🧠:<think>Alright, so I have these two differential equation problems to solve. Let me take them one at a time.Starting with the first problem: Professor Demir is modeling the erosion of fairy chimneys in Cappadocia using the differential equation dh/dt = -k√h. The initial condition is h(0) = h₀. I need to find h(t) in terms of t, k, and h₀.Hmm, okay. This looks like a separable differential equation. So, I can rewrite it as dh/dt = -k√h. To separate the variables, I can divide both sides by √h and multiply both sides by dt. That gives me:(1/√h) dh = -k dtNow, I can integrate both sides. The left side with respect to h and the right side with respect to t.Integrating the left side: ∫(1/√h) dh. I remember that the integral of h^(-1/2) is 2√h. Let me check: d/dh [2√h] = 2*(1/(2√h)) = 1/√h. Yep, that's correct.So, integrating the left side gives 2√h + C₁, where C₁ is the constant of integration.Integrating the right side: ∫-k dt. That's straightforward; it's -k t + C₂, where C₂ is another constant.Putting it all together:2√h = -k t + CWhere I've combined the constants C₁ and C₂ into a single constant C.Now, I need to apply the initial condition to find C. At t = 0, h = h₀. Plugging that in:2√h₀ = -k*0 + C => C = 2√h₀So, substituting back into the equation:2√h = -k t + 2√h₀I can solve for √h:√h = (-k t + 2√h₀)/2 = √h₀ - (k t)/2Then, squaring both sides to solve for h:h(t) = [√h₀ - (k t)/2]^2Let me expand that:h(t) = (√h₀)^2 - 2*(√h₀)*(k t)/2 + (k t / 2)^2Simplifying each term:(√h₀)^2 = h₀The middle term: 2*(√h₀)*(k t)/2 = √h₀ * k tThe last term: (k t / 2)^2 = (k² t²)/4So, putting it all together:h(t) = h₀ - √h₀ * k t + (k² t²)/4Hmm, wait a second. Let me check if this makes sense. If I plug t = 0, I get h(0) = h₀, which is correct. Also, as t increases, h(t) decreases, which aligns with the erosion model. The quadratic term in t suggests that the height decreases quadratically over time, which seems plausible given the square root in the original differential equation.Alternatively, I can write h(t) as [√h₀ - (k t)/2]^2, which is a more compact form. Both forms are correct, but maybe the squared form is preferable for simplicity.So, I think that's the solution for the first problem.Moving on to the second problem: Professor Demir models the spread of volcanic ash deposits with the partial differential equation ∂A/∂t = D(∂²A/∂x² + ∂²A/∂y²). The initial condition is A(r, 0) = A₀ e^{-r²}, where r is the radial distance. I need to solve for A(r, t) in terms of r, t, D, and A₀.Alright, this is a partial differential equation, specifically the heat equation in two dimensions, given the radial symmetry. Since the problem mentions radial symmetry, I can switch to polar coordinates, where the equation simplifies.In polar coordinates, the Laplacian operator ∇² becomes (1/r) ∂/∂r (r ∂A/∂r). So, the PDE becomes:∂A/∂t = D [ (1/r) ∂/∂r (r ∂A/∂r) ]This is the radial heat equation. The initial condition is given as A(r, 0) = A₀ e^{-r²}.I remember that for such equations, the solution can often be found using separation of variables or by recognizing it as a Gaussian function, which is its own Fourier transform. Given the initial condition is a Gaussian, I suspect the solution remains Gaussian over time, but with a time-dependent variance.Let me try to recall the general solution for the heat equation in radial coordinates. The solution for the heat equation with radial symmetry and an initial Gaussian condition is another Gaussian that spreads over time.The general form is A(r, t) = (A₀ / sqrt(1 + 4 D t)) e^{-r² / (1 + 4 D t)}.Wait, let me verify that. Alternatively, sometimes the solution has a factor involving sqrt(4π D t) in the denominator, but since the initial condition here is A₀ e^{-r²}, which is a Gaussian without the normalization factor, perhaps the solution will also not have that.Let me think step by step.First, the heat equation in radial coordinates is:∂A/∂t = D [ (1/r) ∂/∂r (r ∂A/∂r) ]Let me make a substitution to simplify this equation. Let me define u(r, t) = A(r, t). Then, the equation is:∂u/∂t = D [ (1/r) ∂/∂r (r ∂u/∂r) ]This is a standard PDE, and for radially symmetric solutions, we can look for solutions of the form u(r, t) = f(r, t), where f depends only on r and t.The initial condition is u(r, 0) = A₀ e^{-r²}.I recall that the fundamental solution to the heat equation in two dimensions is:u(r, t) = (1/(4 π D t)) e^{-r² / (4 D t)}But this is the solution when the initial condition is a delta function. However, our initial condition is a Gaussian, which is similar but not exactly a delta function.Wait, actually, if the initial condition is a Gaussian, the solution remains Gaussian but with a time-dependent width. Let me see.Suppose we assume a solution of the form:u(r, t) = (A₀ / sqrt(1 + 4 D t)) e^{- r² / (1 + 4 D t)}Let me check if this satisfies the heat equation.First, compute the time derivative ∂u/∂t.∂u/∂t = A₀ * [ (-1/2) (4 D) / (1 + 4 D t)^{3/2} ) ] e^{- r² / (1 + 4 D t)} + A₀ / sqrt(1 + 4 D t) * e^{- r² / (1 + 4 D t)} * (r² / (1 + 4 D t)^2) * 4 DWait, that seems complicated. Maybe it's better to compute it step by step.Let me denote s = 1 + 4 D t. Then, u(r, t) = A₀ / sqrt(s) e^{- r² / s}.Compute ∂u/∂t:du/dt = A₀ * [ (-1/(2 s^{3/2})) * 4 D ] e^{- r² / s} + A₀ / sqrt(s) * e^{- r² / s} * (r² / s²) * 4 DSimplify:= A₀ * (-2 D / s^{3/2}) e^{- r² / s} + A₀ * (4 D r² / s^{5/2}) e^{- r² / s}Factor out A₀ e^{- r² / s} / s^{5/2}:= A₀ e^{- r² / s} / s^{5/2} [ -2 D s + 4 D r² ]Now, compute the spatial derivatives.First, compute ∂u/∂r:∂u/∂r = A₀ / sqrt(s) * e^{- r² / s} * (-2 r / s)= -2 A₀ r e^{- r² / s} / (s^{3/2})Then, compute ∂/∂r (r ∂u/∂r):First, r ∂u/∂r = -2 A₀ r² e^{- r² / s} / (s^{3/2})Then, ∂/∂r [ -2 A₀ r² e^{- r² / s} / (s^{3/2}) ] =-2 A₀ / s^{3/2} [ 2 r e^{- r² / s} + r² * (-2 r / s) e^{- r² / s} ]= -2 A₀ / s^{3/2} [ 2 r e^{- r² / s} - 2 r³ e^{- r² / s} / s ]= -4 A₀ r e^{- r² / s} / s^{3/2} + 4 A₀ r³ e^{- r² / s} / s^{5/2}Now, multiply by (1/r):(1/r) ∂/∂r (r ∂u/∂r) = -4 A₀ e^{- r² / s} / s^{3/2} + 4 A₀ r² e^{- r² / s} / s^{5/2}So, the Laplacian term is D times that:D [ -4 A₀ e^{- r² / s} / s^{3/2} + 4 A₀ r² e^{- r² / s} / s^{5/2} ]Compare this with the time derivative we computed earlier:∂u/∂t = A₀ e^{- r² / s} / s^{5/2} [ -2 D s + 4 D r² ]Let me write both expressions:∂u/∂t = A₀ e^{- r² / s} / s^{5/2} ( -2 D s + 4 D r² )Laplacian term: D [ -4 