🧠:<think>Alright, so I have this problem about an event coordinator planning a storytelling session. The mom is participating, and they need to set up the seating in a specific polygonal shape for good acoustics. The hall can be divided into smaller sections, and they want a closed polygon with an even number of sides. Each side needs to have an equal number of seats, with each side accommodating at least 10 seats but no more than 60 seats. The total seating capacity is 480.First, I need to figure out the possible number of sides ( n ) the polygon can have. Since it's a polygon, ( n ) must be an integer greater than or equal to 3. But the problem specifies it's an even number, so ( n ) must be 4, 6, 8, etc. Also, each side must have between 10 and 60 seats, inclusive.So, the total number of seats is 480, and each side has ( s ) seats. Therefore, the number of sides ( n ) multiplied by the number of seats per side ( s ) must equal 480. So, ( n times s = 480 ). Given that ( s ) must be between 10 and 60, inclusive, we can write that as ( 10 leq s leq 60 ). Since ( s = frac{480}{n} ), substituting that in gives ( 10 leq frac{480}{n} leq 60 ).Let me solve the inequalities:First, ( frac{480}{n} geq 10 ) implies ( n leq frac{480}{10} = 48 ).Second, ( frac{480}{n} leq 60 ) implies ( n geq frac{480}{60} = 8 ).So, ( n ) must be between 8 and 48, inclusive. But since ( n ) must be an even number, the possible values are 8, 10, 12, ..., up to 48.Wait, but hold on. The problem says the polygon must have an even number of sides, so ( n ) must be even. So, starting from 8, 10, 12, ..., 48.But let me check if 48 is feasible. If ( n = 48 ), then each side would have ( s = 480 / 48 = 10 ) seats, which is the minimum. Similarly, if ( n = 8 ), each side would have ( s = 480 / 8 = 60 ) seats, which is the maximum. So, both 8 and 48 are acceptable.Therefore, the possible values of ( n ) are all even integers from 8 to 48, inclusive. Let me list them out:8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48.So, that's 21 possible values of ( n ). But wait, let me count: from 8 to 48 inclusive, stepping by 2. The formula for the number of terms is ((last - first)/step) + 1. So, ((48 - 8)/2) + 1 = (40/2) + 1 = 20 + 1 = 21. Yep, 21 values.So, that answers the first part. Now, moving on to the second part.The event coordinator wants to maximize the use of the hall's seating capacity while keeping the polygonal shape. Hmm, but the total seating capacity is fixed at 480. So, if they want to maximize the use, they must use all 480 seats. So, the number of sides ( n ) must be such that 480 is divisible by ( n ), which it is, as we've already considered.But wait, the second part says "after determining the possible polygonal arrangements," which we have, and now they want to calculate the total cost for each feasible polygonal arrangement, where the cost is given by ( C(n) = 5n^3 - 30n^2 + 45n ). Then, identify which arrangement minimizes the cost.So, for each possible ( n ) (even numbers from 8 to 48), we need to compute ( C(n) ) and find the ( n ) that gives the minimum cost.But calculating ( C(n) ) for each ( n ) from 8 to 48 with step 2 would be tedious. Maybe we can find a way to find the minimum without computing all of them.Alternatively, perhaps we can treat ( C(n) ) as a function and find its minimum. Since ( n ) must be an integer, but for the sake of analysis, let's consider it as a continuous function.So, ( C(n) = 5n^3 - 30n^2 + 45n ). To find its minimum, we can take the derivative with respect to ( n ) and set it equal to zero.The derivative ( C'(n) = 15n^2 - 60n + 45 ).Setting ( C'(n) = 0 ):( 15n^2 - 60n + 45 = 0 ).Divide both sides by 15:( n^2 - 4n + 3 = 0 ).Factor:( (n - 1)(n - 3) = 0 ).So, critical points at ( n = 1 ) and ( n = 3 ).But our ( n ) starts at 8, so these critical points are outside our domain. Therefore, the function ( C(n) ) is either increasing or decreasing throughout our interval.Let me check the second derivative to see the concavity.Second derivative ( C''(n) = 30n - 60 ).At ( n = 8 ), ( C''(8) = 240 - 60 = 180 > 0 ), so the function is concave upward at ( n = 8 ). Since the function is a cubic with positive leading coefficient, it tends to infinity as ( n ) increases. So, in our interval from 8 to 48, the function is increasing because the derivative at ( n = 8 ) is:( C'(8) = 15*(8)^2 - 60*(8) + 45 = 15*64 - 480 + 45 = 960 - 480 + 45 = 525 > 0 ).So, the function is increasing on our interval. Therefore, the minimum cost occurs at the smallest ( n ), which is 8.Wait, but let me verify that. If the function is increasing, then yes, the minimum is at the leftmost point, which is ( n = 8 ).But let me compute ( C(n) ) for ( n = 8 ) and ( n = 10 ) to see if it's indeed increasing.Compute ( C(8) = 5*(8)^3 - 30*(8)^2 + 45*(8) ).Calculate step by step:( 8^3 = 512 ), so 5*512 = 2560.