🧠:<think>Okay, so I have this problem where a lawyer is analyzing the impact of government-imposed fees on her firm. There are two parts to this problem. Let me try to tackle them step by step.Starting with the first part: The firm currently handles ( C(t) = 50 + 5sin(pi t/6) ) cases per month. The government imposes a fee ( F ), and for each 100 increase in fee, the number of cases decreases by 0.1C(t). The fee is set at 200, so I need to find the new function ( C_F(t) ).Hmm, okay. So, the fee is 200, which is two times 100. For each 100, the cases decrease by 0.1C(t). So, for 200, it should decrease by 0.2C(t). That makes sense.So, the decrease in cases is 0.2 times the original case load. Therefore, the new case load should be the original case load minus 0.2 times the original case load. Which is 0.8 times the original case load.So, ( C_F(t) = C(t) - 0.2C(t) = 0.8C(t) ).Substituting the given ( C(t) ):( C_F(t) = 0.8(50 + 5sin(pi t/6)) ).Let me compute that:First, 0.8 times 50 is 40, and 0.8 times 5 is 4. So, ( C_F(t) = 40 + 4sin(pi t/6) ).Wait, that seems straightforward. So, the new case load function is 40 plus 4 sine of pi t over 6.Let me just verify that. If the fee is 200, which is two increments of 100, each causing a 10% decrease. So, 10% decrease twice would be a 20% decrease, so 80% of the original. So, yes, 0.8 times the original function. That seems correct.Alright, so part 1 is done. Now, moving on to part 2.The revenue ( R(t) ) is initially given by ( R(t) = 2000C(t) - 100C(t)^2 ). After the fee is imposed, the revenue function must account for the reduced case load and the additional 200 fee per case. So, I need to formulate the new revenue function ( R_F(t) ) and find the maximum monthly revenue.Okay, so initially, revenue is 2000C(t) minus 100C(t)^2. So, that's a quadratic in terms of C(t). After the fee, the number of cases is reduced, so we have ( C_F(t) ) instead of C(t). Also, there is an additional fee of 200 per case. So, does that mean the firm has to pay 200 per case, which reduces their revenue? Or does it mean that clients have to pay an additional 200, which would increase revenue? Hmm, the problem says "the revenue function must account for the reduced case load and additional 200 fee per case." So, probably, the fee is a cost for the firm, so it reduces their revenue.Wait, but the original revenue is 2000C(t) - 100C(t)^2. So, 2000C(t) is probably the income from clients, and 100C(t)^2 is the cost or something else? Hmm, maybe not. Let me think.Wait, actually, revenue is usually income minus costs. So, if the firm is paying a fee per case, that would be a cost. So, the original revenue is 2000C(t) - 100C(t)^2, which could be income minus some costs. Now, with the fee, they have an additional cost of 200 per case, so the new revenue would be 2000C_F(t) - 100C_F(t)^2 - 200C_F(t). Because for each case, they have to pay an extra 200, so total additional cost is 200C_F(t).Alternatively, if the fee is paid by the clients, then the firm's revenue would increase by 200C_F(t). But the problem says "the revenue function must account for the reduced case load and additional 200 fee per case." So, it's more likely that the fee is a cost for the firm, so subtracting 200C_F(t).So, putting it together, the new revenue function is:( R_F(t) = 2000C_F(t) - 100C_F(t)^2 - 200C_F(t) ).Simplify that:Combine the terms with C_F(t):2000C_F(t) - 200C_F(t) = 1800C_F(t).So, ( R_F(t) = 1800C_F(t) - 100C_F(t)^2 ).Alternatively, factor out 100:( R_F(t) = 100(18C_F(t) - C_F(t)^2) ).But maybe it's better to keep it as is.So, ( R_F(t) = 1800C_F(t) - 100C_F(t)^2 ).We already have ( C_F(t) = 40 + 4sin(pi t/6) ).So, substitute that in:( R_F(t) = 1800(40 + 4sin(pi t/6)) - 100(40 + 4sin(pi t/6))^2 ).Now, to find the maximum monthly revenue, we need to find the maximum value of ( R_F(t) ). Since this is a function of t, which is time in months, we can treat it as a function of a single variable and find its maximum.But before that, let me compute ( R_F(t) ) in terms of sine function.First, compute 1800*(40 + 4 sin(pi t /6)):1800*40 = 72,0001800*4 = 7,200So, 72,000 + 7,200 sin(pi t /6)Now, compute 100*(40 + 4 sin(pi t /6))^2:First, square (40 + 4 sin(pi t /6)):= 40^2 + 2*40*4 sin(pi t /6) + (4 sin(pi t /6))^2= 1600 + 320 sin(pi t /6) + 16 sin^2(pi t /6)Multiply by 100:= 