A₀ e^{- r² / s} / s^{3/2} + 4 A₀ r² e^{- r² / s} / s^{5/2} ]Factor out D A₀ e^{- r² / s} / s^{5/2}:= D A₀ e^{- r² / s} / s^{5/2} [ -4 s² + 4 r² ]Wait, let me check:Wait, the first term is -4 A₀ e^{- r² / s} / s^{3/2} which is -4 A₀ e^{- r² / s} s^{-3/2} = -4 A₀ e^{- r² / s} s^{-3/2} * s² / s² = -4 A₀ e^{- r² / s} s^{1/2} / s²Wait, perhaps it's better to express both terms with the same denominator.Wait, ∂u/∂t:= A₀ e^{- r² / s} / s^{5/2} ( -2 D s + 4 D r² )= A₀ e^{- r² / s} / s^{5/2} * D ( -2 s + 4 r² )Laplacian term:= D [ -4 A₀ e^{- r² / s} / s^{3/2} + 4 A₀ r² e^{- r² / s} / s^{5/2} ]= D A₀ e^{- r² / s} [ -4 / s^{3/2} + 4 r² / s^{5/2} ]= D A₀ e^{- r² / s} / s^{5/2} [ -4 s + 4 r² ]So, the Laplacian term is D A₀ e^{- r² / s} / s^{5/2} ( -4 s + 4 r² )Comparing to ∂u/∂t:∂u/∂t = D A₀ e^{- r² / s} / s^{5/2} ( -2 s + 4 r² )So, unless -2 s + 4 r² equals -4 s + 4 r², which would require -2 s = -4 s, which is only true if s = 0, which isn't the case, they are not equal.Hmm, so my initial guess for the solution doesn't satisfy the PDE. Maybe I made a mistake in the form.Wait, perhaps I need to include a factor of 4 D t in the exponent. Let me think again.Alternatively, maybe the solution is:A(r, t) = (A₀ / (1 + 4 D t)) e^{- r² / (1 + 4 D t)}Wait, let me try this form.Let me define s = 1 + 4 D t.Then, A(r, t) = A₀ / s e^{- r² / s}Compute ∂A/∂t:= A₀ [ - (4 D) / s² e^{- r² / s} + (4 D r² / s³) e^{- r² / s} ]= (4 D A₀ / s³) e^{- r² / s} ( -s + r² )Now, compute the Laplacian term:First, compute ∂A/∂r:= A₀ / s e^{- r² / s} * (-2 r / s )= -2 A₀ r e^{- r² / s} / s²Then, compute ∂/∂r (r ∂A/∂r):= ∂/∂r [ -2 A₀ r² e^{- r² / s} / s² ]= -2 A₀ / s² [ 2 r e^{- r² / s} + r² (-2 r / s) e^{- r² / s} ]= -4 A₀ r e^{- r² / s} / s² + 4 A₀ r³ e^{- r² / s} / s³Multiply by (1/r):= -4 A₀ e^{- r² / s} / s² + 4 A₀ r² e^{- r² / s} / s³Multiply by D:= D [ -4 A₀ e^{- r² / s} / s² + 4 A₀ r² e^{- r² / s} / s³ ]= (4 D A₀ / s³) e^{- r² / s} ( -s + r² )Which is exactly equal to ∂A/∂t. So, this form satisfies the PDE.Therefore, the solution is:A(r, t) = (A₀ / (1 + 4 D t)) e^{- r² / (1 + 4 D t)}So, that's the solution.Alternatively, sometimes the solution is written with a factor of sqrt(π) or something, but since the initial condition doesn't have that, I think this is the correct form.Let me double-check the initial condition: at t = 0, s = 1, so A(r, 0) = A₀ / 1 e^{- r² / 1} = A₀ e^{- r²}, which matches the given initial condition. Good.Also, as t increases, the denominator 1 + 4 D t increases, so the amplitude decreases and the exponent's denominator increases, meaning the Gaussian spreads out, which makes sense for diffusion.So, I think that's the correct solution.Final Answer1. The height of the fairy chimney as a function of time is boxed{h(t) = left( sqrt{h_0} - frac{k t}{2} right)^2}.2. The area covered by volcanic ash deposits as a function of radial distance and time is boxed{A(r, t) = frac{A_0}{1 + 4 D t} e^{-frac{r^2}{1 + 4 D t}}}.</think>

🧠:<think>Alright, so I have this problem about a rugby player's performance being influenced by encouragement and opponent pressure. It's split into two parts. Let me try to tackle them one by one.Starting with part 1: They give me the performance function as ( P(t) = e^{kt} ). They mention that the initial performance at ( t = 0 ) is ( P_0 ). Hmm, so when ( t = 0 ), ( P(0) = e^{k*0} = e^0 = 1 ). But they say it's ( P_0 ). Wait, that means ( P(t) = P_0 e^{kt} ). Maybe I missed that in the problem statement. Let me check again.Oh, the problem says "the player's performance level at time ( t ) is represented by the function ( P(t) = e^{kt} )". So, actually, it's just ( e^{kt} ), not multiplied by ( P_0 ). But then they mention the initial performance is ( P_0 ). That seems contradictory. Wait, maybe ( P_0 ) is the initial value, so ( P(0) = P_0 ). But according to the function, ( P(0) = e^{0} = 1 ). So perhaps ( P(t) = P_0 e^{kt} ). Maybe the problem statement was a bit unclear.Let me assume that ( P(t) = P_0 e^{kt} ). That makes more sense because then ( P(0) = P_0 ). So, given that, after 15 minutes, the performance is ( 2P_0 ). So, at ( t = 15 ), ( P(15) = 2P_0 ).So, substituting into the equation: ( 2P_0 = P_0 e^{15k} ). If I divide both sides by ( P_0 ), assuming ( P_0 neq 0 ), which it isn't because it's a performance level, I get ( 2 = e^{15k} ).To solve for ( k ), I can take the natural logarithm of both sides. So, ( ln(2) = 15k ). Therefore, ( k = ln(2)/15 ). Let me compute that. ( ln(2) ) is approximately 0.6931, so ( k approx 0.6931/15 approx 0.0462 ). But since the problem doesn't specify rounding, I can just leave it as ( ln(2)/15 ).Wait, is that correct? Let me verify. If ( P(t) = P_0 e^{kt} ), then at ( t = 15 ), it's ( P_0 e^{15k} = 2P_0 ). Dividing both sides by ( P_0 ) gives ( e^{15k} = 2 ), so yes, ( 15k = ln(2) ), so ( k = ln(2)/15 ). That seems right.Moving on to part 2: The net performance ( N(t) = P(t) - Q(t) ), where ( Q(t) = mt + c ). They say that at ( t = 30 ) minutes, ( N(t) ) reaches a maximum. So, we need to find the relationship between ( m ), ( c ), and ( k ).Since ( N(t) ) is differentiable and has a maximum at ( t = 30 ), the derivative ( N'(t) ) should be zero at ( t = 30 ).First, let's write out ( N(t) ):( N(t) = P(t) - Q(t) = P_0 e^{kt} - (mt + c) ).Wait, but in part 1, we found ( k ) assuming ( P(t) = P_0 e^{kt} ). But in the problem statement, it was given as ( P(t) = e^{kt} ). Hmm, so maybe in part 2, ( P(t) ) is still ( e^{kt} ) without the ( P_0 ). Let me check.Looking back at the problem: "the player's performance level at time ( t ) is represented by the function ( P(t) = e^{kt} )". So, it's just ( e^{kt} ), not multiplied by ( P_0 ). But in part 1, they mention the initial performance is ( P_0 ). So, perhaps ( P(t) = P_0 e^{kt} ). Maybe the problem statement had a typo or something. Because otherwise, if ( P(t) = e^{kt} ), then ( P(0) = 1 ), but they say it's ( P_0 ). So, I think it's safe to assume that ( P(t) = P_0 e^{kt} ).So, going back, ( N(t) = P_0 e^{kt} - (mt + c) ).Now, to find the maximum at ( t = 30 ), we take the derivative of ( N(t) ) with respect to ( t ) and set it equal to zero at ( t = 30 ).So, ( N'(t) = d/dt [P_0 e^{kt} - mt - c] = P_0 k e^{kt} - m ).Setting ( N'(30) = 0 ):( P_0 k e^{30k} - m = 0 ).So, ( m = P_0 k e^{30k} ).But wait, in part 1, we found ( k = ln(2)/15 ). So, we can substitute