( 8^2 = 64 ), so 30*64 = 1920.45*8 = 360.So, ( C(8) = 2560 - 1920 + 360 = (2560 - 1920) + 360 = 640 + 360 = 1000 ).Now, ( C(10) = 5*(10)^3 - 30*(10)^2 + 45*(10) ).10^3 = 1000, so 5*1000 = 5000.10^2 = 100, so 30*100 = 3000.45*10 = 450.Thus, ( C(10) = 5000 - 3000 + 450 = 2000 + 450 = 2450 ).So, ( C(10) = 2450 ), which is higher than ( C(8) = 1000 ). So, indeed, as ( n ) increases, the cost increases.Therefore, the minimum cost occurs at ( n = 8 ).But wait, let me compute ( C(8) ) again to make sure I didn't make a mistake.5*(8)^3 = 5*512 = 2560.30*(8)^2 = 30*64 = 1920.45*8 = 360.So, 2560 - 1920 = 640. 640 + 360 = 1000. Yep, correct.Similarly, ( C(10) = 5*1000 - 30*100 + 450 = 5000 - 3000 + 450 = 2450 ).So, yes, it's increasing. Therefore, the minimal cost is at ( n = 8 ).But just to be thorough, let me compute ( C(12) ) as well.( C(12) = 5*(12)^3 - 30*(12)^2 + 45*(12) ).12^3 = 1728, so 5*1728 = 8640.12^2 = 144, so 30*144 = 4320.45*12 = 540.Thus, ( C(12) = 8640 - 4320 + 540 = (8640 - 4320) + 540 = 4320 + 540 = 4860 ).Which is even higher. So, the trend is clear: as ( n ) increases, ( C(n) ) increases.Therefore, the minimal cost is achieved when ( n = 8 ).So, summarizing:1. The possible values of ( n ) are all even integers from 8 to 48, inclusive.2. The cost function ( C(n) ) is minimized when ( n = 8 ), with a total cost of 1000.Final Answer1. The possible values of ( n ) are all even integers from 8 to 48. Thus, the possible values are boxed{8, 10, 12, ldots, 48}.2. The arrangement that minimizes the cost is when ( n = 8 ), resulting in a total cost of boxed{1000}.</think>

🧠:<think>Okay, so I have this problem about a recycling facility's machine that compresses plastic waste. The efficiency of the machine decreases over time because it's inversely proportional to the square root of the time it's been operating. Initially, it can compress 500 kg per hour. First, I need to derive the function E(t) that represents the efficiency after t hours. The problem says that E(t) is inversely proportional to the square root of t. Hmm, inversely proportional means that E(t) = k / sqrt(t), where k is the constant of proportionality. They also give me that E(1) = 500 kg/hour. So when t = 1, E(1) = 500. Plugging that into the equation: 500 = k / sqrt(1). Since sqrt(1) is 1, that means k must be 500. So the function E(t) is 500 / sqrt(t). Wait, let me double-check that. If t is 1, then E(1) is 500, which matches the given information. If t increases, say t = 4, then E(4) would be 500 / 2 = 250 kg/hour. That makes sense because as time increases, efficiency decreases, which aligns with the problem statement. Okay, so part 1 seems straightforward.Now, moving on to part 2. I need to calculate the total amount of plastic waste the machine can compress over a 10-hour shift. Since the efficiency is changing over time, I can't just multiply 500 kg/hour by 10 hours. Instead, I need to integrate the efficiency function over the 10-hour period to find the total amount.So, the total amount, let's call it A, is the integral from t = 0 to t = 10 of E(t) dt. That is, A = ∫₀¹⁰ E(t) dt. Substituting E(t) from part 1, that becomes A = ∫₀¹⁰ (500 / sqrt(t)) dt.Let me write that out: A = 500 ∫₀¹⁰ t^(-1/2) dt. The integral of t^(-1/2) is 2t^(1/2), so plugging that in, we get A = 500 [2t^(1/2)] from 0 to 10.Calculating the definite integral: at t = 10, it's 2*sqrt(10), and at t = 0, it's 2*sqrt(0) = 0. So A = 500 * (2*sqrt(10) - 0) = 1000*sqrt(10).Hmm, sqrt(10) is approximately 3.1623, so 1000*3.1623 is about 3162.3 kg. But since the problem doesn't specify rounding, maybe I should leave it in exact form.Wait, let me make sure I did the integral correctly. The integral of t^(-1/2) is indeed 2t^(1/2). So, yes, that part is correct. And evaluating from 0 to 10, so 2*sqrt(10) minus 0. So 500 times 2*sqrt(10) is 1000*sqrt(10). That seems right.Is there another way to think about this? Maybe using substitution or another method? Let's see. Alternatively, I could write the integral as 500 ∫₀¹⁰ t^(-1/2) dt, which is 500*(2t^(1/2)) evaluated from 0 to 10. Yep, same result. So I think that's solid.Just to recap: since efficiency decreases over time, the machine isn't working at 500 kg/hour the whole time. It starts at 500 and gradually slows down. So the total amount is the area under the curve of E(t) from 0 to 10, which is the integral. And that gives us 1000*sqrt(10) kg.I wonder if there's a way to check this without calculus. Maybe approximate the integral with a Riemann sum? Let's try a simple approximation with a few intervals to see if it's in the ballpark.Suppose we divide the 10 hours into 10 intervals of 1 hour each. Then, the efficiency at each hour is E(t) = 500 / sqrt(t). So at t=0, it's undefined, but approaching infinity. Hmm, that's a problem. Maybe start at t=1 instead. But actually, the integral from 0 to 10 includes t=0, which is a point of discontinuity. But since it's an integral, it's okay because the function is integrable despite the singularity at t=0.But for approximation, let's say we take the average of the left and right Riemann sums. For t=1 to t=10, each interval is 1 hour. The left sum would be sum from t=1 to t=10 of E(t-1), but E(0) is infinity, which complicates things. Maybe it's better to use the trapezoidal rule or something else, but this might get too involved.Alternatively, since we know the exact integral is 1000*sqrt(10), which is approximately 3162.3 kg, and if I think about the average efficiency over 10 hours, maybe it's around half of the initial efficiency? 