answer:To solve this problem, we need to follow these steps: Step 1: Derive the Differential Equation for ( P(t) )The revenue ( R(t) ) is given by:[R(t) = D(t) cdot P(t)]where ( D(t) = 1000 cdot e^{-0.1t} + 150 sinleft(frac{pi}{12}tright) ).We want to maximize the rate of change of revenue, (frac{dR}{dt}). Using the product rule, we have:[frac{dR}{dt} = D(t) cdot frac{dP}{dt} + P(t) cdot frac{dD}{dt}]To maximize (frac{dR}{dt}), set its derivative with respect to ( t ) equal to zero:[frac{dR}{dt} = 0 quad Rightarrow quad D(t) cdot frac{dP}{dt} + P(t) cdot frac{dD}{dt} = 0]Rearranging gives:[frac{dP}{dt} = -frac{P(t) cdot frac{dD}{dt}}{D(t)}]Next, we need to compute (frac{dD}{dt}):[frac{dD}{dt} = frac{d}{dt} left(1000 e^{-0.1t} + 150 sinleft(frac{pi}{12}tright)right)][= -100 cdot e^{-0.1t} + 150 cdot frac{pi}{12} cosleft(frac{pi}{12}tright)]Substitute this into the equation for (frac{dP}{dt}):[frac{dP}{dt} = -frac{P(t) left(-100 cdot e^{-0.1t} + frac{150pi}{12} cosleft(frac{pi}{12}tright)right)}{1000 cdot e^{-0.1t} + 150 sinleft(frac{pi}{12}tright)}] Step 2: Determine When Demand Reaches CapacityWe need to solve for ( t ) when ( D(t) = 20000 ):[1000 cdot e^{-0.1t} + 150 sinleft(frac{pi}{12}tright) = 20000]This is a transcendental equation and may need numerical methods to solve. For illustration, let's assume we solve it numerically and find that ( t approx t^* ) is the solution. Step 3: Calculate Corresponding Ticket Price ( P(t) )If ( t = t^* ), substitute ( t^* ) back into the demand function ( D(t) ) and the derived differential equation for ( P(t) ) to find ( P(t^*) ).If ( P(t) ) was initially 120 at ( t = 0 ), use the differential equation to integrate and find ( P(t^*) ).Since exact computation here is complex due to the transcendental nature and numerical integration required, a precise ( P(t) ) value would be deduced via computational assistance. Final AnswerDue to the complexity of solving the transcendental equation and the numerical nature of the integration, the final answer depends on computational solutions, which are typically done with software tools.Final Answer: The final answer is computationally derived. I hope it is correct.

answer:To solve the given problem, we will follow the steps outlined in the problem statement. Part 1: Calculate the total annual donation revenueThe revenue function for the organization is given by:[ R(t) = 5000 + 3000 sinleft( frac{2pi t}{12} right) ]We need to integrate ( R(t) ) from ( t = 1 ) to ( t = 12 ) to find the total annual donation revenue.The integral of ( R(t) ) over the interval ([1, 12]) is:[ int_{1}^{12} R(t) , dt = int_{1}^{12} left( 5000 + 3000 sinleft( frac{2pi t}{12} right) right) dt ]First, we separate the integral into two parts:[ int_{1}^{12} R(t) , dt = int_{1}^{12} 5000 , dt + int_{1}^{12} 3000 sinleft( frac{2pi t}{12} right) , dt ]We solve each integral separately.1. For the first integral:[ int_{1}^{12} 5000 , dt = 5000 int_{1}^{12} 1 , dt = 5000 [t]_{1}^{12} = 5000 (12 - 1) = 5000 times 11 = 55000 ]2. For the second integral, let ( u = frac{2pi t}{12} ), then ( du = frac{2pi}{12} dt ) or ( dt = frac{12}{2pi} du ):[ int_{1}^{12} 3000 sinleft( frac{2pi t}{12} right) , dt = 3000 int_{1}^{12} sinleft( frac{2pi t}{12} right) , dt ]Changing the limits of integration:When ( t = 1 ), ( u = frac{2pi cdot 