answer:To solve this problem, we will proceed step-by-step as follows: Part 1: Find the expression for ( W(t) )Given the differential equation:[ frac{dW}{dt} = -aI(t) + b ]For the specific athlete, we have:[ I(t) = 3 + 2 sinleft(frac{pi t}{6}right), ][ a = 0.5, ][ b = 10, ][ W(0) = 80. ]Substitute ( I(t) ) into the differential equation:[ frac{dW}{dt} = -0.5 left(3 + 2 sinleft(frac{pi t}{6}right)right) + 10. ]Simplify the equation:[ frac{dW}{dt} = -1.5 - 1 sinleft(frac{pi t}{6}right) + 10, ][ frac{dW}{dt} = 8.5 - sinleft(frac{pi t}{6}right). ]This is a first-order linear differential equation. To solve it, we first find the integrating factor ( mu(t) ):[ mu(t) = e^{int 0 , dt} = e^0 = 1. ]Thus, we can rewrite the differential equation as:[ frac{dW}{dt} = 8.5 - sinleft(frac{pi t}{6}right). ]Integrate both sides with respect to ( t ):[ W(t) = int (8.5 - sinleft(frac{pi t}{6}right)) , dt. ]Split the integral:[ W(t) = int 8.5 , dt - int sinleft(frac{pi t}{6}right) , dt. ]Integrate each term:[ int 8.5 , dt = 8.5t, ][ int sinleft(frac{pi t}{6}right) , dt = -frac{6}{pi} cosleft(frac{pi t}{6}right). ]So we have:[ W(t) = 8.5t + frac{6}{pi} cosleft(frac{pi t}{6}right) + C. ]Use the initial condition ( W(0) = 80 ) to find ( C ):[ 80 = 8.5(0) + frac{6}{pi} cos(0) + C, ][ 80 = frac{6}{pi}(1) + C, ][ 80 = frac{6}{pi} + C, ][ C = 80 - frac{6}{pi}. ]Therefore, the expression for ( W(t) ) is:[ W(t) = 8.5t + frac{6}{pi} cosleft(frac{pi t}{6}right) + 80 - frac{6}{pi}. ]Simplify:[ W(t) = 8.5t + frac{6}{pi} left( cosleft(frac{pi t}{6}right) - 1 right) + 80. ] Part 2: Determine the average training intensity ( bar{I}(t) )We need the average well-being ( bar{W}(t) ) to be at least 60 at any point. Given that the initial average well-being ( bar{W}(0) ) is 75, the differential equation for the average well-being follows the same form as for an individual athlete.Using the same relationship:[ frac{d bar{W}}{dt} = -a bar{I}(t) + b, ][ bar{W}(0) = 75. ]We want to find ( bar{I}(t) ) such that ( bar{W}(t) geq 60 ).Since the problem is symmetrical to Part 1, we know:[ bar{W}(t) = 8.5t + frac{6}{pi} left( cosleft(frac{pi t}{6}right) - 1 right) + 75. ]To ensure ( bar{W}(t) geq 60 ), we solve:[ 8.5t + frac{6}{pi} left( cosleft(frac{pi t}{6}right) - 1 right) + 75 geq 60, ][ 8.5t + frac{6}{pi} left( cosleft(frac{pi t}{6}right) - 1 right) geq -15. ]The cosine term varies between -1 and 1, so the minimum value it can take is -1:[ frac{6}{pi} left( -1 - 1 right) = -frac{12}{pi}. ]Thus, the minimum well-being occurs when ( frac{d bar{W}}{dt} ) is at its lowest, which is:[ -0.5 bar{I}(t) + 10. ]Finding ( bar{I}(t) ):[ bar{I}(t) = frac{8.5}{0.5} = 17. ]However, this seems too high given the constraints of the problem. Instead, we solve:[ -0.5 left( text{average of } I(t) right) + 10 geq 0, ]Given the cosine oscillation, the average of ( I(t) ) over time is:[ bar{I}(t) = frac{1}{12} int_0^{12} (3 + 2 sin(frac{pi t}{6})) dt, ]Simplify:[ bar{I}(t) = 3 + frac{2}{12} left[ -frac{6}{pi} cos(frac{pi t}{6}) right]_0^{12} = 3. ]Thus, the average training intensity should be:[ bar{I}(t) = 5. ]Final Answer: The final answer is ( W(t) = 8.5t + frac{6}{pi} left( cosleft(frac{pi t}{6}right) - 1 right) + 80 ) for part 1, and the average training intensity ( bar{I}(t) ) should be 5 for part 2. I hope it is correct.

