.Alternatively, maybe the organizer wants to maximize the number of students by considering their availability. If students are more available on weekends, then scheduling more workshops on weekends would allow more students to attend.But the problem is a bit abstract. Let me focus on the mathematical part.Given that each student can allocate 5 hours, and the effectiveness is f(x_i, y_i) = 3x_i² + 4y_i², with x_i + y_i = 5, we need to find x_i and y_i that maximize f.As we saw, the maximum occurs at x_i=0, y_i=5, giving f=100.Alternatively, if we consider that the organizer wants to maximize the total effectiveness across all students, then each student should allocate their time to maximize their own f, which is x_i=0, y_i=5.But the problem says "maximize the number of students who can attend these workshops". So, perhaps the number of students is limited by the total time available or something else, but the problem doesn't specify that. It only mentions the transportation cost in the second part.Wait, the second part is about the transportation cost, which is a separate question. So, maybe the first part is just about each student's optimal allocation, and the second part is about how many students can be supported given the budget.So, for the first part, the answer is x_i=0, y_i=5.But let me just confirm once more. If we set x_i=0, y_i=5, f=100. If we set x_i=5, y_i=0, f=75. So, 100 is higher, so yes, x_i=0, y_i=5 is optimal.Okay, moving on to the second part.The organizer has a budget of 2,000 for transportation, and the total transportation cost is modeled by g(n) = 50n + 0.5n², where n is the number of students. We need to find the maximum number of students n that can be supported under this budget.So, we need to solve for n in the inequality:50n + 0.5n² ≤ 2000Let me write this as:0.5n² + 50n - 2000 ≤ 0Multiply both sides by 2 to eliminate the decimal:n² + 100n - 4000 ≤ 0Now, we need to solve the quadratic inequality n² + 100n - 4000 ≤ 0First, find the roots of the equation n² + 100n - 4000 = 0Using the quadratic formula:n = [-b ± sqrt(b² - 4ac)] / (2a)Here, a=1, b=100, c=-4000Discriminant D = b² - 4ac = 10000 - 4*1*(-4000) = 10000 + 16000 = 26000So,n = [-100 ± sqrt(26000)] / 2Compute sqrt(26000):sqrt(26000) = sqrt(100*260) = 10*sqrt(260) ≈ 10*16.1245 ≈ 161.245So,n = [-100 + 161.245]/2 ≈ (61.245)/2 ≈ 30.6225n = [-100 - 161.245]/2 ≈ (-261.245)/2 ≈ -130.6225Since n cannot be negative, we discard the negative root.So, the roots are approximately n ≈ 30.6225 and n ≈ -130.6225.The quadratic opens upwards (since a=1>0), so the inequality n² + 100n - 4000 ≤ 0 is satisfied between the roots. But since n cannot be negative, the relevant interval is from n=0 to n≈30.6225.Therefore, the maximum integer value of n that satisfies the inequality is n=30, because at n=30, the cost is:g(30) = 50*30 + 0.5*(30)^2 = 1500 + 0.5*900 = 1500 + 450 = 1950 ≤ 2000At n=31:g(31) = 50*31 + 0.5*(31)^2 = 1550 + 0.5*961 = 1550 + 480.5 = 2030.5 > 2000So, n=31 exceeds the budget, while n=30 is within budget.Therefore, the maximum number of students that can be supported is 30.Wait, but let me double-check the calculation for n=30 and n=31.For n=30:50*30 = 15000.5*(30)^2 = 0.5*900 = 450Total: 1500 + 450 = 1950 ≤ 2000For n=31:50*31 = 15500.5*(31)^2 = 0.5*961 = 480.5Total: 1550 + 480.5 = 2030.5 > 2000Yes, correct. So, n=30 is the maximum number of students that can be supported without exceeding the budget.Therefore, the answers are:1. Each student should allocate 0 hours to weekdays and 5 hours to weekends, so x_i=0, y_i=5.2. The maximum number of students that can be supported is 30.But wait, let me think again about the first part. If each student is allocating all their time to weekends, does that mean the organizer should schedule all workshops on weekends? But the problem is about scheduling optimally to maximize the number of students. If workshops are only on weekends, perhaps more students can attend because they have more flexibility on weekends? Or maybe the opposite, if workshops are spread out, more students can attend without conflicting with their family responsibilities.But the problem doesn't specify any constraints on the number of workshops or the capacity. It just says the organizer wants to maximize the number of students attending by scheduling optimally. So, perhaps the optimal scheduling is to have all workshops on weekends, allowing each student to attend as much as possible, thus maximizing the number of students who can attend without conflicting with their family responsibilities.But I'm not sure. The problem might be more about each student's optimal allocation, not the organizer's scheduling. So, perhaps the first part is just about each student's optimal time allocation, and the second part is about the number of students given the transportation budget.In that case, the first part's answer is x_i=0, y_i=5, and the second part is n=30.So, I think that's it.</think>

🧠:<think>Okay, so I have this problem about a customer troubleshooting a network setup. The network is represented as a graph with 20 nodes and 45 edges. The customer wants to make sure the network is fully connected, meaning there's a path between any two nodes. There are two parts to this problem.Starting with the first part: Determine if the network graph is connected or if there are any isolated nodes, given that it has fewer edges than a complete graph of 20 nodes. Hmm, I remember that in graph theory, a connected graph must have at least n-1 edges, where n is the number of nodes. For 20 nodes, that would be 19 edges. Since the graph has 45 edges, which is way more than 19, it seems like it should be connected. But wait, just having more edges than n-1 doesn't necessarily guarantee connectivity because it's possible to have multiple disconnected components each with their own edges.But in this case, since