500 / 2 = 250 kg/hour, times 10 hours would be 2500 kg. But 3162 is higher than that, which makes sense because the efficiency doesn't drop linearly; it drops as 1/sqrt(t), which is a slower decrease. So the area under the curve is more than the average of the start and end.Wait, let's compute the average efficiency. The total amount is 1000*sqrt(10), so the average efficiency would be total amount divided by time, which is (1000*sqrt(10))/10 = 100*sqrt(10) ≈ 316.23 kg/hour. That seems reasonable because it's less than the initial 500 but not too low.Alternatively, if I compute the average of E(t) over [0,10], it's (1/10)*∫₀¹⁰ E(t) dt = (1/10)*1000*sqrt(10) = 100*sqrt(10), same as above. So that checks out.I think I'm confident with the integral approach. So the total amount is 1000*sqrt(10) kg, which is approximately 3162.3 kg. But since the problem doesn't specify rounding, I should present the exact value.Wait, just to make sure, let me write the integral again:A = ∫₀¹⁰ (500 / sqrt(t)) dt = 500 ∫₀¹⁰ t^(-1/2) dt = 500 [2t^(1/2)] from 0 to 10 = 500*(2*sqrt(10) - 0) = 1000*sqrt(10).Yes, that's correct. So the total amount is 1000*sqrt(10) kg.I think that's all. I don't see any mistakes in my reasoning. The function for efficiency is inversely proportional to sqrt(t), so E(t) = 500 / sqrt(t). The total amount is the integral of that from 0 to 10, which gives 1000*sqrt(10). That should be the answer.Final Answer1. The efficiency function is ( boxed{E(t) = dfrac{500}{sqrt{t}}} ).2. The total amount of plastic waste compressed over a 10-hour shift is ( boxed{1000sqrt{10}} ) kg.</think>

🧠:<think>Alright, so I've got this problem about a neuroscientist studying the impact of cognitive exercises on a child's well-being. The problem has two parts, and I need to tackle them step by step. Let me start with the first part.Problem 1: Determining Steady-State ValuesThe system of differential equations given is:[frac{dT}{dt} = k_1 (T_{text{max}} - T) - k_2 S][frac{dS}{dt} = k_3 T - k_4 S]I need to find the steady-state values of ( T ) and ( S ). Steady-state means that the derivatives are zero because the system isn't changing anymore. So, I can set both (frac{dT}{dt}) and (frac{dS}{dt}) to zero and solve for ( T ) and ( S ).Starting with the first equation:[0 = k_1 (T_{text{max}} - T) - k_2 S]Let me rearrange this:[k_1 (T_{text{max}} - T) = k_2 S][S = frac{k_1}{k_2} (T_{text{max}} - T)]Okay, so that's an expression for ( S ) in terms of ( T ). Now, let's plug this into the second equation.The second equation is:[0 = k_3 T - k_4 S]Substituting ( S ) from above:[0 = k_3 T - k_4 left( frac{k_1}{k_2} (T_{text{max}} - T) right)]Let me simplify this:[0 = k_3 T - frac{k_1 k_4}{k_2} T_{text{max}} + frac{k_1 k_4}{k_2} T]Combine like terms:[0 = left( k_3 + frac{k_1 k_4}{k_2} right) T - frac{k_1 k_4}{k_2} T_{text{max}}]Let me factor out ( T ):[left( k_3 + frac{k_1 k_4}{k_2} right) T = frac{k_1 k_4}{k_2} T_{text{max}}]Solving for ( T ):[T = frac{ frac{k_1 k_4}{k_2} T_{text{max}} }{ k_3 + frac{k_1 k_4}{k_2} }]Let me simplify the denominator by combining the terms:[T = frac{ k_1 k_4 T_{text{max}} }{ k_2 k_3 + k_1 k_4 }]So, that's the steady-state value for ( T ). Now, let's find ( S ) using the expression we had earlier:[S = frac{k_1}{k_2} (T_{text{max}} - T)]Substituting ( T ):[S = frac{k_1}{k_2} left( T_{text{max}} - frac{ k_1 k_4 T_{text{max}} }{ k_2 k_3 + k_1 k_4 } right )]Factor out ( T_{text{max}} ):[S = frac{k_1 T_{text{max}}}{k_2} left( 1 - frac{ k_1 k_4 }{ k_2 k_3 + k_1 k_4 } right )]Simplify the term in the parentheses:[1 - frac{ k_1 k_4 }{ k_2 k_3 + k_1 k_4 } = frac{ k_2 k_3 + k_1 k_4 - k_1 k_4 }{ k_2 k_3 + k_1 k_4 } = frac{ k_2 k_3 }{ k_2 k_3 + k_1 k_4 }]So, substituting back:[S = frac{k_1 T_{text{max}}}{k_2} cdot frac{ k_2 k_3 }{ k_2 k_3 + k_1 k_4 } = frac{ k_1 k_3 T_{text{max}} }{ k_2 k_3 + k_1 k_4 }]Therefore, the steady-state values are:[T_{ss} = frac{ k_1 k_4 T_{text{max}} }{ k_2 k_3 + k_1 k_4 }][S_{ss} = frac{ k_1 k_3 T_{text{max}} }{ k_2 k_3 + k_1 k_4 }]Hmm, let me double-check my algebra to make sure I didn't make a mistake. Starting from the substitution into the second equation, I think it's correct. The expressions for ( T_{ss} ) and ( S_{ss} ) seem to make sense dimensionally as well. The denominators are sums of products of constants, which is consistent with the units.Problem 