🧠:<think>Alright, so I have this project with six tasks, each with specific times and dependencies. I need to figure out the minimum total time to complete all tasks, both when working alone and when collaborating with a senior executive. Let me break this down step by step.First, let's list out all the tasks with their times:- Task A: 3 hours- Task B: 2 hours- Task C: 4 hours- Task D: 1 hour- Task E: 5 hours- Task F: 6 hoursNow, the dependencies are:1. Task A must be done before B and C.2. Task B must be done before D.3. Task C must be done before E.4. Both D and E must be done before F.So, visualizing this as a DAG (Directed Acyclic Graph), the tasks form a structure where A is the root, leading to B and C. B leads to D, and C leads to E. Then, both D and E lead to F.To find the minimum total time when working alone, I need to determine the critical path, which is the longest path from start to finish. The critical path determines the minimum time required because any delay on this path will delay the entire project.Let's map out the possible paths:1. A -> B -> D -> F2. A -> C -> E -> FCalculating the durations:1. A (3) + B (2) + D (1) + F (6) = 3 + 2 + 1 + 6 = 12 hours2. A (3) + C (4) + E (5) + F (6) = 3 + 4 + 5 + 6 = 18 hoursSo, the critical path is the second one, which takes 18 hours. Therefore, the minimum total time required when working alone is 18 hours.Now, moving on to the second part where the senior executive and I can work on tasks in parallel. We can start on different independent tasks simultaneously, but we still have to respect the dependencies. Each of us can only work on one task at a time.I need to figure out how to distribute the tasks between the two of us to minimize the total time. Let's see the dependencies again:- A must be done before B and C.- B must be done before D.- C must be done before E.- D and E must be done before F.So, the tasks can be grouped into levels based on dependencies:Level 1: ALevel 2: B, CLevel 3: D, ELevel 4: FWe can work on Level 1 first, then Level 2, then Level 3, and finally Level 4. However, since we can work in parallel, we can assign tasks in each level to either me or the executive as long as dependencies are satisfied.Let's plan the tasks step by step.First, Level 1: Only Task A. So, one of us has to do Task A. Let's say I take Task A, which takes 3 hours. The executive is free during this time.After 3 hours, Task A is done. Now, Level 2 tasks B and C can start. Since both B and C are independent, we can assign each to one person.Let me take Task B (2 hours) and the executive takes Task C (4 hours). So, I finish Task B in 2 hours (total time so far: 3 + 2 = 5 hours). The executive is still working on Task C, which will take until 3 + 4 = 7 hours.So, at 5 hours, I'm free, and the executive is still busy until 7 hours.Now, Level 3 tasks D and E depend on B and C respectively. Since B is done at 5 hours and C is done at 7 hours, D can start at 5 hours, and E can start at 7 hours.I can take Task D (1 hour) starting at 5 hours, finishing at 6 hours. The executive, who finishes Task C at 7 hours, can take Task E (5 hours), finishing at 7 + 5 = 12 hours.Now, Level 4: Task F depends on both D and E. D finishes at 6 hours, E finishes at 12 hours. So, Task F can start at 12 hours.Now, who is free at 12 hours? I finished Task D at 6 hours and have been idle since then. The executive finished Task E at 12 hours. So, I can start Task F at 12 hours, which takes 6 hours, finishing at 18 hours.Wait, but the executive is free at 12 hours as well. Maybe we can assign Task F to the executive instead? Let me check.If I finish D at 6 hours, I'm idle until 12 hours. The executive finishes E at 12 hours. So, at 12 hours, both of us are free. We can assign Task F to either of us. If I take it, it'll take 6 hours, finishing at 18 hours. If the executive takes it, same thing. So, regardless, the total time is 18 hours.But wait, is there a way to overlap more? Let me think.After Task A is done at 3 hours, I take B (2 hours) until 5 hours, and the executive takes C (4 hours) until 7 hours.At 5 hours, I'm free. The executive is still working on C until 7 hours.At 5 hours, I can start Task D (1 hour), finishing at 6 hours.At 7 hours, the executive is free and can start Task E (5 hours), finishing at 12 hours.At 6 hours, I'm free again. But Task E isn't done until 12 hours. So, I can't start Task F until 12 hours because both D and E need to be done.So, from 6 to 12 hours, I'm idle. Then, I can start Task F at 12 hours, taking 6 hours, finishing at 18 hours.Alternatively, maybe assign Task E to me earlier? Let's see.Wait, after Task A is done at 3 hours, if I take Task B (2 hours) until 5 hours, and the executive takes Task C (4 hours) until 7 hours.At 5 hours, I'm free. The executive is still busy until 7 hours.At 5 hours, I can start Task D (1 hour), finishing at 6 hours.At 7 hours, the executive is free. They can start Task E (5 hours), finishing at 12 hours.Alternatively, at 7 hours, the executive could start Task E, but I'm free from 6 to 7 hours. Maybe I can help with Task E earlier? But Task E depends on Task C, which is only done at 7 hours. So, I can't start Task E before 7 hours.So, the earliest Task E can start is at 7 hours, regardless.Therefore, the total time remains 18 hours.Wait, but maybe if we assign tasks differently. For example, after Task A is done, maybe the executive takes Task B and I take Task C.Let me try that.Task A: 3 hours (me).At 3 hours, I'm free. The executive can take Task B (2 hours), finishing at 5 hours. I can take Task C (4 hours), finishing at 7 hours.At 5 hours, the executive is free. They can take Task D (1 hour), finishing at 6 hours.At 7 hours, I'm free. I can take Task E (5 hours), finishing at 12 hours.Now, Task F depends on D and E. D is done at 6 hours, E at 12 hours. So, Task F can start at 12 hours.At 12 hours, the executive is free (they finished D at 6 hours and were idle until 12). I finish E at 12 hours. So, we can assign Task F to either.If the executive takes Task F, they start at 12, finish at 18. If I take it, same thing.So, total time is still 18 hours.Hmm, same result.Is there a way to make it faster?Wait, let's think about the critical path again. The critical path is A -> C -> E -> F, which is 3 + 4 + 5 + 6 = 18 hours.But when working in parallel, maybe we can reduce this.Wait, but the critical path is still the same because even if we parallelize, the dependencies still require that C and E are done before F, and A must be done before C.But perhaps, if we can work on some tasks in parallel that are not on the critical path, we might reduce the overall time.Wait, let's see. The critical path is 18 hours, but maybe the other path (A -> B -> D -> F) is 12 hours, which is shorter. So, if we can make sure that the longer path is minimized.But since the critical path is 18 hours, that's the minimum time regardless of parallelization because the critical path can't be shortened by parallel work.Wait, but actually, when working in parallel, maybe we can overlap some tasks.Wait, let me think again.After Task A is done, we can have both B and C being worked on in parallel. So, if I do B and the executive does C, then B finishes at 2 hours after A, which is 5 hours total, and C finishes at 4 hours after A, which is 7 hours total.Then, D can start at 5 hours, and E can start at 7 hours.If I do D (1 hour) starting at 5 hours, finishing at 6 hours.The executive can do E (5 hours) starting at 7 hours, finishing at 12 hours.Then, F can start at 12 hours, taking 6 hours, finishing at 18 hours.Alternatively, if the executive does D and I do E.After A is done at 3 hours:I take B (2 hours), finishing at 5 hours.Executive takes C (4 hours), finishing at 7 hours.At 5 hours, I'm free. I can take D (1 hour), finishing at 6 hours.Executive is busy until 7 hours.At 7 hours, executive is free. They can take E (5 hours), finishing at 12 hours.Then, F starts at 12 hours, taking 6 hours.Same result.Alternatively, after A is done, executive takes B, I take C.A: 3 hours (me).At 3 hours, I'm free. Executive takes B (2 hours), finishing at 5 hours. I take C (4 hours), finishing at 7 hours.At 5 hours, executive is free. They can take D (1 hour), finishing at 6 hours.At 7 hours, I'm free. I can take E (5 hours), finishing at 12 hours.Then, F starts at 12 hours, taking 6 hours.Same total time.Wait, but what if we try to do some tasks in a different order?For example, after A is done, I do B, executive does C.I finish B at 5 hours, executive finishes C at 7 hours.Then, I can do E (5 hours) starting at 7 hours, finishing at 12 hours.Executive can do D (1 hour) starting at 5 hours, finishing at 6 hours.Then, F can start at 12 hours, taking 6 hours.Same result.Alternatively, after A is done, I do C, executive does B.I finish C at 7 hours, executive finishes B at 5 hours.Then, I can do E (5 hours) starting at 7 hours, finishing at 12 hours.Executive can do D (1 hour) starting at 5 hours, finishing at 6 hours.Then, F starts at 12 hours, taking 6 hours.Same total time.So, regardless of how we assign B and C, the total time remains 18 hours.Wait, but is there a way to overlap Task E and Task D more?After A is done, if I do B and executive does C.I finish B at 5 hours, executive finishes C at 7 hours.Then, I can do D (1 hour) at 5 hours, finishing at 6 hours.Executive can do E (5 hours) starting at 7 hours, finishing at 12 hours.Then, F starts at 12 hours.Alternatively, after I finish D at 6 hours, can I help with E?But E depends on C, which is done at 7 hours. So, I can't start E before 7 hours.So, from 6 to 7 hours, I'm idle. Then, at 7 hours, I can start E, but the executive is also free at 7 hours.Wait, if I start E at 7 hours, taking 5 hours, finishing at 12 hours.The executive, who finished C at 7 hours, can do something else? But after C, the only task is E, which I'm starting. So, the executive is idle from 7 to 12 hours, except for E.Wait, no, the executive can't do E because I'm already doing it. So, the executive is idle from 7 to 12 hours.Alternatively, maybe the executive can help with F earlier? But F can't start until both D and E are done.D is done at 6 hours, E at 12 hours. So, F can't start before 12 hours.Therefore, the earliest F can start is 12 hours, regardless.So, the total time is 18 hours.Wait, but what if we assign E to the executive earlier?After A is done at 3 hours, I take B (2 hours) until 5 hours.Executive takes C (4 hours) until 7 hours.At 5 hours, I'm free. I can take D (1 hour) until 6 hours.At 7 hours, executive is free. They can take E (5 hours) until 12 hours.Then, F starts at 12 hours, taking 6 hours.Same result.Alternatively, after A is done, I take C (4 hours) until 7 hours.Executive takes B (2 hours) until 5 hours.At 5 hours, executive is free. They can take D (1 hour) until 6 hours.At 7 hours, I'm free. I can take E (5 hours) until 12 hours.Then, F starts at 12 hours, taking 6 hours.Same total time.So, it seems that regardless of how we assign the tasks, the total time remains 18 hours because the critical path is still 18 hours, and we can't shorten it by parallelizing since the dependencies require that certain tasks must be done in sequence.Wait, but let me think again. Maybe there's a way to interleave tasks differently.After A is done at 3 hours, I take B (2 hours) until 5 hours.Executive takes C (4 hours) until 7 hours.At 5 hours, I'm free. I can take E (5 hours), but E depends on C, which is done at 7 hours. So, I can't start E until 7 hours.Alternatively, at 5 hours, I can take D (1 hour), finishing at 6 hours.Then, at 7 hours, I can take E (5 hours), finishing at 12 hours.Meanwhile, the executive is busy until 7 hours.At 7 hours, the executive is free. They can take F (6 hours), but F depends on D and E. D is done at 6 hours, E at 12 hours. So, F can't start until 12 hours.So, the executive is idle from 7 to 12 hours.Alternatively, the executive can help with E, but E is already being done by me from 7 to 12 hours.Wait, but if both of us work on E, can we finish it faster? But the problem states that each person can only work on one task at a time. So, we can't split tasks; each task must be done by one person entirely.Therefore, E must be done by one person, taking 5 hours.So, regardless, E takes 5 hours, and since it can't start before 7 hours, it finishes at 12 hours.Then, F starts at 12 hours, taking 6 hours.Total time: 18 hours.Therefore, even with parallel work, the minimum total time remains 18 hours because the critical path can't be shortened.Wait, but let me check if there's any way to overlap tasks more cleverly.Suppose after A is done at 3 hours, I take B (2 hours) until 5 hours.Executive takes C (4 hours) until 7 hours.At 5 hours, I'm free. I can take E (5 hours), but E depends on C, which is done at 7 hours. So, I can't start E until 7 hours.Alternatively, at 5 hours, I can take D (1 hour), finishing at 6 hours.Then, at 7 hours, I can take E (5 hours), finishing at 12 hours.Meanwhile, the executive is busy until 7 hours.At 7 hours, the executive is free. They can take F (6 hours), but F can't start until E is done at 12 hours.So, the executive is idle from 7 to 12 hours.Alternatively, the executive can take E at 7 hours, but I'm already free at 7 hours. So, maybe I can take E and the executive can do something else? But after C is done, the only task is E, which is dependent on C.Wait, no, after C is done, E is the only task, so only one person can do