🧠:<think>Okay, so I have this problem about Jane's performance efficiency, E(t), which is modeled by a differential equation. Dr. Anderson is helping her manage stress, and the equation given is dE/dt = -kE + S(t). I need to solve two parts: first, find E(t) given the initial condition E(0) = E0, and second, figure out the optimal S(t) to keep E(t) constant at Ec.Starting with part 1. The differential equation is linear, right? It looks like a first-order linear ordinary differential equation. The standard form is dE/dt + P(t)E = Q(t). In this case, P(t) would be k, since it's -kE, so moving the -kE to the left side gives dE/dt + kE = S(t). So, yes, it's linear.To solve this, I can use an integrating factor. The integrating factor, μ(t), is e^(∫P(t)dt). Here, P(t) is k, so μ(t) = e^(∫k dt) = e^(kt). Multiplying both sides of the equation by μ(t):e^(kt) dE/dt + k e^(kt) E = S(t) e^(kt)The left side is the derivative of (E(t) e^(kt)) with respect to t. So, integrating both sides from 0 to t:∫₀ᵗ d/dτ [E(τ) e^(kτ)] dτ = ∫₀ᵗ S(τ) e^(kτ) dτThis simplifies to:E(t) e^(kt) - E(0) = ∫₀ᵗ S(τ) e^(kτ) dτWe know E(0) is E0, so:E(t) e^(kt) = E0 + ∫₀ᵗ S(τ) e^(kτ) dτTherefore, E(t) = e^(-kt) [E0 + ∫₀ᵗ S(τ) e^(kτ) dτ]Now, S(t) is given as S0 e^(-α t). So substituting that into the integral:E(t) = e^(-kt) [E0 + ∫₀ᵗ S0 e^(-α τ) e^(k τ) dτ]Simplify the exponent in the integral: e^(-α τ + k τ) = e^{(k - α) τ}So the integral becomes S0 ∫₀ᵗ e^{(k - α) τ} dτIf k ≠ α, the integral is S0 [ (e^{(k - α) τ} ) / (k - α) ) ] from 0 to t.Which is S0 [ (e^{(k - α) t} - 1) / (k - α) ]So putting it all together:E(t) = e^(-kt) [ E0 + S0 (e^{(k - α) t} - 1)/(k - α) ]Simplify that expression:Let me factor out e^(-kt):E(t) = E0 e^(-kt) + S0 e^(-kt) (e^{(k - α) t} - 1)/(k - α)Simplify the second term:e^(-kt) * e^{(k - α) t} = e^{-α t}So the second term becomes S0 (e^{-α t} - e^{-kt}) / (k - α)Therefore, E(t) = E0 e^(-kt) + S0 (e^{-α t} - e^{-kt}) / (k - α)Alternatively, we can write it as:E(t) = E0 e^(-kt) + [S0 / (k - α)] (e^{-α t} - e^{-kt})That should be the solution for part 1.Wait, let me check if k = α. If k = α, then the integral becomes S0 ∫₀ᵗ e^{0 τ} dτ = S0 t. So in that case, E(t) = e^(-kt) [E0 + S0 t]. So we need to consider that separately. But since the problem says S(t) = S0 e^{-α t}, and S0 and α are positive constants, but doesn't specify if α equals k. So perhaps in the general case, we can leave it as is, noting that if k = α, the solution is different.But since the problem doesn't specify, maybe we can just present the solution assuming k ≠ α, and mention the case when k = α separately.But for now, let's proceed with k ≠ α.So, summarizing part 1, the solution is:E(t) = E0 e^{-kt} + [S0 / (k - α)] (e^{-α t} - e^{-kt})Alright, moving on to part 2. Dr. Anderson wants to optimize S(t) so that Jane's performance efficiency is sustained at a constant level Ec over time. So, E(t) = Ec for all t.If E(t) is constant, then dE/dt = 0. So plugging into the differential equation:0 = -k Ec + S(t)Therefore, S(t) = k EcSo, S(t) must be a constant function equal to k Ec. But in the problem, S(t) is given as S0 e^{-α t}. So to have S(t) constant, we need S0 e^{-α t} = k Ec for all t.But S0 e^{-α t} is only constant if α = 0, because otherwise, it's an exponential decay. So if α = 0, then S(t) = S0, which is a constant. Therefore, to have S(t) constant, set α = 0 and S0 = k Ec.Wait, but in the problem statement, S(t) is given as S0 e^{-α t}, with S0 and α positive constants. So if we set α = 0, S(t) becomes S0, which is a constant. So, to have S(t) = k Ec, we must have S0 = k Ec and α = 0.But wait, α is a positive constant. If we set α = 0, is that allowed? The problem says α is a positive constant, so α = 0 is not positive. Hmm, that's a problem.Alternatively, maybe we can think of S(t) as a function that needs to be adjusted so that E(t) remains constant. So, maybe S(t) is not necessarily of the form S0 e^{-α t}, but in part 2, we are to find the optimal S(t) that makes E(t) constant.Wait, let me reread the problem."Suppose Dr. Anderson wants to optimize S(t) to sustain Jane's performance efficiency at a constant level Ec over time. Find the optimal function S(t) and the corresponding constants S0 and α that achieve this goal."Hmm, so the question is, given that S(t) is of the form S0 e^{-α t}, find S0 and α such that E(t) = Ec for all t.But from part 1, we have E(t) expressed in terms of S(t). So if E(t) is to be constant, then dE/dt = 0, which gives S(t) = k Ec. But S(t) is given as S0 e^{-α t}, so setting S0 e^{-α t} = k Ec for all t.But the only way this can be true for all t is if α = 0 and S0 = k Ec. However, α is supposed to be a positive constant, so α = 0 is not allowed.Wait, that suggests that it's impossible to have E(t) constant if S(t) is of the form S0 e^{-α t} with α > 0. Because S(t) would decay exponentially, while to have E(t) constant, S(t) must be a constant function. So unless α = 0, which is not allowed, it's not possible.But the problem says "find the optimal function S(t)" and "the corresponding constants S0 and α". So maybe we need to reinterpret.Alternatively, perhaps the question is to find S(t) such that E(t) approaches a constant as t approaches infinity, rather than being exactly constant for all t.But the problem says "sustain Jane's performance efficiency at a constant level Ec over time", which sounds like E(t) should be Ec for all t, not just asymptotically.Hmm, maybe I need to think differently. If E(t) is to be constant, then from the differential equation, S(t) must equal k Ec. So S(t) must be a constant function equal to k Ec. But since S(t) is given as S0 e^{-α t}, the only way for S(t) to be constant is if α = 0, but α must be positive. So perhaps the problem is to find S(t) such that E(t) approaches Ec as t approaches infinity.Wait, let me check. If we set S(t) = k Ec, then E(t) would be Ec for all t, but S(t) must be S0 e^{-α t}. So unless S0 e^{-α t} is constant, which requires α = 0, which is not allowed, it's impossible. So maybe the problem is misinterpreted.Alternatively, maybe the question is to find S(t) such that E(t) tends to Ec as t approaches infinity. So, in that case, we can have S(t) = k Ec e^{-α t}, but then as t approaches infinity, S(t) approaches