Appearance
question:A food industry executive is working on optimizing nutrient retention in a new line of vegetable-based products. The executive must balance the cooking time and temperature to ensure maximum retention of two key vitamins, Vitamin A and Vitamin C. The retention rates of these vitamins are modeled by the following functions based on cooking time ( t ) (in minutes) and temperature ( T ) (in degrees Celsius):[ R_A(t, T) = 100 - 0.02t^2 - 0.05(T - 50)^2 ][ R_C(t, T) = 100 - 0.03t^2 - 0.07(T - 50)^2 ]where ( R_A(t, T) ) and ( R_C(t, T) ) represent the retention rates of Vitamin A and Vitamin C, respectively, as percentages of their original amounts.Sub-problem 1:Determine the optimal cooking time ( t ) and temperature ( T ) that maximize the combined retention rate ( R(t, T) = R_A(t, T) + R_C(t, T) ). Show all necessary steps, including the use of partial derivatives and any critical points.Sub-problem 2:Given that the cooking process can only be conducted within the range of 30 to 60 degrees Celsius and 0 to 30 minutes, find the maximum possible combined retention rate within these constraints. Use appropriate optimization techniques to account for boundary conditions.
answer:To solve the given problem, we will follow the specified steps and use calculus to find the optimal cooking time ( t ) and temperature ( T ) for maximizing the combined retention rate of Vitamin A and Vitamin C. Sub-problem 1: Find the Critical Points# Step 1: Define the combined retention rate functionThe combined retention rate function ( R(t, T) ) is given by:[ R(t, T) = R_A(t, T) + R_C(t, T) ][ R(t, T) = left(100 - 0.02t^2 - 0.05(T - 50)^2right) + left(100 - 0.03t^2 - 0.07(T - 50)^2right) ][ R(t, T) = 200 - 0.05t^2 - 0.12(T - 50)^2 ]# Step 2: Compute the partial derivativesTo find the critical points, we need to set the partial derivatives of ( R(t, T) ) with respect to ( t ) and ( T ) to zero.Compute ( frac{partial R}{partial t} ):[ frac{partial R}{partial t} = -0.1t ]Compute ( frac{partial R}{partial T} ):[ frac{partial R}{partial T} = -0.24(T - 50) ]# Step 3: Set the partial derivatives to zero and solve for ( t ) and ( T )[ frac{partial R}{partial t} = -0.1t = 0 ][ t = 0 ][ frac{partial R}{partial T} = -0.24(T - 50) = 0 ][ T = 50 ]So, the critical point is at ( (t, T) = (0, 50) ).# Step 4: Verify if this critical point is a maximumTo determine if this critical point is a maximum, we need to check the second partial derivatives and use the second derivative test.Compute the second partial derivatives:[ frac{partial^2 R}{partial t^2} = -0.1 ][ frac{partial^2 R}{partial T^2} = -0.24 ][ frac{partial^2 R}{partial t partial T} = frac{partial^2 R}{partial T partial t} = 0 ]The Hessian determinant ( H ) is given by:[ H = left(frac{partial^2 R}{partial t^2}right)left(frac{partial^2 R}{partial T^2}right) - left(frac{partial^2 R}{partial t partial T}right)^2 ][ H = (-0.1)(-0.24) - (0)^2 ][ H = 0.024 ]Since ( H > 0 ) and ( frac{partial^2 R}{partial t^2} < 0 ), the critical point ( (0, 50) ) is a local maximum. Sub-problem 2: Maximize within ConstraintsGiven the constraints:[ 30 leq T leq 60 ][ 0 leq t leq 30 ]We need to evaluate the function ( R(t, T) ) at the boundaries.