E.Therefore, regardless, E takes 5 hours, starting at 7 hours, finishing at 12 hours.Then, F starts at 12 hours, taking 6 hours.Total time: 18 hours.So, it seems that no matter how we assign the tasks, the total time remains 18 hours because the critical path is too long and can't be shortened by parallel work.Wait, but let me think about the total work involved. The total work is 3+2+4+1+5+6 = 21 hours. Since there are two people, the theoretical minimum time is ceil(21/2) = 11 hours. But due to dependencies, we can't reach that.But in reality, the dependencies force us to have a critical path of 18 hours, so the actual minimum time is 18 hours.Therefore, the answer for part 1 is 18 hours, and for part 2, it's also 18 hours.Wait, but that seems counterintuitive. Usually, parallel work should reduce the time, but in this case, the dependencies are so strict that we can't overlap the critical path tasks.Yes, because the critical path is A -> C -> E -> F, which is 3 + 4 + 5 + 6 = 18 hours. Since these tasks are dependent, they can't be overlapped. So, even with two people, the critical path remains the same.Therefore, the minimum total time required is 18 hours in both cases.But wait, in the second part, maybe we can do some tasks in a different order to reduce the time.Wait, let's see. After A is done, can we do E before C? No, because C must be done before E.Similarly, B must be done before D.So, the dependencies are fixed.Therefore, the critical path remains the same, and the total time can't be reduced below 18 hours.So, the answer is 18 hours for both parts.Wait, but let me double-check.In the first part, working alone, the critical path is 18 hours.In the second part, with two people, the critical path is still 18 hours because the dependencies can't be overlapped.Therefore, the minimum total time is 18 hours in both cases.But wait, is that correct? Because in some cases, parallel work can reduce the time even with dependencies.For example, if two tasks on the critical path can be done in parallel, but in this case, the critical path is a single sequence.Wait, the critical path is A -> C -> E -> F. These are all dependent, so they can't be parallelized.Therefore, even with two people, these tasks must be done in sequence, each taking their respective times.So, A (3) + C (4) + E (5) + F (6) = 18 hours.Therefore, the total time is 18 hours.So, the answer is 18 hours for both parts.But wait, in the second part, maybe we can do some tasks in parallel that are not on the critical path, thus allowing the critical path to be completed faster.Wait, for example, after A is done, we can do B and C in parallel.B takes 2 hours, C takes 4 hours.So, B is done at 5 hours, C at 7 hours.Then, D can be done at 5 hours (after B), and E at 7 hours (after C).So, D is done at 6 hours, E at 12 hours.Then, F can be done at 12 hours, taking 6 hours.So, the total time is 18 hours.But if we can do D and E in parallel, but D is done at 6 hours, E at 12 hours, so F can't start until 12 hours.Alternatively, if we can do D and E in parallel, but E depends on C, which is done at 7 hours, so E can start at 7 hours.But D is done at 6 hours, so E can start at 7 hours, and F can start when both D and E are done.So, E starts at 7 hours, takes 5 hours, finishes at 12 hours.D is done at 6 hours.So, F starts at 12 hours, takes 6 hours.Total time: 18 hours.Therefore, even with parallelization, the total time remains 18 hours because the critical path is too long.So, the answer is 18 hours for both parts.Wait, but let me think again. Maybe if we do some tasks in a different order.After A is done at 3 hours, I take C (4 hours) until 7 hours.Executive takes B (2 hours) until 5 hours.At 5 hours, executive is free. They can take D (1 hour) until 6 hours.At 7 hours, I'm free. I can take E (5 hours) until 12 hours.Then, F starts at 12 hours, taking 6 hours.Total time: 18 hours.Same result.Alternatively, after A is done, I take B (2 hours) until 5 hours.Executive takes C (4 hours) until 7 hours.At 5 hours, I take D (1 hour) until 6 hours.At 7 hours, executive takes E (5 hours) until 12 hours.Then, F starts at 12 hours, taking 6 hours.Same total time.So, regardless of the order, the total time is 18 hours.Therefore, the minimum total time required is 18 hours in both cases.Wait, but that seems a bit strange because usually, parallel work should help. But in this case, the dependencies are such that the critical path can't be shortened.Therefore, the answer is 18 hours for both parts.</think>

🧠:<think>Okay, so I have this problem about a philanthropist organizing a charity event. They’re using a tiered donation strategy with Gold, Silver, and Bronze tiers. Each tier has a different model for the expected number of donors based on the number of promotional activities, x. Sub-problem 1 is asking me to find the expression for the total expected donation amount, D(x), and then determine the number of promotional activities x that maximizes D(x). The models given are quadratic for Gold, linear for Silver, and exponential for Bronze. Let me break this down. The total donation D(x) is a combination of the donations from each tier. Each tier has its own function: G(x) for Gold, S(x) for Silver, and B(x) for Bronze. The total donation is calculated by multiplying the number of donors in each tier by the average donation amount per donor in that tier. So, D(x) = p*G(x) + q*S(x) + r*B(x), where p, q, r are the average donations per donor.Given that G(x) is quadratic: G(x) = a x² + b x + c. So, p*G(x) would be p*(a x² + b x + c) = p a x² + p b x + p c.S(x) is linear: S(x) = d x + e. So, q*S(x) is q*(d x + e) = q d x + q e.B(x) is exponential: B(x) = f * g^x. So, r*B(x) is r*f*g^x.Putting it all together, D(x) = p a x² + p b x + p c + q d x + q e + r f g^x.I can combine like terms. The x² term is p a x². The x terms are p b x + q d x, which can be written as (p b + q d) x. The constant terms are p c + q e. Then, the exponential term is r f g^x.So, D(x) = p a x² + (p b + q d) x + (p c + q e) + r f g^x.Now, to find the x that maximizes D(x). Since D(x) is a combination of a quadratic and an exponential function, it's going to be a bit tricky. The quadratic term is a x², which is a parabola opening upwards if a is positive, but since all constants are positive, a is positive, so the quadratic part is convex. The exponential term, r f g^x, is also convex because the exponential function is convex. So, the sum of convex functions is convex, meaning D(x) is a convex function. Wait, but convex functions have a minimum, not a maximum. Hmm, that's confusing. If D(x) is convex, then it doesn't have a maximum unless it's bounded above. But since the exponential term will dominate as x increases, D(x) will