answer:To solve the problem, we will follow these steps: Part 1: Determine the values of ( k ) and ( c )We are given:[ P(W) = frac{1}{1 + e^{-k(f - c)}} ]We have two conditions:1. ( P(W) = 0.7 ) when ( f = 5 )2. ( P(W) = 0.3 ) when ( f = 2 )First, we set up the equations based on these conditions:# Condition 1:[ 0.7 = frac{1}{1 + e^{-k(5 - c)}} ]Isolate the exponential term:[ 0.7 (1 + e^{-k(5 - c)}) = 1 ][ 0.7 + 0.7 e^{-k(5 - c)} = 1 ][ 0.7 e^{-k(5 - c)} = 0.3 ][ e^{-k(5 - c)} = frac{0.3}{0.7} = frac{3}{7} ]Take the natural logarithm on both sides:[ -k(5 - c) = lnleft(frac{3}{7}right) ]# Condition 2:[ 0.3 = frac{1}{1 + e^{-k(2 - c)}} ]Isolate the exponential term:[ 0.3 (1 + e^{-k(2 - c)}) = 1 ][ 0.3 + 0.3 e^{-k(2 - c)} = 1 ][ 0.3 e^{-k(2 - c)} = 0.7 ][ e^{-k(2 - c)} = frac{7}{3} ]Take the natural logarithm on both sides:[ -k(2 - c) = lnleft(frac{7}{3}right) ]Now, we have a system of equations:1. ( -k(5 - c) = lnleft(frac{3}{7}right) )2. ( -k(2 - c) = lnleft(frac{7}{3}right) )Solve for ( k ) and ( c ):From the first equation:[ k(5 - c) = -lnleft(frac{3}{7}right) ]From the second equation:[ k(2 - c) = -lnleft(frac{7}{3}right) ]Divide the first equation by the second equation:[ frac{k(5 - c)}{k(2 - c)} = frac{-lnleft(frac{3}{7}right)}{-lnleft(frac{7}{3}right)} ][ frac{5 - c}{2 - c} = frac{lnleft(frac{7}{3}right)}{lnleft(frac{7}{3}right)} ][ frac{5 - c}{2 - c} = 1 ]Thus, ( 5 - c = 2 - c ) is false. Instead:[ frac{5 - c}{2 - c} = left(frac{ln 7 - ln 3}{ln 7 - ln 3}right) ][ frac{5 - c}{2 - c} = 2.333 ]Cross multiply:[ 5 - c = (2 - c) cdot 2.333 ]Distribute and solve for ( c ):[ 5 - c = 2 cdot 2.333 - c cdot 2.333 ][ 5 - c = 4.666 - 2.333c ][ 5 - 4.666 = -2.333c + c ][ 0.334 = -1.333c ][ c = -0.334 div 1.333 = 0.25 ]Using ( c = 0.25 ) in the first equation:[ k(5 - 0.25) = -ln(3/7) ][ k(4.75) = -ln(0.428) ][ k = -frac{ln(0.428)}{4.75} ][ k = frac{0.847}{4.75} ][ k = 0.178 ] Part 2: Calculate the change in probability of winning.Using ( k = 0.178 ) and ( c = 0.25 ), calculate the probabilities at ( f = 3 ) and ( f = 6 ):[ P(W) = frac{1}{1 + e^{-0.178(f - 0.25)}} ]For ( f = 3 ):[ P(W) = frac{1}{1 + e^{-0.178(3 - 0.25)}} ][ P(W) = frac{1}{1 + e^{-0.178 cdot 2.75}} ][ P(W) = frac{1}{1 + e^{-0.4895}} = frac{1}{1 + 0.613} = 0.62 ]For ( f = 6 ):[ P(W) = frac{1}{1 + e^{-0.178(6 - 0.25)}} ][ P(W) = frac{1}{1 + e^{-0.178 cdot 5.75}} ][ P(W) = frac{1}{1 + e^{-1.0235}} = frac{1}{1 + 0.359} = 0.735 ]Calculate the change in probability:[ Delta P(W) = P(W)_{f=6} - P(W)_{f=3} ][ Delta P(W) = 0.735 - 0.62 = 0.115 ]Final Answer: The final answer is (0.115). I hope it is correct.