# Evaluate at the boundaries1. At ( t = 0 ) and ( 30 leq T leq 60 ):[ R(0, T) = 200 - 0.12(T - 50)^2 ]The maximum will be at ( T = 50 ) (as found earlier):[ R(0, 50) = 200 - 0.12(50 - 50)^2 = 200 ]2. At ( t = 30 ) and ( 30 leq T leq 60 ):[ R(30, T) = 200 - 0.05(30)^2 - 0.12(T - 50)^2 ][ R(30, T) = 200 - 45 - 0.12(T - 50)^2 ][ R(30, T) = 155 - 0.12(T - 50)^2 ]The maximum will be at ( T = 50 ):[ R(30, 50) = 155 ]3. At ( 0 leq t leq 30 ) and ( T = 30 ):[ R(t, 30) = 200 - 0.05t^2 - 0.12(30 - 50)^2 ][ R(t, 30) = 200 - 0.05t^2 - 0.12(20)^2 ][ R(t, 30) = 200 - 0.05t^2 - 48 ][ R(t, 30) = 152 - 0.05t^2 ]The maximum will be at ( t = 0 ):[ R(0, 30) = 152 ]4. At ( 0 leq t leq 30 ) and ( T = 60 ):[ R(t, 60) = 200 - 0.05t^2 - 0.12(60 - 50)^2 ][ R(t, 60) = 200 - 0.05t^2 - 0.12(10)^2 ][ R(t, 60) = 200 - 0.05t^2 - 12 ][ R(t, 60) = 188 - 0.05t^2 ]The maximum will be at ( t = 0 ):[ R(0, 60) = 188 ]# ConclusionThe maximum combined retention rate within the given constraints is at ( t = 0 ) and ( T = 50 ):[ R(0, 50) = 200 ]Final Answer: The final answer is ( 200 ). I hope it is correct.
question:A school bus driver, Alex, plans his route based on real-time traffic updates from a traffic news presenter. The traffic updates are provided every 15 minutes and predict the average speed of vehicles on different segments of the route. Alex needs to determine the optimal departure time and route to minimize travel time while ensuring that he arrives at the school exactly at 8:00 AM. Alex has two route options:- Route A: 12 miles long with changing traffic conditions.- Route B: 15 miles long with more stable traffic conditions.Based on historical data provided by the traffic news presenter, the average speeds for each route segment (in miles per hour) are as follows:Route A:- Segment 1 (3 miles): Speed varies between 20 mph and 30 mph, updated every 15 minutes.- Segment 2 (5 miles): Speed varies between 15 mph and 25 mph, updated every 15 minutes.- Segment 3 (4 miles): Speed varies between 25 mph and 35 mph, updated every 15 minutes.Route B:- Segment 1 (5 miles): Speed is consistently 25 mph.- Segment 2 (10 miles): Speed varies between 20 mph and 30 mph, updated every 15 minutes.Sub-problems:1. Given the traffic update times and the speed variations, calculate the expected travel time for Route A and Route B if Alex departs at 7:00 AM. Assume the speeds change randomly within the given ranges for each segment.2. Determine the latest possible departure time for Alex to ensure he arrives at the school exactly at 8:00 AM, considering the worst-case traffic conditions for Route A and Route B.
answer:Let's break down the problem step-by-step to provide a solution. Sub-problem 1: Expected Travel Time for Route A and Route B if Alex departs at 7:00 AM# Route A Travel Time Calculation:We need to calculate the expected travel time for each segment of Route A.Segment 1 (3 miles):- Speed varies between 20 mph and 30 mph.- Average speed = (20 mph + 30 mph) / 2 = 25 mph- Travel time = Distance / Average Speed = 3 miles / 25 mph = 0.12 hours = 7.2 minutesSegment 2 (5 miles):- Speed varies between 15 mph and 25 mph.- Average speed = (15 mph + 25 mph) / 2 = 20 mph- Travel time = Distance / Average Speed = 5 miles / 20 mph = 0.25 hours = 15 minutesSegment 3 (4 miles):- Speed varies between 25 mph and 35 mph.- Average speed = (25 mph + 35 mph) / 2 = 30 mph- Travel time = Distance / Average Speed = 4 miles / 30 mph = 0.1333 hours ≈ 8 minutesTotal expected travel time for Route A:- 7.2 minutes + 15 minutes + 8 minutes = 30.2 minutes# Route B Travel Time Calculation:We need to calculate the expected travel time for each segment of Route B.Segment 1 (5 miles):- Speed is consistently 25 mph.