go to infinity as x increases. So, actually, D(x) doesn't have a maximum; it increases without bound as x increases. But that can't be right because in reality, you can't have an infinite number of promotional activities. Wait, but in the problem statement, for Sub-problem 1, it just says to find the number of promotional activities x that maximizes D(x). Maybe I need to consider if D(x) actually has a maximum or if it's unbounded. Let me think again. The quadratic term is p a x², which is positive, so as x increases, that term grows quadratically. The exponential term is r f g^x, which grows even faster, exponentially. So, as x increases, both terms go to infinity. Therefore, D(x) tends to infinity as x increases. So, there is no finite maximum; D(x) can be made as large as desired by increasing x. But that seems counterintuitive because in reality, there might be constraints on x, like time, money, or other resources. But in Sub-problem 1, it doesn't mention any constraints. So, mathematically, without constraints, D(x) doesn't have a maximum; it's unbounded above. Wait, but maybe I made a mistake. Let me double-check. The functions G(x), S(x), B(x) are all increasing functions of x? Because a, b, c, d, e, f, g are positive constants. So, G(x) is quadratic, so it will eventually increase as x increases. S(x) is linear, so it increases with x. B(x) is exponential, which also increases with x. So, all three components of D(x) are increasing functions of x, meaning D(x) is also increasing with x. Therefore, as x increases, D(x) increases without bound. So, in that case, there is no maximum; D(x) can be made as large as possible by increasing x. But that seems odd because in reality, you can't have an infinite number of promotional activities. Maybe the problem is expecting me to consider the derivative and set it to zero to find a critical point, but if the function is always increasing, the derivative will always be positive, so there's no critical point where the function changes direction.Wait, let me compute the derivative of D(x). D'(x) = derivative of p a x² + (p b + q d) x + (p c + q e) + r f g^x.So, D'(x) = 2 p a x + (p b + q d) + r f ln(g) g^x.Since all constants are positive, 2 p a x is positive, (p b + q d) is positive, and r f ln(g) g^x is positive if g > 1. If g < 1, ln(g) is negative, but since g is a base for an exponential model of donors, it's likely that g > 1, meaning the number of Bronze donors increases with x. So, assuming g > 1, all terms in D'(x) are positive. Therefore, D'(x) is always positive, meaning D(x) is strictly increasing. Therefore, D(x) doesn't have a maximum; it just keeps increasing as x increases.But the problem says to determine the number of promotional activities x that maximizes D(x). So, maybe I'm missing something. Perhaps the functions G(x), S(x), B(x) have different behaviors? Let me check the models again.G(x) is quadratic, so it's a parabola. Depending on the coefficients, it could open upwards or downwards, but since a is positive, it opens upwards. So, G(x) has a minimum at x = -b/(2a). But since x represents the number of promotional activities, it's non-negative. So, if the minimum is at a negative x, then G(x) is increasing for x >=0. So, G(x) is increasing for x >=0.Similarly, S(x) is linear with positive slope d, so it's increasing.B(x) is exponential. If g > 1, it's increasing; if 0 < g < 1, it's decreasing. But since it's modeling the number of donors, which should increase with more promotional activities, I think g > 1.Therefore, all three functions are increasing in x, so their combination D(x) is also increasing in x. Therefore, D(x) has no maximum; it increases without bound as x increases.But the problem is asking to find the x that maximizes D(x). Maybe I need to consider if there's a point where the marginal increase in D(x) starts to decrease, but since D'(x) is always increasing (because the second derivative is positive), D(x) is convex, so it's always increasing at an increasing rate.Wait, the second derivative of D(x) is D''(x) = 2 p a + r f (ln g)^2 g^x, which is positive because all constants are positive. So, D(x) is convex, meaning it curves upward. Therefore, D(x) is increasing and its rate of increase is also increasing. So, there's no maximum; the more x you do, the higher D(x) gets.Therefore, in Sub-problem 1, without any constraints, the optimal x is infinity. But since x can't be infinity in reality, maybe the problem expects us to say that D(x) is unbounded and thus there's no finite maximum. But the problem says to "determine the number of promotional activities x that maximizes the total expected donation amount." So, perhaps the answer is that there is no maximum; D(x) increases indefinitely with x.But maybe I'm missing something. Let me think again. Perhaps the functions G(x), S(x), B(x) have different behaviors. For example, maybe G(x) is a quadratic that peaks and then decreases, but since a is positive, it's a parabola opening upwards, so it has a minimum, not a maximum. So, G(x) is increasing for x > -b/(2a). Since x is non-negative, if -b/(2a) is negative, which it is because a and b are positive, then G(x) is increasing for all x >=0.Similarly, S(x) is linear increasing, and B(x) is exponential increasing. So, all three are increasing, so D(x) is increasing. Therefore, no maximum.But the problem is asking to find x that maximizes D(x). So, perhaps the answer is that D(x) is unbounded above, so there's no finite x that maximizes it; it can be made arbitrarily large by increasing x.Alternatively, maybe the problem expects us to find the critical point, but since D'(x) is always positive, there's no critical point where D'(x)=0. So, the function doesn't have a maximum.Therefore, for Sub-problem 1, the expression for D(x) is D(x) = p a x² + (p b + q d) x + (p c + q e) + r f g^x, and there is no finite x that maximizes D(x); it increases without bound as x increases.But maybe I should write that the function is always increasing, so the maximum is achieved as x approaches infinity, but in practical terms, x can't be infinite, so the philanthropist should conduct as many promotional activities as possible.But the problem says to "determine the number of promotional activities x that maximizes the total expected donation amount." So, perhaps the answer is that there is no maximum; D(x) increases indefinitely with x.Alternatively, maybe I made a mistake in interpreting the functions. Let me check the models again.G(x) = a x² + b x + c. Since a, b, c are positive, G(x) is increasing for x >=0.S(x) = d x + e. Increasing.B(x) = f g^x. If g >1, increasing; if g=1, constant; if g<1, decreasing. But since it's modeling donors increasing with promotional