that into this equation.First, let's compute ( e^{30k} ). Since ( k = ln(2)/15 ), then ( 30k = 30*(ln2)/15 = 2 ln2 ). So, ( e^{30k} = e^{2 ln2} = (e^{ln2})^2 = 2^2 = 4 ).Therefore, ( m = P_0 k * 4 ).But from part 1, we have ( k = ln2 /15 ), so substituting:( m = P_0 * (ln2 /15) * 4 = (4 P_0 ln2)/15 ).But wait, the problem asks for the relationship between ( m ), ( c ), and ( k ). So, in the equation ( m = P_0 k e^{30k} ), we can express ( P_0 ) in terms of ( k ) if needed.Wait, from part 1, we have ( P(t) = P_0 e^{kt} ). At ( t = 15 ), ( P(15) = 2P_0 ). So, ( 2P_0 = P_0 e^{15k} ), which gives ( e^{15k} = 2 ), so ( P_0 = P_0 ). Wait, that doesn't help. Alternatively, maybe we can express ( P_0 ) in terms of ( k ) from the initial condition.Wait, actually, from ( P(0) = P_0 ), which is given, so ( P_0 ) is just a constant. So, in the expression ( m = P_0 k e^{30k} ), we have ( m ) in terms of ( P_0 ), ( k ), and ( e^{30k} ).But since ( e^{30k} = 4 ) as we found earlier, because ( 30k = 2 ln2 ), so ( e^{30k} = 4 ). Therefore, ( m = P_0 k *4 ).But we can also express ( P_0 ) in terms of ( k ) from part 1. Wait, no, because ( P_0 ) is just the initial performance level, which is given as a constant. So, unless we can relate ( P_0 ) to ( k ), which we can't directly, because ( k ) is determined from the doubling time.Wait, but in part 1, we found ( k = ln2 /15 ), which is a constant, independent of ( P_0 ). So, ( P_0 ) is just a constant multiplier. So, in the expression ( m = 4 P_0 k ), we can write ( m = 4 k P_0 ).But the problem asks for the relationship between ( m ), ( c ), and ( k ). So, in the equation ( m = 4 k P_0 ), we have ( m ) in terms of ( k ) and ( P_0 ). But we need to relate ( m ), ( c ), and ( k ). So, perhaps we need another equation.Wait, but in the expression for ( N(t) ), ( c ) is just a constant term. So, does ( c ) play any role in the derivative? No, because the derivative of ( c ) is zero. So, the maximum condition only gives us a relationship between ( m ) and ( k ), but not involving ( c ).Wait, but the problem says "find the relationship between ( m ), ( c ), and ( k )". Hmm, maybe I'm missing something. Let me think again.We have ( N(t) = P(t) - Q(t) = P_0 e^{kt} - (mt + c) ).We found that ( N'(t) = P_0 k e^{kt} - m ), and at ( t = 30 ), ( N'(30) = 0 ), so ( P_0 k e^{30k} = m ).So, ( m = P_0 k e^{30k} ).But from part 1, we have ( k = ln2 /15 ), so ( e^{30k} = e^{2 ln2} = 4 ). So, ( m = 4 P_0 k ).But the problem is asking for the relationship between ( m ), ( c ), and ( k ). So, unless there's another condition, perhaps at ( t = 30 ), the value of ( N(t) ) is maximum, but we don't know its value, so we can't get another equation involving ( c ).Wait, unless we consider that ( N(t) ) has a maximum at ( t = 30 ), so the second derivative at ( t = 30 ) should be negative. Let's compute the second derivative.( N''(t) = d/dt [P_0 k e^{kt} - m] = P_0 k^2 e^{kt} ).At ( t = 30 ), ( N''(30) = P_0 k^2 e^{30k} ). Since ( P_0 ), ( k ), and ( e^{30k} ) are all positive, ( N''(30) ) is positive, which would mean it's a minimum, not a maximum. Wait, that contradicts the problem statement which says it's a maximum.Hmm, that's a problem. Did I make a mistake in computing the derivatives?Wait, ( N(t) = P(t) - Q(t) = P_0 e^{kt} - (mt + c) ).First derivative: ( N'(t) = P_0 k e^{kt} - m ).Second derivative: ( N''(t) = P_0 k^2 e^{kt} ).Since ( P_0 > 0 ), ( k > 0 ), ( e^{kt} > 0 ), so ( N''(t) > 0 ) for all ( t ). That means the function ( N(t) ) is concave upward everywhere, so any critical point would be a minimum, not a maximum. But the problem says that ( N(t) ) reaches a maximum at ( t = 30 ). That seems contradictory.Wait, maybe I made a wrong assumption about ( P(t) ). Let me double-check.The problem says: "the player's performance level at time ( t ) is represented by the function ( P(t) = e^{kt} )". So, ( P(t) = e^{kt} ), not ( P_0 e^{kt} ). So, perhaps I was wrong earlier. Let me correct that.So, ( P(t) = e^{kt} ), so ( P(0) = e^{0} = 1 ). But the problem says the initial performance is ( P_0 ). So, that suggests that ( P(t) = P_0 e^{kt} ). So, maybe the problem statement had a typo, or perhaps ( P_0 = 1 ). Hmm, this is confusing.Alternatively, maybe ( P(t) = e^{kt} ) and ( P_0 ) is just a label for ( P(0) ), which is 1. So, in that case, ( P_0 = 1 ). So, in part 1, when they say the initial performance is ( P_0 ), it's just 1, and after 15 minutes, it's ( 2P_0 = 2 ). So, ( P(15) = 2 ). So, ( e^{15k} = 2 ), so ( k = ln2 /15 ). So, that part is consistent.Then, in part 2, ( N(t) = P(t) - Q(t) = e^{kt} - (mt + c) ).So, the initial performance is ( P_0 = 1 ), but in the net performance, it's ( N(0) = 1 - c ). But the problem doesn't specify anything about ( N(0) ), so maybe ( c ) is just another constant.So, going back, ( N(t) = e^{kt} - (mt + c) ).Taking the derivative: ( N'(t) = k e^{kt} - m ).Setting ( N'(30) = 0 ):( k e^{30k} - m = 0 ).So, ( m = k e^{30k} ).But from part 1, ( k = ln2 /15 ), so ( 30k = 2 ln2 ), so ( e^{30k} = e^{2 ln2} = 4 ). Therefore, ( m = k *4 ).So, ( m = 4k ).But the problem asks for the relationship between ( m ), ( c ), and ( k ). So, in the equation ( m = 4k ), we have ( m ) in terms of ( k ). But ( c ) isn't involved in the derivative condition. So, is there another condition?Wait, perhaps the maximum value of ( N(t) ) at ( t = 30 ) is also given, but the problem doesn't specify the value, just that it's a maximum. So, without knowing the value, we can't get another equation involving ( c ). So, maybe the relationship is only between ( m ) and ( k ), which is ( m = 4k ), and ( c ) is arbitrary? But the problem says "find the relationship between ( m ), ( c ), and ( k )", so perhaps I'm missing something.Wait, maybe the maximum value of ( N(t) ) is achieved at ( t = 30 ), so ( N(30) ) is the maximum. But without knowing the value, we can't relate ( c ). Alternatively, maybe the maximum value is zero? But that's not stated.Wait, perhaps I need to consider the second derivative. Earlier, I thought that ( N''(t) = k^2 e^{kt} ), which is positive, meaning it's a minimum. But the problem says it's a maximum. That suggests a contradiction, which means my assumption about ( P(t) ) might be wrong.Wait, if ( P(t) = e^{kt} ), then ( N(t) = e^{kt} - (mt + c) ). The derivative is ( N'(t) = k e^{kt} - m ). Setting to zero at ( t = 30 ): ( k e^{30k} = m ). So, ( m = k e^{30k} ). As before, ( e^{30k} = 4 ), so ( m = 4k ).But the second derivative is ( N''(t) = k^2 e^{kt} ), which is positive, so it's a minimum, not a maximum. That's a problem because the question says it's a maximum. So, perhaps the function ( N(t) ) is concave down at ( t = 30 ), meaning ( N''(30) < 0 ). But ( N''(t) = k^2 e^{kt} ), which is always positive. So, that's impossible.Wait, maybe I made a mistake in the derivative. Let me check again.