- Travel time = Distance / Speed = 5 miles / 25 mph = 0.2 hours = 12 minutesSegment 2 (10 miles):- Speed varies between 20 mph and 30 mph.- Average speed = (20 mph + 30 mph) / 2 = 25 mph- Travel time = Distance / Average Speed = 10 miles / 25 mph = 0.4 hours = 24 minutesTotal expected travel time for Route B:- 12 minutes + 24 minutes = 36 minutes Sub-problem 2: Latest Possible Departure Time# Route A Worst-case Travel Time Calculation:We need to determine the worst-case travel time for Route A.Segment 1 (3 miles):- Worst-case speed = 20 mph- Travel time = Distance / Speed = 3 miles / 20 mph = 0.15 hours = 9 minutesSegment 2 (5 miles):- Worst-case speed = 15 mph- Travel time = Distance / Speed = 5 miles / 15 mph = 0.3333 hours ≈ 20 minutesSegment 3 (4 miles):- Worst-case speed = 25 mph- Travel time = Distance / Speed = 4 miles / 25 mph = 0.16 hours ≈ 10 minutesTotal worst-case travel time for Route A:- 9 minutes + 20 minutes + 10 minutes = 39 minutes# Route B Worst-case Travel Time Calculation:We need to determine the worst-case travel time for Route B.Segment 1 (5 miles):- Speed is consistently 25 mph.- Travel time = Distance / Speed = 5 miles / 25 mph = 0.2 hours = 12 minutesSegment 2 (10 miles):- Worst-case speed = 20 mph- Travel time = Distance / Speed = 10 miles / 20 mph = 0.5 hours = 30 minutesTotal worst-case travel time for Route B:- 12 minutes + 30 minutes = 42 minutes Latest Possible Departure Time Calculation:To ensure arrival at 8:00 AM, we need to subtract the worst-case travel times from 8:00 AM.For Route A:- Departure time = 8:00 AM - 39 minutes = 7:21 AMFor Route B:- Departure time = 8:00 AM - 42 minutes = 7:18 AM Final Answer:The final answer is:1. Expected travel time if Alex departs at 7:00 AM: - Route A: 30.2 minutes - Route B: 36 minutes2. Latest possible departure time to arrive at 8:00 AM: - Route A: 7:21 AM - Route B: 7:18 AMI hope it is correct.
question:The nostalgic descendant of prominent families in Fauquier County, Virginia, is interested in the historical land division of their ancestral estate. The estate, originally spanning 500 acres, was divided among three siblings in a ratio reflecting their birth order: 1:2:3. Over the centuries, the land has experienced various transactions, and the descendant wants to understand the current value and configuration of the estate.Sub-problem 1:The estate was initially divided such that the first sibling received 1 part, the second sibling received 2 parts, and the third sibling received 3 parts of the total 500 acres. Given that the land value has appreciated exponentially according to the function ( V(t) = 10,000 cdot e^{0.05t} ) dollars per acre, where ( t ) is the number of years since the initial division (considering the division occurred 200 years ago), calculate the current value of the land each sibling originally received.Sub-problem 2:Over time, portions of the land were sold and merged. The first sibling's descendants sold 50 acres, the second sibling's descendants sold 100 acres, and the third sibling's descendants sold 150 acres. The remaining land is now owned by the nostalgic descendant. Determine the total current value of the land owned by the descendant based on the exponential land value function provided.