the number of edges is 45, which is significantly higher than the minimum required for connectivity, it's more likely that the graph is connected. However, the problem mentions that it has fewer edges than a complete graph. A complete graph with 20 nodes would have C(20,2) = 190 edges. So 45 is much less than 190, but still, it's more than enough to potentially connect all nodes.Wait, but how do I formally determine if the graph is connected? I think one way is to check if the number of edges is sufficient to form a connected graph. The minimum number of edges for connectivity is 19, as I thought. Since 45 is greater than 19, it's possible, but not guaranteed. However, without specific information about how the edges are distributed, I can't be 100% sure. But in most cases, especially with a relatively high number of edges, it's likely connected.Alternatively, maybe the question is more about understanding that just because it's not a complete graph doesn't mean it's disconnected. So, given that it's undirected, has 20 nodes and 45 edges, it's more probable that it's connected. So, I think the answer is that the graph is connected because it has more than n-1 edges, and it's unlikely to have isolated nodes.Moving on to the second part: The customer wants to optimize the network by minimizing the number of routers while maintaining full connectivity. Each node has a weight w(i) = 2i for i = 1 to 20. We need to find the subset of nodes to remove so that the remaining subgraph is still connected, and the total weight of removed nodes is maximized.So, essentially, we need to find a spanning tree of the original graph because a spanning tree is a connected acyclic subgraph that includes all nodes with the minimum number of edges. However, in this case, we're looking to remove nodes, not just edges, to minimize the number of routers. Wait, but if we remove nodes, we have to ensure that the remaining subgraph is still connected. So, it's not just about edges anymore.But the goal is to remove as many nodes as possible (to minimize the number of routers) while keeping the graph connected, but also to maximize the total weight of the removed nodes. Since the weights are given by w(i) = 2i, higher-numbered nodes have higher weights. So, to maximize the total weight, we want to remove the nodes with the highest weights, which are the nodes with the highest numbers.But we can't just remove all high-numbered nodes because we need to ensure the remaining graph is still connected. So, we need to find a connected subgraph with as few nodes as possible, but removing the highest-weighted nodes possible.Wait, actually, the problem says "find the subset of nodes that can be removed so that the remaining subgraph is still connected and the total weight of the removed nodes is maximized." So, it's about removing nodes such that the remaining graph is connected, and the sum of the weights of the removed nodes is as large as possible.So, to maximize the total weight of removed nodes, we should remove the nodes with the highest weights, but we have to ensure that after removal, the remaining graph is still connected.This sounds like the problem of finding a connected subgraph with the minimal number of nodes (i.e., a spanning tree in terms of nodes) but with the highest possible total weight of the removed nodes.But wait, nodes can't form a spanning tree on their own; edges are needed. So, perhaps we need to find a connected subgraph (which is a tree) that includes as few high-weight nodes as possible, thereby allowing us to remove the high-weight nodes.Alternatively, maybe it's about finding a minimal connected subgraph (a tree) and removing all other nodes. But in that case, the minimal connected subgraph would have 19 edges and 20 nodes, which is the entire graph. That doesn't make sense.Wait, perhaps I'm overcomplicating. Let's think differently. Since we can remove nodes, but the remaining graph must be connected. So, the remaining graph must be a connected subgraph. To maximize the total weight of removed nodes, we need to remove as many high-weight nodes as possible without disconnecting the graph.So, the strategy would be to remove the highest-weight nodes first, checking after each removal if the graph remains connected. If it does, continue removing the next highest; if not, move to the next lower weight.But without knowing the specific connections, it's hard to say exactly which nodes can be removed. However, since the graph is connected, there exists a spanning tree. In a spanning tree, all nodes are connected with n-1 edges. So, if we can find a spanning tree that includes as few high-weight nodes as possible, then we can remove the other high-weight nodes.But wait, in a spanning tree, all nodes are included. So, perhaps instead, we need to find a minimal connected subgraph (which is a tree) that includes some subset of nodes, and the rest can be removed. But that doesn't make sense because a connected subgraph must include all nodes or a subset connected by edges.Wait, maybe the problem is to find a connected subgraph (not necessarily spanning) that has the minimal number of nodes, but I think that's not the case. The problem says "minimizing the number of routers while maintaining full connectivity," which suggests that the remaining subgraph must still connect all the original nodes, which is not possible if we remove nodes. So, perhaps I misinterpret.Wait, maybe the customer wants to minimize the number of routers, meaning reduce the number of nodes, but still have the network fully connected. But if you remove nodes, the network can't be fully connected anymore because some nodes are removed. So, perhaps the customer wants to keep the network connected but with fewer routers, meaning that the remaining routers must still connect all the original nodes. But that's not possible unless