160,000 + 32,000 sin(pi t /6) + 1,600 sin^2(pi t /6)So, putting it all together:( R_F(t) = [72,000 + 7,200 sin(pi t /6)] - [160,000 + 32,000 sin(pi t /6) + 1,600 sin^2(pi t /6)] )Simplify term by term:72,000 - 160,000 = -88,0007,200 sin(pi t /6) - 32,000 sin(pi t /6) = (-24,800) sin(pi t /6)And then minus 1,600 sin^2(pi t /6)So, ( R_F(t) = -88,000 - 24,800 sin(pi t /6) - 1,600 sin^2(pi t /6) )Hmm, that seems a bit messy, but maybe we can write it as:( R_F(t) = -1,600 sin^2(pi t /6) - 24,800 sin(pi t /6) - 88,000 )Alternatively, factor out -1,600:( R_F(t) = -1,600 left( sin^2(pi t /6) + 15.5 sin(pi t /6) + 55 right) )But I'm not sure if that helps. Alternatively, maybe we can express this quadratic in sin(pi t /6) and find its maximum.Let me denote ( x = sin(pi t /6) ). Then, the revenue function becomes:( R_F(t) = -1,600 x^2 - 24,800 x - 88,000 )But since x is between -1 and 1, because sine function ranges between -1 and 1.Wait, but in the original case load, C(t) = 50 + 5 sin(pi t /6). So, sin(pi t /6) ranges between -1 and 1, so C(t) ranges between 45 and 55. After the fee, C_F(t) = 40 + 4 sin(pi t /6), so sin(pi t /6) still ranges between -1 and 1, so C_F(t) ranges between 36 and 44.But in the revenue function, we have x = sin(pi t /6), which is still between -1 and 1.So, our quadratic in x is:( R_F = -1,600 x^2 - 24,800 x - 88,000 )We can treat this as a quadratic function in x, where x ∈ [-1, 1]. The maximum of this quadratic will occur either at the vertex or at the endpoints.But since the coefficient of x^2 is negative (-1,600), the parabola opens downward, so the vertex is the maximum point.The vertex of a quadratic ( ax^2 + bx + c ) is at x = -b/(2a).So, here, a = -1,600, b = -24,800.So, x = -(-24,800)/(2*(-1,600)) = 24,800 / (-3,200) = -7.75Wait, that's x = -7.75, but x is sin(pi t /6), which can only be between -1 and 1. So, the vertex is outside the domain of x.Therefore, the maximum must occur at one of the endpoints, x = -1 or x = 1.So, we can compute R_F at x = -1 and x = 1.Compute R_F at x = -1:( R_F = -1,600 (-1)^2 - 24,800 (-1) - 88,000 )= -1,600 (1) + 24,800 - 88,000= -1,600 + 24,800 - 88,000= (24,800 - 1,600) - 88,000= 23,200 - 88,000= -64,800Now, at x = 1:( R_F = -1,600 (1)^2 - 24,800 (1) - 88,000 )= -1,600 - 24,800 - 88,000= (-1,600 - 24,800) - 88,000= -26,400 - 88,000= -114,400So, comparing R_F at x = -1 is -64,800 and at x =1 is -114,400. So, the maximum revenue is at x = -1, which is -64,800.But wait, revenue can't be negative, right? Or is this in dollars? Wait, the original revenue function was 2000C(t) - 100C(t)^2. Let me check what that evaluates to.At C(t) = 50, R(t) = 2000*50 - 100*2500 = 100,000 - 250,000 = -150,000. Hmm, negative? That doesn't make sense. Maybe I misunderstood the revenue function.Wait, maybe the revenue function is different. Let me think again.Revenue is usually income minus costs. So, perhaps 2000C(t) is the income, and 100C(t)^2 is the cost. So, if the cost is 100C(t)^2, then the revenue is 2000C(t) - 100C(t)^2.But when the fee is imposed, the firm has to pay an additional 200 per case, so the cost becomes 100C(t)^2 + 200C(t). Therefore, the new revenue is 2000C_F(t) - (100C_F(t)^2 + 200C_F(t)).Which is 2000C_F(t) - 100C_F(t)^2 - 200C_F(t) = 1800C_F(t) - 100C_F(t)^2.So, that's consistent with what I did earlier.But then, when I plug in C_F(t) = 40 + 4 sin(pi t /6), the revenue becomes negative. That seems odd because revenue shouldn't be negative, unless the costs exceed the income.Wait, maybe the original revenue function is already considering some costs, and the fee adds another cost. So, perhaps the firm is losing money because the costs are too high.But in any case, the problem asks for the maximum monthly revenue after the fee is imposed, so even if it's negative, we have to find the maximum, which is the least negative, i.e., -64,800.But let me double-check my calculations because getting a negative revenue seems counterintuitive.Wait, let's compute R_F(t) when x = -1:C_F(t) = 40 + 4*(-1) = 40 -4 = 36.So, R_F(t) = 1800*36 - 100*(36)^2.Compute 1800*36: 1800*30=54,000; 1800*6=10,800; total=64,800.Compute 100*(36)^2: 36^2=1,296; 100*1,296=129,600.So, R_F(t) = 64,800 - 129,600 = -64,800.Similarly, at x=1:C_F(t)=40 +4*1=44.R_F(t)=1800*44 -100*(44)^2.1800*44: 1800*40=72,000; 1800*4=7,200; total=79,200.100*(44)^2=100*1,936=193,600.So, R_F(t)=79,200 -193,600= -114,400.So, yes, the calculations are correct. The revenue is negative in both cases, but the maximum (least negative) is -64,800.But wait, is there a way for the revenue to be positive? Let me check the original revenue function.Original revenue: R(t)=2000C(t)-100C(t)^2.At C(t)=50, R(t)=2000*50 -100*2500=100,000 -250,000=-150,000.At C(t)=45, R(t)=2000*45 -100*2025=90,000 -202,500=-112,500.At C(t)=55, R(t)=2000*55 -100*3025=110,000 -302,500=-192,500.So, the original revenue is also negative. So, the firm is actually losing money regardless of the number of cases? That seems odd, but maybe the cost structure is such that it's a loss-making operation.So, after the fee, the maximum revenue is -64,800, which is better than before.But wait, maybe I made a mistake in interpreting the fee. Maybe the fee is a cost per case, so the firm's total cost is 100C(t)^2 + 200C(t). So, the revenue is income minus costs, which is 2000C(t) - (100C(t)^2 + 200C(t)).But in that case, the revenue function would be 2000C(t) -100C(t)^2 -200C(t)=1800C(t)-100C(t)^2, which is what I have.Alternatively, maybe the fee is a fixed fee, but the problem says "additional 200 fee per case," so it's variable cost.Alternatively, perhaps the fee is a cost for the clients, so the firm's revenue increases by 200 per case. But the problem says "the revenue function must account for the reduced case load and additional 200 fee per case." So, it's more likely that the fee is a cost for the firm.Alternatively, maybe the fee is a reduction in the firm's income. For example, clients have to pay an extra 200, but the firm's revenue is reduced by that amount because it's a fee they have to pay to the government. So, the firm's revenue is 2000C(t) -200C(t) -100C(t)^2.Which is again 1800C(t) -100C(t)^2.So, I think my approach is correct. Therefore, the maximum revenue is -64,800, which is the least negative, so the maximum.But let me think again. Maybe the fee is a cost that the firm has to pay per case, so it's subtracted from the revenue. So, the original revenue is 2000C(t) -100C(t)^2, which is negative, as we saw.After the fee, the revenue is 2000C_F(t) -100C_F(t)^2 -200C_F(t). So, that's 1800C_F(t) -100C_F(t)^2.Given that, and C_F(t)=40 +4 sin(pi t /6), the maximum occurs when sin(pi t /6) is -1, giving C_F(t)=36, and R_F(t)= -64,800.Alternatively, maybe I should consider that the fee is a one-time fee, but the problem says "additional 200 fee per case," so it's per case, so it's variable.Alternatively, maybe the fee is a reduction in the firm's income per case. So, instead of getting 2000 per case, they get 2000 - 200 = 1800 per case. So, the revenue would be 1800C_F(t) -100C_F(t)^2, which is the same as before.So, regardless, the revenue function is 1800C_F(t) -100C_F(t)^2.So, given that, and C_F(t)=40 +4 sin(pi t /6), the maximum occurs at the minimum of C_F(t), which is 36, giving R_F(t)= -64,800.But wait, is that correct? Because when C_F(t) is lower, the quadratic term is smaller, but the linear term is also smaller. Let me see.Wait, the revenue function is quadratic in C_F(t): R_F = -100C_F^2 + 1800C_F.This is a quadratic function opening downward, so its maximum is at the vertex.The vertex is at C_F = -b/(2a) = -1800/(2*(-100)) = 1800/200 = 9.Wait, that's interesting. So, the maximum revenue occurs when C_F(t)=9.But wait, C_F(t) is 40 +4 sin(pi t /6), which ranges from 36 to 44. So, 9 is way below that range. So, the maximum of the quadratic is at C_F=9, but our C_F(t) can't reach that. So, the maximum of R_F(t) occurs at the highest possible C_F(t), which is 44.Wait, but earlier, when I substituted x=1 (which gives C_F=44), I got R_F(t)= -114,400, which is lower than at x=-1 (C_F=36, R_F=-64,800).Wait, that contradicts. So, perhaps I made a mistake in interpreting the quadratic.Wait, the revenue function is R_F = -100C_F^2 + 1800C_F.This is a quadratic in C_F, which opens downward, so the maximum is at C_F=9. But since C_F(t) ranges from 36 to 44, which is much higher than 9, the function is decreasing in that interval. Therefore, the maximum occurs at the smallest C_F(t), which is 36, giving R_F(t)= -64,800.Wait, that makes sense. Because the quadratic peaks at C_F=9, but our C_F(t) is always above 36, so in the interval [36,44], the function is decreasing. Therefore, the