activities, g>1.Therefore, D(x) is increasing for all x >=0, so no maximum.Therefore, the answer is that D(x) is unbounded above, so there is no finite x that maximizes it; it can be made as large as desired by increasing x.But the problem might expect a different approach. Maybe I need to consider the derivative and set it to zero, but since D'(x) is always positive, there's no solution. So, the maximum is at infinity.Alternatively, perhaps the problem expects me to consider that while D(x) is increasing, the rate of increase might slow down, but since D''(x) is positive, the rate of increase is actually accelerating.So, in conclusion, for Sub-problem 1, the expression for D(x) is D(x) = p a x² + (p b + q d) x + (p c + q e) + r f g^x, and there is no finite x that maximizes D(x); it increases without bound as x increases.Now, moving on to Sub-problem 2. It says that due to logistical constraints, x cannot exceed a certain limit L. So, we need to formulate a constraint optimization problem to maximize D(x) subject to x <= L, and solve for the optimal x within the constraint.So, the problem is to maximize D(x) = p a x² + (p b + q d) x + (p c + q e) + r f g^x, subject to x <= L, and x >=0 (since you can't have negative promotional activities).Since D(x) is increasing in x, as established in Sub-problem 1, the maximum D(x) occurs at the upper bound of x, which is x = L.Therefore, the optimal number of promotional activities is x = L.But let me think again. If D(x) is increasing, then yes, the maximum within x <= L is at x = L. So, the optimal x is L.Alternatively, if D(x) had a maximum within the feasible region, we would have to check if the critical point is within [0, L]. But since D(x) is always increasing, the maximum is at x = L.Therefore, the optimal x is L.So, summarizing:Sub-problem 1: D(x) = p a x² + (p b + q d) x + (p c + q e) + r f g^x, and there is no finite x that maximizes D(x); it increases indefinitely with x.Sub-problem 2: With x <= L, the optimal x is L.But wait, in Sub-problem 1, the problem says to "determine the number of promotional activities x that maximizes the total expected donation amount." So, if there's no maximum, perhaps the answer is that the maximum is achieved as x approaches infinity, but in practical terms, the philanthropist should conduct as many promotional activities as possible.But in Sub-problem 2, with x <= L, the maximum is at x = L.So, I think that's the conclusion.</think>

🧠:<think>Okay, so I have this problem about a theater actor who is trying to perfect his monologue delivery. He's using a function to model his speech rate, which is given by ( f(t) = A sin(Bt + C) + D ). The function models his speech rate in words per minute (wpm) over time ( t ) in minutes. There are two parts to this problem. Let me tackle them one by one.Problem 1: Derive a relationship between the constants ( A, B, C, ) and ( D ) given that the average speech rate over the entire monologue is 150 wpm.Hmm, okay. So the average speech rate over the entire monologue is 150 wpm. The monologue lasts exactly 10 minutes, so the average would be the integral of the speech rate function over 0 to 10 minutes divided by 10. Mathematically, the average value ( bar{f} ) of a function ( f(t) ) over the interval ([a, b]) is given by:[bar{f} = frac{1}{b - a} int_{a}^{b} f(t) dt]In this case, ( a = 0 ), ( b = 10 ), and ( bar{f} = 150 ). So,[150 = frac{1}{10 - 0} int_{0}^{10} [A sin(Bt + C) + D] dt]Simplify the equation:[150 = frac{1}{10} int_{0}^{10} [A sin(Bt + C) + D] dt]Let me compute the integral. The integral of ( A sin(Bt + C) ) with respect to ( t ) is:[int A sin(Bt + C) dt = -frac{A}{B} cos(Bt + C) + K]And the integral of ( D ) with respect to ( t ) is:[int D dt = D t + K]So putting it all together, the integral from 0 to 10 is:[left[ -frac{A}{B} cos(Bt + C) + D t right]_0^{10}]Which is:[left( -frac{A}{B} cos(10B + C) + 10D right) - left( -frac{A}{B} cos(C) + 0 right)]Simplify this expression:[-frac{A}{B} cos(10B + C) + 10D + frac{A}{B} cos(C)]So the integral becomes:[frac{1}{10} left( -frac{A}{B} cos(10B + C) + 10D + frac{A}{B} cos(C) right) = 150]Multiply both sides by 10:[-frac{A}{B} cos(10B + C) + 10D + frac{A}{B} cos(C) = 1500]Hmm, that seems a bit complicated. But wait, I remember that the average value of a sinusoidal function over a full period is equal to its vertical shift ( D ). Because the sine function oscillates symmetrically around its midline, so the average cancels out the oscillation part. Is the interval from 0 to 10 minutes a multiple of the period of the sine function? Let's check.The period ( T ) of ( sin(Bt + C) ) is ( frac{2pi}{B} ). So unless ( 10 ) is a multiple of ( frac{2pi}{B} ), the integral won't necessarily simplify to just ( 10D ). But in the problem statement, it's not specified whether the function completes an integer number of periods over the 10-minute interval. So, perhaps we can't assume that. Wait, but the average speech rate is given as 150 wpm. If the function ( f(t) ) is sinusoidal, its average over any interval is equal to its vertical shift ( D ) only if the interval is a multiple of the period. Otherwise, the average might be different. But the problem says the average over the entire monologue is 150 wpm, which is 10 minutes. So unless ( 10 ) is a multiple of the period, the average might not just be ( D ). Hmm, so perhaps I need to consider whether the integral of the sine term over 0 to 10 is zero. For that to happen, the interval 0 to 10 must be an integer multiple of the period. Let me think. If ( 10 = n times frac{2pi}{B} ) for some integer ( n ), then the sine function would complete ( n ) full periods, and the integral of the sine term over that interval would be zero. But since we don't know ( B ), we can't be sure. So perhaps the problem expects us to use the average value formula regardless, leading to the equation above.But that equation is:[-frac{A}{B} cos(10B + C) + 10D + frac{A}{B} cos(C) = 1500]But this seems complicated because it involves ( B ) and ( C ), which are constants we don't know. Wait, maybe the problem is expecting a simpler relationship. Since the average of a sine function over its period is zero, perhaps the average over any interval that is a multiple of the period is zero. But if the interval is not a multiple, it's not zero.But without knowing ( B ) and ( C ), perhaps we can't derive a specific relationship. Wait, but the problem says "derive a relationship between the constants ( A, B, C, ) and ( D )", so maybe it's expecting an equation involving these constants.So, from the integral, we have:[-frac{A}{B} [cos(10B + C) - cos(C)] + 10D = 1500]So, that's the relationship.But let me check my steps again.1. The average is 150, so:[frac{1}{10} int_{0}^{10} [A