zero, so E(t) would approach E0 e^{-kt}, which would go to zero unless E0 is also adjusted.Wait, perhaps I need to think again.Wait, if we want E(t) to approach Ec as t approaches infinity, then we can set the steady-state solution. From part 1, the solution is E(t) = E0 e^{-kt} + [S0 / (k - α)] (e^{-α t} - e^{-kt})As t approaches infinity, e^{-kt} and e^{-α t} go to zero if k and α are positive. So E(t) approaches [S0 / (k - α)] (0 - 0) = 0. So that's not helpful.Wait, unless k = α, in which case the solution is E(t) = E0 e^{-kt} + S0 t e^{-kt}As t approaches infinity, if k > 0, e^{-kt} goes to zero, so E(t) approaches zero. So that's not helpful either.Wait, maybe I need to have S(t) such that E(t) is Ec for all t. So, from the differential equation, 0 = -k Ec + S(t). So S(t) must be equal to k Ec for all t. So if S(t) is a constant function, then S(t) = k Ec. But in the problem, S(t) is given as S0 e^{-α t}, which is only constant if α = 0, which is not allowed. Therefore, it's impossible to have E(t) constant for all t with S(t) of the form S0 e^{-α t} where α > 0.Therefore, perhaps the problem is misstated, or I'm misinterpreting it. Alternatively, maybe the question is to find S(t) such that E(t) is constant for all t, which would require S(t) = k Ec, but since S(t) is given as S0 e^{-α t}, the only way is to set α = 0 and S0 = k Ec, but α must be positive. So perhaps the answer is that it's impossible, but the problem says "find the optimal function S(t)", so maybe I'm missing something.Wait, perhaps the question is to find S(t) such that E(t) is constant, regardless of the form. So, in that case, S(t) must be k Ec, which is a constant function. So the optimal S(t) is a constant function S(t) = k Ec, which would correspond to S0 = k Ec and α = 0. But since α must be positive, maybe we can't do that. Alternatively, perhaps the problem allows α = 0, even though it's stated as positive. Maybe it's a typo.Alternatively, perhaps the question is to find S(t) such that E(t) approaches Ec as t approaches infinity. In that case, we can set the steady-state solution. From the differential equation, as t approaches infinity, dE/dt approaches zero, so -k Ec + S(t) approaches zero. Therefore, S(t) must approach k Ec as t approaches infinity. But S(t) is S0 e^{-α t}, which approaches zero as t approaches infinity, unless S0 = 0, which would make S(t) zero, but then E(t) would decay to zero. So that's not helpful.Wait, perhaps I need to think differently. Maybe the question is to have E(t) = Ec for all t, which requires S(t) = k Ec, but since S(t) is given as S0 e^{-α t}, we can't have that unless α = 0. So perhaps the answer is that it's impossible, but the problem says "find the optimal function S(t)", so maybe I'm missing something.Alternatively, perhaps the question is to have E(t) approach Ec as t approaches infinity, but as I saw earlier, that would require S(t) to approach k Ec, but S(t) approaches zero, so that's not possible unless S0 is infinite, which is not practical.Wait, maybe I need to reconsider the differential equation. If E(t) is to be constant, then dE/dt = 0, so S(t) must equal k E(t). So if E(t) is Ec, then S(t) must be k Ec. So S(t) must be a constant function. But since S(t) is given as S0 e^{-α t}, the only way for S(t) to be constant is if α = 0, which is not allowed. Therefore, it's impossible to have E(t) constant for all t with S(t) of the form S0 e^{-α t} where α > 0.Therefore, the optimal function S(t) would have to be a constant function, which would require α = 0, but since α must be positive, perhaps the problem is to find S(t) such that E(t) is as close as possible to Ec, but that's a different optimization problem.Alternatively, perhaps the question is to have E(t) approach Ec as t approaches infinity, but as I saw earlier, that's not possible because S(t) decays to zero, making E(t) decay to zero as well.Wait, maybe I made a mistake in the solution of part 1. Let me double-check.The differential equation is dE/dt = -k E + S(t), with S(t) = S0 e^{-α t}.The integrating factor is e^{∫k dt} = e^{kt}.Multiplying both sides:e^{kt} dE/dt + k e^{kt} E = S0 e^{(k - α) t}Integrate both sides:E(t) e^{kt} = E0 + ∫₀ᵗ S0 e^{(k - α) τ} dτIf k ≠ α, the integral is S0/(k - α) (e^{(k - α) t} - 1)So E(t) = e^{-kt} [E0 + S0/(k - α) (e^{(k - α) t} - 1)]Which simplifies to E0 e^{-kt} + S0/(k - α) (e^{-α t} - e^{-kt})If k = α, the integral becomes S0 t, so E(t) = e^{-kt} (E0 + S0 t)So that's correct.Now, for part 2, if we want E(t) = Ec for all t, then:Ec = E0 e^{-kt} + S0/(k - α) (e^{-α t} - e^{-kt})This must hold for all t. Let's rearrange:Ec = E0 e^{-kt} + S0/(k - α) e^{-α t} - S0/(k - α) e^{-kt}Group the e^{-kt} terms:Ec = [E0 - S0/(k - α)] e^{-kt} + S0/(k - α) e^{-α t}For this to hold for all t, the coefficients of e^{-kt} and e^{-α t} must be zero, and the constant term must be Ec.But wait, e^{-kt} and e^{-α t} are not constants unless k = α, but if k = α, then the equation becomes:Ec = [E0 - S0/(0)] e^{-kt} + S0/(0) e^{-kt}Which is undefined because we have division by zero. So, if k ≠ α, the only way for the equation to hold for all t is if the coefficients of e^{-kt} and e^{-α t} are zero, and the constant term is Ec.But in our equation, there is no constant term; all terms are exponential functions. Therefore, the only way for Ec to be a constant is if the coefficients of e^{-kt} and e^{-α t} are zero, and the remaining term is Ec.But looking at the equation:Ec = [E0 - S0/(k - α)] e^{-kt} + S0/(k - α) e^{-α t}To have this equal to Ec for all t, the coefficients of e^{-kt} and e^{-α t} must be zero, and the remaining term must be Ec. But there is no constant term, so the only way is if:E0 - S0/(k - α) = 0andS0/(k - α) = 0But S0/(k - α) = 0 implies S0 = 0, which would make E(t) = E0 e^{-kt}, which is not constant unless k = 0, which is not the case.Therefore, it's impossible to have E(t) = Ec for all t with S(t) = S0 e^{-α t} where α > 0.Therefore, the conclusion is that it's impossible to sustain E(t) at a constant level Ec using S(t) of the form S0 e^{-α t} with α > 0. The only way to have E(t) constant is if S(t) is a constant function, which would require α = 0, but α must be positive.Therefore, perhaps the problem is misstated, or I'm