maximum is at C_F=36.So, that's consistent with my earlier calculation.Therefore, the maximum monthly revenue is -64,800.But wait, that's a negative number. Is that possible? The firm is losing money. Maybe, given the cost structure, it's unavoidable.Alternatively, perhaps I made a mistake in the sign when substituting.Wait, let me re-express the revenue function:R_F(t) = 1800C_F(t) - 100C_F(t)^2.So, factor out 100:R_F(t) = 100*(18C_F(t) - C_F(t)^2).Which is 100*(-C_F(t)^2 +18C_F(t)).So, that's a quadratic in C_F(t), opening downward, with vertex at C_F=9, as before.So, in the interval [36,44], the function is decreasing, so maximum at 36.So, R_F(t)=100*(-36^2 +18*36)=100*(-1296 +648)=100*(-648)= -64,800.Yes, that's correct.So, the maximum monthly revenue is -64,800 dollars, which is a loss of 64,800.But the problem says "determine the maximum monthly revenue the firm can achieve after the fee is imposed." So, even though it's a loss, it's the maximum, meaning the least loss.Alternatively, maybe the firm should shut down? But the problem doesn't mention that, so I think we have to go with the maximum as per the function.Therefore, the new revenue function is ( R_F(t) = 1800C_F(t) - 100C_F(t)^2 ), and the maximum monthly revenue is -64,800.But let me check if I can express this in terms of t.Wait, the maximum occurs when C_F(t) is minimized, which is 36, which occurs when sin(pi t /6) = -1, which is when pi t /6 = 3pi/2, so t/6 = 3/2, so t=9 months.So, at t=9 months, the revenue is -64,800, which is the maximum.Alternatively, maybe I should present the maximum revenue as 64,800, but since it's negative, it's a loss.Wait, but in the problem statement, the original revenue function is 2000C(t) -100C(t)^2, which is also negative. So, the firm is already making a loss, and after the fee, the loss is reduced to 64,800, which is better.So, in conclusion, the maximum monthly revenue after the fee is imposed is -64,800 dollars, which is a loss of 64,800.But let me make sure I didn't make any calculation errors.Compute R_F(t) at C_F=36:1800*36=64,800100*(36)^2=129,600So, 64,800 -129,600= -64,800.Yes, that's correct.Similarly, at C_F=44:1800*44=79,200100*(44)^2=193,60079,200 -193,600= -114,400.So, yes, the maximum is at C_F=36, which is -64,800.Therefore, the answers are:1. ( C_F(t) = 40 + 4sin(pi t/6) )2. ( R_F(t) = 1800C_F(t) - 100C_F(t)^2 ), and the maximum monthly revenue is -64,800.But wait, the problem says "determine the maximum monthly revenue the firm can achieve after the fee is imposed." So, maybe they expect a positive number, but given the calculations, it's negative. Alternatively, perhaps I made a mistake in the sign when considering the fee.Wait, let me think again about the fee. If the fee is a cost, then it's subtracted. If it's a reduction in income, it's also subtracted. But maybe the fee is a cost that the firm has to pay, so the revenue is 2000C_F(t) -100C_F(t)^2 -200C_F(t). So, that's correct.Alternatively, maybe the fee is a reduction in the firm's income per case, so instead of getting 2000 per case, they get 2000 -200=1800 per case. So, the revenue is 1800C_F(t) -100C_F(t)^2, which is what I have.So, I think my calculations are correct, even though the result is negative.Therefore, the final answers are:1. ( C_F(t) = 40 + 4sinleft(frac{pi t}{6}right) )2. The new revenue function is ( R_F(t) = 1800C_F(t) - 100C_F(t)^2 ), and the maximum monthly revenue is -64,800 dollars.But since the problem might expect a positive number, maybe I should express it as a loss, but I think in terms of revenue, it's just a negative number.Alternatively, maybe I should write the maximum revenue as 64,800, but that would be incorrect because it's a loss.Wait, but in the original revenue function, at C(t)=50, R(t)= -150,000, which is worse than -64,800. So, the firm is better off with the fee, even though they are still losing money.So, I think the answer is correct.Final Answer1. The new case load function is boxed{40 + 4sinleft(frac{pi t}{6}right)}.2. The maximum monthly revenue after the fee is imposed is boxed{-64800} dollars.</think>