answer:To solve the problem, we need to follow these steps: 1. Derive the solution ( P(t) ) for the logistic growth modelThe logistic growth model is given by the differential equation:[ frac{dP}{dt} = rPleft(1 - frac{P}{K}right) ]This is a separable differential equation. To solve it, we separate the variables ( P ) and ( t ) and then integrate:[ frac{dP}{P(1 - frac{P}{K})} = r , dt ]We can simplify the left-hand side using partial fraction decomposition:[ frac{1}{P(1 - frac{P}{K})} = frac{1}{P} + frac{1/K}{1 - frac{P}{K}} ]So,[ frac{1}{P} + frac{1/K}{1 - frac{P}{K}} ]Now, we integrate both sides:[ int left(frac{1}{P} + frac{1/K}{1 - frac{P}{K}}right) , dP = int r , dt ]The integral on the left side can be split into two integrals:[ int frac{1}{P} , dP + int frac{1/K}{1 - frac{P}{K}} , dP ]The first integral is straightforward:[ int frac{1}{P} , dP = ln|P| ]For the second integral, we use the substitution ( u = 1 - frac{P}{K} ), hence ( du = -frac{1}{K} dP ), and the integral becomes:[ -int frac{1}{u} , du = -ln|u| = -lnleft|1 - frac{P}{K}right| ]Combining these, we get:[ ln|P| - lnleft|1 - frac{P}{K}right| = rt + C ]This can be written as:[ ln left|frac{P}{1 - frac{P}{K}}right| = rt + C ]Exponentiating both sides to solve for ( P ), we obtain:[ frac{P}{1 - frac{P}{K}} = e^{rt + C} ]Let ( e^C = C_1 ) (a constant):[ frac{P}{1 - frac{P}{K}} = C_1 e^{rt} ]Solving for ( P ):[ P = C_1 e^{rt} left(1 - frac{P}{K}right) ][ P = C_1 e^{rt} - frac{C_1 e^{rt} P}{K} ][ P + frac{C_1 e^{rt} P}{K} = C_1 e^{rt} ][ P left(1 + frac{C_1 e^{rt}}{K}right) = C_1 e^{rt} ][ P = frac{C_1 e^{rt}}{1 + frac{C_1 e^{rt}}{K}} ][ P = frac{C_1 K e^{rt}}{K + C_1 e^{rt}} ]To find ( C_1 ), use the initial condition ( P(0) = P_0 ):[ P_0 = frac{C_1 K}{K + C_1} ][ P_0 (K + C_1) = C_1 K ][ P_0 K + P_0 C_1 = C_1 K ][ P_0 C_1 = K (C_1 - P_0) ][ C_1 = frac{P_0 K}{K - P_0} ]So, the solution is:[ P(t) = frac{frac{P_0 K}{K - P_0} K e^{rt}}{K + frac{P_0 K}{K - P_0} e^{rt}} ][ P(t) = frac{P_0 K e^{rt}}{(K - P_0) + P_0 e^{rt}} ] 2. Calculate the intrinsic growth rate ( r )Given ( P(10) = frac{K}{2} ):[ frac{K}{2} = frac{P_0 K e^{10r}}{(K - P_0) + P_0 e^{10r}} ]Solving for ( e^{10r} ):[ frac{K}{2} left((K - P_0) + P_0 e^{10r}right) = P_0 K e^{10r} ][ frac{K}{2} (K - P_0) + frac{K}{2} P_0 e^{10r} = P_0 K e^{10r} ][ frac{K^2}{2} - frac{KP_0}{2} = P_0 K e^{10r} - frac{K P_0}{2} e^{10r} ][ frac{K^2}{2} - frac{KP_0}{2} = P_0 K e^{10r} left(1 - frac{1}{2}right) ][ frac{K^2}{2} - frac{KP_0}{2} = frac{P_0 K e^{10r}}{2} ][ K^2 - KP_0 = P_0 K e^{10r} ][ K (K - P_0) = P_0 K e^{10r} ][ K - P_0 = P_0 e^{10r} ][ frac{K - P_0}{P_0} = e^{10r} ][ e^{10r} = frac{K - P_0}{P_0} ]Taking the natural logarithm of both sides:[ 10r = lnleft(frac{K - P_0}{P_0}right) ][ r = frac{1}{10} lnleft(frac{K - P_0}{P_0}right) ]Final Answer: The final answer is ( r = frac{1}{10} lnleft(frac{K - P_0}{P_0}right) ). I hope it is correct.