misinterpreting it. Alternatively, maybe the question is to find S(t) such that E(t) approaches Ec as t approaches infinity, but as I saw earlier, that's not possible because S(t) decays to zero, making E(t) decay to zero as well.Alternatively, perhaps the question is to find S(t) such that E(t) is constant for all t, which would require S(t) = k Ec, but since S(t) is given as S0 e^{-α t}, the only way is to set α = 0 and S0 = k Ec. So, even though α is supposed to be positive, perhaps the optimal function is S(t) = k Ec, which corresponds to S0 = k Ec and α = 0.But since α must be positive, maybe the answer is that it's impossible, but the problem says "find the optimal function S(t)", so perhaps the answer is that S(t) must be a constant function, which would require α = 0, but since α must be positive, it's impossible. Therefore, the optimal S(t) is a constant function, but it's not possible with the given form.Alternatively, perhaps the problem allows α = 0, even though it's stated as positive. In that case, the optimal S(t) is S(t) = k Ec, which corresponds to S0 = k Ec and α = 0.But since the problem states that α is a positive constant, I think the correct answer is that it's impossible to sustain E(t) at a constant level Ec with S(t) of the form S0 e^{-α t} where α > 0. Therefore, the optimal function S(t) would have to be a constant function, which is not possible with the given form.But the problem says "find the optimal function S(t)", so perhaps the answer is that S(t) must be a constant function, which would require α = 0, but since α must be positive, it's impossible. Therefore, the optimal S(t) is a constant function, but it's not possible with the given form.Alternatively, perhaps the problem is to find S(t) such that E(t) is constant for all t, which requires S(t) = k Ec, so the optimal function is S(t) = k Ec, which is a constant function, corresponding to S0 = k Ec and α = 0.But since α must be positive, maybe the answer is that it's impossible, but the problem says "find the optimal function S(t)", so perhaps the answer is S(t) = k Ec, which is a constant function, with S0 = k Ec and α = 0, even though α is supposed to be positive.Alternatively, perhaps the problem allows α = 0, so the answer is S(t) = k Ec, with S0 = k Ec and α = 0.But I'm not sure. Maybe I should proceed with that.So, in summary:1. The solution for E(t) is E(t) = E0 e^{-kt} + [S0 / (k - α)] (e^{-α t} - e^{-kt})2. To have E(t) constant at Ec, S(t) must be k Ec, which is a constant function. Therefore, S(t) = k Ec, which corresponds to S0 = k Ec and α = 0.But since α must be positive, perhaps the answer is that it's impossible, but the problem says "find the optimal function S(t)", so perhaps the answer is S(t) = k Ec, with S0 = k Ec and α = 0.Alternatively, perhaps the problem allows α = 0, so the answer is S(t) = k Ec, with S0 = k Ec and α = 0.I think that's the best I can do.</think>

🧠:<think>Okay, so I have this problem where a novice TV show reviewer is trying to figure out the relationship between the ratings given by three experienced reviewers and the overall average rating of each episode. There are two parts to this problem. Let me try to work through each part step by step.Starting with part 1: The novice wants to create a weighted average model using the ratings from Reviewer A, B, and C. The weights are ( w_A, w_B, w_C ) respectively, and they must sum up to 1. The weighted average rating ( R_i ) for each episode ( i ) is given by ( R_i = w_A cdot a_i + w_B cdot b_i + w_C cdot c_i ). The overall average rating for each episode ( i ) is known as ( r_i ). So, the goal is to set up a system of equations to find the weights ( w_A, w_B, w_C ).Hmm, okay. So, for each episode, the weighted average should equal the known overall average. Since there are 5 episodes, that means we have 5 equations. Each equation will correspond to one episode. Let me write that out.For episode 1:( w_A cdot a_1 + w_B cdot b_1 + w_C cdot c_1 = r_1 )For episode 2:( w_A cdot a_2 + w_B cdot b_2 + w_C cdot c_2 = r_2 )Similarly, for episodes 3, 4, and 5:( w_A cdot a_3 + w_B cdot b_3 + w_C cdot c_3 = r_3 )( w_A cdot a_4 + w_B cdot b_4 + w_C cdot c_4 = r_4 )( w_A cdot a_5 + w_B cdot b_5 + w_C cdot c_5 = r_5 )So, that's five equations. But we have only three unknowns: ( w_A, w_B, w_C ). Wait, so this is an overdetermined system because there are more equations than unknowns. That means we might not have an exact solution, but we can find a least squares solution or something like that. But the problem just asks to formulate the system of equations, not necessarily solve it. So, I think I just need to write these five equations.But let me make sure. The problem says, "formulate the system of equations that the novice reviewer needs to solve to find the weights ( w_A, w_B, ) and ( w_C )." So, yes, it's just setting up the equations. So, the system is as I wrote above.But wait, also, the sum of the weights must equal 1. So, that's another equation: ( w_A + w_B + w_C = 1 ). So, actually, the system has six equations? Wait, no. Wait, the five equations are from the five episodes, each equating the weighted average to the known average. Then, the sixth equation is the constraint on the weights. So, in total, we have six equations with three unknowns. That seems even more overdetermined. Hmm.But actually, no. Wait, the five equations are each linear equations in terms of ( w_A, w_B, w_C ). So, each equation is of the form ( a_i w_A + b_i w_B + c_i w_C = r_i ). So, that's five equations, and the sixth equation is ( w_A + w_B + w_C = 1 ). So, total six equations for three variables. That's a lot. So, the system is overdetermined, as I thought.But since the problem is just asking to formulate the system, I think that's all. So, the system is:1. ( a_1 w_A + b_1 w_B + c_1 w_C = r_1 )2. ( a_2 w_A + b_2 w_B + c_2 w_C = r_2 )3. ( a_3 w_A + b_3 w_B + c_3 w_C = r_3 )4. ( a_4 w_A + b_4 w_B + c_4 w_C = r_4 )5. ( a_5 w_A + b_5 w_B + c_5 w_C = r_5 )6. ( w_A + w_B + w_C = 1 )So, that's the system. I think that's part 1 done.Moving on to part 2: The novice wants to analyze the variance in ratings among the three reviewers for each episode. The variance ( D_i ) is given by ( D_i = frac{1}{3} left( (a_i - overline{r_i})^2 + (b_i - overline{r_i})^2 + (c_i - overline{r_i})^2 right) ), where ( overline{r_i} ) is the mean rating for episode ( i ). The task is to express ( D_i ) in terms of ( a_i, b_i, c_i ) and then calculate the average variance ( overline{D} ) over the 5 episodes.Alright, so first, express ( D_i ) in terms of ( a_i, b_i, c_i ). Currently, it's expressed in terms of ( overline{r_i} ). So, I need to substitute ( overline{r_i} ) with an expression involving ( a_i, b_i, c_i ).Wait, ( overline{r_i} ) is the mean rating for episode ( i ). But wait, is that the mean of the three reviewers? Or is it the overall average? Hmm, the problem says, "the overall average rating for each episode ( i ) is known to be ( r_i )". So, in part 1, ( r_i ) is the known average. But in part 2, ( overline{r_i} ) is the mean rating for episode ( i ). Wait, that might be confusing.Wait, in part 1, ( r_i ) is the overall average, which is known. So, in part 2, ( D_i ) is defined as ( frac{1}{3} left( (a_i - overline{r_i})^2 + (b_i - overline{r_i})^2 + (c_i - overline{r_i})^2 right) ). So, ( overline{r_i} ) is the mean of the three reviewers' ratings for episode ( i ). So, ( overline{r_i} = frac{a_i + b_i + c_i}{3} ).Therefore, ( D_i ) can be expressed in terms of ( a_i, b_i, c_i ) by substituting ( overline{r_i} ) with ( frac{a_i + b_i + c_i}{3} ).Let me write that out:( D_i = frac{1}{3} left( left(a_i - frac{a_i + b_i + c_i}{3}right)^2 + left(b_i - frac{a_i + b_i + c_i}{3}right)^2 + left(c_i - frac{a_i + b_i + c_i}{3}right)^2 right) )Simplify each term inside the parentheses.Let me compute ( a_i - frac{a_i + b_i + c_i}{3} ):That's equal to ( frac{3a_i - a_i - b_i - c_i}{3} = frac{2a_i - b_i - c_i}{3} )Similarly, ( b_i - frac{a_i + b_i + c_i}{3} = frac{2b_i - a_i - c_i}{3} )And ( c_i - frac{a_i + b_i + c_i}{3} = frac{2c_i - a_i - b_i}{3} )So, substituting back into ( D_i ):( D_i = frac{1}{3} left( left( frac{2a_i - b_i - c_i}{3} right)^2 + left( frac{2b_i - a_i - c_i}{3} right)^2 + left( frac{2c_i - a_i - b_i}{3} right)^2 right) )Let me compute each squared term:First term: ( left( frac{2a_i - b_i - c_i}{3} right)^2 = frac{(2a_i - b_i - c_i)^2}{9} )Second term: ( left( frac{2b_i - a_i - c_i}{3} right)^2 = frac{(2b_i - a_i - c_i)^2}{9} )Third term: ( left( frac{2c_i - a_i - b_i}{3} right)^2 = frac{(2c_i - a_i - b_i)^2}{9} )So, putting it all together:( D_i = frac{1}{3} left( frac{(2a_i - b_i - c_i)^2 + (2b_i - a_i - c_i)^2 + (2c_i - a_i - b_i)^2}{9} right) )Simplify the constants:( D_i = frac{1}{27} left( (2a_i - b_i - c_i)^2 + (2b_i - a_i - c_i)^2 + (2c_i - a_i - b_i)^2 right) )Alternatively, that can be written as:( D_i = frac{(2a_i - b_i - c_i)^2 + (2b_i - a_i - c_i)^2 + (2c_i - a_i - b_i)^2}{27} )Is there a way to simplify this expression further? Let me see.Alternatively, perhaps we can express ( D_i ) in terms of the sum of squares and cross products.Wait, another approach: Variance can also be expressed as ( frac{1}{n} sum (x_j - mu)^2 ), where ( mu ) is the mean. So, in this case, ( D_i ) is the variance of the three ratings ( a_i, b_i, c_i ). So, another formula for variance is ( frac{1}{n} sum x_j^2 - mu^2 ). So, maybe we can use that.Let me compute ( D_i ) using that formula.First, ( mu = frac{a_i + b_i + c_i}{3} )Then, ( D_i = frac{1}{3}(a_i^2 + b_i^2 + c_i^2) - mu^2 )Substituting ( mu ):( D_i = frac{1}{3}(a_i^2 + b_i^2 + c_i^2) - left( frac{a_i + b_i + c_i}{3} right)^2 )Let me compute that:First term: ( frac{1}{3}(a_i^2 + b_i^2 + c_i^2) )Second term: ( frac{(a_i + b_i + c_i)^2}{9} )So, ( D_i = frac{1}{3}(a_i^2 + b_i^2 + c_i^2) - frac{(a_i + b_i + c_i)^2}{9} )Let me combine these terms over a common denominator:Multiply the first term by 3/3 to get ( frac{3(a_i^2 + b_i^2 + c_i^2)}{9} )So, ( D_i = frac{3(a_i^2 + b_i^2 + c_i^2) - (a_i + b_i + c_i)^2}{9} )Now, expand ( (a_i + b_i + c_i)^2 ):( (a_i + b_i + c_i)^2 = a_i^2 + b_i^2 + c_i^2 + 2a_i b_i + 2a_i c_i + 2b_i c_i )So, substituting back:( D_i = frac{3(a_i^2 + b_i^2 + c_i^2) - (a_i^2 + b_i^2 + c_i^2 + 2a_i b_i + 2a_i c_i + 2b_i c_i)}{9} )Simplify numerator:( 3a_i^2 + 3b_i^2 + 3c_i^2 - a_i^2 - b_i^2 - c_i^2 - 2a_i b_i - 2a_i c_i - 2b_i c_i )Which simplifies to:( 2a_i^2 + 2b_i^2 + 2c_i^2 - 2a_i b_i - 2a_i c_i - 2b_i c_i )Factor out 2:( 2(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) )So, numerator is ( 2(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) )Therefore, ( D_i = frac{2(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i)}{9} )Which can be written as:( D_i = frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) )Alternatively, that's another expression for ( D_i ). So, this might be a simpler way to express it.So, either expression is correct. The first one I had was:( D_i = frac{(2a_i - b_i - c_i)^2 + (2b_i - a_i - c_i)^2 + (2c_i - a_i - b_i)^2}{27} )And the second one is:( D_i = frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) )Either is acceptable, but perhaps the second one is simpler.So, now, the problem asks to express ( D_i ) in terms of ( a_i, b_i, c_i ). So, either of these expressions is correct. I think the second one is more straightforward.So, ( D_i = frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) )Now, the next part is to calculate the average variance ( overline{D} ) over the 5 episodes. So, that would be the average of ( D_1, D_2, D_3, D_4, D_5 ).So, ( overline{D} = frac{1}{5} sum_{i=1}^{5} D_i )Substituting the expression for each ( D_i ):( overline{D} = frac{1}{5} sum_{i=1}^{5} frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) )We can factor out the constants:( overline{D} = frac{2}{45} sum_{i=1}^{5} (a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) )Alternatively, we can write this as:( overline{D} = frac{2}{45} left( sum_{i=1}^{5} a_i^2 + sum_{i=1}^{5} b_i^2 + sum_{i=1}^{5} c_i^2 - sum_{i=1}^{5} a_i b_i - sum_{i=1}^{5} a_i c_i - sum_{i=1}^{5} b_i c_i right) )So, that's the expression for the average variance.Wait, but is there a way to express this in terms of the overall averages or something else? Or is this sufficient? The problem just says to express ( D_i ) in terms of ( a_i, b_i, c_i ) and then calculate the average variance. So, I think this is sufficient.Alternatively, if we use the first expression for ( D_i ), we can write:( overline{D} = frac{1}{5} sum_{i=1}^{5} frac{(2a_i - b_i - c_i)^2 + (2b_i - a_i - c_i)^2 + (2c_i - a_i - b_i)^2}{27} )Which simplifies to:( overline{D} = frac{1}{135} sum_{i=1}^{5} left( (2a_i - b_i - c_i)^2 + (2b_i - a_i - c_i)^2 + (2c_i - a_i - b_i)^2 right) )But that seems more complicated. So, I think the second expression is better.So, to recap, ( D_i ) can be expressed as ( frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) ), and the average variance ( overline{D} ) is ( frac{2}{45} ) times the sum over all episodes of ( a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i ).So, that's part 2.Wait, let me just double-check my calculations for ( D_i ). I used the formula for variance as ( frac{1}{n} sum x_j^2 - mu^2 ). Let me confirm that.Yes, variance is ( E[X^2] - (E[X])^2 ). So, in this case, ( D_i = frac{1}{3}(a_i^2 + b_i^2 + c_i^2) - left( frac{a_i + b_i + c_i}{3} right)^2 ). That's correct.So, expanding that, we get:( frac{1}{3}(a_i^2 + b_i^2 + c_i^2) - frac{(a_i + b_i + c_i)^2}{9} )Which simplifies to:( frac{3(a_i^2 + b_i^2 + c_i^2) - (a_i^2 + b_i^2 + c_i^2 + 2a_i b_i + 2a_i c_i + 2b_i c_i)}{9} )Which is:( frac{2(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i)}{9} )Yes, that's correct.So, I think I did that correctly.Therefore, the expressions are correct.So, summarizing:1. The system of equations is five equations from the weighted average equaling the known average for each episode, plus the constraint that the weights sum to 1.2. The variance ( D_i ) is expressed as ( frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i) ), and the average variance is the average of these over the five episodes.I think that's all. I don't see any mistakes in my reasoning, so I think I'm done.Final Answer1. The system of equations is:[begin{cases}a_1 w_A + b_1 w_B + c_1 w_C = r_1 a_2 w_A + b_2 w_B + c_2 w_C = r_2 a_3 w_A + b_3 w_B + c_3 w_C = r_3 a_4 w_A + b_4 w_B + c_4 w_C = r_4 a_5 w_A + b_5 w_B + c_5 w_C = r_5 w_A + w_B + w_C = 1end{cases}]So, the final answer is boxed{begin{cases} a_1 w_A + b_1 w_B + c_1 w_C = r_1 a_2 w_A + b_2 w_B + c_2 w_C = r_2 a_3 w_A + b_3 w_B + c_3 w_C = r_3 a_4 w_A + b_4 w_B + c_4 w_C = r_4 a_5 w_A + b_5 w_B + c_5 w_C = r_5 w_A + w_B + w_C = 1 end{cases}}.2. The variance ( D_i ) is expressed as:[D_i = frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i)]The average variance ( overline{D} ) over the 5 episodes is:[overline{D} = frac{2}{45} sum_{i=1}^{5} (a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i)]So, the final answers are:[D_i = boxed{frac{2}{9}(a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i)}]and[overline{D} = boxed{frac{2}{45} sum_{i=1}^{5} (a_i^2 + b_i^2 + c_i^2 - a_i b_i - a_i c_i - b_i c_i)}]</think>

🧠:<think>Okay, so I have these two probability questions to solve about a customer's spending at Quickmart. Let me try to figure them out step by step.Starting with the first question: The customer visits Quickmart 3 times a week, and each visit's expenditure is normally distributed with a mean of 1200 KES and a variance of 300 squared, which is 90,000. I need to find the probability that the customer spends more than 1,500 KES in at least one of the visits in a given week.Hmm, okay. So each visit is independent, right? So the customer makes three visits, each with the same distribution. I need the probability that at least one of these three visits exceeds 1,500 KES.I remember that for such problems, it's often easier to calculate the complement probability and then subtract it from 1. The complement here would be the probability that all three visits are less than or equal to 1,500 KES.So, let me denote X as the expenditure on a single visit. Then, X ~ N(1200, 300²). I need P(X > 1500). But since I want the probability that all three are <= 1500, it's [P(X <= 1500)]^3. Then, subtract that from 1 to get the desired probability.First, let me compute P(X > 1500). To do that, I can standardize X. The Z-score is (1500 - 1200)/300 = 300/300 = 1. So, Z = 1.Looking up the standard normal distribution table, P(Z <= 1) is approximately 0.8413. Therefore, P(X <= 1500) is 0.8413, so P(X > 1500) is 1 - 0.8413 = 0.1587.But wait, no. Wait, actually, for the first part, I need P(at least one visit > 1500). So, as I thought earlier, it's 1 - P(all visits <= 1500). Since each visit is independent, P(all <= 1500) is [P(X <= 1500)]^3.So, P(X <= 1500) is 0.8413, so [0.8413]^3 is approximately... let me calculate that. 0.8413 * 0.8413 is about 0.7079, then multiplied by 0.8413 again is approximately 0.7079 * 0.8413 ≈ 0.596.Therefore, the probability that all three visits are <= 1500 is approximately 0.596. So, the probability that at least one visit exceeds 1500 is 1 - 0.596 = 0.404, or 40.4%.Wait, let me double-check the calculations because sometimes I make arithmetic errors.First, Z = (1500 - 1200)/300 = 1. So, P(X <= 1500) is indeed 0.8413. Then, [0.8413]^3. Let me compute it more accurately.0.8413 * 0.8413: 0.8 * 0.8 = 0.64, 0.8 * 0.0413 = ~0.033, 0.0413 * 0.8 = ~0.033, and 0.0413 * 0.0413 ≈ 0.0017. Adding these up: 0.64 + 0.033 + 0.033 + 0.0017 ≈ 0.7077. So, approximately 0.7077.Then, 0.7077 * 0.8413: Let's compute 0.7 * 0.8 = 0.56, 0.7 * 0.0413 ≈ 0.0289, 0.0077 * 0.8 ≈ 0.00616, and 0.0077 * 0.0413 ≈ 0.000317. Adding these: 0.56 + 0.0289 + 0.00616 + 0.000317 ≈ 0.595377. So, approximately 0.5954.Thus, 1 - 0.5954 ≈ 0.4046, so about 40.46%. So, approximately 40.5% chance.Okay, that seems reasonable.Moving on to the second question: The customer has a budget constraint such that the total expenditure over any 4-week period should not exceed 60,000 KES. Each visit is still normally distributed with mean 1200 and variance 300². I need to find the probability that the customer will exceed their budget in a 4-week period.First, let me figure out how many visits the customer makes in 4 weeks. Since they visit 3 times a week, over 4 weeks, that's 3 * 4 = 12 visits.So, the total expenditure over 4 weeks is the sum of 12 independent normal variables, each with mean 1200 and variance 90,000.I know that the sum of independent normal variables is also normal. So, the total expenditure, let's denote it as S, will be N(12*1200, 12*90,000).Calculating the mean: 12 * 1200 = 14,400 KES.Calculating the variance: 12 * 90,000 = 1,080,000. Therefore, the standard deviation is sqrt(1,080,000). Let me compute that.sqrt(1,080,000) = sqrt(1,080 * 1000) = sqrt(1,080) * sqrt(1000). sqrt(1000) is approximately 31.6227766. sqrt(1,080): Let's see, 32^2 = 1024, 33^2=1089. So sqrt(1080) is between 32 and 33. 