🧠:<think>Alright, so I have this problem about a missionary setting up a health clinic in a remote area. The goal is to figure out how many patients the clinic can treat without exceeding their budget and also considering their operational hours. Let me try to break this down step by step.Starting with the first part: the clinic has a budget of 500,000. Out of this, 200,000 is fixed for infrastructure and medical equipment. The remaining money will be spent on variable costs, which are 50 per patient for medical supplies and medications. The clinic wants to treat at least 3,000 patients in the first year, but we need to find the maximum number they can handle without going over the budget.Okay, so let's think about the costs. The total budget is fixed at 500,000. The fixed costs are 200,000, so the variable costs will be the total budget minus the fixed costs. That would be 500,000 - 200,000 = 300,000. This 300,000 is what's left for the variable costs per patient.Each patient costs 50 in variable costs. So, if we let x be the number of patients, then the total variable cost is 50x. We need this to be less than or equal to 300,000. So, the inequality would be 50x ≤ 300,000.To solve for x, we can divide both sides by 50. That gives x ≤ 6,000. So, the maximum number of patients the clinic can treat without exceeding the budget is 6,000. But wait, the clinic aims to treat at least 3,000 patients. So, 6,000 is more than enough, and it's within the budget. So, that's the first part done.Moving on to the second part: the clinic also provides faith-based counseling sessions. Each patient needs 1 hour of medical treatment and 0.5 hours of counseling. The clinic operates 8 hours a day, 5 days a week. There are 2 doctors and 3 counselors available. Each doctor can see one patient at a time, and each counselor can conduct one session at a time. We need to find the maximum number of patients that can be treated in a week.Hmm, okay. So, this is a bit more complex. It involves both the medical staff and the counseling staff. Let me try to model this.First, let's figure out how much time each department (medical and counseling) can allocate per week.For the doctors: There are 2 doctors, each working 8 hours a day, 5 days a week. So, each doctor can work 8*5 = 40 hours per week. With 2 doctors, that's 2*40 = 80 hours per week.Similarly, for the counselors: There are 3 counselors, each working 8 hours a day, 5 days a week. So, each counselor can work 40 hours per week. With 3 counselors, that's 3*40 = 120 hours per week.Now, each patient requires 1 hour of medical treatment and 0.5 hours of counseling. So, if we let x be the number of patients treated in a week, then the total medical time needed is 1*x = x hours, and the total counseling time needed is 0.5*x hours.But we also have constraints on the total time available. The medical department can only provide 80 hours, and the counseling department can provide 120 hours.So, setting up the inequalities:For medical: x ≤ 80For counseling: 0.5x ≤ 120Wait, but actually, each patient requires both medical and counseling time. So, we need to make sure that both constraints are satisfied.So, the maximum number of patients is limited by both the medical staff and the counseling staff. Let's solve both inequalities.First, from the medical side: x ≤ 80From the counseling side: 0.5x ≤ 120 => x ≤ 240So, the medical staff can handle up to 80 patients, while the counseling staff can handle up to 240 patients. Since both are required, the limiting factor is the medical staff. Therefore, the maximum number of patients that can be treated in a week is 80.But wait, let me double-check. Each doctor can see one patient at a time, so with 2 doctors, they can see 2 patients simultaneously. The total time per week is 80 hours. Each patient takes 1 hour, so 80 hours / 1 hour per patient = 80 patients. Similarly, for counseling, each counselor can handle one session at a time, so 3 counselors can handle 3 sessions simultaneously. Total counseling time is 120 hours, each session is 0.5 hours. So, 120 / 0.5 = 240 sessions. But since each patient needs one session, the maximum is 240. But since we only have 80 patients due to the medical constraint, the counseling capacity isn't fully utilized.Therefore, the maximum number of patients is 80 per week.Wait, but is that the case? Let me think again. Maybe I should model it as a system of equations.Let me denote x as the number of patients.Each patient requires 1 hour of medical time and 0.5 hours of counseling.Total medical time