2: Modifying the System with a Perturbation and Analyzing StabilityNow, the second part is about modifying the system to include an external stress factor ( E(t) = E_0 cos(omega t) ) as a perturbation in the equation for ( frac{dS}{dt} ). The perturbation is ( epsilon E(t) ), where ( epsilon ) is a small parameter.So, the original system is:[frac{dT}{dt} = k_1 (T_{text{max}} - T) - k_2 S][frac{dS}{dt} = k_3 T - k_4 S]We need to add the perturbation ( epsilon E(t) ) to the second equation. So, the modified system becomes:[frac{dT}{dt} = k_1 (T_{text{max}} - T) - k_2 S][frac{dS}{dt} = k_3 T - k_4 S + epsilon E(t)]Where ( E(t) = E_0 cos(omega t) ).Now, the task is to analyze the stability of the modified system's steady-state solution using linearization techniques.First, I need to find the steady-state solution of the modified system. However, since ( E(t) ) is a time-dependent perturbation, the steady-state might not be constant anymore. But perhaps we can consider the steady-state in the presence of the perturbation or analyze the stability around the original steady-state.Wait, the question says "analyze the stability of the modified system's steady-state solution." So, maybe the steady-state is still the same as before, but now with a perturbation. Or perhaps the steady-state shifts slightly.But since ( E(t) ) is a periodic function, it might not settle to a steady-state but instead lead to oscillations or other behaviors. Hmm, maybe I need to consider the system near the original steady-state and see how the perturbation affects it.Let me think. In linear stability analysis, we linearize the system around the steady-state and then analyze the eigenvalues of the resulting linear system. If the real parts of the eigenvalues are negative, the steady-state is stable; otherwise, it's unstable.So, let's denote the original steady-state as ( (T_{ss}, S_{ss}) ). We can write the perturbed system as:[frac{dT}{dt} = k_1 (T_{text{max}} - T) - k_2 S][frac{dS}{dt} = k_3 T - k_4 S + epsilon E(t)]To linearize, we consider small deviations from the steady-state. Let me define:[T = T_{ss} + delta T][S = S_{ss} + delta S]Where ( delta T ) and ( delta S ) are small perturbations. Substituting these into the equations:First equation:[frac{d}{dt}(T_{ss} + delta T) = k_1 (T_{text{max}} - (T_{ss} + delta T)) - k_2 (S_{ss} + delta S)]But since ( T_{ss} ) is a steady-state, its derivative is zero:[frac{d}{dt}(T_{ss}) = 0]So, the equation becomes:[frac{d (delta T)}{dt} = -k_1 delta T - k_2 delta S]Similarly, for the second equation:[frac{d}{dt}(S_{ss} + delta S) = k_3 (T_{ss} + delta T) - k_4 (S_{ss} + delta S) + epsilon E(t)]Again, since ( S_{ss} ) is a steady-state, its derivative is zero:[frac{d}{dt}(S_{ss}) = 0]So, the equation becomes:[frac{d (delta S)}{dt} = k_3 delta T - k_4 delta S + epsilon E(t)]Therefore, the linearized system is:[frac{d (delta T)}{dt} = -k_1 delta T - k_2 delta S][frac{d (delta S)}{dt} = k_3 delta T - k_4 delta S + epsilon E(t)]Now, to analyze the stability, we can write this system in matrix form:[begin{pmatrix}frac{d (delta T)}{dt} frac{d (delta S)}{dt}end{pmatrix}=begin{pmatrix}- k_1 & - k_2 k_3 & - k_4end{pmatrix}begin{pmatrix}delta T delta Send{pmatrix}+begin{pmatrix}0 epsilon E(t)end{pmatrix}]This is a linear nonhomogeneous system. To analyze the stability, we can look at the homogeneous part first, which is the linear system without the perturbation ( epsilon E(t) ).The stability of the homogeneous system is determined by the eigenvalues of the matrix:[A = begin{pmatrix}- k_1 & - k_2 k_3 & - k_4end{pmatrix}]The characteristic equation is:[lambda^2 - text{tr}(A) lambda + det(A) = 0]Where ( text{tr}(A) = -k_1 - k_4 ) and ( det(A) = (-k_1)(-k_4) - (-k_2)(k_3) = k_1 k_4 + k_2 k_3 ).So, the characteristic equation is:[lambda^2 + (k_1 + k_4) lambda + (k_1 k_4 + k_2 k_3) = 0]The roots (eigenvalues) are:[lambda = frac{ - (k_1 + k_4) pm sqrt{(k_1 + k_4)^2 - 4 (k_1 k_4 + k_2 k_3)} }{2}]Simplify the discriminant:[D = (k_1 + k_4)^2 - 4 (k_1 k_4 + k_2 k_3) = k_1^2 + 2 k_1 k_4 + k_4^2 - 4 k_1 k_4 - 4 k_2 k_3][D = k_1^2 - 2 k_1 k_4 + k_4^2 - 4 k_2 k_3 = (k_1 - k_4)^2 - 4 k_2 k_3]Now, the nature of the eigenvalues depends on the discriminant ( D ).If ( D > 0 ), we have two real eigenvalues.If ( D = 0 ), we have a repeated real eigenvalue.If ( D < 0 ), we have complex conjugate eigenvalues.Given that all constants ( k_1, k_2, k_3, k_4 ) are positive, let's analyze ( D ):( D = (k_1 - k_4)^2 - 4 k_2 k_3 )Since ( (k_1 - k_4)^2 ) is non-negative, and ( 4 k_2 k_3 ) is positive, the sign of ( D ) depends on whether ( (k_1 - k_4)^2 ) is greater than ( 4 k_2 k_3 ).If ( (k_1 - k_4)^2 > 4 k_2 k_3 ), then ( D > 0 ), leading to two real eigenvalues.If ( (k_1 - k_4)^2 < 4 k_2 k_3 ), then ( D < 0 ), leading to complex eigenvalues.In either case, the real parts of the eigenvalues are determined by the coefficients in the characteristic equation.The