answer:Let's solve the problem step-by-step. Sub-problem 1: Initial Division and Current Value CalculationThe estate was divided among three siblings in a ratio of 1:2:3.1. Calculate the parts each sibling received: Total parts = 1 + 2 + 3 = 6 parts. - First sibling's share: (frac{1}{6} times 500 = frac{500}{6} = 83.overline{3} text{ acres}) - Second sibling's share: (frac{2}{6} times 500 = frac{1000}{6} = 166.overline{6} text{ acres}) - Third sibling's share: (frac{3}{6} times 500 = frac{1500}{6} = 250 text{ acres})2. Calculate the current value of the land each sibling received: The value function is given by ( V(t) = 10,000 cdot e^{0.05t} ) dollars per acre, where ( t = 200 ) years. Plugging ( t = 200 ) into the function: [ V(200) = 10,000 cdot e^{0.05 times 200} = 10,000 cdot e^{10} ] Using the approximate value ( e^{10} approx 22026.46579 ): [ V(200) approx 10,000 cdot 22026.46579 = 220,264,657.9 text{ dollars per acre} ] - First sibling's land value: [ 83.overline{3} text{ acres} times 220,264,657.9 text{ dollars per acre} = 18,355,438,158.67 text{ dollars} ] - Second sibling's land value: [ 166.overline{6} text{ acres} times 220,264,657.9 text{ dollars per acre} = 36,710,876,317.33 text{ dollars} ] - Third sibling's land value: [ 250 text{ acres} times 220,264,657.9 text{ dollars per acre} = 55,066,164,475 text{ dollars} ] Sub-problem 2: Current Value of the Remaining Land1. Calculate the remaining land after sales: - First sibling's descendants sold 50 acres: [ 83.overline{3} - 50 = 33.overline{3} text{ acres remaining} ] - Second sibling's descendants sold 100 acres: [ 166.overline{6} - 100 = 66.overline{6} text{ acres remaining} ] - Third sibling's descendants sold 150 acres: [ 250 - 150 = 100 text{ acres remaining} ]2. Calculate the total current value of the remaining land: - First sibling's remaining land value: [ 33.overline{3} text{ acres} times 220,264,657.9 text{ dollars per acre} = 7,342,175,263 text{ dollars} ] - Second sibling's remaining land value: [ 66.overline{6} text{ acres} times 220,264,657.9 text{ dollars per acre} = 14,684,350,526 text{ dollars} ] - Third sibling's remaining land value: [ 100 text{ acres} times 220,264,657.9 text{ dollars per acre} = 22,026,465,790 text{ dollars} ] - Total current value of the remaining land: [ 7,342,175,263 + 14,684,350,526 + 22,026,465,790 = 44,053,991,579 text{ dollars} ] Final Answer:The final answer is 44,053,991,579. I hope it is correct.
question:A business analyst is studying the relationship between the cement industry and global macroeconomic trends. They have collected data on the annual cement production (in millions of tons) ( C(t) ) and the global GDP (in trillion USD) ( G(t) ) over the past 20 years, where ( t ) represents time in years.1. Given the following differential equations that describe the interaction between cement production and global GDP: [ frac{dC(t)}{dt} = k_1 C(t) left(1 - frac{C(t)}{K}right) + k_2 G(t) ] [ frac{dG(t)}{dt} = k_3 G(t) left(1 - frac{G(t)}{M}right) + k_4 C(t) ] where ( k_1, k_2, k_3, k_4, K, ) and ( M ) are positive constants, find the steady-state solutions ( C^* ) and ( G^* ) for cement production and global GDP, respectively.2. Assume the constants have the following values: ( k_1 = 0.02 ), ( k_2 = 0.01 ), ( k_3 = 0.03 ), ( k_4 = 0.005 ), ( K = 500 ) million tons, and ( M = 100 ) trillion USD. Using these values, determine if the steady-state solutions found in sub-problem 1 are stable or unstable by analyzing the Jacobian matrix at the steady-state points.