available: 2 doctors * 8 hours/day * 5 days = 80 hours.Total counseling time available: 3 counselors * 8 hours/day * 5 days = 120 hours.So, the constraints are:1x ≤ 80 (medical)0.5x ≤ 120 (counseling)So, solving these:From medical: x ≤ 80From counseling: x ≤ 240So, the maximum x is the smaller of these two, which is 80.Therefore, the maximum number of patients is 80 per week.But wait, is there a way to utilize the counseling time more effectively? For example, can we schedule counseling sessions while medical treatments are happening? Since the doctors and counselors are different people, perhaps the time can overlap.Wait, but each patient needs both medical treatment and counseling. So, for each patient, they need 1 hour with a doctor and 0.5 hours with a counselor. But these can happen at different times, right? So, perhaps the total time isn't additive per patient, but rather, the total time required is the sum of all medical hours and all counseling hours.But in terms of scheduling, the doctors can only handle so many patients per week, and the counselors can handle more. So, the bottleneck is the doctors.Therefore, even though the counselors can handle more, the number of patients is limited by the number of medical hours available.So, I think my initial conclusion is correct: the maximum number of patients per week is 80.But let me think about it another way. Suppose we have 2 doctors, each can see a patient in 1 hour. So, in 8 hours a day, each doctor can see 8 patients. With 2 doctors, that's 16 patients per day. Over 5 days, that's 16*5 = 80 patients per week. So, that's consistent.For the counselors, each can conduct a session in 0.5 hours. So, in 8 hours a day, each counselor can do 16 sessions. With 3 counselors, that's 48 sessions per day. Over 5 days, that's 240 sessions per week. So, 240 counseling sessions can be done, but since each patient only needs one session, the maximum number of patients is 240. But since the doctors can only handle 80, the number of patients is limited by the doctors.Therefore, the maximum number of patients is 80 per week.Wait, but is there a way to stagger the appointments or something? For example, can a patient have their counseling session on a different day than their medical treatment? If so, maybe the total time isn't as restrictive.But in reality, each patient needs both services, but the services can be provided on different days. So, the total number of patients is limited by the total medical hours and the total counseling hours, but since they can be scheduled separately, perhaps the maximum number is determined by the minimum of (total medical hours / medical time per patient) and (total counseling hours / counseling time per patient). Which is exactly what I did before: 80 and 240, so 80 is the limit.Alternatively, if we consider that each patient requires both services, but the services can be provided at different times, the total number of patients isn't necessarily limited by the sum of the times, but rather by the individual constraints.So, I think my conclusion is correct.Therefore, summarizing:1. The maximum number of patients without exceeding the budget is 6,000.2. The maximum number of patients that can be treated in a week, considering both medical and counseling staff, is 80.Final Answer1. The maximum number of patients the clinic can treat without exceeding the budget is boxed{6000}.2. The maximum number of patients that can be treated in a week is boxed{80}.</think>

🧠:<think>Okay, so I have this problem about two plants in Trinidad and Tobago, and I need to analyze their growth patterns. Let me take it step by step.Starting with Plant A. The problem says that the height of Plant A, H_A(t), is modeled by a differential equation:dH_A/dt = k_A * H_A * (1 - H_A / L_A)They give me k_A = 0.05 per week and L_A = 150 cm. I need to find the general solution for H_A(t).Hmm, this looks like a logistic growth model. I remember that the logistic equation is a common model for population growth, but it can also apply to things like plant growth where there's a carrying capacity. The general form is dP/dt = rP(1 - P/K), where P is the population, r is the growth rate, and K is the carrying capacity.In this case, H_A is analogous to the population, k_A is the growth rate, and L_A is the carrying capacity or maximum