( N(t) = e^{kt} - mt - c ).First derivative: ( N'(t) = k e^{kt} - m ).Second derivative: ( N''(t) = k^2 e^{kt} ).Yes, that's correct. So, ( N''(t) > 0 ) always, meaning any critical point is a minimum. So, the problem statement must have a mistake, or perhaps I misunderstood the problem.Alternatively, maybe the net performance is ( N(t) = Q(t) - P(t) ), so ( N(t) = mt + c - e^{kt} ). Then, the derivative would be ( N'(t) = m - k e^{kt} ). Setting to zero at ( t = 30 ): ( m = k e^{30k} ), same as before. Then, the second derivative would be ( N''(t) = -k^2 e^{kt} ), which is negative, so it's a maximum. That would make sense.Wait, the problem says "the net performance of the player, considering both encouragement and pressure, is given by ( N(t) = P(t) - Q(t) )". So, it's ( P(t) - Q(t) ). So, if ( N(t) ) is supposed to have a maximum, but with ( N''(t) > 0 ), which contradicts, perhaps the problem meant ( Q(t) - P(t) ). Alternatively, maybe the function is ( N(t) = Q(t) - P(t) ), but the problem says ( P(t) - Q(t) ).Alternatively, maybe the problem is correct, and I need to consider that ( N(t) ) has a maximum despite the second derivative being positive. But that's impossible because a positive second derivative implies convexity, meaning any critical point is a minimum.Wait, unless ( k ) is negative. If ( k ) is negative, then ( e^{kt} ) decreases over time, and ( N(t) = e^{kt} - (mt + c) ) would have a maximum somewhere. Let me consider that.If ( k ) is negative, then ( e^{kt} ) decreases, so ( N(t) ) would be a decreasing exponential minus a linear function. So, the derivative ( N'(t) = k e^{kt} - m ). If ( k ) is negative, then ( k e^{kt} ) is negative and decreasing. So, ( N'(t) ) starts negative and becomes more negative, but if ( m ) is positive, it might cross zero from above to below, creating a maximum.Wait, let me think. If ( k ) is negative, say ( k = -|k| ), then ( N'(t) = -|k| e^{-|k| t} - m ). So, as ( t ) increases, ( e^{-|k| t} ) decreases, so ( -|k| e^{-|k| t} ) increases towards zero. So, ( N'(t) ) starts at ( -|k| - m ) and approaches ( -m ). So, if ( -|k| - m < 0 ) and ( -m < 0 ), then ( N'(t) ) is always negative, so no maximum.Alternatively, if ( k ) is positive, as we had before, ( N'(t) = k e^{kt} - m ), which is increasing because ( k e^{kt} ) grows exponentially. So, ( N'(t) ) starts at ( k - m ) and increases to infinity. So, if ( k - m < 0 ), then ( N'(t) ) crosses zero from below to above, meaning a minimum. If ( k - m > 0 ), then ( N'(t) ) is always positive, so no critical point.Wait, but the problem says that ( N(t) ) reaches a maximum at ( t = 30 ). So, perhaps the function ( N(t) ) is concave down, meaning ( N''(t) < 0 ). But in our case, ( N''(t) = k^2 e^{kt} ), which is always positive if ( k ) is real. So, unless ( k ) is imaginary, which it's not, we can't have ( N''(t) < 0 ).This is confusing. Maybe the problem has a typo, and ( N(t) ) is supposed to be ( Q(t) - P(t) ). Let me assume that for a moment.If ( N(t) = Q(t) - P(t) = mt + c - e^{kt} ), then ( N'(t) = m - k e^{kt} ), and ( N''(t) = -k^2 e^{kt} ), which is negative, so any critical point is a maximum.So, setting ( N'(30) = 0 ):( m - k e^{30k} = 0 ), so ( m = k e^{30k} ).From part 1, ( k = ln2 /15 ), so ( e^{30k} = 4 ), so ( m = 4k ).So, the relationship is ( m = 4k ). But the problem asks for the relationship between ( m ), ( c ), and ( k ). So, unless ( c ) is related somehow, but in this case, ( c ) doesn't affect the derivative, so it can be any constant. So, maybe the relationship is ( m = 4k ), and ( c ) is arbitrary.But the problem says "find the relationship between ( m ), ( c ), and ( k )", so perhaps I need to express ( c ) in terms of ( m ) and ( k ). But without another condition, I can't do that. So, maybe the answer is just ( m = 4k ), and ( c ) is unrelated.Alternatively, maybe I need to consider the value of ( N(t) ) at ( t = 30 ), but since it's a maximum, we don't know its value. So, perhaps the only relationship is ( m = 4k ).Wait, but the problem says "the net performance of the player, considering both encouragement and pressure, is given by ( N(t) = P(t) - Q(t) )". So, it's ( P(t) - Q(t) ), not the other way around. So, unless the problem is wrong, or I'm misunderstanding something.Alternatively, maybe the function ( N(t) ) is defined as ( Q(t) - P(t) ), but the problem says ( P(t) - Q(t) ). So, perhaps the problem is correct, and I need to reconcile the fact that ( N(t) ) has a maximum despite ( N''(t) > 0 ). That seems impossible, so maybe the problem is intended to have ( N(t) = Q(t) - P(t) ), which would make sense.Given that, I think the intended answer is ( m = 4k ), and ( c ) is arbitrary, so the relationship is ( m = 4k ). But since the problem mentions ( c ), maybe I need to express ( c ) in terms of ( m ) and ( k ). But without another condition, I can't do that. So, perhaps the answer is ( m = 4k ), and ( c ) is a constant, so the relationship is ( m = 4k ) and ( c ) is arbitrary.But the problem says "find the relationship between ( m ), ( c ), and ( k )", so maybe it's just ( m = 4k ), and ( c ) is not related. Alternatively, perhaps ( c ) is related through the value of ( N(t) ) at ( t = 30 ), but since we don't know the value, we can't express ( c ) in terms of ( m ) and ( k ).Wait, maybe I can express ( c ) in terms of ( m ) and ( k ) using the fact that ( N(t) ) has a maximum at ( t = 30 ). So, ( N(30) = e^{30k} - (30m + c) ). But since it's a maximum, we don't know its value, so we can't solve for ( c ). So, I think ( c ) remains arbitrary, and the only relationship is ( m = 4k ).Therefore, the relationship is ( m = 4k ), and ( c ) is a constant that can be any value. But the problem asks for the relationship between all three, so maybe it's just ( m = 4k ), and ( c ) is independent.Alternatively, perhaps the problem expects ( c ) to be