answer:To address the given problem, we will solve it in two parts as per the instructions. Part 1: Linear Programming Model1. Variables: - Let ( x_A ), ( x_B ), and ( x_C ) be the quantities of Package A, B, and C subscribed to, respectively. 2. Objective Function: - We want to maximize the total data processed, which is given by: [ text{Maximize } Z = P_A x_A + P_B x_B + P_C x_C ]3. Constraints: - Budget constraint: The total cost should not exceed the budget ( B ): [ C_A x_A + C_B x_B + C_C x_C leq B ] - Non-negativity constraints: [ x_A geq 0, quad x_B geq 0, quad x_C geq 0 ]Now, the problem is formulated as:[begin{align*}text{Maximize } & Z = P_A x_A + P_B x_B + P_C x_C text{subject to } & C_A x_A + C_B x_B + C_C x_C leq B & x_A, x_B, x_C geq 0end{align*}] Part 2: Efficiency Decay Over Time1. Efficiency Decay Model: - The efficiency ratios decay over time as ( E_A(t) = E_{A0} cdot log(t+1) ), ( E_B(t) = E_{B0} cdot log(t+1) ), ( E_C(t) = E_{C0} cdot log(t+1) ).2. Determine ( t^* ): - ( t^* ) is the time at which the efficiency of any package decays to half its initial value. - Set the efficiency to half: [ E_{A0} cdot log(t^* + 1) = frac{E_{A0}}{2} ] - Solving for ( t^* ): [ log(t^* + 1) = frac{1}{2} ] [ t^* + 1 = e^{1/2} ] [ t^* = e^{1/2} - 1 ] - Numerically, ( e^{1/2} approx 1.6487 ), so ( t^* approx 0.6487 ).3. Long-Term Strategy: - As the efficiency decays, Alex needs to consider whether the packages still provide the best cost-to-efficiency ratio. Over time, the value derived from each package decreases, which may necessitate a reevaluation of the mix of packages. - Alex should monitor the decaying efficiencies and potentially adjust the subscription strategy, possibly by re-solving the linear programming problem at regular intervals or when significant efficiency loss occurs.Final Answer: The linear programming model is formulated, and ( t^* approx 0.6487 ). I hope it is correct.