height. So, the equation is similar to the logistic equation.I think the solution to the logistic equation is:H_A(t) = L_A / (1 + (L_A / H_A_0 - 1) * e^{-k_A t})Where H_A_0 is the initial height at time t=0.Wait, let me make sure. The standard solution is:H_A(t) = L_A / (1 + (L_A / H_A_0 - 1) * e^{-k_A t})Yes, that seems right. So, if I can express it in terms of the initial condition, that would be the general solution.But the problem just asks for the general solution, not a particular solution with initial conditions. So, maybe I can leave it in terms of H_A_0.Alternatively, sometimes it's written as:H_A(t) = L_A / (1 + C * e^{-k_A t})Where C is a constant determined by the initial condition. So, if H_A(0) = H_A_0, then:H_A(0) = L_A / (1 + C) = H_A_0So, solving for C:1 + C = L_A / H_A_0 => C = (L_A / H_A_0) - 1Therefore, substituting back:H_A(t) = L_A / (1 + (L_A / H_A_0 - 1) * e^{-k_A t})So, that's the general solution. Since the problem doesn't specify an initial condition, I think this is acceptable as the general solution.Let me write that down:H_A(t) = 150 / (1 + (150 / H_A_0 - 1) * e^{-0.05 t})But maybe I can write it in a more compact form. Alternatively, if I let C = (150 / H_A_0 - 1), then it's:H_A(t) = 150 / (1 + C * e^{-0.05 t})Either way, I think that's the general solution.Moving on to Plant B. The height is given by:H_B(t) = M_B / (1 + e^{-k_B(t - t_0)})They give M_B = 120 cm, k_B = 0.1 per week, and t_0 = 10 weeks. I need to find the height at t = 5 weeks and t = 15 weeks.So, plugging in the values, the equation becomes:H_B(t) = 120 / (1 + e^{-0.1(t - 10)})I need to compute H_B(5) and H_B(15).Let me compute H_B(5) first.At t = 5:H_B(5) = 120 / (1 + e^{-0.1(5 - 10)}) = 120 / (1 + e^{-0.1*(-5)}) = 120 / (1 + e^{0.5})Compute e^{0.5}. I know that e^0.5 is approximately 1.6487.So, H_B(5) = 120 / (1 + 1.6487) = 120 / (2.6487) ≈ 120 / 2.6487 ≈ 45.3 cm.Wait, let me check that calculation again.Wait, 1 + e^{0.5} ≈ 1 + 1.6487 ≈ 2.6487.120 divided by 2.6487. Let me compute that.2.6487 * 45 = 119.1915, which is close to 120. So, 45.3 cm is approximately correct.Now, for t = 15 weeks.H_B(15) = 120 / (1 + e^{-0.1(15 - 10)}) = 120 / (1 + e^{-0.1*5}) = 120 / (1 + e^{-0.5})Compute e^{-0.5} ≈ 0.6065.So, 1 + 0.6065 ≈ 1.6065.Therefore, H_B(15) = 120 / 1.6065 ≈ 74.7 cm.Wait, let me verify that division.1.6065 * 74 = 118.871, which is close to 120. So, 74.7 cm is approximately correct.Alternatively, maybe I can compute it more precisely.For t = 5:e^{0.5} ≈ 1.64872So, 1 + 1.64872 = 2.64872120 / 2.64872 ≈ 45.32 cmSimilarly, for t = 15:e^{-0.5} ≈ 0.606531 + 0.60653 = 1.60653120 / 1.60653 ≈ 74.68 cmSo, rounding to two decimal places, approximately 45.32 cm and 74.68 cm.But maybe the problem expects exact expressions or more precise decimal places? Let me see.Alternatively, I can express it in terms of e^{0.5} and e^{-0.5}, but since they are transcendental numbers, it's better to compute their approximate decimal values.So, summarizing:At t = 5 weeks, H_B ≈ 45.32 cmAt t = 15 weeks, H_B ≈ 74.68 cmI think that's it.Wait, let me double-check the calculations.For t = 5:Exponent: -0.1*(5 - 10) = -0.1*(-5) = 0.5So, e^{0.5} ≈ 1.64872Denominator: 1 + 1.64872 ≈ 2.64872120 / 2.64872 ≈ 45.32 cmYes.For t = 15:Exponent: -0.1*(15 - 10) = -0.1*5 = -0.5e^{-0.5} ≈ 0.60653Denominator: 1 + 0.60653 ≈ 1.60653120 / 1.60653 ≈ 74.68 cmYes, that seems correct.So, I think I have the solutions for both parts.Final Answer1. The general solution for Plant A's height is boxed{H_A(t) = dfrac{150}{1 + left(dfrac{150}{H_A(0)} - 1right) e^{-0.05t}}}.2. The height of Plant B at ( t = 5 ) weeks is approximately boxed{45.32} cm and at ( t = 15 ) weeks is approximately boxed{74.68} cm.</think>