answer:Let's solve the problem step-by-step. Step 1: Finding the Steady-State SolutionsThe steady-state solutions ( C^* ) and ( G^* ) occur when the time derivatives ( frac{dC(t)}{dt} ) and ( frac{dG(t)}{dt} ) are both zero. This gives us the following system of equations:[frac{dC(t)}{dt} = k_1 C(t) left(1 - frac{C(t)}{K}right) + k_2 G(t) = 0][frac{dG(t)}{dt} = k_3 G(t) left(1 - frac{G(t)}{M}right) + k_4 C(t) = 0]We will solve this system of equations for ( C^* ) and ( G^* ).# Equation 1:[k_1 C left(1 - frac{C}{K}right) + k_2 G = 0]Rearranging terms,[k_1 C - frac{k_1 C^2}{K} + k_2 G = 0]# Equation 2:[k_3 G left(1 - frac{G}{M}right) + k_4 C = 0]Rearranging terms,[k_3 G - frac{k_3 G^2}{M} + k_4 C = 0]We have two equations:1. ( k_1 C - frac{k_1 C^2}{K} + k_2 G = 0 )2. ( k_3 G - frac{k_3 G^2}{M} + k_4 C = 0 )Let's solve these equations simultaneously. Step 2: Solving for Steady-State ValuesFirst, solve for ( C ) in terms of ( G ) from Equation 1:[k_1 C - frac{k_1 C^2}{K} + k_2 G = 0][k_1 C left(1 - frac{C}{K}right) = -k_2 G][C left(1 - frac{C}{K}right) = -frac{k_2 G}{k_1}][C - frac{C^2}{K} = -frac{k_2 G}{k_1}]This is a quadratic equation in ( C ):[C^2 - KC + frac{K k_2 G}{k_1} = 0]Solving this quadratic equation using the quadratic formula ( C = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[C = frac{K pm sqrt{K^2 - 4 cdot 1 cdot frac{K k_2 G}{k_1}}}{2}][C = frac{K pm sqrt{K^2 - frac{4 K k_2 G}{k_1}}}{2}][C = frac{K pm sqrt{K(K - frac{4 k_2 G}{k_1})}}{2}]For the steady-state, we need to consider the biologically plausible solution (i.e., positive values).Next, solve for ( G ) in terms of ( C ) from Equation 2:[k_3 G - frac{k_3 G^2}{M} + k_4 C = 0][G left(1 - frac{G}{M}right) = -frac{k_4 C}{k_3}][G - frac{G^2}{M} = -frac{k_4 C}{k_3}]This is a quadratic equation in ( G ):[G^2 - MG + frac{M k_4 C}{k_3} = 0]Solving this quadratic equation using the quadratic formula ( G = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[G = frac{M pm sqrt{M^2 - 4 cdot 1 cdot frac{M k_4 C}{k_3}}}{2}][G = frac{M pm sqrt{M(M - frac{4 k_4 C}{k_3})}}{2}]For the steady-state, we need to consider the biologically plausible solution (i.e., positive values). Step 3: Stability Analysis Using Jacobian MatrixThe Jacobian matrix ( J ) of the system is given by the partial derivatives of the right-hand side of the differential equations with respect to ( C ) and ( G ):[J = begin{pmatrix}frac{partial}{partial C} left( k_1 C left(1 - frac{C}{K}right) + k_2 G right) & frac{partial}{partial G} left( k_1 C left(1 - frac{C}{K}right) + k_2 G right) frac{partial}{partial C} left( k_3 G left(1 - frac{G}{M}right) + k_4 C right) & frac{partial}{partial G} left( k_3 G left(1 - frac{G}{M}right) + k_4 C right)end{pmatrix}]Calculate the partial derivatives:[frac{partial}{partial C} left( k_1 C left(1 - frac{C}{K}right) + k_2 G right) = k_1 left(1 - frac{2C}{K}right)][frac{partial}{partial G} left( k_1 C left(1 - frac{C}{K}right) + k_2 G right) = k_2][frac{partial}{partial C} left( k_3 G left(1 - frac{G}{M}right) + k_4 C right) = k_4][frac{partial}{partial G} left( k_3 G left(1 - frac{G}{M}right) + k_4 C right) = k_3 left(1 - frac{2G}{M}right)]Thus, the Jacobian matrix ( J ) at the steady-state ((C^*, G^*)) is:[J = begin{pmatrix}k_1 left(1 - frac{2C^*}{K}right) & k_2 k_4 & k_3 left(1 - frac{2G^*}{M}right)end{pmatrix}] Stability AnalysisThe stability of the steady state can be determined by the eigenvalues of the Jacobian matrix ( J ). The eigenvalues ( lambda ) are found by solving the characteristic equation:[text{det}(J - lambda I) = 0][begin{vmatrix}k_1 left(1 - frac{2C^*}{K}right) - lambda & k_2 k_4 & k_3 left(1 - frac{2G^*}{M}right) - lambdaend{vmatrix} = 0][left( k_1 left(1 - frac{2C^*}{K}right) - lambda right) left( k_3 left(1 - frac{2G^*}{M}right) - lambda right) - k_2 k_4 = 0]The eigenvalues ( lambda ) are the roots of the characteristic polynomial. The nature of the eigenvalues (real or complex, positive or negative) will determine the stability of the steady state.To determine specific eigenvalues, we would need the values of ( C^* ) and ( G^* ). For now, let's assume ( C^* ) and ( G^* ) are such that the eigenvalues are real and negative, indicating a stable steady state.Final Answer: The final answer is that the steady-state solutions are ( C^* ) and ( G^* ), and their stability depends on the eigenvalues of the Jacobian matrix evaluated at ( (C^*, G^*) ). Given the constants, further analysis would be required to determine the specific stability. I hope it is correct.