expressed in terms of ( m ) and ( k ) using the maximum condition, but without knowing the value of ( N(30) ), we can't do that. So, I think the answer is ( m = 4k ), and ( c ) is arbitrary.But to be thorough, let me write down the equations:From part 1: ( k = ln2 /15 ).From part 2: ( m = k e^{30k} = k *4 = 4k ).So, ( m = 4k ).And ( c ) is just a constant in ( Q(t) = mt + c ), which doesn't affect the derivative, so it's arbitrary.Therefore, the relationship is ( m = 4k ), and ( c ) can be any constant.But the problem says "find the relationship between ( m ), ( c ), and ( k )", so maybe it's just ( m = 4k ), and ( c ) is unrelated. Alternatively, if they expect a single equation involving all three, but without another condition, I can't see how.Wait, perhaps the maximum value of ( N(t) ) is zero, meaning ( N(30) = 0 ). Then, we could write ( e^{30k} - (30m + c) = 0 ). Substituting ( m = 4k ), we get ( 4 - (30*4k + c) = 0 ). But ( k = ln2 /15 ), so ( 30*4k = 120k = 120*(ln2)/15 = 8 ln2 ). So, ( 4 - (8 ln2 + c) = 0 ), so ( c = 4 - 8 ln2 ). But the problem doesn't state that ( N(30) = 0 ), so I can't assume that.Therefore, I think the only relationship we can establish is ( m = 4k ), and ( c ) is arbitrary. So, the relationship between ( m ), ( c ), and ( k ) is ( m = 4k ), with ( c ) being any constant.But the problem might expect a different approach. Let me think again.Wait, maybe I need to consider the entire function ( N(t) = e^{kt} - (mt + c) ) and find the condition for a maximum at ( t = 30 ). So, we have two conditions:1. ( N'(30) = 0 ): ( k e^{30k} - m = 0 ) => ( m = k e^{30k} = 4k ).2. ( N''(30) < 0 ): But ( N''(t) = k^2 e^{kt} > 0 ), which contradicts the maximum. So, unless ( k ) is negative, but then ( e^{kt} ) would decrease, and ( N(t) ) would be a decreasing exponential minus a linear function, which might have a maximum.Wait, if ( k ) is negative, let's say ( k = -|k| ), then ( N(t) = e^{-|k| t} - (mt + c) ).Then, ( N'(t) = -|k| e^{-|k| t} - m ).Setting ( N'(30) = 0 ):( -|k| e^{-30|k|} - m = 0 ) => ( m = -|k| e^{-30|k|} ).But ( m ) is a constant in the linear function ( Q(t) = mt + c ). If ( m ) is negative, that would mean the pressure decreases over time, which might not make sense in the context of a rugby match where pressure typically increases. So, perhaps ( k ) is positive, and the problem is intended to have ( N(t) = Q(t) - P(t) ), making the second derivative negative.Given that, I think the intended answer is ( m = 4k ), and ( c ) is arbitrary. So, the relationship is ( m = 4k ).Therefore, summarizing:1. ( k = ln2 /15 ).2. ( m = 4k ), and ( c ) is arbitrary.But since the problem asks for the relationship between ( m ), ( c ), and ( k ), and we can't express ( c ) in terms of ( m ) and ( k ) without additional information, the only relationship is ( m = 4k ).Alternatively, if we consider that ( c ) can be expressed in terms of ( m ) and ( k ) using the value of ( N(t) ) at ( t = 30 ), but since we don't know that value, we can't do so. Therefore, the relationship is ( m = 4k ), and ( c ) is a constant.So, to answer the question:1. ( k = ln2 /15 ).2. ( m = 4k ), and ( c ) is arbitrary.But the problem might expect a single equation involving all three, but without another condition, I can't see how. So, I think the answer is ( m = 4k ), and ( c ) is arbitrary.Wait, but the problem says "find the relationship between ( m ), ( c ), and ( k )", so maybe it's just ( m = 4k ), and ( c ) is unrelated. Alternatively, if they expect ( c ) to be expressed in terms of ( m ) and ( k ), but without another condition, it's impossible.Therefore, I think the answer is ( m = 4k ), and ( c ) is arbitrary. So, the relationship is ( m = 4k ).</think>

🧠:<think>Okay, so I have this problem about calculating the probability that at least 25 articles are published by six different writing groups in a given year. Each group follows a Poisson distribution with different average rates. Let me try to break this down step by step.First, I remember that the Poisson distribution is used to model the number of events happening in a fixed interval of time or space. In this case, each writing group's number of publications per year is modeled by a Poisson distribution with its own lambda (λ_i). The first part of the problem asks for the probability that the total number of articles published by all six groups in a year is at least 25. So, I need to find P(X₁ + X₂ + X₃ + X₄ + X₅ + X₆ ≥ 25), where each X_i is Poisson distributed with λ_i given.I recall that the sum of independent Poisson random variables is also a Poisson random variable with the parameter being the sum of the individual lambdas. So, if I sum up all the λ_i, that should give me the lambda for the total publications.Let me calculate the total lambda first:λ_total = λ₁ + λ₂ + λ₃ + λ₄ + λ₅ + λ₆= 3 + 5 + 4 + 2 + 6 + 7= 3 + 5 is 8, plus 4 is 12, plus 2 is 14, plus 6 is 20, plus 7 is 27.So, the total lambda is 27. That means the total number of articles published in a year by all groups follows a Poisson distribution with λ = 27.Now, I need to find P(X ≥ 25), where X ~ Poisson(27). Calculating this directly might be a bit tricky because Poisson probabilities can get cumbersome for large λ. But maybe I can use the complement rule: P(X ≥ 25) = 1 - P(X ≤ 24).Calculating P(X ≤ 24) would require summing up the probabilities from X=0 to X=24. That's a lot of terms, but perhaps there's a better way. Alternatively, I remember that for large λ, the Poisson distribution can be approximated by a normal distribution with mean μ = λ and variance σ² = λ. So, maybe I can use the normal approximation here.Let me check if the normal approximation is appropriate. The rule of thumb is that both μ and σ should be sufficiently large, usually μ ≥ 10 and σ ≥ 3. Here, μ = 27 and σ = sqrt(27) ≈ 5.196, which are both well above those thresholds. So, the normal approximation should be reasonable.So, I can model X as approximately N(27, 27). To find P(X ≥ 25), I can standardize this and use the Z-score.But wait, since we're dealing with a discrete distribution (Poisson) and approximating it with a continuous distribution (normal), I should apply the continuity correction. That means I should adjust the boundary by 0.5. So, instead of P(X ≥ 25), I should calculate P(X ≥ 24.5).Let me write that down:P(X ≥ 25) ≈ P(X ≥ 24.5) using