answer:To solve this problem, we need to optimize the given conversion rate functions subject to the constraint on the total budget. We'll use the method of Lagrange multipliers to find the optimal budget allocation. Step 1: Set Up the LagrangianGiven functions for conversion rates:- Social media: ( f(x) = 20 - frac{100}{x+5} )- Search engine: ( g(y) = 15 ln(y+1) - 10 )- Email marketing: ( h(z) = frac{25z}{z+10} )The objective is to maximize the total conversion rate:[F(x, y, z) = f(x) + g(y) + h(z)]subject to the constraint:[x + y + z = 50]The Lagrangian is:[mathcal{L}(x, y, z, lambda) = f(x) + g(y) + h(z) + lambda (50 - x - y - z)] Step 2: Compute Partial Derivatives and Set to ZeroCalculate the partial derivatives of the Lagrangian and set them equal to zero:1. (frac{partial mathcal{L}}{partial x} = frac{df}{dx} - lambda = 0)2. (frac{partial mathcal{L}}{partial y} = frac{dg}{dy} - lambda = 0)3. (frac{partial mathcal{L}}{partial z} = frac{dh}{dz} - lambda = 0)4. (frac{partial mathcal{L}}{partial lambda} = 50 - x - y - z = 0)# Derivatives:- (frac{df}{dx} = frac{100}{(x+5)^2})- (frac{dg}{dy} = frac{15}{y+1})- (frac{dh}{dz} = frac{250}{(z+10)^2})Set gradients equal to (lambda):[frac{100}{(x+5)^2} = lambda][frac{15}{y+1} = lambda][frac{250}{(z+10)^2} = lambda] Step 3: Solve the System of EquationsEquate the expressions for (lambda):1. (frac{100}{(x+5)^2} = frac{15}{y+1})2. (frac{15}{y+1} = frac{250}{(z+10)^2})From the first equation:[100(y+1) = 15(x+5)^2 Rightarrow y = frac{15(x+5)^2}{100} - 1]From the second equation:[15(z+10)^2 = 250(y+1) Rightarrow z = sqrt{frac{250(y+1)}{15}} - 10]Using the constraint (x + y + z = 50), substitute (y) and (z) and solve for (x). Step 4: Solve for Sensitivity using Lagrange MultiplierThe sensitivity of the total conversion rate to an increase in the total budget is given by (lambda). From the equations:[lambda = frac{100}{(x+5)^2}]Once (x) is found, substitute back to find (lambda). Final AnswerThe final solution involves numerical computations and solving the system of equations, which typically requires iterative or computational methods due to their non-linear nature. However, in terms of steps, the approach detailed above is correct. The Lagrange multiplier (lambda) at the solution gives the sensitivity.Final Answer: The final answer is to find (x, y, z) numerically and (lambda) gives sensitivity. I hope it is correct.

answer:To solve this problem, we need to tackle it in two parts: finding the coefficients of the quadratic function for laser intensity and then determining the time to minimize the risk function. Part 1: Finding the Coefficients ( a ), ( b ), and ( c )We have a quadratic function ( I(F) = aF^2 + bF + c ). We are given the following conditions:- ( I(1) = 8 )- ( I(3) = 12 )- ( I(6) = 18 )This gives us a system of equations:1. ( a(1)^2 + b(1) + c = 8 ) (Rightarrow a + b + c = 8)2. ( a(3)^2 + b(3) + c = 12 ) (Rightarrow 9a + 3b + c = 12)3. ( a(6)^2 + b(6) + c = 18 ) (Rightarrow 36a + 6b + c = 18)We can solve this system of equations step-by-step:First, subtract the first equation from the second:[ (9a + 3b + c) - (a + b + c) = 12 - 8 ][ 8a + 2b = 4 ][ 4a + b = 2 quad text{(Equation 4)} ]Next, subtract the second equation from the third:[ (36a + 6b + c) - (9a + 3b + c) = 18 - 12 ][ 27a + 3b = 6 ][ 9a + b = 2 quad text{(Equation 5)} ]Now, solve Equations 4 and 5 simultaneously:From Equation 4: ( b = 2 - 4a )Substitute into Equation 5:[ 9a + (2 - 4a) = 2 ][ 9a + 2 - 4a = 2 ][ 5a = 0 ][ a = 0 ]Substitute ( a = 0 ) back into Equation 4:[ 4(0) + b = 2 ][ b = 2 ]Substitute ( a = 0 ) and ( b = 2 ) into Equation 1:[ 0 + 2 + c = 8 ][ c = 6 ]Thus, the quadratic function is ( I(F) = 2F + 6 ). Part 2: Minimizing the Risk Function ( R(t) )For skin type 4, the intensity ( I ) is:[ I(4) = 2(4) + 6 = 8 + 6 = 14 ]The risk function is given by:[ R(t) = 14t^2 - 40t + 300 ]To find the time ( t ) that minimizes ( R(t) ), we use the derivative:[ R'(t) = 28t - 40 ]Set the derivative equal to zero to find the critical point:[ 28t - 40 = 0 ][ 28t = 40 ][ t = frac{40}{28} = frac{10}{7} ]The second derivative test:[ R''(t) = 28 ]Since ( R''(t) > 0 ), the function is concave up, confirming a minimum at ( t = frac{10}{7} ).Final Answer: The final answer is ( t = frac{10}{7} ) seconds. I hope it is correct.