1080 - 1024 = 56, so approximately 32 + 56/(2*32) = 32 + 56/64 = 32 + 0.875 = 32.875. So, sqrt(1080) ≈ 32.875.Therefore, sqrt(1,080,000) ≈ 32.875 * 31.6227766 ≈ Let's compute that.32 * 31.6227766 ≈ 1011.9288480.875 * 31.6227766 ≈ 27.643366Adding them together: 1011.928848 + 27.643366 ≈ 1039.572214. So, approximately 1039.57 KES.So, the total expenditure S ~ N(14,400, (1039.57)^2).The customer's budget is 60,000 KES. So, we need to find P(S > 60,000).Wait, hold on. Wait, 12 visits, each averaging 1200, so total expected expenditure is 14,400. But the budget is 60,000? That seems way higher than the expected total.Wait, is that correct? Wait, 60,000 KES over 4 weeks, with 3 visits per week, so 12 visits. So, 60,000 / 12 = 5,000 KES per visit on average. But the mean per visit is 1,200, so 5,000 is way higher.Wait, that seems like a very high budget. So, the probability that the total expenditure exceeds 60,000 KES is going to be extremely low, practically zero.But let me verify. Maybe I misread the problem. Wait, the customer has a budget constraint such that the total expenditure over any 4-week period should not exceed 60,000 KES. So, 60,000 is the maximum they want to spend. So, we need P(S > 60,000). But since the mean is 14,400, which is much lower than 60,000, the probability should be almost zero.But maybe I made a mistake in the total number of visits. Let me check.Wait, the customer visits 3 times a week, over 4 weeks, so 12 visits. Each visit is 1,200 on average, so total is 14,400. So, 60,000 is way higher. So, the probability of exceeding 60,000 is practically zero.But maybe the question is the other way around? Maybe the customer wants the total expenditure not to exceed 60,000, but given that they spend on average 1,200 per visit, 12 visits would be 14,400, so 60,000 is much higher. So, the probability of exceeding 60,000 is almost zero.But perhaps I misread the problem. Let me read again."the total expenditure over any 4-week period should not exceed KES 60,000."So, the customer wants to make sure that in any 4-week period, they don't spend more than 60,000. So, we need the probability that their total spending exceeds 60,000.But as the mean is 14,400, 60,000 is way above. So, the probability is almost zero.But maybe the question is actually about not exceeding 60,000, so P(S <= 60,000). But the wording says "exceed their budget", so P(S > 60,000). So, it's a very small probability.But let me compute it formally.First, S ~ N(14,400, 1,080,000). So, to find P(S > 60,000), we can standardize it.Z = (60,000 - 14,400)/1039.57 ≈ (45,600)/1039.57 ≈ Let's compute that.45,600 / 1039.57 ≈ 45,600 / 1000 ≈ 45.6, but since denominator is 1039.57, it's slightly less. Let me compute 45,600 / 1039.57.Let me compute 1039.57 * 44 = 1039.57 * 40 = 41,582.8; 1039.57 * 4 = 4,158.28; so total 41,582.8 + 4,158.28 = 45,741.08. That's more than 45,600.So, 44 * 1039.57 = 45,741.08, which is higher than 45,600. So, 44 - (45,741.08 - 45,600)/1039.57 ≈ 44 - (141.08)/1039.57 ≈ 44 - 0.1357 ≈ 43.8643.So, approximately 43.86.So, Z ≈ 43.86.Looking up Z = 43.86 in the standard normal table. But wait, standard normal tables usually go up to about Z = 3 or 4, beyond that, the probability is practically zero.In reality, for Z = 43.86, the probability that Z > 43.86 is essentially zero. So, P(S > 60,000) ≈ 0.But let me think again. Maybe I misread the budget. Maybe the budget is 60,000 per week, not per 4 weeks? Let me check.The question says: "the total expenditure over any 4-week period should not exceed KES 60,000." So, no, it's 60,000 over 4 weeks, which is 15,000 per week. Wait, 60,000 over 4 weeks is 15,000 per week, but the customer is spending on average 1,200 per visit, 3 times a week, so 3,600 per week. So, 15,000 per week is still higher than their average.Wait, 3 visits per week at 1,200 each is 3,600 per week. So, 4 weeks would be 14,400. So, 60,000 is way higher. So, the probability of exceeding 60,000 is practically zero.Alternatively, maybe the budget is 60,000 in total, so 15,000 per week. Wait, but the customer is only spending 3,600 per week on average, so 15,000 is still higher.Wait, perhaps I made a mistake in the number of visits. Let me check again.The customer visits 3 times a week, so in 4 weeks, it's 12 visits. Each visit is 1,200 on average, so total is 14,400. So, 60,000 is way higher. So, the probability is practically zero.But maybe the question is about not exceeding 60,000, so P(S <= 60,000). But the wording says "exceed their budget", so it's P(S > 60,000). So, it's a very small probability.But perhaps I need to compute it formally.So, Z = (60,000 - 14,400)/1039.57 ≈ 45,600 / 1039.57 ≈ 43.86.Looking up Z = 43.86, which is way beyond the typical Z-table range. The probability that Z > 43.86 is effectively zero. So, the probability is approximately zero.But maybe I should express it in terms of the standard normal distribution's tail probability. For such a high Z-score, the probability is negligible, like 10^(-something).Alternatively, maybe I made a mistake in the variance calculation. Let me double-check.Each visit has variance 300² = 90,000. So, for 12 visits, variance is 12 * 90,000 = 1,080,000. So, standard deviation is sqrt(1,080,000) ≈ 1039.57. That seems correct.So, yes, the Z-score is about 43.86, which is extremely high. So, the probability is practically zero.But maybe the question intended the budget to be 60,000 over 4 weeks, which is 15,000 per week, but the customer is only spending 3,600 per week on average, so 15,000 is still much higher. So, the probability is still practically zero.Wait, maybe I misread the budget. Let me read again: "the total expenditure over any 4-week period should not exceed KES 60,000." So, 60,000 is the maximum they want to spend in 4 weeks. So, if their average is 14,400, 60,000 is way above. So, the probability of exceeding 60,000 is practically zero.Alternatively, maybe the budget is 60,000 per week, which would be 240,000 over 4 weeks. But that would be even higher. Wait, no, the question says "over any 4-week period", so 60,000 is the total for 4 weeks.So, in conclusion, the probability is practically zero.But let me think again. Maybe the question is about not exceeding 60,000 in total, but the customer is concerned about exceeding it. So, the probability is P(S > 60,000). But given the mean is 14,400, and 60,000 is way higher, the probability is almost zero.Alternatively, maybe I made a mistake in the total number of visits. Wait, 3 times a week, 4 weeks: 3*4=12 visits. Each visit is 1,200, so total is 14,400. So, 60,000 is 4.1667 times the mean. So, 4.1667 standard deviations above the mean. Wait, no, 60,000 - 14,400 = 45,600. 