1}{12} = frac{2pi}{12} = frac{pi}{6} )When ( t = 12 ), ( u = frac{2pi cdot 12}{12} = 2pi )Thus,[ 3000 int_{pi/6}^{2pi} sin(u) frac{12}{2pi} , du = frac{3000 times 12}{2pi} int_{pi/6}^{2pi} sin(u) , du ][ = frac{36000}{2pi} left[ -cos(u) right]_{pi/6}^{2pi} ][ = frac{36000}{2pi} left( -cos(2pi) + cosleft( frac{pi}{6} right) right) ]Since (cos(2pi) = 1) and (cosleft( frac{pi}{6} right) = frac{sqrt{3}}{2} ),[ = frac{36000}{2pi} left( -1 + frac{sqrt{3}}{2} right) ][ = frac{36000}{2pi} left( frac{sqrt{3} - 2}{2} right) ][ = frac{36000}{4pi} (sqrt{3} - 2) ][ = frac{9000 (sqrt{3} - 2)}{pi} ]Combining both results:[ int_{1}^{12} R(t) , dt = 55000 + frac{9000 (sqrt{3} - 2)}{pi} ] Part 2: Calculate the new annual donation revenue after switching to .com domainAfter switching to a .com domain, the monthly donation revenue decreases by 15%. The new revenue function is:[ R_{text{new}}(t) = 0.85 R(t) = 0.85 left( 5000 + 3000 sinleft( frac{2pi t}{12} right) right) ]We integrate ( R_{text{new}}(t) ) over the interval ([1, 12]):[ int_{1}^{12} R_{text{new}}(t) , dt = int_{1}^{12} 0.85 left( 5000 + 3000 sinleft( frac{2pi t}{12} right) right) dt ][ = 0.85 int_{1}^{12} R(t) , dt ]From Part 1:[ int_{1}^{12} R(t) , dt = 55000 + frac{9000 (sqrt{3} - 2)}{pi} ]Thus,[ int_{1}^{12} R_{text{new}}(t) , dt = 0.85 left( 55000 + frac{9000 (sqrt{3} - 2)}{pi} right) ][ = 0.85 times 55000 + 0.85 times frac{9000 (sqrt{3} - 2)}{pi} ][ = 46750 + frac{0.85 times 9000 (sqrt{3} - 2)}{pi} ][ = 46750 + frac{7650 (sqrt{3} - 2)}{pi} ]Final Answer: The final answer is ( 46750 + frac{7650 (sqrt{3} - 2)}{pi} ). I hope it is correct.

answer:Let's solve the problems step-by-step. Problem 1: Derive an expression for the maximum number of lawsuits ( n ) that can be handled within the budget ( B ).Given:- Budget ( B )- Cost per lawsuit ( C )- Number of lawsuits ( n )The total cost to handle ( n ) lawsuits is ( n times C ). This total cost must not exceed the budget ( B ). Therefore, we have the inequality:[n times C leq B]To find the maximum number of lawsuits ( n ) that can be handled, solve for ( n ):[n leq frac{B}{C}]Since ( n ) must be an integer, the maximum number ( n ) is:[n = leftlfloor frac{B}{C} rightrfloor] Problem 2: Express the expected number of lawsuits won ( E ) in terms of ( n ) and ( p ).Given:- Number of lawsuits ( n )- Probability of winning a single lawsuit ( p )The expected value ( E ) of the total number of lawsuits won is:[E = n times p] Problem 3: Find the minimum probability ( p ) required to ensure that the expected number of lawsuits won is at least ( k ).Given:- Expected number of lawsuits won ( E geq k )We already expressed ( E ) as:[E = n times p]To ensure that the expected number of lawsuits won is at least ( k ), we set up the inequality:[n times p geq k]Solving for ( p ):[p geq frac{k}{n}] Final Answers:1. The maximum number of lawsuits ( n ) that can be handled within the budget ( B ) is:[n = leftlfloor frac{B}{C} rightrfloor]2. The expected number of lawsuits won ( E ) in terms of ( n ) and ( p ) is:[E = n times p]3. The minimum probability ( p ) required to ensure that the expected number of lawsuits won is at least ( k ) is:[p geq frac{k}{n}]Final Answer: The final answers are ( n = leftlfloor frac{B}{C} rightrfloor ), ( E = n times p ), and ( p geq frac{k}{n} ). I hope it is correct.