sum of the eigenvalues is ( - (k_1 + k_4) ), which is negative because ( k_1 ) and ( k_4 ) are positive. The product of the eigenvalues is ( k_1 k_4 + k_2 k_3 ), which is positive.Therefore, regardless of whether the eigenvalues are real or complex, their real parts are negative (since for complex eigenvalues, the real part is ( - (k_1 + k_4)/2 ), which is negative). This implies that the homogeneous system is asymptotically stable; small perturbations decay over time.However, in our case, we have a nonhomogeneous term ( epsilon E(t) ), which is a periodic forcing function. So, the system is subject to a periodic perturbation. The stability in the presence of this perturbation can be analyzed using the concept of forced oscillations and the method of harmonic balance or by looking at the system's response to the periodic input.But since the question asks to analyze the stability using linearization techniques, perhaps we can consider the effect of the perturbation on the stability.In linear systems, the presence of a periodic forcing term can lead to resonances if the forcing frequency matches the system's natural frequency. However, since the homogeneous system is stable, the response to the periodic perturbation will not grow without bound but will instead reach a steady oscillation.To analyze this, we can consider the system's response to the perturbation ( epsilon E(t) ). The solution to the nonhomogeneous system will be the sum of the homogeneous solution (which decays to zero due to stability) and a particular solution that represents the steady-state response to the perturbation.The particular solution can be found using methods like the method of undetermined coefficients, assuming a solution of the form ( delta T_p = A cos(omega t) + B sin(omega t) ) and similarly for ( delta S_p ).However, since the question is about stability, and the homogeneous system is already stable, the addition of a bounded periodic perturbation will not make the system unstable in the sense of Lyapunov. The system will remain stable, but the perturbations will cause oscillations around the steady-state.Therefore, the steady-state solution of the modified system is stable because the homogeneous system is asymptotically stable, and the perturbation only introduces bounded oscillations without causing the deviations to grow indefinitely.Alternatively, if we consider the full system with the perturbation, the steady-state might not be a fixed point anymore but a periodic solution. However, the stability analysis around the original steady-state still holds because the perturbation is small (since ( epsilon ) is a small parameter). Therefore, the system remains stable, and the deviations from the steady-state remain bounded.So, putting it all together, the modified system's steady-state is stable because the eigenvalues of the linearized system have negative real parts, ensuring that any perturbations decay over time, and the periodic forcing only introduces bounded oscillations.Final Answer1. The steady-state values are (boxed{T_{ss} = dfrac{k_1 k_4 T_{text{max}}}{k_2 k_3 + k_1 k_4}}) and (boxed{S_{ss} = dfrac{k_1 k_3 T_{text{max}}}{k_2 k_3 + k_1 k_4}}).2. The modified system's steady-state solution is stable because the eigenvalues of the linearized system have negative real parts, ensuring that perturbations decay over time, and the periodic forcing introduces only bounded oscillations.Final Answer1. The steady-state values are (boxed{T_{ss} = dfrac{k_1 k_4 T_{text{max}}}{k_2 k_3 + k_1 k_4}}) and (boxed{S_{ss} = dfrac{k_1 k_3 T_{text{max}}}{k_2 k_3 + k_1 k_4}}).2. The modified system's steady-state solution is stable.</think>

🧠:<think>Okay, so I need to help Amina figure out how to maximize her profits by optimizing the number of sarees she produces and sells. There are two parts to this problem. Let me tackle them one by one.Problem 1: Budget Constraint and Maximum ProductionFirst, Amina has a budget of BDT 20,000 per month for raw materials and labor. The cost function is given as ( C(n) = an + bn^2 ), where ( n ) is the number of sarees produced, and ( a ) and ( b ) are positive constants. I need to find the maximum number of sarees she can produce within her budget.So, the total cost ( C(n) ) must be less than or equal to 20,000 BDT. That gives the inequality:[ an + bn^2 leq 20,000 ]This is a quadratic inequality in terms of ( n ). To find the maximum ( n ), I can solve the quadratic equation:[ bn^2 + an - 20,000 = 0 ]Using the quadratic formula, ( n = frac{-a pm sqrt{a^2 + 80,000b}}{2b} ). Since ( n ) can't be negative, we take the positive root:[ n = frac{-a + sqrt{a^2 + 80,000b}}{2b} ]This will give the maximum number of sarees Amina can produce without exceeding her budget. The conditions for this to be valid are that the discriminant ( a^2 + 80,000b ) must be positive, which it is since ( a ) and ( b ) are positive. Also, ( n ) must be an integer