45,600 / 1039.57 ≈ 43.86 standard deviations above the mean. That's an enormous number.So, the probability is effectively zero. So, the answer is approximately zero.But maybe I should express it as a very small number, like 10^(-something). But without a calculator, it's hard to compute, but it's safe to say it's practically zero.Wait, but let me think again. Maybe the budget is 60,000 per week, not per 4 weeks. Let me check the question again."the total expenditure over any 4-week period should not exceed KES 60,000."So, no, it's 60,000 over 4 weeks. So, 15,000 per week. But the customer is only spending 3,600 per week on average. So, 15,000 is still way higher.Wait, maybe the question is about not exceeding 60,000 in total, but the customer is worried about exceeding it. So, the probability is P(S > 60,000). But as we saw, it's practically zero.Alternatively, maybe the question is about the total expenditure over 4 weeks not exceeding 60,000, so P(S <= 60,000). But the wording says "exceed their budget", so it's P(S > 60,000).So, in conclusion, the probability is practically zero.But to be thorough, let me compute the Z-score again.Z = (60,000 - 14,400)/1039.57 ≈ 45,600 / 1039.57 ≈ 43.86.Looking up Z = 43.86 in standard normal table. Since standard normal tables typically don't go beyond Z = 3 or 4, we can use the approximation for the tail probability.The formula for the tail probability for large Z is approximately (1/(Z * sqrt(2π))) * e^(-Z²/2). So, let's compute that.First, Z = 43.86.Compute Z² = 43.86² ≈ 1,923. So, Z²/2 ≈ 961.5.Compute e^(-961.5). That's an extremely small number, practically zero.So, the probability is approximately (1/(43.86 * sqrt(2π))) * e^(-961.5). Since e^(-961.5) is like 10^(-417), which is unimaginably small.Therefore, the probability is effectively zero.So, summarizing:1. The probability of spending more than 1,500 in at least one visit in a week is approximately 40.5%.2. The probability of exceeding 60,000 KES in a 4-week period is practically zero.But let me just make sure I didn't make any mistakes in the first part.In the first part, I calculated P(at least one visit > 1500) = 1 - [P(X <= 1500)]^3 ≈ 1 - (0.8413)^3 ≈ 1 - 0.595 ≈ 0.405, which is 40.5%.Yes, that seems correct.So, final answers:1. Approximately 40.5%2. Practically zero, which we can write as 0.But in terms of probability, it's better to write it as approximately 0.Alternatively, if we need to express it more formally, we can say it's less than 0.0000000001 or something, but in the context of the question, it's safe to say the probability is practically zero.So, I think that's it.</think>

🧠:<think>Okay, so I have this problem about a middle-aged woman with diabetes, and her blood glucose level is modeled by this piecewise function. There are two parts to the problem: calculating the total variation of her blood glucose over 24 hours and determining how much time her glucose is outside the recommended range. Let me tackle them one by one.Starting with part 1: Calculate the total variation of her blood glucose level over 24 hours. The total variation is given by the integral of the absolute value of the derivative of G(t) over [0, 24]. Hmm, okay, so I need to find G'(t) for each piece, take the absolute value, and integrate over their respective intervals.First, let's write down the function again:G(t) = 150 + 10 sin(πt/12) for 0 ≤ t < 12andG(t) = 180 + 5 cos(π(t - 12)/6) for 12 ≤ t < 24.So, I need to find the derivatives for both pieces.For the first part, 0 ≤ t < 12:G(t) = 150 + 10 sin(πt/12)The derivative G'(t) is the derivative of 150, which is 0, plus the derivative of 10 sin(πt/12). The derivative of sin(u) is cos(u) times u', so:G'(t) = 10 * (π/12) cos(πt/12) = (10π/12) cos(πt/12) = (5π/6) cos(πt/12)So, G'(t) = (5π/6) cos(πt/12) for 0 ≤ t < 12.Now, for the second part, 12 ≤ t < 24:G(t) = 180 + 5 cos(π(t - 12)/6)Let me simplify the argument of the cosine:π(t - 12)/6 = πt/6 - 2πBut since cosine is periodic with period 2π, cos(πt/6 - 2π) = cos(πt/6). So, actually, G(t) simplifies to 180 + 5 cos(πt/6) for 12 ≤ t < 24.Wait, is that right? Let me check:cos(π(t - 12)/6) = cos(πt/6 - 2π) = cos(πt/6) because cosine is 2π periodic. So yes, that's correct. So, G(t) can be rewritten as 180 + 5 cos(πt/6) for 12 ≤ t < 24.So, the derivative G'(t) is the derivative of 180, which is 0, plus the derivative of 5 cos(πt/6). The derivative of cos(u) is -sin(u) times u', so:G'(t) = 5 * (-π/6) sin(πt/6) = (-5π/6) sin(πt/6)So, G'(t) = (-5π/6) sin(πt/6) for 12 ≤ t < 24.Now, to compute the total variation, I need to integrate |G'(t)| from 0 to 24. Since G(t) is piecewise, I can split the integral into two parts: from 0 to 12 and from 12 to 24.So, total variation = ∫₀²⁴ |G'(t)| dt = ∫₀¹² |(5π/6) cos(πt/12)| dt + ∫₁₂²⁴ |(-5π/6) sin(πt/6)| dtSimplify the absolute values:First integral: |(5π/6) cos(πt/12)| = (5π/6) |cos(πt/12)|Second integral: |(-5π/6) sin(πt/6)| = (5π/6) |sin(πt/6)|So, total variation = (5π/6) ∫₀¹² |cos(πt/12)| dt + (5π/6) ∫₁₂²⁴ |sin(πt/6)| dtLet me compute each integral separately.First integral: I1 = ∫₀¹² |cos(πt/12)| dtLet me make a substitution: let u = πt/12, so du = π/12 dt, which means dt = (12/π) du.When t = 0, u = 0; when t = 12, u = π.So, I1 = ∫₀^π |cos(u)| * (12/π) du = (12/π) ∫₀^π |cos(u)| duWe know that cos(u) is positive from 0 to π/2 and negative from π/2 to π. So, |cos(u)| = cos(u) for 0 ≤ u ≤ π/2 and -cos(u) for π/2 ≤ u ≤ π.Thus, I1 = (12/π) [ ∫₀^(π/2) cos(u) du + ∫_(π/2)^π (-cos(u)) du ]Compute the integrals:∫ cos(u) du = sin(u)First integral: sin(π/2) - sin(0) = 1 - 0 = 1Second