sin(Bt + C) + D] dt = 150]2. The integral becomes:[frac{1}{10} left[ -frac{A}{B} cos(Bt + C) + D t right]_0^{10}]3. Evaluating from 0 to 10:[frac{1}{10} left( -frac{A}{B} cos(10B + C) + 10D + frac{A}{B} cos(C) right) = 150]4. Multiply both sides by 10:[-frac{A}{B} cos(10B + C) + 10D + frac{A}{B} cos(C) = 1500]So yes, that's the relationship. So, the equation is:[10D + frac{A}{B} [cos(C) - cos(10B + C)] = 1500]Alternatively, factoring:[10D + frac{A}{B} [cos(C) - cos(10B + C)] = 1500]I think that's the relationship. It involves all four constants ( A, B, C, D ). But maybe we can simplify it further? Let's see.Using the trigonometric identity:[cos theta - cos phi = -2 sin left( frac{theta + phi}{2} right) sin left( frac{theta - phi}{2} right)]So, let ( theta = C ) and ( phi = 10B + C ). Then,[cos(C) - cos(10B + C) = -2 sin left( frac{C + 10B + C}{2} right) sin left( frac{C - (10B + C)}{2} right)]Simplify:[= -2 sin left( frac{2C + 10B}{2} right) sin left( frac{-10B}{2} right)][= -2 sin(C + 5B) sin(-5B)][= -2 sin(C + 5B) (-sin(5B)) quad text{since } sin(-x) = -sin x][= 2 sin(C + 5B) sin(5B)]So, substituting back into the equation:[10D + frac{A}{B} times 2 sin(C + 5B) sin(5B) = 1500]So,[10D + frac{2A}{B} sin(C + 5B) sin(5B) = 1500]I don't know if this is helpful, but it's another form of the relationship. Alternatively, perhaps the problem expects a simpler relationship, assuming that the average of the sine function over the interval is zero. But that would only be the case if the interval is a multiple of the period. Wait, if the function ( f(t) ) is sinusoidal, its average over a full period is ( D ). So, if the monologue duration is a multiple of the period, then the average speech rate would be ( D ). But the monologue is 10 minutes long. So, if ( 10 ) is a multiple of ( frac{2pi}{B} ), then the average would be ( D ). But since we don't know ( B ), perhaps the problem is expecting us to consider that the average is ( D ), regardless of the interval. But that's not necessarily true unless the interval is a multiple of the period.Wait, maybe the problem is designed so that the average is ( D ), so ( D = 150 ). But that might not be the case unless the integral of the sine term is zero.Wait, let me think again. If the function is ( A sin(Bt + C) + D ), then over a full period, the average is ( D ). But over an interval that's not a multiple of the period, the average could be different.But the problem says the average over the entire monologue is 150 wpm. So unless the monologue duration is a multiple of the period, the average isn't necessarily ( D ). But perhaps the problem is designed in such a way that the average is ( D ), so ( D = 150 ). Maybe the function is set up so that the sine term averages out to zero over the 10-minute interval. Wait, but without knowing ( B ) and ( C ), we can't be sure. So, perhaps the answer is simply ( D = 150 ). But that might not be accurate unless the integral of the sine term is zero.Wait, let me test this. Suppose ( D = 150 ). Then, the average would be 150 only if the integral of ( A sin(Bt + C) ) over 0 to 10 is zero. So, if:[int_{0}^{10} A sin(Bt + C) dt = 0]Then,[-frac{A}{B} [cos(10B + C) - cos(C)] = 0]Which implies:[cos(10B + C) = cos(C)]Which would be true if ( 10B ) is a multiple of ( 2pi ), i.e., ( 10B = 2pi n ) for some integer ( n ). So, if ( B = frac{pi n}{5} ), then ( 10B = 2pi n ), and the cosine terms would be equal, making the integral zero. Therefore, if ( B ) is chosen such that ( 10B ) is a multiple of ( 2pi ), then the average speech rate is ( D ), which is 150. But since the problem doesn't specify anything about ( B ) or ( C ), maybe it's expecting us to assume that the average is ( D ), so ( D = 150 ). But I'm not entirely sure. The problem says "derive a relationship between the constants ( A, B, C, ) and ( D )", so perhaps it's expecting the equation I derived earlier:[10D + frac{A}{B} [cos(C) - cos(10B + C)] = 1500]Yes, that seems to be the correct relationship, involving all four constants. Problem 2: Determine the values of ( A ) and ( D ) that will satisfy the conditions of a maximum rate of 180 wpm and a minimum rate of 120 wpm.Okay, so the function is ( f(t) = A sin(Bt + C) + D ). The maximum value of a sine function is 1, and the minimum is -1. So, the maximum of ( f(t) ) is ( A(1) + D = A + D ), and the minimum is ( A(-1) + D = -A + D ).Given that the maximum rate is 180 wpm and the minimum is 120 wpm, we can set up the following equations:1. ( A + D = 180 )2. ( -A + D = 120 )So, we have a system of two equations:1. ( A + D = 180 )2. ( -A + D = 120 )Let me solve this system.First, subtract the second equation from the first:[(A + D) - (-A + D) = 180 - 120][A + D + A - D = 60][2A = 60][A = 30]Now, substitute ( A = 30 ) into the first equation:[30 + D = 180][D = 180 - 30][D = 150]So, ( A = 30 ) and ( D = 150 ).Wait, but in Problem 1, we had an equation involving ( A, B, C, D ). If ( D = 150 ), then from Problem 1's equation:[10D + frac{A}{B} [cos(C) - cos(10B + C)] = 1500]Substituting ( D = 150 ):[10 times 150 + frac{30}{B} [cos(C) - cos(10B + C)] = 1500][1500 + frac{30}{B} [cos(C) - cos(10B + C)] = 1500]Subtract 1500 from both sides:[frac{30}{B} [cos(C) - cos(10B + C)] = 0]Which implies:[cos(C) - cos(10B + C) = 0]So,[cos(C) = cos(10B + C)]Which means that ( 10B + C = 2pi n pm C ) for some integer ( n ).So,Case 1: ( 10B + C = 2pi n + C )Subtract ( C ):[10B = 2pi n][B = frac{pi n}{5}]Case 2: ( 10B + C = 2pi n - C )Add ( C ):[10B + 2C = 2pi n][5B + C = pi n]So, either ( B = frac{pi n}{5} ) or ( 5B + C = pi n ).But since ( B ) and ( C ) are constants, and we're only asked to find ( A ) and ( D ), perhaps we don't need to worry about ( B ) and ( C ) here. But it's interesting that for the average speech rate to be 150 wpm, given that ( D = 150 ), the function must satisfy ( cos(C) = cos(10B + C) ), which constrains ( B ) and ( C ) as above.But since the problem only asks for ( A ) and ( D ), and we've found ( A = 30 ) and ( D = 150 ), that should be sufficient.Let me just verify.If ( A = 30 ) and ( D = 150 ), then the function is ( f(t) = 30 sin(Bt + C) + 150 ).The maximum value is ( 30 times 1 + 150 = 180 ), and the minimum is ( 30 times (-1) + 150 = 120 ). Perfect, that matches the given maximum and minimum rates.So, yeah, ( A = 30 ) and ( D = 150 ).Final Answer1. The relationship is ( 10D + frac{A}{B} [cos(C) - cos(10B + C)] = 1500 ).2. The values are ( A = boxed{30} ) and ( D = boxed{150} ).</think>