because you can't produce a fraction of a saree, so we'll take the floor of the result if necessary.Problem 2: Maximizing ProfitNow, Amina's profit function is given by:[ P(n) = n left( p(n) - frac{C(n)}{n} right) ]Where the selling price per saree is ( p(n) = 1500 - 5n ). Let's substitute ( p(n) ) and ( C(n) ) into the profit function:First, compute ( frac{C(n)}{n} ):[ frac{C(n)}{n} = frac{an + bn^2}{n} = a + bn ]So, the profit function becomes:[ P(n) = n left( (1500 - 5n) - (a + bn) right) ][ P(n) = n left( 1500 - 5n - a - bn right) ][ P(n) = n left( 1500 - a - (5 + b)n right) ][ P(n) = (1500 - a)n - (5 + b)n^2 ]So, ( P(n) ) is a quadratic function in terms of ( n ), and since the coefficient of ( n^2 ) is negative (because ( 5 + b > 0 )), the parabola opens downward, meaning it has a maximum point.To find the value of ( n ) that maximizes profit, we can take the derivative of ( P(n) ) with respect to ( n ) and set it to zero.But since it's a quadratic function, we can also use the vertex formula. The vertex occurs at ( n = -frac{B}{2A} ) for a quadratic ( An^2 + Bn + C ).In this case, ( A = -(5 + b) ) and ( B = 1500 - a ). So,[ n = -frac{1500 - a}{2 times -(5 + b)} ][ n = frac{1500 - a}{2(5 + b)} ]This gives the number of sarees that will maximize profit. However, we need to ensure that this number is within the feasible production range determined by the budget constraint in Problem 1.So, we have two expressions:1. Maximum production ( n_{max} = frac{-a + sqrt{a^2 + 80,000b}}{2b} )2. Profit maximizing production ( n_{opt} = frac{1500 - a}{2(5 + b)} )We need to check if ( n_{opt} leq n_{max} ). If it is, then Amina should produce ( n_{opt} ) sarees. If not, she should produce as many as her budget allows, which is ( n_{max} ).To analyze the behavior of the profit function, since it's a downward opening parabola, it will have a single maximum point. Therefore, as long as ( n_{opt} ) is within the feasible range (i.e., less than or equal to ( n_{max} )), the profit will be maximized at ( n_{opt} ). If ( n_{opt} ) exceeds ( n_{max} ), the maximum profit within the budget constraint will be at ( n_{max} ).But without specific values for ( a ) and ( b ), we can't compute exact numbers. However, we can express the conditions in terms of ( a ) and ( b ).Wait, actually, the problem doesn't provide specific values for ( a ) and ( b ). So, perhaps we need to express the answer in terms of ( a ) and ( b ), or maybe there's more information I missed.Looking back at the problem statement, it says "given her budget constraint from the first sub-problem." So, in the second part, we need to use the maximum ( n ) from the first part as the upper limit for ( n ) in the profit function.Therefore, to find the optimal ( n ), we need to compare ( n_{opt} ) with ( n_{max} ). If ( n_{opt} leq n_{max} ), then ( n_{opt} ) is the solution. Otherwise, ( n_{max} ) is the solution.But without knowing ( a ) and ( b ), we can't numerically solve for ( n ). Maybe the problem expects an expression in terms of ( a ) and ( b ), or perhaps it's implied that ( n_{opt} ) is within the budget constraint.Alternatively, maybe I need to consider that the cost function is ( C(n) = an + bn^2 ), and the budget is 20,000, so ( an + bn^2 leq 20,000 ). Therefore, when maximizing profit, we have to consider both the profit function and the budget constraint.So, perhaps the optimal ( n ) is the minimum of ( n_{opt} ) and ( n_{max} ). But since ( n_{opt} ) is derived from the profit function without considering the budget, and ( n_{max} ) is the maximum possible due to budget, the optimal production is the smaller of the two.But again, without specific values, we can't compute exact numbers. Maybe the problem expects us to set up the equations and discuss the conditions.Alternatively, perhaps I can express the optimal ( n ) in terms of ( a ) and ( b ), ensuring that ( n leq n_{max} ).Wait, maybe I can relate ( n_{opt} ) and ( n_{max} ) by substituting ( n_{opt} ) into the budget constraint.Let me try that.From the budget constraint:[ an + bn^2 leq 20,000 ]Substitute ( n = n_{opt} = frac{1500 - a}{2(5 + b)} ):[ a left( frac{1500 - a}{2(5 + b)} right) + b left( frac{1500 - a}{2(5 + b)} right)^2 leq 20,000 ]This would give a condition on ( a ) and ( b ) whether ( n_{opt} ) is within the budget. If this inequality holds, then ( n_{opt} ) is feasible. Otherwise, Amina should produce ( n_{max} ).But without specific values, this is as far as we can go.Alternatively, maybe the problem expects us to assume that ( n_{opt} ) is within the budget, so we can proceed with that.Wait, perhaps I'm overcomplicating. Let me re-express the profit function and see if I can find the optimal ( n ) in terms of ( a ) and ( b ), and then relate it to the budget.Given:[ P(n) = (1500 - a)n - (5 + b)n^2 ]Taking derivative:[ P'(n) = 1500 - a - 2(5 + b)n ]Set to zero:[ 1500 - a - 2(5 + b)n = 0 ][ 2(5 + b)n = 1500 - a ][ n = frac{1500 - a}{2(5 + b)} ]So, this is the critical point. To ensure it's a maximum, the second derivative is:[ P''(n) = -2(5 + b) ]Which is negative since ( 5 + b > 0 ), confirming it's a maximum.Therefore, the optimal ( n ) is ( frac{1500 - a}{2(5 + b)} ), provided this ( n ) is less than or equal to ( n_{max} ).So, the conditions are:1. ( frac{1500 - a}{2(5 + b)} leq frac{-a + sqrt{a^2 + 80,000b}}{2b} )If this holds, then ( n = frac{1500 - a}{2(5 + b)} ) is the optimal. Otherwise, ( n = frac{-a + sqrt{a^2 + 80,000b}}{2b} ).But without specific values for ( a ) and ( b ), we can't simplify further. So, the answer is expressed in terms of ( a ) and ( b ).Alternatively, maybe the problem expects us to assume that ( n_{opt} ) is within the budget, so we can present ( n_{opt} ) as the solution.But to be thorough, I think the answer should consider both possibilities. So, the optimal number of sarees Amina should produce is the minimum of ( frac{1500 - a}{2(5 + b)} ) and ( frac{-a + sqrt{a^2 + 80,000b}}{2b} ).However, since ( a ) and ( b ) are positive, let's see if ( n_{opt} ) is necessarily less than ( n_{max} ).Let me compare the two expressions:( n_{opt} = frac{1500 - a}{2(5 + b)} )( n_{max} = frac{-a + sqrt{a^2 + 80,000b}}{2b} )We can see that ( n_{max} ) is the solution to ( bn^2 + an = 20,000 ), which is a quadratic in ( n ). The term ( sqrt{a^2 + 80,000b} ) is greater than ( a ), so ( n_{max} ) is positive.Now, to see if ( n_{opt} leq n_{max} ), let's substitute ( n_{opt} ) into the budget constraint:[ a n_{opt} + b n_{opt}^2 leq 20,000 ]Substitute ( n_{opt} = frac{1500 - a}{2(5 + b)} ):[ a left( frac{1500 - a}{2(5 + b)} right) + b left( frac{1500 - a}{2(5 + b)} right)^2 leq 20,000 ]This inequality must hold for ( n_{opt} ) to be feasible. If it does, then ( n_{opt} ) is the optimal. Otherwise, ( n_{max} ) is the optimal.But without knowing ( a ) and ( b ), we can't determine this. Therefore, the answer is that Amina should produce ( n = frac{1500 - a}{2(5 + b)} ) sarees, provided this number does not exceed her budget constraint. If it does, she should produce the maximum number allowed by her budget, which is ( n_{max} = frac{-a + sqrt{a^2 + 80,000b}}{2b} ).Alternatively, if we assume that the optimal ( n ) is within the budget, then the answer is simply ( n = frac{1500 - a}{2(5 + b)} ).But since the problem mentions "given her budget constraint from the first sub-problem," it implies that we need to consider the budget when determining the optimal ( n ). Therefore, the optimal ( n ) is the minimum of ( n_{opt} ) and ( n_{max} ).However, without specific values, we can't compute the exact number. So, the answer is expressed in terms of ( a ) and ( b ).Wait, maybe I can express the condition for ( n_{opt} leq n_{max} ):[ frac{1500 - a}{2(5 + b)} leq frac{-a + sqrt{a^2 + 80,000b}}{2b} ]Multiply both sides by 2:[ frac{1500 - a}{5 + b} leq frac{-a + sqrt{a^2 + 80,000b}}{b} ]Multiply both sides by ( b(5 + b) ) (which is positive since ( b > 0 )):[ b(1500 - a) leq (-a + sqrt{a^2 + 80,000b})(5 + b) ]This is a complex inequality, but perhaps we can square both sides to eliminate the square root, but that might complicate things further.Alternatively, maybe we can assume that ( n_{opt} ) is less than ( n_{max} ), so the optimal is ( n_{opt} ). But without knowing ( a ) and ( b ), it's safer to say that the optimal ( n ) is the smaller of the two.But since the problem asks to "determine the number of sarees Amina should produce and sell to maximize her profit, given her budget constraint," it implies that we need to express it in terms of ( a ) and ( b ), considering the budget.Therefore, the optimal number is:[ n = minleft( frac{1500 - a}{2(5 + b)}, frac{-a + sqrt{a^2 + 80,000b}}{2b} right) ]But to present it as a single expression, perhaps we can write it as:[ n = begin{cases} frac{1500 - a}{2(5 + b)} & text{if } frac{1500 - a}{2(5 + b)} leq frac{-a + sqrt{a^2 + 80,000b}}{2b} frac{-a + sqrt{a^2 + 80,000b}}{2b} & text{otherwise}end{cases} ]But since the problem doesn't provide specific values, this is the most precise answer we can give.Alternatively, if we consider that the profit function's optimal point is within the budget, then the answer is simply ( n = frac{1500 - a}{2(5 + b)} ).But to be thorough, I think the answer should account for the budget constraint, so the optimal ( n ) is the minimum of the two expressions.However, in many optimization problems, especially in economics, the profit-maximizing quantity is often within the feasible region defined by constraints, so perhaps we can assume that ( n_{opt} leq n_{max} ), making ( n_{opt} ) the solution.But without specific values, it's safer to present both possibilities.Wait, perhaps I can find a relationship between ( a ) and ( b ) that would make ( n_{opt} leq n_{max} ).Let me set ( n_{opt} leq n_{max} ):[ frac{1500 - a}{2(5 + b)} leq frac{-a + sqrt{a^2 + 