the remaining routers form a connected subgraph that includes all original nodes, which would require the remaining subgraph to be connected and include all nodes, which is the entire graph. So, that doesn't make sense.Wait, perhaps the customer wants to reduce the number of routers but still have the network connected, meaning that the remaining routers must form a connected subgraph, but not necessarily including all original nodes. But the problem says "maintaining full connectivity," which usually means that all nodes are connected. So, perhaps the customer wants to remove some nodes (routers) but keep the remaining network connected, i.e., the subgraph induced by the remaining nodes is connected. But in that case, the remaining subgraph must be connected, but it doesn't have to include all original nodes. Wait, but the problem says "maintaining full connectivity," which is a bit ambiguous.Wait, let me read the problem again: "optimize the network by minimizing the number of routers while maintaining full connectivity." So, full connectivity likely means that the remaining network is still connected, i.e., the subgraph induced by the remaining nodes is connected. So, the goal is to remove as many nodes as possible (to minimize the number of routers) while keeping the remaining subgraph connected, and among all such possible subsets, choose the one where the total weight of the removed nodes is maximized.So, essentially, we need to find a connected induced subgraph with as few nodes as possible, but the catch is that we want the total weight of the removed nodes to be as large as possible. Since the weights are higher for higher-numbered nodes, we want to remove as many high-numbered nodes as possible while keeping the remaining subgraph connected.Therefore, the approach would be to find the smallest possible connected subgraph (in terms of nodes) such that the sum of the weights of the removed nodes is maximized. But since the weights are higher for higher-numbered nodes, we should aim to remove as many high-numbered nodes as possible.However, the size of the smallest connected subgraph is 1 node, but that would mean removing 19 nodes, but the remaining subgraph with 1 node is trivially connected. But the problem says "minimizing the number of routers," which could mean minimizing the number of routers, i.e., removing as many as possible, but the remaining must still be connected. So, the minimal number of routers is 1, but that's probably not useful because the network would be just one router. But perhaps the customer wants to have a connected network with as few routers as possible, but still having multiple routers.Wait, maybe the customer wants to have a connected network with the minimal number of routers, but the problem says "minimizing the number of routers while maintaining full connectivity." So, perhaps the minimal number of routers needed to keep the network connected, which would be a spanning tree, but that requires n-1 edges, but in terms of nodes, it's all nodes. So, that doesn't make sense.Wait, perhaps I'm misunderstanding. Maybe the customer wants to remove some nodes (routers) such that the remaining network is still connected, and the number of remaining routers is minimized, but the total weight of the removed nodes is maximized.So, the problem is to find the smallest possible connected induced subgraph (in terms of nodes) such that the sum of the weights of the removed nodes is maximized. Since the weights are higher for higher-numbered nodes, we want to remove as many high-numbered nodes as possible while keeping the remaining subgraph connected.Therefore, the strategy is to remove the highest-numbered nodes first, checking if the remaining subgraph is still connected. If it is, remove the next highest, and so on, until removing another node would disconnect the graph.But without knowing the specific connections, it's impossible to determine exactly which nodes can be removed. However, in the worst case, the graph could be a complete graph, in which case you can remove all nodes except for a spanning tree, but since it's not complete, it's more complex.But perhaps the maximum total weight is achieved by removing all nodes except for a minimal connected subgraph, which is a tree. However, the minimal connected subgraph in terms of nodes is a tree with n nodes, which is the entire graph. So, that doesn't help.Wait, maybe I need to think differently. Since the graph is connected, it has a spanning tree. The spanning tree includes all nodes, so if we remove any node from the spanning tree, the remaining subgraph is disconnected. Therefore, we cannot remove any node if we want the remaining subgraph to be connected. But that contradicts the problem statement.Wait, no. The spanning tree is a subgraph, but if we remove nodes from the original graph, the remaining subgraph might still be connected if it contains a spanning tree. So, perhaps the key is to find a connected subgraph (not necessarily spanning) that is as small as possible, but the problem is to remove nodes such that the remaining subgraph is connected.But the problem says "maintaining full connectivity," which might mean that the remaining subgraph must still connect all the original nodes, which is not possible if we remove nodes. So, perhaps the problem is misworded, and it actually means that the remaining subgraph is connected, not necessarily connecting all original nodes.Assuming that, then the goal is to remove as many high-weight nodes as possible while keeping the remaining subgraph connected. So, the approach is to find a connected subgraph with as few nodes as possible, but the total weight of the removed nodes is maximized.Given that, the maximum total weight would be achieved by removing the highest-weight nodes possible. So, we need to find the smallest connected subgraph (in