integral: ∫ (-cos(u)) du = -sin(u) evaluated from π/2 to π: -sin(π) + sin(π/2) = -0 + 1 = 1So, I1 = (12/π)(1 + 1) = (12/π)(2) = 24/πOkay, so the first integral is 24/π.Now, the second integral: I2 = ∫₁₂²⁴ |sin(πt/6)| dtAgain, let me make a substitution: let v = πt/6, so dv = π/6 dt, so dt = (6/π) dv.When t = 12, v = π*12/6 = 2π; when t = 24, v = π*24/6 = 4π.So, I2 = ∫_{2π}^{4π} |sin(v)| * (6/π) dv = (6/π) ∫_{2π}^{4π} |sin(v)| dvWe know that |sin(v)| has a period of π, and over each interval of length π, the integral of |sin(v)| is 2. So, from 2π to 4π is two periods.Thus, ∫_{2π}^{4π} |sin(v)| dv = 2 * 2 = 4Therefore, I2 = (6/π) * 4 = 24/πSo, both integrals I1 and I2 are equal to 24/π.Therefore, total variation = (5π/6)(24/π) + (5π/6)(24/π) = (5π/6)(24/π) * 2Simplify:(5π/6)(24/π) = (5/6)(24) = 20So, total variation = 20 + 20 = 40Wait, hold on. Let me double-check that:(5π/6)(24/π) = (5/6)(24) = 20, yes. So, each integral contributes 20, so total variation is 40.So, the total variation over 24 hours is 40 mg/dL.Wait, that seems low? Let me think again.Wait, the integral of |G'(t)| over [0,24] is the total variation. So, if G(t) is oscillating, the total variation would be the sum of all the rises and falls.But in this case, the first part is a sine wave, so over 0 to 12 hours, it completes half a period (since the period is 24 hours for sin(πt/12)), but wait, sin(πt/12) has period 24, so from 0 to 12 is half a period.Similarly, the second part is a cosine wave with argument π(t - 12)/6, which simplifies to cos(πt/6 - 2π) = cos(πt/6). So, cos(πt/6) has period 12 hours. So, from 12 to 24 is one full period.Wait, so the first part is half a sine wave, and the second part is a full cosine wave.But when we took the derivatives, we had for the first part, G'(t) = (5π/6) cos(πt/12), and for the second part, G'(t) = (-5π/6) sin(πt/6). Then, when we took absolute values, we integrated over their absolute values.Wait, but when I calculated I1, it was 24/π, and I2 was 24/π as well. Then, each integral was multiplied by (5π/6), so (5π/6)*(24/π) = 20 for each, so total variation is 40.That seems correct. So, 40 mg/dL is the total variation.Okay, moving on to part 2: Determine the total duration within a 24-hour period during which her blood glucose level is outside the recommended range [70, 180] mg/dL.So, I need to find the total time when G(t) < 70 or G(t) > 180.Looking at the function G(t):For 0 ≤ t < 12, G(t) = 150 + 10 sin(πt/12)For 12 ≤ t < 24, G(t) = 180 + 5 cos(π(t - 12)/6) = 180 + 5 cos(πt/6 - 2π) = 180 + 5 cos(πt/6)So, let me analyze each interval.First interval: 0 ≤ t < 12G(t) = 150 + 10 sin(πt/12)We need to find when G(t) < 70 or G(t) > 180.Let's check the maximum and minimum of G(t) in this interval.The sine function varies between -1 and 1, so G(t) varies between 150 - 10 = 140 and 150 + 10 = 160.So, in the first 12 hours, G(t) is between 140 and 160 mg/dL.The recommended range is [70, 180], so 140 to 160 is entirely within [70, 180]. Therefore, in the first 12 hours, G(t) is always within the recommended range. So, no time is spent outside the range here.Now, the second interval: 12 ≤ t < 24G(t) = 180 + 5 cos(πt/6)Again, let's find when G(t) is outside [70, 180].First, find the maximum and minimum of G(t):cos(πt/6) varies between -1 and 1, so G(t) varies between 180 - 5 = 175 and 180 + 5 = 185.So, G(t) ranges from 175 to 185 mg/dL in this interval.The upper limit is 185, which is above 180, and the lower limit is 175, which is within [70, 180].So, the glucose level is above 180 when G(t) > 180, which occurs when cos(πt/6) > 0.Wait, let's solve G(t) > 180:180 + 5 cos(πt/6) > 180Subtract 180: 5 cos(πt/6) > 0Divide by 5: cos(πt/6) > 0So, cos(πt/6) > 0.Similarly, G(t) < 70 would be 180 + 5 cos(πt/6) < 70, which would require 5 cos(πt/6) < -110, but since cos(πt/6) is between -1 and 1, 5 cos(πt/6) is between -5 and 5. So, 180 + 5 cos(πt/6) is between 175 and 185, which is always above 70. So, G(t) never goes below 175, which is above 70. Therefore, G(t) is never below 70 in this interval.Therefore, the only time G(t) is outside the recommended range is when G(t) > 180, which is when cos(πt/6) > 0.So, let's solve cos(πt/6) > 0 for t in [12, 24).First, let me find the values of t where cos(πt/6) = 0.cos(πt/6) = 0 when πt/6 = π/2 + kπ, where k is integer.So, πt/6 = π/2 + kπ => t/6 = 1/2 + k => t = 3 + 6k.Within t ∈ [12, 24), let's find k such that t = 3 + 6k is in [12, 24).So, 3 + 6k ≥ 12 => 6k ≥ 9 => k ≥ 1.5And 3 + 6k < 24 => 6k < 21 => k < 3.5Since k is integer, k = 2, 3.So, t = 3 + 6*2 = 15, and t = 3 + 6*3 = 21.So, cos(πt/6) is positive in the intervals where t is between the zeros. Since cosine is positive in the intervals ( -π/2 + 2πk, π/2 + 2πk ). Translating back to t:cos(πt/6) > 0 when πt/6 ∈ (-π/2 + 2πk, π/2 + 2πk) for some integer k.But since t is in [12, 24), let's find the corresponding intervals.First, let's find the period of cos(πt/6). The period is 2π / (π/6) = 12 hours. So, every 12 hours, the cosine function repeats.Within [12, 24), which is one full period, the function cos(πt/6) starts at t=12:At t=12, cos(π*12/6) = cos(2π) = 1.It goes down to cos(π*18/6) = cos(3π) = -1 at t=18, and back to cos(π*24/6) = cos(4π) = 1 at t=24.Wait, so in the interval [12, 24), cos(πt/6) starts at 1, goes down to -1 at t=18, then back to 1 at t=24.So, the function is positive from t=12 to t=15 (where it crosses zero), negative from t=15 to t=21, and positive again from t=21 to t=24.Wait, but earlier I found zeros at t=15 and t=21. So, cos(πt/6) is positive in [12,15) and (21,24), and negative in (15,21).Therefore, cos(πt/6) > 0 when t ∈ [12,15) ∪ (21,24).Therefore, G(t) > 180 when t ∈ [12,15) ∪ (21,24).So, the duration when G(t) is above 180 is (15 - 12) + (24 - 21) = 3 + 3 = 6 hours.Since G(t) never goes below 70, the total duration outside the recommended range is 6 hours.Wait, let me confirm.From t=12 to t=15, G(t) is above 180, and from t=21 to t=24, G(t) is above 180. Each interval is 3 hours, so total 6 hours.Yes, that seems correct.So, summarizing:1. Total variation is 40 mg/dL.2. Total duration outside the recommended range is 6 hours.Final Answer1. The total variation is boxed{40} mg/dL.2. The total duration outside the recommended range is boxed{6} hours.</think>