🧠:<think>Okay, so I have this problem where a journalism student is looking at how a famous critic's score affects a film's box office revenue. They're using a cubic polynomial to model this relationship. The function is given as ( R(S) = aS^3 + bS^2 + cS + d ). I need to find the coefficients ( a, b, c, ) and ( d ) using the given data points. Then, in part 2, I have to find the score ( S ) that maximizes the revenue by taking the derivative of ( R(S) ) and finding its critical points.Alright, let's start with part 1. I have four data points: when ( S = 20 ), ( R = 15 ); when ( S = 40 ), ( R = 50 ); when ( S = 60 ), ( R = 140 ); and when ( S = 80 ), ( R = 300 ). Since it's a cubic polynomial, and I have four points, I can set up a system of four equations to solve for the four unknowns ( a, b, c, ) and ( d ).So, plugging each data point into the equation:1. When ( S = 20 ), ( R = 15 ): ( a(20)^3 + b(20)^2 + c(20) + d = 15 ) Simplify: ( 8000a + 400b + 20c + d = 15 ) [Equation 1]2. When ( S = 40 ), ( R = 50 ): ( a(40)^3 + b(40)^2 + c(40) + d = 50 ) Simplify: ( 64000a + 1600b + 40c + d = 50 ) [Equation 2]3. When ( S = 60 ), ( R = 140 ): ( a(60)^3 + b(60)^2 + c(60) + d = 140 ) Simplify: ( 216000a + 3600b + 60c + d = 140 ) [Equation 3]4. When ( S = 80 ), ( R = 300 ): ( a(80)^3 + b(80)^2 + c(80) + d = 300 ) Simplify: ( 512000a + 6400b + 80c + d = 300 ) [Equation 4]Now, I have four equations:1. ( 8000a + 400b + 20c + d = 15 )2. ( 64000a + 1600b + 40c + d = 50 )3. ( 216000a + 3600b + 60c + d = 140 )4. ( 512000a + 6400b + 80c + d = 300 )I need to solve this system for ( a, b, c, d ). Since all equations have ( d ), I can subtract each equation from the next one to eliminate ( d ). Let's do that step by step.First, subtract Equation 1 from Equation 2:Equation 2 - Equation 1:( (64000a - 8000a) + (1600b - 400b) + (40c - 20c) + (d - d) = 50 - 15 )Simplify:( 56000a + 1200b + 20c = 35 ) [Equation 5]Next, subtract Equation 2 from Equation 3:Equation 3 - Equation 2:( (216000a - 64000a) + (3600b - 1600b) + (60c - 40c) + (d - d) = 140 - 50 )Simplify:( 152000a + 2000b + 20c = 90 ) [Equation 6]Then, subtract Equation 3 from Equation 4:Equation 4 - Equation 3:( (512000a - 216000a) + (6400b - 3600b) + (80c - 60c) + (d - d) = 300 - 140 )Simplify:( 296000a + 2800b + 20c = 160 ) [Equation 7]Now, I have three new equations (5, 6, 7) without ( d ):5. ( 56000a + 1200b + 20c = 35 )6. ( 152000a + 2000b + 20c = 90 )7. ( 296000a + 2800b + 20c = 160 )Next, I can subtract Equation 5 from Equation 6 and Equation 6 from Equation 7 to eliminate ( c ).First, subtract Equation 5 from Equation 6:Equation 6 - Equation 5:( (152000a - 56000a) + (2000b - 1200b) + (20c - 20c) = 90 - 35 )Simplify:( 96000a + 800b = 55 ) [Equation 8]Next, subtract Equation 6 from Equation 7:Equation 7 - Equation 6:( (296000a - 152000a) + (2800b - 2000b) + (20c - 20c) = 160 - 90 )Simplify:( 144000a + 800b = 70 ) [Equation 9]Now, Equations 8 and 9 are:8. ( 96000a + 800b = 55 )9. ( 144000a + 800b = 70 )Subtract Equation 8 from Equation 9 to eliminate ( b ):Equation 9 - Equation 8:( (144000a - 96000a) + (800b - 800b) = 70 - 55 )Simplify:( 48000a = 15 )So, ( a = 15 / 48000 = 1 / 3200 ). Let me compute that: 15 divided by 48000 is 0.0003125, which is 1/3200. So, ( a = 1/3200 ).Now, plug ( a = 1/3200 ) into Equation 8 to find ( b ):Equation 8: ( 96000*(1/3200) + 800b = 55 )Calculate ( 96000 / 3200 = 30 ). So, 30 + 800b = 55Subtract 30: 800b = 25Thus, ( b = 25 / 800 = 5/160 = 1/32 ). So, ( b = 1/32 ).Now, with ( a = 1/3200 ) and ( b = 1/32 ), go back to Equation 5 to find ( c ):Equation 5: ( 56000*(1/3200) + 1200*(1/32) + 20c = 35 )Compute each term:56000 / 3200 = 17.51200 / 32 = 37.5So, 17.5 + 37.5 + 20c = 35Add 17.5 and 37.5: 55 + 20c = 35Subtract 55: 20c = -20Thus, ( c = -1 ).Now, with ( a = 1/3200 ), ( b = 1/32 ), and ( c = -1 ), plug into Equation 1 to find ( d ):Equation 1: ( 8000*(1/3200) + 400*(1/32) + 20*(-1) + d = 15 )Calculate each term:8000 / 3200 = 2.5400 / 32 = 12.520*(-1) = -20So, 2.5 + 12.5 - 20 + d = 15Add 2.5 and 12.5: 15 - 20 + d = 15So, -5 + d = 15Thus, ( d = 20 ).So, the coefficients are:( a = 1/3200 ), ( b = 1/32 ), ( c = -1 ), ( d = 20 ).Let me write the polynomial:( R(S) = (1/3200)S^3 + (1/32)S^2 - S + 20 )Wait, let me double-check these calculations because it's easy to make arithmetic errors.First, checking ( a = 1/3200 ). Then, in Equation 8: 96000*(1/3200) = 96000 / 3200 = 30. Correct. Then 30 + 800b = 55, so 800b = 25, so b = 25/800 = 1/32. Correct.In Equation 5: 56000*(1/3200) = 56000 / 3200 = 17.5. 1200*(1/32) = 1200 / 32 = 37.5. So, 17.5 + 37.5 = 55. Then, 55 + 20c = 35, so 20c = -20, so c = -1. Correct.Then, Equation 1: 8000*(1/3200) = 2.5, 400*(1/32) = 12.5, 20*(-1) = -20. So, 2.5 + 12.5 - 20 = -5. Then, -5 + d = 15, so d = 20. Correct.So, the coefficients seem correct.Now, moving on to part 2: finding the critic's score ( S ) that maximizes the box office revenue. To do this, I need to find the critical points of ( R(S) ) by taking its derivative and setting it equal to zero.First, let's write the derivative ( R'(S) ):( R'(S) = 3aS^2 + 2bS + c )Plugging in the values of ( a, b, c ):( R'(S) = 3*(1/3200)S^2 + 2*(1/32)S + (-1) )Simplify each term:3*(1/3200) = 3/32002*(1/32) = 2/32 = 1/16So, ( R'(S) = (3/3200)S^2 + (1/16)S - 1 )To find critical points, set ( R'(S) = 0 ):( (3/3200)S^2 + (1/16)S - 1 = 0 )This is a quadratic equation in terms of ( S ). Let's write it as:( (3/3200)S^2 + (1/16)S - 1 = 0 )To make it easier, multiply all terms by 3200 to eliminate denominators:3200*(3/3200)S^2 + 3200*(1/16)S - 3200*1 = 0Simplify:3S^2 + 200S - 3200 = 0So, the quadratic equation is:3S^2 + 200S - 3200 = 0Let me write that as:3S² + 200S - 3200 = 0We can solve this using the quadratic formula:( S = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Where ( a = 3 ), ( b = 200 ), ( c = -3200 ).Compute discriminant ( D = b² - 4ac ):D = (200)^2 - 4*3*(-3200) = 40000 + 38400 = 78400Square root of D: sqrt(78400) = 280So, solutions:( S = frac{-200 pm 280}{2*3} )Compute both possibilities:First solution: ( (-200 + 280)/6 = 80/6 = 40/3 ≈ 13.333 )Second solution: ( (-200 - 280)/6 = (-480)/6 = -80 )Since the score ( S ) is on a scale from 0 to 100, we can ignore the negative solution. So, the critical point is at ( S = 40/3 ≈ 13.333 ).But wait, the data points given start at ( S = 20 ). So, 13.333 is below the lowest score in the data. Hmm, that seems odd. Maybe I made a mistake in the derivative or the calculations.Wait, let me double-check the derivative.Original function: ( R(S) = (1/3200)S³ + (1/32)S² - S + 20 )Derivative: ( R'(S) = 3*(1/3200)S² + 2*(1/32)S - 1 )Which is ( (3/3200)S² + (1/16)S - 1 ). Correct.Then, setting equal to zero:( (3/3200)S² + (1/16)S - 1 = 0 )Multiply by 3200:3S² + 200S - 3200 = 0Quadratic formula: ( S = [-200 ± sqrt(200² - 4*3*(-3200))]/(2*3) )Compute discriminant:200² = 400004*3*3200 = 12*3200 = 38400So, D = 40000 + 38400 = 78400sqrt(78400) = 280Thus, ( S = (-200 ± 280)/6 )So, positive solution: (80)/6 ≈13.333, negative