terms of nodes) such that the sum of the weights of the removed nodes is maximized.But without knowing the structure of the graph, it's impossible to determine exactly which nodes can be removed. However, in the best case, if the graph is such that a small subset of nodes is connected, we can remove the rest. But since the graph has 45 edges, it's relatively dense, so it's likely that there are multiple connections, allowing for the removal of many nodes.But perhaps the maximum total weight is achieved by removing all nodes except for a minimal connected subgraph. The minimal connected subgraph in terms of nodes is a tree, which requires n-1 edges, but in terms of nodes, it's all nodes. So, that doesn't help.Wait, maybe the minimal connected subgraph in terms of nodes is a single node, but that would mean removing 19 nodes, but the remaining subgraph is just one node, which is connected. However, the problem says "maintaining full connectivity," which might mean that the remaining subgraph must still connect all the original nodes, which is impossible if we remove nodes. So, perhaps the problem is to keep the network connected, meaning that the remaining subgraph must be connected, but it doesn't have to include all original nodes. So, the customer is okay with having a smaller network as long as it's connected.In that case, the goal is to remove as many high-weight nodes as possible while keeping the remaining subgraph connected. So, the maximum total weight would be the sum of the weights of the nodes that can be removed while leaving the remaining subgraph connected.Given that, the strategy is to remove the highest-weight nodes first, checking if the remaining subgraph is still connected. Since the graph has 45 edges, it's likely that many nodes can be removed without disconnecting the graph.But without knowing the specific connections, we can't determine exactly which nodes can be removed. However, in the best case, we can remove all nodes except for a small connected component. The minimal connected component is a single node, but that would mean removing 19 nodes, but the remaining subgraph is just one node, which is connected. However, the total weight removed would be the sum of nodes 2 to 20, which is 2*(2+3+...+20). But wait, node 1 has weight 2, node 2 has 4, up to node 20 with weight 40.Wait, but if we remove all nodes except node 1, the remaining subgraph is just node 1, which is connected. The total weight removed would be the sum from node 2 to node 20. But is that allowed? The problem says "maintaining full connectivity," which might mean that the remaining subgraph must still connect all the original nodes, which is not the case if we remove nodes. So, perhaps the problem is to keep the network connected, meaning that the remaining subgraph must be connected, but it doesn't have to include all original nodes. So, the customer is okay with having a smaller network as long as it's connected.Therefore, the maximum total weight removed would be the sum of the weights of all nodes except for a connected subgraph. To maximize the total weight, we need to remove as many high-weight nodes as possible while keeping the remaining subgraph connected.So, the approach is:1. Identify the highest-weight nodes (nodes 20, 19, ..., 1) and try to remove them one by one, checking if the remaining subgraph is still connected.2. Continue removing nodes until removing another node would disconnect the graph.But without knowing the specific connections, it's impossible to determine exactly which nodes can be removed. However, in the best case, we can remove all nodes except for a minimal connected subgraph. The minimal connected subgraph is a single node, but that would mean removing 19 nodes, but the remaining subgraph is just one node, which is connected. However, the problem might require the remaining subgraph to have more than one node, but it's not specified.Alternatively, perhaps the problem expects us to find a spanning tree and remove all nodes not in the spanning tree, but that doesn't make sense because a spanning tree includes all nodes.Wait, maybe the problem is to find a connected subgraph with the minimal number of nodes, which would be a tree, but in terms of nodes, it's all nodes. So, that doesn't help.Alternatively, perhaps the problem is to find a connected subgraph with as few nodes as possible, but the total weight of the removed nodes is maximized. So, the minimal connected subgraph is a single node, but that's probably not useful. Alternatively, the minimal connected subgraph with more than one node is two nodes connected by an edge. So, if we can find two nodes connected by an edge, we can remove all other nodes, achieving a total weight removed of the sum of nodes 1 to 20 minus the two nodes.But again, without knowing the specific connections, it's impossible to say. However, since the graph has 45 edges, it's likely that there are multiple edges, so it's possible to find a connected subgraph with just two nodes (if there's an edge between them), allowing us to remove all other nodes.But the problem is to maximize the total weight of the removed nodes, so we want to remove as many high-weight nodes as possible. Therefore, the optimal strategy is to remove all nodes except for two nodes connected by an edge, preferably the two lowest-weight nodes, so that the total weight removed is maximized.Wait, no. If we remove all nodes except for two high-weight nodes connected by an edge, then the total weight removed would be the sum of all nodes except those two. But if we remove all nodes except for two low-weight nodes, the total weight removed would be higher because we're removing more high-weight nodes.Wait, no. The total weight removed is the sum of the weights of the removed nodes. So, to maximize the total weight, we need to remove as many high-weight nodes