🧠:<think>Okay, so I have this problem where an artist and a chemistry teacher are working together on an interactive art installation. The sculpture's height changes over time based on the concentration of a reactant A in a chemical reaction. The reaction is first-order, which I remember means the rate depends linearly on the concentration of one reactant.The problem has two parts. Let me start with the first one.1. Derive the expression for [A](t) and calculate the concentration after 10 minutes.Alright, the differential equation given is:[ frac{d[A](t)}{dt} = -k[A](t) ]This is a first-order linear differential equation. I think the general solution for such an equation is an exponential function. Let me recall. For a first-order reaction, the concentration over time is given by:[ [A](t) = [A]_0 e^{-kt} ]Where [A]_0 is the initial concentration, k is the rate constant, and t is time.Given that [A]_0 is 2 mol/L and k is 0.3 min⁻¹. So plugging these values in:[ [A](t) = 2 e^{-0.3 t} ]To find the concentration after 10 minutes, I substitute t = 10.[ [A](10) = 2 e^{-0.3 times 10} ]Calculating the exponent first:-0.3 * 10 = -3So,[ [A](10) = 2 e^{-3} ]I need to compute e⁻³. I remember that e³ is approximately 20.0855, so e⁻³ is 1/20.0855 ≈ 0.0498.Therefore,[ [A](10) ≈ 2 * 0.0498 ≈ 0.0996 , text{mol/L} ]So, approximately 0.0996 mol/L after 10 minutes.Wait, let me double-check my calculations. Maybe I should use a calculator for e⁻³ to be precise.But since I don't have a calculator here, I can recall that e⁻³ is roughly 0.0498, so 2 times that is about 0.0996. That seems correct.2. Express H(t) as a function of time and determine the height after 10 minutes.The height H(t) is proportional to the square root of the concentration of A. The proportionality constant is 5. So,[ H(t) = 5 sqrt{[A](t)} ]We already have [A](t) from part 1:[ [A](t) = 2 e^{-0.3 t} ]So,[ H(t) = 5 sqrt{2 e^{-0.3 t}} ]Simplify that:First, square root of 2 is √2, and square root of e^{-0.3 t} is e^{-0.15 t}.So,[ H(t) = 5 sqrt{2} e^{-0.15 t} ]Alternatively, I can write it as:[ H(t) = 5 times sqrt{2} times e^{-0.15 t} ]But maybe it's better to keep it in terms of the square root of [A](t).Alternatively, since [A](t) is 2 e^{-0.3 t}, then sqrt([A](t)) is sqrt(2) e^{-0.15 t}, so H(t) is 5 times that.So, either way, the expression is correct.Now, to find H(10):First, compute [A](10), which we found to be approximately 0.0996 mol/L.Then,[ H(10) = 5 sqrt{0.0996} ]Compute sqrt(0.0996). Let me think, sqrt(0.1) is approximately 0.3162, and 0.0996 is just slightly less than 0.1, so sqrt(0.0996) ≈ 0.3156.Therefore,[ H(10) ≈ 5 * 0.3156 ≈ 1.578 ]So, approximately 1.578 units of height. Since the problem doesn't specify the units, I assume it's just a numerical value.Alternatively, I can compute it more precisely.Given that [A](10) = 2 e^{-3} ≈ 2 * 0.049787 ≈ 0.099574 mol/L.Then sqrt(0.099574) ≈ sqrt(0.099574). Let me compute this more accurately.Let me note that 0.315^2 = 0.099225, and 0.316^2 = 0.099856.Since 0.099574 is between 0.099225 and 0.099856, so sqrt(0.099574) is between 0.315 and 0.316.Compute 0.315^2 = 0.0992250.3155^2 = ?0.3155^2 = (0.315 + 0.0005)^2 = 0.315² + 2*0.315*0.0005 + 0.0005² = 0.099225 + 0.000315 + 0.00000025 ≈ 0.09954025That's very close to 0.099574.So, 0.3155^2 ≈ 0.09954025Difference between 0.099574 and 0.09954025 is about 0.00003375.Each increment of 0.0001 in x adds approximately 2*0.3155*0.0001 + (0.0001)^2 ≈ 0.0000631 to x².So, to get an additional 0.00003375, we need approximately 0.00003375 / 0.0000631 ≈ 0.535 of 0.0001, which is about 0.0000535.So, sqrt(0.099574) ≈ 0.3155 + 0.0000535 ≈ 0.3155535.Therefore, approximately 0.31555.Thus,H(10) = 5 * 0.31555 ≈ 1.57775.So, approximately 1.578 units.Alternatively, if I use exact expressions:H(t) = 5 * sqrt(2 e^{-0.3 t}) = 5 * sqrt(2) * e^{-0.15 t}So, H(10) = 5 * sqrt(2) * e^{-1.5}Compute e^{-1.5} ≈ 0.22313sqrt(2) ≈ 1.41421So,H(10) ≈ 5 * 1.41421 * 0.22313First, 1.41421 * 0.22313 ≈ Let's compute that:1.41421 * 0.2 = 0.2828421.41421 * 0.02313 ≈ Approximately 1.41421 * 0.02 = 0.0282842 and 1.41421 * 0.00313 ≈ 0.004423So, total ≈ 0.0282842 + 0.004423 ≈ 0.032707Thus, total 0.282842 + 0.032707 ≈ 0.31555Then, 5 * 0.31555 ≈ 1.57775, which is the same as before.So, H(10) ≈ 1.57775, which is approximately 1.578.So, rounding to three decimal places, 1.578.Alternatively, if we want to present it as a fraction or something, but probably decimal is fine.Wait, but the problem says the proportionality constant is 5. It doesn't specify units, so maybe the height is in some arbitrary units, like centimeters or meters? The problem doesn't specify, so I think just the numerical value is okay.So, summarizing:1. [A](t) = 2 e^{-0.3 t}, and [A](10) ≈ 0.0996 mol/L.2. H(t) = 5 sqrt(2) e^{-0.15 t}, and H(10) ≈ 1.578.I think that's it.Final Answer1. The concentration of A after 10 minutes is boxed{0.0996} mol/L.2. The height of the sculpture after 10 minutes is boxed{1.578} units.</think>