solution: (-480)/6 = -80.So, the critical point is at S ≈13.333, which is below the given data points. That suggests that the maximum revenue might occur at the lowest score in the data, which is 20. But wait, let's check the behavior of the derivative.Since the leading coefficient of the cubic polynomial is positive (a = 1/3200 > 0), the cubic function tends to infinity as S increases. However, the derivative is a quadratic with a positive leading coefficient (3/3200 > 0), so it opens upwards. Therefore, the critical point at S ≈13.333 is a local minimum, not a maximum.Wait, that's important. So, the derivative has a minimum at S ≈13.333, which is a local minimum for R(S). Therefore, the function R(S) is decreasing before S ≈13.333 and increasing after that. But since our data starts at S=20, which is after the local minimum, the function is increasing from S=20 onwards.Therefore, in the domain of S from 20 to 80, the function is increasing. So, the maximum revenue would occur at the highest S, which is 80. But wait, let's check the derivative's sign in the interval [20,80].Compute R'(20):( R'(20) = (3/3200)*(400) + (1/16)*(20) - 1 )Calculate each term:(3/3200)*400 = (3*400)/3200 = 1200/3200 = 0.375(1/16)*20 = 20/16 = 1.25So, R'(20) = 0.375 + 1.25 - 1 = 0.625 > 0So, the derivative is positive at S=20, meaning the function is increasing at S=20. Since the derivative is a quadratic opening upwards, and the only critical point is at S≈13.333, which is a local minimum, the function is increasing for all S >13.333. Therefore, in the interval [20,80], the function is always increasing, meaning the maximum revenue occurs at S=80.But wait, let's check R'(80):( R'(80) = (3/3200)*(6400) + (1/16)*(80) - 1 )Calculate each term:(3/3200)*6400 = (3*6400)/3200 = 19200/3200 = 6(1/16)*80 = 5So, R'(80) = 6 + 5 -1 = 10 >0So, derivative is still positive at S=80, meaning the function is still increasing beyond S=80. Therefore, in the given data range, the function is increasing throughout, so the maximum revenue in the given data is at S=80, which is 300 million.But wait, the question is asking for the critic's score S that maximizes the box office revenue. If the function is increasing on [20,80], then the maximum is at S=80. However, if we consider beyond S=80, the function would continue to increase, but since the critic's score is capped at 100, maybe the maximum is at S=100.But the problem doesn't specify the domain beyond the given data. It just says S is from 0 to 100. So, technically, the function is increasing for all S >13.333, so the maximum would be at S=100.But let's check the derivative at S=100:( R'(100) = (3/3200)*(10000) + (1/16)*(100) -1 )Calculate:(3/3200)*10000 = 30000/3200 = 9.375(1/16)*100 = 6.25So, R'(100) = 9.375 + 6.25 -1 = 14.625 >0So, derivative is still positive at S=100, meaning the function is still increasing. Therefore, the maximum revenue would be at S=100. However, since the problem only provides data up to S=80, and the function is increasing throughout, the maximum in the given data is at S=80, but theoretically, it's unbounded as S increases beyond 100, but since S is capped at 100, the maximum would be at S=100.But wait, the question is a bit ambiguous. It says "the critic's score S that maximizes the box office revenue." If we consider the entire domain of S from 0 to 100, then the maximum is at S=100. However, if we consider only the given data points, the maximum is at S=80.But since the polynomial is defined for all S, and it's increasing beyond S=80, the maximum would be at S=100. However, let's check the second derivative to confirm the concavity.Wait, actually, since the derivative is always positive beyond S=13.333, the function is monotonically increasing in that interval, so the maximum occurs at the upper limit, which is S=100.But let me think again. The problem says "using the polynomial function R(S) derived in sub-problem 1, calculate the critic's score S that maximizes the box office revenue." So, it's not restricted to the given data points. It's asking for the score S where R(S) is maximized, which, given that R(S) is a cubic with positive leading coefficient, tends to infinity as S increases. However, since S is bounded between 0 and 100, the maximum would be at S=100.But wait, let's compute R(100):( R(100) = (1/3200)*(100)^3 + (1/32)*(100)^2 -100 +20 )Calculate each term:(1/3200)*1,000,000 = 1,000,000 / 3200 = 312.5(1/32)*10,000 = 10,000 /32 ≈312.5So, R(100) = 312.5 + 312.5 -100 +20 = 312.5 + 312.5 = 625; 625 -100 = 525; 525 +20 = 545.So, R(100) = 545 million.But let's check R(80) which is given as 300 million. So, R(100) is higher, as expected.But wait, the problem is about a critic's score, which is from 0 to 100. So, the maximum revenue would be at S=100, giving R=545 million.However, the question is whether the function actually has a maximum at S=100 or if it's increasing beyond that. Since S can't go beyond 100, the maximum is at S=100.But wait, let me think again. The derivative is positive at S=100, meaning the function is still increasing at S=100. So, if S could go beyond 100, the revenue would keep increasing. But since S is capped at 100, the maximum is at S=100.Therefore, the score S that maximizes R is 100.But wait, let me double-check the derivative. We found that the critical point is at S≈13.333, which is a local minimum. So, the function decreases until S≈13.333, then increases thereafter. Therefore, in the domain S ≥13.333, the function is increasing. Since our data starts at S=20, which is after the local minimum, the function is increasing throughout the given data range and beyond. Therefore, the maximum occurs at the highest possible S, which is 100.Therefore, the answer is S=100.But wait, let me check if the function actually has a maximum at S=100 or if it's just the highest point in the domain. Since the function is a cubic with positive leading coefficient, it tends to infinity as S increases, so the maximum is indeed at S=100 within the domain [0,100].So, to summarize:1. The coefficients are ( a = 1/3200 ), ( b = 1/32 ), ( c = -1 ), ( d = 20 ).2. The score S that maximizes R is 100.But wait, let me make sure I didn't make a mistake in interpreting the critical points. The derivative has only one real critical point at S≈13.333, which is a local minimum. Therefore, the function is decreasing before that and increasing after that. So, in the interval [20,80], the function is increasing, and beyond that, it continues to increase. Therefore, the maximum revenue occurs at the highest possible S, which is 100.Yes, that seems correct.So, the final answers are:1. ( a = frac{1}{3200} ), ( b = frac{1}{32} ), ( c = -1 ), ( d = 20 ).2. The score S that maximizes revenue is 100.</think>