continuity correction.Now, converting 24.5 to a Z-score:Z = (24.5 - μ) / σ= (24.5 - 27) / sqrt(27)= (-2.5) / 5.196≈ -0.481So, Z ≈ -0.481. Now, I need to find the probability that Z is greater than or equal to -0.481. Looking at standard normal distribution tables, the area to the left of Z = -0.48 is approximately 0.3156, and the area to the left of Z = -0.481 is roughly the same. Therefore, the area to the right (which is what we need) is 1 - 0.3156 = 0.6844.Wait, but hold on. If Z is -0.481, the area to the left is 0.3156, so the area to the right is 1 - 0.3156 = 0.6844. So, P(X ≥ 24.5) ≈ 0.6844.But wait, that seems high. Let me double-check. If the mean is 27, then 24.5 is just slightly below the mean. So, the probability of being above 24.5 should be more than 0.5, which 0.6844 is. Hmm, that seems plausible.Alternatively, maybe I can compute the exact probability using the Poisson formula. But calculating the sum from 25 to infinity is tedious. Maybe I can compute it using the cumulative distribution function (CDF) for Poisson.But without a calculator or software, it's difficult. Alternatively, I can use the normal approximation as I did before. Let me see if I can get a better approximation.Alternatively, maybe I can use the Poisson CDF formula:P(X ≤ k) = e^{-λ} * Σ_{i=0}^{k} (λ^i / i!)But calculating this up to k=24 with λ=27 is going to be time-consuming. Maybe I can use some properties or approximations.Alternatively, I remember that for Poisson distributions, the median is approximately equal to the mean, so 27. So, P(X ≤ 27) is about 0.5. But we're looking for P(X ≥ 25), which is more than half.Wait, but 25 is just 2 less than 27. So, maybe the probability is more than 0.5. The normal approximation gave me 0.6844, which seems reasonable.Alternatively, maybe I can use the fact that for Poisson, the probability mass function is symmetric around the mean in some way, but I don't think that's exactly true. Poisson is skewed, especially for smaller λ, but for larger λ, it becomes more symmetric.Given that λ=27 is reasonably large, the normal approximation should be decent. So, I think 0.6844 is a reasonable approximation.But let me see if I can get a better estimate. Maybe using the exact Poisson CDF. Let me try to compute P(X ≤ 24) using the Poisson formula.But calculating this manually would take a lot of time. Alternatively, I can use the recursive formula for Poisson probabilities:P(X = k) = (λ / k) * P(X = k - 1)Starting from P(X=0) = e^{-27} ≈ 0.00000005 (since e^{-27} is a very small number). Then, P(X=1) = 27 * P(X=0) ≈ 27 * 0.00000005 ≈ 0.00000135.But even P(X=1) is extremely small. Wait, that can't be right. Wait, no, actually, for Poisson(27), the probabilities are spread out, but the peak is around 27. So, the probabilities for k=0,1,...,24 are actually not that small, but the sum up to 24 is still a significant portion.Wait, but calculating each term up to 24 is going to be time-consuming. Maybe I can use the relationship between Poisson and chi-squared distributions? I remember that the sum of independent Poisson variables can be related to chi-squared, but I'm not sure.Alternatively, maybe I can use the fact that the Poisson CDF can be expressed in terms of the incomplete gamma function:P(X ≤ k) = Γ(k + 1, λ) / k!But without computational tools, this is difficult.Alternatively, perhaps using the normal approximation with continuity correction is the best approach here.So, going back, I had Z ≈ -0.481, leading to P(X ≥ 24.5) ≈ 0.6844.But let me verify this with another method. Maybe using the Central Limit Theorem, since we're summing multiple Poisson variables. Wait, but in this case, we already have the sum as a single Poisson variable with λ=27, so the CLT would suggest that it's approximately normal, which is what I used.Alternatively, maybe I can use the exact Poisson probability for X=25 and see how it compares.But again, without computational tools, it's difficult.Alternatively, maybe I can use the fact that for Poisson(λ), the probability P(X ≥ λ + k) can be approximated using the normal distribution.But I think I've spent enough time on this. The normal approximation with continuity correction gives me approximately 0.6844, so I'll go with that.Wait, but let me check the Z-score again. I had:Z = (24.5 - 27) / sqrt(27) ≈ (-2.5)/5.196 ≈ -0.481Looking up Z=-0.48 in standard normal tables, the cumulative probability is approximately 0.3156. So, P(Z ≥ -0.481) = 1 - 0.3156 = 0.6844.Yes, that seems correct.So, the probability that at least 25 articles are published is approximately 0.6844, or 68.44%.But wait, let me think again. If the mean is 27, then 25 is just slightly below the mean. So, the probability of being above 25 should be more than 50%, which 68% is. That seems reasonable.Alternatively, maybe I can use the exact Poisson CDF. Let me try to compute it approximately.I know that for Poisson(λ), the CDF can be approximated using the normal distribution, but for better accuracy, maybe I can use the fact that the Poisson CDF can be expressed in terms of the gamma function, but without computational tools, it's hard.Alternatively, I can use the fact that the Poisson distribution is approximately normal for large λ, so the normal approximation is acceptable.Therefore, I think the answer is approximately 0.6844, or 68.44%.But let me check if I can get a better approximation. Maybe using the Poisson CDF formula with some terms.Alternatively, I can use the relationship between Poisson and chi-squared. The sum of independent Poisson variables is Poisson, and the CDF can be related to the chi-squared distribution.Wait, I think there's a formula that relates the Poisson CDF to the chi-squared CDF. Specifically, P(X ≤ k) = Γ(k + 1, λ) / k! = 1 - Γc(k + 1, λ) / k!, where Γc is the complementary incomplete gamma function.But without computational tools, I can't compute this exactly. Alternatively, I can use an approximation for the incomplete gamma function.Alternatively, maybe I can use the fact that for Poisson(λ), the CDF can be approximated using the normal distribution with continuity correction, which is what I did earlier.Given that, I think the normal approximation is the way to go here.So, to summarize:Total λ = 27We need P(X ≥ 25) ≈ P(X ≥ 24.5) using continuity correction.Z = (24.5 - 27)/sqrt(27) ≈ -0.481P(Z ≥ -0.481) ≈ 0.6844Therefore, the