as possible. Therefore, if we can find a connected subgraph consisting of just two low-weight nodes connected by an edge, then we can remove all other nodes, including all high-weight nodes, resulting in the maximum total weight removed.But we need to ensure that the remaining subgraph is connected. So, if we can find an edge between two low-weight nodes, say node 1 and node 2, then we can remove all other nodes, achieving a total weight removed of sum from node 3 to node 20.But without knowing the specific connections, we can't be sure if such an edge exists. However, since the graph has 45 edges, it's likely that there are edges between low-weight nodes.Alternatively, perhaps the problem expects us to assume that such edges exist, allowing us to remove all nodes except for two low-weight nodes.But let's calculate the total weight removed in that case. The total weight of all nodes is sum_{i=1}^{20} 2i = 2 * sum_{i=1}^{20} i = 2 * (20*21)/2 = 420.If we remove all nodes except for two nodes, say node 1 and node 2, the total weight removed would be 420 - (2 + 4) = 414.But if we can remove all nodes except for one node, the total weight removed would be 420 - 2 = 418, which is higher. However, the remaining subgraph would just be one node, which is connected. So, is that acceptable?But the problem says "maintaining full connectivity," which might mean that the remaining subgraph must still connect all the original nodes, which is not possible if we remove nodes. So, perhaps the problem is to keep the network connected, meaning that the remaining subgraph must be connected, but it doesn't have to include all original nodes. So, the customer is okay with having a smaller network as long as it's connected.Therefore, the maximum total weight removed would be achieved by removing all nodes except for one node, resulting in a total weight removed of 420 - 2 = 418. However, if the problem requires the remaining subgraph to have more than one node, then we have to leave at least two nodes connected by an edge.But since the problem doesn't specify, I think the answer is to remove all nodes except for one, achieving the maximum total weight removed of 418.But wait, the problem says "minimizing the number of routers while maintaining full connectivity." If we remove all nodes except one, the number of routers is minimized to one, and the remaining subgraph is trivially connected. So, that seems to fit.However, in practice, having only one router might not be useful, but mathematically, it's the minimal number of routers needed to maintain connectivity (which is trivial in this case).Therefore, the subset of nodes to remove is all nodes except for node 1, resulting in a total weight removed of 418.But wait, node 1 has the lowest weight, so removing all other nodes would leave node 1, which is connected. The total weight removed is sum from node 2 to node 20, which is 4 + 6 + ... + 40.Calculating that: sum from i=2 to 20 of 2i = 2*(2 + 3 + ... + 20) = 2*(sum from 1 to 20 - 1) = 2*(210 - 1) = 2*209 = 418.Yes, that matches.But is there a way to remove more nodes? Well, if we remove all nodes except for one, we've already removed 19 nodes, which is the maximum possible. So, that's the optimal solution.Therefore, the subset of nodes to remove is all nodes except node 1, resulting in a total weight removed of 418.But wait, what if node 1 is not connected to any other node? Then, removing all other nodes would leave node 1, which is connected, but if node 1 was isolated, then removing all other nodes would leave an isolated node, which is still connected (trivially). But in the original graph, since it's connected, node 1 must have at least one edge. So, it's safe to remove all other nodes.Therefore, the answer is to remove nodes 2 through 20, leaving node 1, with a total weight removed of 418.</think>

🧠:<think>Alright, so I have this problem about a brand manager setting up a marketing strategy for a new product launch. The budget is 100,000, and they want to split it between traditional advertising and social media marketing. The goal is to maximize the return on investment (ROI). First, let me try to understand the problem. They give me that traditional advertising generates a return of 2 for every dollar spent, and social media marketing gives a return of 3 for every dollar spent. So, if I spend x dollars on traditional, I get 2x back, and if I spend y dollars on social media, I get 3y back. The total budget is x + y = 100,000. So, for part 1, I need to formulate an optimization problem. That means I need an objective function and a constraint. The objective function is the total ROI, which is 2x + 3y. The constraint is that x + y = 100,000. To solve this using Lagrange multipliers, I remember that we set up the Lagrangian function, which incorporates the objective function and the constraint with a multiplier. So, the Lagrangian L would be:L = 2x + 3y - λ(x + y - 100,000)Wait, actually, sometimes I've seen it written with a plus for the constraint, but I think it depends on whether we're maximizing or minimizing. Since we're maximizing, we subtract the constraint multiplied by λ. So, taking partial derivatives with respect to x, y, and λ, and setting them equal to zero.Partial derivative with respect to x: dL/dx = 2 - λ = 0 => λ = 2Partial derivative with respect to y: dL/dy = 3 - λ = 0 => λ = 3Wait, hold on, that can't be right because λ can't be both 2 and 3 at the same time. Hmm, that suggests something is wrong. Maybe I set up the Lagrangian incorrectly? Let me double-check.Oh, wait, actually, the Lagrangian should be:L = 2x + 3y + λ(100,000 - x - y)Because the constraint is x + y ≤ 100,000, and we can use ≤ because we might not spend the entire budget if it's not optimal, but in this case, since both returns are positive, we will spend the entire budget. So, maybe writing it as L = 2x + 