question:The owner of a small convenience store along the trunk road, known for their friendly smile and local recommendations, has noticed that the number of customers entering the store throughout the day follows a Poisson distribution. The average number of customers per hour is 10. 1. What is the probability that exactly 15 customers will enter the store in a given hour?Additionally, the store owner is planning to optimize their inventory based on customer purchase patterns. They have observed that on average, 60% of the customers buy a particular popular snack. If 15 customers enter the store in a given hour:2. What is the probability that at least 10 customers will buy the popular snack during that hour?These calculations will help the owner make better recommendations and improve the store's service.
answer:To solve the given problem, let's break it down into two parts: Part 1: Poisson DistributionThe number of customers entering the store follows a Poisson distribution with an average rate (λ) of 10 customers per hour.We need to find the probability that exactly 15 customers will enter the store in a given hour. The probability mass function of a Poisson distribution is given by:[ P(X = k) = frac{e^{-lambda} lambda^k}{k!} ]where:- ( lambda = 10 )- ( k = 15 )- ( e ) is the base of the natural logarithm (approximately equal to 2.71828)Let's plug in the values:[ P(X = 15) = frac{e^{-10} cdot 10^{15}}{15!} ]First, calculate ( e^{-10} ):[ e^{-10} approx 4.53999 times 10^{-5} ]Next, calculate ( 10^{15} ):[ 10^{15} = 1,000,000,000,000,000 ]Now, calculate ( 15! ) (15 factorial):[ 15! = 1 times 2 times 3 times ldots times 15 = 1,307,674,368,000 ]Now, combine these results:[ P(X = 15) = frac{(4.53999 times 10^{-5}) times 1,000,000,000,000,000}{1,307,674,368,000} ]Calculate the numerator:[ (4.53999 times 10^{-5}) times 1,000,000,000,000,000 = 4.53999 times 10^{10} ]Now, divide by the denominator:[ P(X = 15) = frac{4.53999 times 10^{10}}{1,307,674,368,000} approx 0.03472 ]So, the probability that exactly 15 customers will enter the store in a given hour is approximately 0.03472. Part 2: Binomial DistributionIf 15 customers enter the store in a given hour, we need to find the probability that at least 10 customers will buy the popular snack. Let ( Y ) be the number of customers who buy the popular snack. ( Y ) follows a Binomial distribution with parameters ( n = 15 ) (number of trials) and ( p = 0.6 ) (probability of success in each trial).We need to find ( P(Y geq 10) ).The probability mass function of a Binomial distribution is given by:[ P(Y = k) = binom{n}{k} p^k (1-p)^{n-k} ]where:- ( n = 15 )- ( p = 0.6 )- ( k ) is the number of successesWe need to sum the probabilities for ( k ) from 10 to 15:[ P(Y geq 10) = sum_{k=10}^{15} binom{15}{k} (0.6)^k (0.4)^{15-k} ]Calculate each term using:[ binom{n}{k} = frac{n!}{k!(n-k)!} ]Let's calculate each probability:[ P(Y = 10) = binom{15}{10} cdot (0.6)^{10} cdot (0.4)^5 = 3003 cdot 0.0060466176 cdot 0.01024 approx 0.1852 ][ P(Y = 11) = binom{15}{11} cdot (0.6)^{11} cdot (0.4)^4 = 1365 cdot 0.00362797056 cdot 0.0256 approx 0.1263 ][ P(Y = 12) = binom{15}{12} cdot (0.6)^{12} cdot (0.4)^3 = 455 cdot 0.002176782336 cdot 0.064 approx 0.0632 ][ P(Y = 13) = binom{15}{13} cdot (0.6)^{13} cdot (0.4)^2 = 105 cdot 0.0013060694016 cdot 0.16 approx 0.0219 ][ P(Y = 14) = binom{15}{14} cdot (0.6)^{14} cdot (0.4)^1 = 15 cdot 0.00078364164096 cdot 0.4 approx 0.0047 ][ P(Y = 15) = binom{15}{15} cdot (0.6)^{15} cdot (0.4)^0 = 1 cdot 0.000470184984576 cdot 1 approx 0.0005 ]Now, sum these probabilities:[ P(Y geq 10) approx 0.1852 + 0.1263 + 0.0632 + 0.0219 + 0.0047 + 0.0005 approx 0.4018 ]Final Answer: The final answer is 0.03472 for part 1 and 0.4018 for part 2. I hope it is correct.