probability is approximately 0.6844, or 68.44%.But let me double-check the Z-score calculation:24.5 - 27 = -2.5sqrt(27) ≈ 5.19615-2.5 / 5.19615 ≈ -0.481Yes, that's correct.Looking up Z=-0.48 in standard normal table:The area to the left of Z=-0.48 is approximately 0.3156, so the area to the right is 1 - 0.3156 = 0.6844.Yes, that's correct.Therefore, the probability is approximately 0.6844.But let me think if there's another way to approach this. Maybe using the moment generating function or something else, but I think the normal approximation is sufficient here.So, for part 1, the answer is approximately 0.6844.Now, moving on to part 2.The history enthusiast wants to predict the total number of articles published over the next 10 years. The publication rates remain constant, and each group's annual publications are independent.We need to determine the expected total number of articles and the standard deviation of this total number.First, the expected total number of articles over 10 years.Since each year, the total number of articles is Poisson(27), as calculated earlier. The expectation of a Poisson variable is λ, so E[X] = 27 per year.Over 10 years, the expected total is 10 * 27 = 270.Now, for the standard deviation.The variance of a Poisson variable is also λ, so Var(X) = 27 per year.Over 10 years, since each year is independent, the total variance is 10 * 27 = 270.Therefore, the standard deviation is sqrt(270).Calculating sqrt(270):270 = 9 * 30, so sqrt(270) = 3 * sqrt(30) ≈ 3 * 5.477 ≈ 16.43.Alternatively, sqrt(270) ≈ 16.4316767.So, the standard deviation is approximately 16.43.Wait, let me verify that.Yes, because for independent variables, the variance adds up. So, over 10 years, the total variance is 10 * 27 = 270, so standard deviation is sqrt(270).Yes, that's correct.Alternatively, since each year's total is Poisson(27), the total over 10 years is Poisson(270), because the sum of independent Poisson variables is Poisson with λ equal to the sum of individual λs.Wait, that's an important point. So, actually, the total number of articles over 10 years is Poisson(270), because each year is Poisson(27), and they're independent, so the sum is Poisson(27*10)=Poisson(270).Therefore, the expected total is 270, and the standard deviation is sqrt(270) ≈ 16.43.So, that's the answer for part 2.But let me think again. Since each year is independent, the total over 10 years is indeed the sum of 10 independent Poisson(27) variables, which is Poisson(270). Therefore, the expectation is 270, and the variance is 270, so standard deviation is sqrt(270).Yes, that's correct.So, to summarize:1. The probability that at least 25 articles are published in a year is approximately 0.6844.2. The expected total number of articles over 10 years is 270, with a standard deviation of approximately 16.43.I think that's it.</think>

🧠:<think>Okay, so I have this problem where a non-profit organization is trying to analyze their fundraising strategies. They have two strategies, A and B, each modeled by different functions. The first part asks me to calculate the expected total donation amount per month if they allocate 60% focus on Strategy A and 40% on Strategy B. The second part introduces a risk function and asks me to find the month that minimizes this risk. Hmm, let's break this down step by step.Starting with part 1. The functions given are:- Strategy A: ( f(x) = 5000 + 3000 sinleft(frac{pi}{6}xright) )- Strategy B: ( g(x) = 4000 + 2000 cosleft(frac{pi}{4}xright) )They want to allocate 60% to A and 40% to B. Since the focus affects the outputs linearly, I think this means we can just take a weighted average of the two functions. So, the expected total donation amount per month, let's call it ( h(x) ), would be:( h(x) = 0.6 times f(x) + 0.4 times g(x) )Plugging in the functions:( h(x) = 0.6 times left(5000 + 3000 sinleft(frac{pi}{6}xright)right) + 0.4 times left(4000 + 2000 cosleft(frac{pi}{4}xright)right) )Let me compute this step by step. First, distribute the 0.6 and 0.4:For Strategy A:- 0.6 * 5000 = 3000- 0.6 * 3000 = 1800, so the sine term becomes 1800 sin(πx/6)For Strategy B:- 0.4 * 4000 = 1600- 0.4 * 2000 = 800, so the cosine term becomes 800 cos(πx/4)Adding these together:( h(x) = 3000 + 1800 sinleft(frac{pi}{6}xright) + 1600 + 800 cosleft(frac{pi}{4}xright) )Combine the constants:3000 + 1600 = 4600So, ( h(x) = 4600 + 1800 sinleft(frac{pi}{6}xright) + 800 cosleft(frac{pi}{4}xright) )Alright, so that's the expected total donation amount per month. It seems straightforward. I think that's the answer for part 1.Moving on to part 2. The risk function is given by:( R(x) = sqrt{(f(x) - bar{f})^2 + (g(x) - bar{g})^2} )Where ( bar{f} ) and ( bar{g} ) are the average monthly donations from Strategies A and B over 12 months. So, I need to compute the average of f(x) and g(x) over a year (12 months) first.Let me recall that the average value of a function over an interval can be found by integrating over that interval and dividing by the length. Since x is the month number from 1 to 12, the interval is from x=1 to x=12.But wait, actually, since these are periodic functions, their average over a full period can be found by integrating over one period. Let's check the periods of the sine and cosine functions.For Strategy A: The sine function has an argument of ( frac{pi}{6}x ). The period of sin(kx) is ( 2pi / k ). So, period is ( 2pi / (pi/6) ) = 12 ). So, the period is 12 months, which makes sense because it's over a year.Similarly, for Strategy B: The cosine function has an argument of ( frac{pi}{4}x ). The period is ( 2pi / (pi/4) ) = 8 ). Wait, 8 months? But we're looking at 12 months. Hmm, so over 12 months, Strategy B's function completes 1.5 periods. That might complicate things a bit.But for the average, since we're integrating over 12 months, which is not an integer multiple of the period for Strategy B, the average might not be as straightforward. Let me compute the average for both f(x) and g(x) over 12 months.Starting with ( bar{f} ):( bar{f} = frac{1}{12} int_{1}^{13} f(x) dx )Wait, actually, since x is the month number starting from January, which is x=1, and we need to compute the average over 12 months, so from x=1 to x=12. So, the integral is from 1 to 12.But integrating from 1 to 12 might be a bit messy, but perhaps we can adjust the variable to make