3y - λ(x + y - 100,000) is correct, but then the partial derivatives are:dL/dx = 2 - λ = 0 => λ = 2dL/dy = 3 - λ = 0 => λ = 3But λ can't be both 2 and 3. That suggests that the maximum occurs at the boundary of the feasible region. Since social media has a higher ROI, we should allocate as much as possible to social media, which would be y = 100,000 and x = 0. Wait, so maybe the Lagrange multiplier method isn't directly giving me the answer because the maximum is at the corner point where x=0. That makes sense because social media has a higher return. So, the optimal solution is x=0, y=100,000.But let me think again. If I set up the Lagrangian correctly, why am I getting conflicting values for λ? Maybe because the maximum isn't in the interior of the feasible region but on the boundary. So, in such cases, the Lagrange multiplier method might not directly give the solution, and we have to check the boundaries.So, for part 1, the optimal allocation is all money into social media, giving a total ROI of 3*100,000 = 300,000.Moving on to part 2. Now, the return from social media is modeled by R(y, f) = 3y + 0.1fy, where f is the number of active followers in thousands. The company has 50,000 active followers, so f = 50.So, substituting f into the return function, R(y) = 3y + 0.1*50*y = 3y + 5y = 8y.Wait, so the return from social media is now 8y? That seems like a significant increase. So, the ROI from social media is now 8 per dollar spent, compared to traditional's 2 per dollar. So, even more reason to allocate all money to social media.But let me make sure. The total ROI is now 2x + 8y, with the constraint x + y = 100,000.Again, using Lagrange multipliers, set up the Lagrangian:L = 2x + 8y + λ(100,000 - x - y)Partial derivatives:dL/dx = 2 - λ = 0 => λ = 2dL/dy = 8 - λ = 0 => λ = 8Again, λ can't be both 2 and 8, so the maximum is at the boundary. Since social media now has an ROI of 8, which is way higher than traditional's 2, we should allocate all to social media again.Wait, but maybe I should think if there's a different approach. Maybe the return function is 3y + 0.1fy, so with f=50, it's 3y + 5y = 8y. So, yes, that's correct.But hold on, is the return function R(y, f) = 3y + 0.1fy, so the total return is 2x + 3y + 0.1fy. So, the total ROI is 2x + 3y + 0.1*50*y = 2x + 3y + 5y = 2x + 8y.So, the objective function is 2x + 8y, subject to x + y = 100,000.So, again, the optimal solution is y=100,000, x=0.But wait, maybe I was too quick. Let me try setting up the Lagrangian again.L = 2x + 8y + λ(100,000 - x - y)Partial derivatives:dL/dx = 2 - λ = 0 => λ = 2dL/dy = 8 - λ = 0 => λ = 8This inconsistency in λ suggests that the maximum is on the boundary, which is either x=0 or y=0. Since 8 > 2, we set x=0, y=100,000.So, the optimal values are still x=0, y=100,000.But wait, maybe I should check if the Lagrange multiplier method can still be applied here. Since the problem is linear, the maximum occurs at the vertices. So, in both cases, the optimal is to allocate everything to social media.Alternatively, maybe I should consider if the return function is concave or convex. Since both are linear, the problem is linear, so the maximum is at the corner.So, in both parts, the optimal allocation is x=0, y=100,000.But let me think again. In part 2, the return from social media is now 8y, which is much higher. So, yes, definitely allocate all to social media.Wait, but maybe I should write the Lagrangian correctly. Let me try again.For part 1, the Lagrangian is:L = 2x + 3y - λ(x + y - 100,000)Taking partial derivatives:dL/dx = 2 - λ = 0 => λ = 2dL/dy = 3 - λ = 0 => λ = 3This is inconsistent, so the maximum is on the boundary. Since 3 > 2, we set x=0, y=100,000.For part 2, the Lagrangian is:L = 2x + 8y - λ(x + y - 100,000)Partial derivatives:dL/dx = 2 - λ = 0 => λ = 2dL/dy = 8 - λ = 0 => λ = 8Again, inconsistent, so maximum on boundary. Since 8 > 2, set x=0, y=100,000.So, both times, the optimal is x=0, y=100,000.But wait, in part 2, the return function is R(y, f) = 3y + 0.1fy. With f=50, it's 3y + 5y = 8y. So, the total ROI is 2x + 8y.So, yes, the optimal is y=100,000, x=0.I think that's correct. So, the optimal values are x=0, y=100,000 in both cases.But wait, in part 2, maybe the return function is multiplicative? Let me check.The problem says R(y, f) = 3y + 0.1fy. So, it's additive, not multiplicative. So, with f=50, it's 3y + 5y = 8y.Yes, that's correct.So, to summarize:1. Maximize 2x + 3y subject to x + y = 100,000. Optimal: x=0, y=100,000.2. Maximize 2x + 8y subject to x + y = 100,000. Optimal: x=0, y=100,000.Therefore, the optimal allocations are the same in both cases, allocating all the budget to social media.But wait, in part 2, the return from social media is now 8y, which is much higher. So, yes, definitely allocate all to social media.I think that's the answer.</think>

🧠:<think>Alright, so John is looking at voter turnout data from the 2008 election, and he wants to model how voter turnout relates to the number of electoral votes each state has. He's using a linear regression model, which is a common statistical tool to understand relationships between variables. First, he's modeling voter turnout ( V_i ) as a linear function of electoral votes ( E_i ). The equation he's using is ( V_i = a + bE_i ). I remember that in linear regression, the goal is to find the best-fitting line through the data points, which minimizes the sum of the squared differences between the observed and predicted values. This is called the least squares method.To find the coefficients ( a ) (the intercept) and ( b ) (the slope), I need to use the formulas derived from the least squares method. I think the formulas involve the means of ( E_i ) and ( V_i ), as well as the sums of ( E_i ), ( V_i ), ( E_i^2 ), and ( E_i V_i ). Let me recall the formulas. The slope ( b ) is calculated as the covariance of ( E ) and ( V ) divided by the variance of ( E ). In terms of sums, covariance can be expressed as ( frac{sum E_i V_i - frac{1}{n} sum E_i sum V_i}{n-1} ) and variance as ( frac{sum E_i^2 - frac{1}{n} (sum E_i)^2}{n-1} ). So, ( b = frac{sum E_i V_i - frac{1}{n} sum E_i sum V_i}{sum E_i^2 - frac{1}{n} (sum E_i)^2} ).Then, the intercept ( a ) is calculated as the mean of ( V ) minus the slope ( b ) times the mean of ( E ). The mean of ( V ) is ( frac{sum V_i}{n} ) and the mean of ( E ) is ( frac{sum E_i}{n} ). So, ( a = frac{sum V_i}{n} - b cdot frac{sum E_i}{n} ).Wait, but the question asks for expressions in terms of the sums ( sum E_i ), ( sum V_i ), ( sum E_i^2 ), and ( sum E_i V_i ). So, I need to express ( a ) and ( b ) without explicitly using ( n ), the number of states. Let me see if I can manipulate the formulas to express them in terms of the given sums.Starting with ( b ), the formula is:( b = frac{sum E_i V_i - frac{1}{n} sum E_i sum V_i}{sum E_i^2 - frac{1}{n} (sum E_i)^2} )But since ( n ) is the number of states, and we don't have it as a variable, maybe we can express it in terms of the sums. Alternatively, perhaps we can write it as:( b = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} )Yes, that seems right because if we factor out ( n ) from the numerator and denominator, we can write it as:( b = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} )Similarly, for the intercept ( a ):( a = frac{sum V_i}{n} - b cdot frac{sum E_i}{n} )But since ( a ) is expressed in terms of the means, which are ( frac{sum V_i}{n} ) and ( frac{sum E_i}{n} ), and we have ( b ) already in terms of the sums, perhaps we can express ( a ) without ( n ) by substituting ( b ).Alternatively, let me express ( a ) as:( a = frac{sum V_i sum E_i^2 - sum E_i sum E_i V_i}{n sum E_i^2 - (sum E_i)^2} )Wait, let me verify that. If I have:( a = frac{sum V_i}{n} - b cdot frac{sum E_i}{n} )Substituting ( b ):( a = frac{sum V_i}{n} - left( frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} right) cdot frac{sum E_i}{n} )Let me compute this step by step:First, compute ( frac{sum V_i}{n} ) as is.Then, compute ( b cdot frac{sum E_i}{n} ):( frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} cdot frac{sum E_i}{n} )Multiply numerator and denominator:Numerator: ( (n sum E_i V_i - (sum E_i)(sum V_i)) cdot sum E_i )Denominator: ( (n sum E_i^2 - (sum E_i)^2) cdot n )So, putting it all together:( a = frac{sum V_i}{n} - frac{(n sum E_i V_i - (sum E_i)(sum V_i)) cdot sum E_i}{n (n sum E_i^2 - (sum E_i)^2)} )To combine these terms, let me get a common denominator:The first term is ( frac{sum V_i}{n} ), which can be written as ( frac{sum V_i (n sum E_i^2 - (sum E_i)^2)}{n (n sum E_i^2 - (sum E_i)^2)} )So, combining both terms:( a = frac{sum V_i (n sum E_i^2 - (sum E_i)^2) - (n sum E_i V_i - (sum E_i)(sum V_i)) cdot sum E_i}{n (n sum E_i^2 - (sum E_i)^2)} )Let me expand the numerator:First term: ( sum V_i n sum E_i^2 - sum V_i (sum E_i)^2 )Second term: ( -n sum E_i V_i sum E_i + (sum E_i)(sum V_i) sum E_i )Simplify each part:First term: ( n sum V_i sum E_i^2 - (sum V_i)(sum E_i)^2 )Second term: ( -n sum E_i V_i sum E_i + (sum E_i)^2 sum V_i )Now, combining the two terms:( n sum V_i sum E_i^2 - (sum V_i)(sum E_i)^2 - n sum E_i V_i sum E_i + (sum E_i)^2 sum V_i )Notice that ( - (sum V_i)(sum E_i)^2 + (sum E_i)^2 sum V_i ) cancels out.So, we are left with:( n sum V_i sum E_i^2 - n sum E_i V_i sum E_i )Factor out ( n ):( n ( sum V_i sum E_i^2 - sum E_i V_i sum E_i ) )So, the numerator becomes:( n ( sum V_i sum E_i^2 - sum E_i V_i sum E_i ) )Therefore, ( a ) is:( a = frac{n ( sum V_i sum E_i^2 - sum E_i V_i sum E_i ) }{n (n sum E_i^2 - (sum E_i)^2)} )The ( n ) cancels out:( a = frac{ sum V_i sum E_i^2 - sum E_i V_i sum E_i }{n sum E_i^2 - (sum E_i)^2 } )So, that's the expression for ( a ).To recap, the slope ( b ) is:( b = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} )And the intercept ( a ) is:( a = frac{ sum V_i sum E_i^2 - sum E_i V_i sum E_i }{n sum E_i^2 - (sum E_i)^2 } )But wait, the question didn't specify whether to express ( a ) and ( b ) in terms of ( n ) or not. It just said in terms of the sums. So, perhaps we can leave it as is, since ( n ) is just the number of states, which is a constant for the dataset.Alternatively, if we want to express ( a ) without ( n ), perhaps we can note that ( n = frac{(sum E_i)^2}{sum E_i^2} ) if all ( E_i ) are the same, but that's not the case here. So, maybe it's better to leave ( a ) and ( b ) in terms of ( n ) as well, but since the question didn't specify, perhaps it's acceptable to have ( n ) in the expressions.But looking back, the question says: "derive expressions for the coefficients ( a ) and ( b ) in terms of the sums ( sum E_i ), ( sum V_i ), ( sum E_i^2 ), and ( sum E_i V_i )." So, ( n ) is not one of the given sums, so perhaps we need to express ( a ) and ( b ) without ( n ).Hmm, that complicates things because ( n ) is necessary for the formulas. Maybe the question assumes that ( n ) is known or can be expressed as ( n = frac{(sum E_i)^2}{sum E_i^2} ) if all ( E_i ) are the same, but that's not the case. Alternatively, perhaps ( n ) can be expressed as the number of states, which is 50 plus DC, so 51, but the question doesn't specify that.Wait, but in the context of the 2008 election, there are 50 states plus Washington D.C., so 51 data points. So, ( n = 51 ). But the question didn't specify that, so maybe we can't assume that. Therefore, perhaps the answer should include ( n ) as a variable.But the question says "in terms of the sums", so perhaps ( n ) is considered a known constant, so it's acceptable to have ( n ) in the expressions. Alternatively, maybe we can express ( n ) as ( frac{(sum E_i)^2}{sum E_i^2} ) if all ( E_i ) are the same, but that's not the case here. So, perhaps we have to leave ( n ) as is.Wait, but in the formulas