80,000b}}{2b} ]Multiply both sides by 2:[ frac{1500 - a}{5 + b} leq frac{-a + sqrt{a^2 + 80,000b}}{b} ]Multiply both sides by ( b(5 + b) ):[ b(1500 - a) leq (-a + sqrt{a^2 + 80,000b})(5 + b) ]Let me expand the right side:[ (-a)(5 + b) + sqrt{a^2 + 80,000b}(5 + b) ][ = -5a - ab + (5 + b)sqrt{a^2 + 80,000b} ]So, the inequality becomes:[ 1500b - ab leq -5a - ab + (5 + b)sqrt{a^2 + 80,000b} ]Simplify both sides by adding ( ab ):[ 1500b leq -5a + (5 + b)sqrt{a^2 + 80,000b} ]Bring ( -5a ) to the left:[ 1500b + 5a leq (5 + b)sqrt{a^2 + 80,000b} ]Now, divide both sides by ( 5 + b ):[ frac{1500b + 5a}{5 + b} leq sqrt{a^2 + 80,000b} ]Square both sides to eliminate the square root:[ left( frac{1500b + 5a}{5 + b} right)^2 leq a^2 + 80,000b ]This is a complex inequality, but perhaps we can simplify it.Let me denote ( k = 5 + b ), so ( b = k - 5 ). Substitute into the inequality:[ left( frac{1500(k - 5) + 5a}{k} right)^2 leq a^2 + 80,000(k - 5) ]Simplify the numerator:[ 1500k - 7500 + 5a ]So,[ left( frac{1500k - 7500 + 5a}{k} right)^2 leq a^2 + 80,000k - 400,000 ]This might not lead us anywhere useful without specific values.Given the complexity, perhaps it's best to conclude that the optimal ( n ) is ( frac{1500 - a}{2(5 + b)} ) provided it does not exceed the budget constraint, otherwise, it's ( n_{max} ).But since the problem doesn't provide specific values for ( a ) and ( b ), we can't compute a numerical answer. Therefore, the answer must be expressed in terms of ( a ) and ( b ).So, summarizing:1. The maximum number of sarees Amina can produce is ( n_{max} = frac{-a + sqrt{a^2 + 80,000b}}{2b} ).2. The optimal number of sarees to maximize profit is ( n_{opt} = frac{1500 - a}{2(5 + b)} ), provided ( n_{opt} leq n_{max} ). If ( n_{opt} > n_{max} ), then Amina should produce ( n_{max} ) sarees.Therefore, the final answer is:Amina should produce ( boxed{frac{1500 - a}{2(5 + b)}} ) sarees to maximize her profit, provided this number does not exceed her budget constraint. If it does, she should produce the maximum number allowed by her budget, which is ( boxed{frac{-a + sqrt{a^2 + 80,000b}}{2b}} ) sarees.However, since the problem asks for a single number, perhaps we need to express it as the minimum of the two. But without specific values, it's not possible. Therefore, the optimal number is ( frac{1500 - a}{2(5 + b)} ), assuming it's within the budget.But to be precise, the answer should consider both. So, the optimal number is the smaller of the two expressions.But since the problem doesn't provide specific values, I think the answer is simply ( frac{1500 - a}{2(5 + b)} ), as it's derived from the profit function, and the budget constraint is considered in the first part.Wait, no, because the budget constraint is a hard limit. So, the optimal ( n ) is the minimum of ( n_{opt} ) and ( n_{max} ).But without knowing which is larger, we can't say for sure. Therefore, the answer is:Amina should produce ( boxed{minleft( frac{1500 - a}{2(5 + b)}, frac{-a + sqrt{a^2 + 80,000b}}{2b} right)} ) sarees to maximize her profit.But since the problem asks to "determine the number," and not necessarily express it in terms of a minimum function, perhaps it's better to present both expressions and state the condition.However, given the structure of the problem, I think the intended answer is to find ( n ) that maximizes profit, considering the budget, which would be ( n_{opt} ) if it's within the budget, else ( n_{max} ).But without specific values, we can't compute it numerically. Therefore, the answer is expressed in terms of ( a ) and ( b ).So, to wrap up:1. The maximum number of sarees Amina can produce is ( frac{-a + sqrt{a^2 + 80,000b}}{2b} ).2. The optimal number to maximize profit is ( frac{1500 - a}{2(5 + b)} ), provided it's less than or equal to the maximum production.Therefore, the final answer is ( boxed{frac{1500 - a}{2(5 + b)}} ) sarees, assuming this number is within her budget constraint. If not, she should produce ( boxed{frac{-a + sqrt{a^2 + 80,000b}}{2b}} ) sarees.But since the problem asks for a single answer, perhaps it's expected to present the profit-maximizing ( n ) without considering the budget, but given the first part, it's better to consider both.Alternatively, maybe the problem expects us to solve for ( n ) in terms of ( a ) and ( b ), so the answer is ( frac{1500 - a}{2(5 + b)} ).I think that's the most straightforward answer, as the budget constraint is already considered in the first part, and the second part is about maximizing profit given that constraint. So, the optimal ( n ) is ( frac{1500 - a}{2(5 + b)} ), provided it doesn't exceed ( n_{max} ).But since the problem doesn't provide specific values, the answer must be in terms of ( a ) and ( b ).Therefore, the final answer is:Amina should produce ( boxed{frac{1500 - a}{2(5 + b)}} ) sarees to maximize her profit, provided this number does not exceed her budget constraint of 20,000 BDT.</think>