answer:Let's solve the problem step-by-step. Step 1: Determine the constants (a), (b), and (c)Given the three data points:1. ( A_1 = 2 ), ( I_1 = 40 )2. ( A_2 = 5 ), ( I_2 = 80 )3. ( A_3 = 7 ), ( I_3 = 85 )We can set up the following system of equations:[begin{cases}4a + 2b + c = 40 & text{(1)} 25a + 5b + c = 80 & text{(2)} 49a + 7b + c = 85 & text{(3)}end{cases}]We solve this system using the method of elimination or substitution.First, subtract equation (1) from equation (2):[(25a + 5b + c) - (4a + 2b + c) = 80 - 40][21a + 3b = 40 quad text{(4)}]Next, subtract equation (2) from equation (3):[(49a + 7b + c) - (25a + 5b + c) = 85 - 80][24a + 2b = 5 quad text{(5)}]Now, we have two simpler equations (4) and (5):[begin{cases}21a + 3b = 40 & text{(4)} 24a + 2b = 5 & text{(5)}end{cases}]We can solve these by elimination. First, we can multiply equation (5) by 3 and equation (4) by 2 to align the coefficients of (b):[63a + 9b = 120 quad text{(6)}][48a + 4b = 10 quad text{(7)}]Now subtract equation (7) from equation (6):[(63a + 9b) - (48a + 4b) = 120 - 10][15a + 5b = 110 Rightarrow 3a + b = 22 quad text{(8)}]From equation (8), we can express (b) in terms of (a):[b = 22 - 3a quad text{(9)}]Substitute equation (9) back into equation (4):[21a + 3(22 - 3a) = 40][21a + 66 - 9a = 40][12a = -26][a = -frac{26}{12} = -frac{13}{6}]Now, substitute (a = -frac{13}{6}) into equation (9):[b = 22 - 3left(-frac{13}{6}right)][b = 22 + frac{39}{6}][b = 22 + 6.5 = 28.5]Finally, substitute both (a) and (b) into equation (1) to find (c):[4left(-frac{13}{6}right) + 2(28.5) + c = 40][-frac{52}{6} + 57 + c = 40][-frac{26}{3} + 57 + c = 40]Convert (-frac{26}{3}) to decimal:[-frac{26}{3} approx -8.67][-8.67 + 57 + c = 40][48.33 + c = 40][c = 40 - 48.33 = -8.33]Thus, the constants are:[a = -frac{13}{6}, quad b = 28.5, quad c = -8.33] Step 2: Calculate the RSSThe RSS is given by:[text{RSS} = sum_{i=4}^{50} (I_i - (aA_i^2 + bA_i + c))^2]For the fourth organization:[A_4 = 3, quad I_4 = 45]Substitute the values of (a), (b), and (c):[I_4 - (aA_4^2 + bA_4 + c) = 45 - left(-frac{13}{6} cdot 3^2 + 28.5 cdot 3 - 8.33right)][= 45 - left(-frac{13}{6} cdot 9 + 85.5 - 8.33right)][= 45 - left(-frac{117}{6} + 85.5 - 8.33right)][= 45 - left(-19.5 + 85.5 - 8.33right)][= 45 - left(57.67right)][= 45 - 57.67 = -12.67]Square the result:[(-12.67)^2 approx 160.49]Thus, the RSS for the fourth organization is:[text{RSS}_4 approx 160.49]Final Answer: The final answer is a = -frac{13}{6}, b = 28.5, c = -8.33, text{RSS}_4 approx 160.49. I hope it is correct.

answer:To solve the given problem, follow these steps: Step 1: Rewrite the Differential EquationThe differential equation given is:[ frac{dE(t)}{dt} = kE(t) left(1 - frac{E(t)}{L} right) - a e^{-bt}. ] Step 2: Simplify the EquationLet’s rewrite it in a more explicit form:[ frac{dE(t)}{dt} = kE(t) - frac{kE(t)^2}{L} - a e^{-bt}. ] Step 3: Solve the Homogeneous PartConsider the homogeneous part of the differential equation:[ frac{dE_h(t)}{dt} = kE_h(t) left(1 - frac{E_h(t)}{L} right). ]This is a separable differential equation. To solve it, rewrite it as:[ frac{dE_h}{E_h left(1 - frac{E_h}{L}right)} = k , dt. ] Step 4: Perform Partial Fraction DecompositionRewrite the left-hand side using partial fractions:[ frac{1}{E_h left(1 - frac{E_h}{L}right)} = frac{1}{E_h} + frac{1}{L - E_h}. ]So, we have:[ left( frac{1}{E_h} + frac{1}{L - E_h} right) , dE_h = k , dt. ]Integrate both sides:[ int left( frac{1}{E_h} + frac{1}{L - E_h} right) , dE_h = int k , dt. ] Step 5: Integrate the EquationIntegrate the left-hand side:[ int frac{1}{E_h} , dE_h + int frac{1}{L - E_h} , dE_h = k int dt. ][ ln|E_h| - ln|L - E_h| = kt + C_1, ]where (C_1) is an integration constant. Combine the logarithms:[ ln left| frac{E_h}{L - E_h} right| = kt + C_1. ]Exponentiate both sides:[ frac{E_h}{L - E_h} = e^{kt + C_1} = C_2 e^{kt}, ]where (C_2 = e^{C_1}). Solve for (E_h):[ E_h = frac{C_2 e^{kt} L}{1 + C_2 e^{kt}}. ] Step 6: Apply Initial ConditionUse the initial condition (E(0) = E_0):[ E_0 = frac{C_2 L}{1 + C_2}, ][ C_2 = frac{E_0}{L - E_0}. ]Thus, the solution to the homogeneous part is:[ E_h(t) = frac{E_0 L e^{kt}}{(L - E_0) + E_0 e^{kt}}. ] Step 7: Solve the Non-Homogeneous PartNow, consider the non-homogeneous part. We use an integrating factor or variation of parameters, but for simplicity, let’s assume a solution of the form:[ E(t) = E_h(t) + E_p(t), ]where (E_p(t)) is a particular solution to the non-homogeneous equation. An ansatz for (E_p(t)) could be proportional to (e^{-bt}), but the exact form depends on the specific (a e^{-bt}) term. Step 8: Determine the Long-Term BehaviorAs (t to infty), the term (a e^{-bt}) approaches zero because (b > 0). The homogeneous solution (E_h(t)) approaches (L) because (e^{kt}) grows exponentially. Thus:[ lim_{t to infty} E(t) = L. ] Final AnswerThe long-term behavior of (E(t)) as (t to infty) is that the economic impact stabilizes at the carrying capacity (L), provided that (k), (L), (a), and (b) are all positive constants. Therefore, the conditions on (a), (b), (k), and (L) are already satisfied as given in the problem.Final Answer: The final answer is that the economic impact (E(t)) stabilizes at the carrying capacity (L) as (t to infty). I hope it is correct.

answer:To solve the given problem, let's tackle each part step by step. Part 1: Derive the General Solution for ( T(t) )We are given the differential equation:[frac{dT}{dt} = aT + bsin(omega t)]This is a non-homogeneous linear first-order differential equation. To solve it, we can use the method of integrating factors.1. Find the Integrating Factor: The standard form of the equation is: [ frac{dT}{dt} - aT = bsin(omega t) ] The integrating factor ( mu(t) ) is given by: [ mu(t) = e^{int -a , dt} = e^{-at} ]2. Multiply through by the Integrating Factor: [ e^{-at} frac{dT}{dt} - ae^{-at}T = b e^{-at} sin(omega t) ] This can be rewritten as: [ frac{d}{dt}(e^{-at} T) = b e^{-at} sin(omega t) ]3. Integrate Both Sides: Integrate with respect to ( t ): [ e^{-at} T = int b e^{-at} sin(omega t) , dt + C ] Using the method of integration by parts or a known integral formula, the integral can be solved (let's assume the solution): [ int e^{-at} sin(omega t) , dt = frac{e^{-at}}{a^2 + omega^2} (a sin(omega t) - omega cos(omega t)) ] Plugging this back, we have: [ e^{-at} T = frac{b}{a^2 + omega^2} e^{-at} (a sin(omega t) - omega cos(omega t)) + C ] Solving for ( T ): [ T(t) = frac{b}{a^2 + omega^2} (a sin(omega t) - omega cos(omega t)) + Ce^{at} ]Thus, the general solution for ( T(t) ) is:[T(t) = frac{b}{a^2 + omega^2} (a sin(omega t) - omega cos(omega t)) + Ce^{at}] Part 2: Express ( T(t) ) in Terms of Fourier Coefficients of ( theta(t) )Given ( theta(t) = ccos(omega t + phi) ), we want to relate it to the torque ( T(t) ).1. Fourier Series Expansion: Since ( theta(t) ) is a simple harmonic function, its Fourier series expansion in terms of sines and cosines will have coefficients ( c_n ) for the cosine terms and ( b_n ) for the sine terms, but since it's a single cosine function, it simplifies to: [ theta(t) = ccos(omega t + phi) = c(cos(phi)cos(omega t) - sin(phi)sin(omega t)) ]2. Relate ( T(t) ) to ( theta(t) ): ( T(t) ) is derived involving sine and cosine terms reflective of ( sin(omega t) ) and (cos(omega t) ) components. By the nature of Fourier series, ( T(t) ) can be expressed as a linear combination of these components: [ T(t) = A cos(omega t) + B sin(omega t) + text{higher order terms} ] where ( A ) and ( B ) depend on the interaction between the driving force (described by ( sin(omega t) )) and the system parameters.3. Conditions for Validity: The approximation using Fourier series is valid under conditions where: - ( T(t) ) can be assumed periodic or approximately periodic. - The system's response bandwidth is not too wide.Final Answer: The general solution for ( T(t) ) is ( T(t) = frac{b}{a^2 + omega^2} (a sin(omega t) - omega cos(omega t)) + Ce^{at} ). Fourier coefficients can model ( T(t) ) under periodic conditions. I hope it is correct.