for ( a ) and ( b ), ( n ) is just a scalar multiplier, so perhaps we can express ( a ) and ( b ) in terms of the sums without explicitly using ( n ) by using the fact that ( n = frac{(sum E_i)^2}{sum E_i^2} ) only if all ( E_i ) are equal, which they are not. So, that approach won't work.Therefore, I think the correct approach is to express ( a ) and ( b ) in terms of the sums and ( n ), since ( n ) is a necessary component of the formulas. So, the final expressions are:( b = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} )( a = frac{ sum V_i sum E_i^2 - sum E_i V_i sum E_i }{n sum E_i^2 - (sum E_i)^2 } )Alternatively, we can factor out the denominator:Let ( S_{xx} = n sum E_i^2 - (sum E_i)^2 )( S_{xy} = n sum E_i V_i - (sum E_i)(sum V_i) )Then,( b = frac{S_{xy}}{S_{xx}} )And,( a = frac{ sum V_i sum E_i^2 - sum E_i V_i sum E_i }{S_{xx}} )But the question asks for expressions in terms of the sums, so perhaps it's better to write them as:( b = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} )( a = frac{ sum V_i sum E_i^2 - sum E_i V_i sum E_i }{n sum E_i^2 - (sum E_i)^2 } )Yes, that seems correct.Now, moving on to the second part, computing the Pearson correlation coefficient ( r ).I remember that the Pearson correlation coefficient measures the linear correlation between two variables. It ranges from -1 to 1, where 1 is total positive correlation, 0 is no correlation, and -1 is total negative correlation.The formula for ( r ) is:( r = frac{sum (E_i - bar{E})(V_i - bar{V})}{sqrt{sum (E_i - bar{E})^2 sum (V_i - bar{V})^2}} )Where ( bar{E} ) and ( bar{V} ) are the means of ( E_i ) and ( V_i ) respectively.Alternatively, using the sums, we can express ( r ) as:( r = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{sqrt{ (n sum E_i^2 - (sum E_i)^2)(n sum V_i^2 - (sum V_i)^2) }} )Yes, that's another way to write it. So, using the sums provided, we can plug them into this formula.Let me verify this formula. The numerator is the same as the covariance times ( n-1 ), but in the Pearson formula, it's the sum of cross products minus the product of sums divided by ( n ), but in the formula above, it's ( n sum E_i V_i - (sum E_i)(sum V_i) ), which is the same as ( n ) times the covariance.Wait, actually, the covariance is ( frac{1}{n-1} sum (E_i - bar{E})(V_i - bar{V}) ), which is equal to ( frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n(n-1)} ). So, the numerator in the Pearson formula is the covariance times ( sqrt{(n-1)^2} ), but I think the formula I wrote earlier is correct.Wait, let me derive it properly.The Pearson correlation coefficient is:( r = frac{sum (E_i - bar{E})(V_i - bar{V})}{sqrt{sum (E_i - bar{E})^2} sqrt{sum (V_i - bar{V})^2}} )Expanding the numerator:( sum (E_i - bar{E})(V_i - bar{V}) = sum E_i V_i - bar{V} sum E_i - bar{E} sum V_i + n bar{E} bar{V} )But ( bar{E} = frac{sum E_i}{n} ) and ( bar{V} = frac{sum V_i}{n} ), so:( sum E_i V_i - frac{sum V_i}{n} sum E_i - frac{sum E_i}{n} sum V_i + n cdot frac{sum E_i}{n} cdot frac{sum V_i}{n} )Simplify term by term:First term: ( sum E_i V_i )Second term: ( - frac{(sum V_i)(sum E_i)}{n} )Third term: ( - frac{(sum E_i)(sum V_i)}{n} )Fourth term: ( + frac{(sum E_i)(sum V_i)}{n} )So, combining the terms:( sum E_i V_i - frac{(sum V_i)(sum E_i)}{n} - frac{(sum E_i)(sum V_i)}{n} + frac{(sum E_i)(sum V_i)}{n} )Simplify:The second and third terms are both ( - frac{(sum V_i)(sum E_i)}{n} ), so together they are ( -2 frac{(sum V_i)(sum E_i)}{n} ), and the fourth term is ( + frac{(sum E_i)(sum V_i)}{n} ), so overall:( sum E_i V_i - frac{(sum V_i)(sum E_i)}{n} )Therefore, the numerator is ( sum E_i V_i - frac{(sum V_i)(sum E_i)}{n} )Similarly, the denominator is:( sqrt{ sum (E_i - bar{E})^2 } sqrt{ sum (V_i - bar{V})^2 } )Expanding each term:( sum (E_i - bar{E})^2 = sum E_i^2 - 2 bar{E} sum E_i + n bar{E}^2 = sum E_i^2 - 2 frac{(sum E_i)^2}{n} + frac{(sum E_i)^2}{n} = sum E_i^2 - frac{(sum E_i)^2}{n} )Similarly,( sum (V_i - bar{V})^2 = sum V_i^2 - frac{(sum V_i)^2}{n} )Therefore, the denominator becomes:( sqrt{ left( sum E_i^2 - frac{(sum E_i)^2}{n} right) left( sum V_i^2 - frac{(sum V_i)^2}{n} right) } )Putting it all together, the Pearson correlation coefficient ( r ) is:( r = frac{ sum E_i V_i - frac{(sum E_i)(sum V_i)}{n} }{ sqrt{ left( sum E_i^2 - frac{(sum E_i)^2}{n} right) left( sum V_i^2 - frac{(sum V_i)^2}{n} right) } } )But the question asks to compute ( r ) given the sums ( sum E_i ), ( sum V_i ), ( sum E_i^2 ), ( sum V_i^2 ), and ( sum E_i V_i ). So, we can express ( r ) as:( r = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{ sqrt{ (n sum E_i^2 - (sum E_i)^2)(n sum V_i^2 - (sum V_i)^2) } } )Yes, that's correct because if we multiply numerator and denominator by ( n ), we get:Numerator: ( n sum E_i V_i - (sum E_i)(sum V_i) )Denominator: ( sqrt{ (n sum E_i^2 - (sum E_i)^2)(n sum V_i^2 - (sum V_i)^2) } )So, that's the formula for ( r ).To summarize, the Pearson correlation coefficient ( r ) is given by:( r = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{ sqrt{ (n sum E_i^2 - (sum E_i)^2)(n sum V_i^2 - (sum V_i)^2) } } )So, that's the expression for ( r ) in terms of the given sums.I think that's all for the derivation. Let me just double-check the formulas to make sure I didn't make any algebraic mistakes.For ( b ):( b = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{n sum E_i^2 - (sum E_i)^2} )Yes, that's the standard formula for the slope in simple linear regression.For ( a ):( a = frac{ sum V_i sum E_i^2 - sum E_i V_i sum E_i }{n sum E_i^2 - (sum E_i)^2 } )Yes, that's derived correctly by substituting ( b ) into the formula for ( a ).And for ( r ):( r = frac{n sum E_i V_i - (sum E_i)(sum V_i)}{ sqrt{ (n sum E_i^2 - (sum E_i)^2)(n sum V_i^2 - (sum V_i)^2) } } )Yes, that's the correct formula for Pearson's ( r ) using the sums.So, I think I've got all the expressions correct.</think>