question:A loyal fan of podcasts, Alex, decided to start their own show to contribute to the podcasting community. Alex plans to release a weekly episode and wants to optimize the production time while maintaining a high quality of content. To measure the quality of each podcast episode, Alex uses a quadratic function ( Q(t) = -2t^2 + 12t + 30 ) where ( t ) is the time in hours spent on production, and ( Q(t) ) is the quality score of the episode.1. Determine the optimal time ( t ) that Alex should spend on production to maximize the quality score of an episode. What is the maximum quality score ( Q(t) )?2. If Alex plans to spend a certain fixed amount of time each week on production, ( t_w ), and wants to maintain an average quality score of at least 70 over 10 weeks, what is the minimum value of ( t_w )? Use calculus to justify your answer.
answer:Let's solve the given problem step-by-step. Part 1: Optimal Production Time to Maximize Quality Score1. Determine the optimal time ( t ) to maximize the quality score ( Q(t) )The quality score ( Q(t) ) is given by the quadratic function: [ Q(t) = -2t^2 + 12t + 30 ]To find the time ( t ) that maximizes ( Q(t) ), we need to find the vertex of the parabola. Since the coefficient of ( t^2 ) is negative, the parabola opens downward, and the vertex represents the maximum point.The vertex form of a quadratic function ( at^2 + bt + c ) has its vertex at ( t = frac{-b}{2a} ).Here, ( a = -2 ), ( b = 12 ), and ( c = 30 ).So, the optimal time ( t ) is:[ t = frac{-b}{2a} = frac{-12}{2(-2)} = frac{-12}{-4} = 3 ]2. Calculate the maximum quality score ( Q(3) )Now we substitute ( t = 3 ) into the original function ( Q(t) ):[ Q(3) = -2(3)^2 + 12(3) + 30 ][ Q(3) = -2(9) + 36 + 30 ][ Q(3) = -18 + 36 + 30 ][ Q(3) = 48 ]So, the maximum quality score is 48 when Alex spends 3 hours on production. Part 2: Minimum Production Time to Maintain an Average Quality Score of at least 70 Over 10 Weeks3. Determine the minimum value of ( t_w )We need to find the minimum ( t_w ) such that the average quality score over 10 weeks is at least 70. This means each week's quality score ( Q(t_w) ) should be at least 70.Set up the inequality:[ Q(t_w) geq 70 ]Substitute the quadratic function:[ -2t_w^2 + 12t_w + 30 geq 70 ]Solve for ( t_w ):[ -2t_w^2 + 12t_w + 30 - 70 geq 0 ][ -2t_w^2 + 12t_w - 40 geq 0 ][ 2t_w^2 - 12t_w + 40 leq 0 ]Divide the inequality by 2:[ t_w^2 - 6t_w + 20 leq 0 ]To solve the quadratic inequality, find the roots of the equality ( t_w^2 - 6t_w + 20 = 0 ).Calculate the discriminant:[ Delta = b^2 - 4ac = (-6)^2 - 4(1)(20) = 36 - 80 = -44 ]Since the discriminant is negative, the quadratic equation has no real roots. This means the quadratic expression ( t_w^2 - 6t_w + 20 ) is always positive because the parabola opens upwards (coefficient of ( t_w^2 ) is positive).Given this, there is no real value of ( t_w ) that satisfies the inequality ( t_w^2 - 6t_w + 20 leq 0 ). Therefore, it is impossible for Alex to maintain an average quality score of 70 with any fixed production time per week.Final Answer: The final answer is t_w text{ cannot be determined because the quadratic inequality has no real solutions}. I hope it is correct.