it from 0 to 12, which might make the integral easier.Let me set t = x - 1, so when x=1, t=0, and x=12, t=11. Hmm, maybe not so helpful. Alternatively, perhaps we can note that the average of a sine function over its period is zero. Since f(x) is 5000 plus a sine function with period 12, the average of the sine term over 12 months would be zero. Similarly, for g(x), which is 4000 plus a cosine function with period 8, but over 12 months, which is 1.5 periods, the average of the cosine term might not be zero.Wait, let's think about it. For f(x):( f(x) = 5000 + 3000 sinleft(frac{pi}{6}xright) )The average of sin(kx) over one period is zero. Since the period is 12, integrating from x=1 to x=13 (which is one full period) would give zero. But we're integrating from x=1 to x=12, which is just one month short of a full period. So, the average might not be exactly zero. Hmm, this complicates things.Similarly, for g(x):( g(x) = 4000 + 2000 cosleft(frac{pi}{4}xright) )The period is 8, so over 12 months, it's 1.5 periods. The average of cosine over 1.5 periods... Hmm, the average of cosine over an integer number of periods is zero, but over 1.5 periods, it might not be.Wait, maybe I can compute the average by integrating over 12 months.Let me compute ( bar{f} ):( bar{f} = frac{1}{12} int_{1}^{12} left[5000 + 3000 sinleft(frac{pi}{6}xright)right] dx )Similarly, ( bar{g} = frac{1}{12} int_{1}^{12} left[4000 + 2000 cosleft(frac{pi}{4}xright)right] dx )Let me compute ( bar{f} ) first.Compute the integral:( int_{1}^{12} 5000 dx = 5000 times (12 - 1) = 5000 times 11 = 55000 )Now, the integral of the sine term:( int_{1}^{12} 3000 sinleft(frac{pi}{6}xright) dx )Let me compute this integral. The integral of sin(ax) dx is -(1/a) cos(ax) + C.So,( 3000 times left[ -frac{6}{pi} cosleft(frac{pi}{6}xright) right] ) evaluated from 1 to 12.Compute at x=12:( -frac{6}{pi} cosleft(frac{pi}{6} times 12right) = -frac{6}{pi} cos(2pi) = -frac{6}{pi} times 1 = -frac{6}{pi} )Compute at x=1:( -frac{6}{pi} cosleft(frac{pi}{6}right) = -frac{6}{pi} times frac{sqrt{3}}{2} = -frac{3sqrt{3}}{pi} )So, the integral is:3000 * [ (-6/π) - (-3√3/π) ] = 3000 * [ (-6 + 3√3)/π ] = 3000 * (3√3 - 6)/πSimplify:3000 * 3(√3 - 2)/π = 9000(√3 - 2)/πSo, the integral of the sine term is 9000(√3 - 2)/πTherefore, the total integral for f(x) is 55000 + 9000(√3 - 2)/πThus, the average ( bar{f} ) is:( frac{55000 + 9000(sqrt{3} - 2)/pi}{12} )Let me compute this numerically to get an approximate value.First, compute 9000(√3 - 2)/π:√3 ≈ 1.732, so √3 - 2 ≈ -0.26799000 * (-0.2679) ≈ -2411.1Divide by π ≈ 3.1416: -2411.1 / 3.1416 ≈ -767.5So, the integral of the sine term is approximately -767.5Therefore, total integral ≈ 55000 - 767.5 ≈ 54232.5Average ( bar{f} ) ≈ 54232.5 / 12 ≈ 4519.375So, approximately 4519.38Now, let's compute ( bar{g} ):( bar{g} = frac{1}{12} int_{1}^{12} left[4000 + 2000 cosleft(frac{pi}{4}xright)right] dx )Compute the integral:( int_{1}^{12} 4000 dx = 4000 times 11 = 44000 )Integral of the cosine term:( int_{1}^{12} 2000 cosleft(frac{pi}{4}xright) dx )Integral of cos(ax) dx is (1/a) sin(ax) + C.So,2000 * [ (4/π) sin(πx/4) ] evaluated from 1 to 12.Compute at x=12:(4/π) sin(3π) = (4/π) * 0 = 0Compute at x=1:(4/π) sin(π/4) = (4/π) * (√2/2) = (2√2)/πSo, the integral is:2000 * [ 0 - (2√2)/π ] = -2000 * (2√2)/π = -4000√2 / πApproximately, √2 ≈ 1.414, so 4000 * 1.414 ≈ 5656So, -5656 / π ≈ -5656 / 3.1416 ≈ -1800.5Therefore, the integral of the cosine term is approximately -1800.5Total integral ≈ 44000 - 1800.5 ≈ 42199.5Average ( bar{g} ) ≈ 42199.5 / 12 ≈ 3516.625So, approximately 3516.63Wait, let me double-check the integral calculations because I might have messed up the signs.For ( bar{f} ):Integral of sine term was:3000 * [ (-6/π) - (-3√3/π) ] = 3000 * [ (-6 + 3√3)/π ]Which is 3000*(3√3 -6)/π ≈ 3000*(5.196 -6)/π ≈ 3000*(-0.804)/π ≈ -2412/π ≈ -767.5, which matches.For ( bar{g} ):Integral of cosine term:2000 * [ (4/π)(sin(3π) - sin(π/4)) ] = 2000*(4/π)(0 - √2/2) = 2000*( -2√2 / π ) ≈ 2000*(-2.828)/π ≈ -5656/π ≈ -1800.5, which is correct.So, the averages are approximately:( bar{f} ≈ 4519.38 )( bar{g} ≈ 3516.63 )Now, the risk function is:( R(x) = sqrt{(f(x) - bar{f})^2 + (g(x) - bar{g})^2} )We need to find the month x (from 1 to 12) that minimizes R(x).This seems a bit complex because R(x) is a function involving both sine and cosine terms. Maybe we can express R(x)^2 to simplify, since the square root is a monotonic function, so minimizing R(x) is equivalent to minimizing R(x)^2.So, let's define:( R(x)^2 = (f(x) - bar{f})^2 + (g(x) - bar{g})^2 )Plugging in f(x) and g(x):( R(x)^2 = left(5000 + 3000 sinleft(frac{pi}{6}xright) - bar{f}right)^2 + left(4000 + 2000 cosleft(frac{pi}{4}xright) - bar{g}right)^2 )We already have approximate values for ( bar{f} ) and ( bar{g} ), so let's plug those in:( R(x)^2 ≈ left(5000 + 3000 sinleft(frac{pi}{6}xright) - 4519.38right)^2 + left(4000 + 2000 cosleft(frac{pi}{4}xright) - 3516.63right)^2 )Simplify the constants:For the first term:5000 - 4519.38 ≈ 480.62So, first term becomes:( (480.62 + 3000 sin(pi x /6))^2 )Second term:4000 - 3516.63 ≈ 483.37So, second term becomes:( (483.37 + 2000 cos(pi x /4))^2 )Thus,( R(x)^2 ≈ (480.62 + 3000 sin(pi x /6))^2 + (483.37 + 2000 cos(pi x /4))^2 )This is a function of x, and we need to find the x (integer from 1 to 12) that minimizes this expression.This seems a bit involved. Maybe we can compute R(x)^2 for each month from 1 to 12 and find the minimum.Alternatively, perhaps we can find the x that makes both terms as small as possible. Since both terms are squares, the minimum occurs when both deviations are minimized.But since the functions are periodic, it's possible that the minimum occurs when both sine and cosine terms are at their average points, but I'm not sure.Alternatively, perhaps we can set the derivatives to zero, but since x is discrete (months 1-12), it's probably easier to compute R(x) for each x and pick the smallest.Let me try that approach.First, let's note the values of sin(πx/6) and cos(πx/4) for x from 1 to 12.Compute sin(πx/6):x | sin(πx/6)1 | sin(π/6) = 0.52 | sin(π/3) ≈ 0.8663 | sin(π/2) = 14 | sin(2π/3) ≈ 0.8665 | sin(5π/6) = 0.56 | sin(π) = 07 | sin(7π/6) = -0.58 | sin(4π/3) ≈ -0.8669 | sin(3π/2) = -110 | sin(5π/3) ≈ -0.86611 | sin(11π/6) = -0.512 | sin(2π) = 0Similarly, compute cos(πx/4):x | cos(πx/4)1 | cos(π/4) ≈ 0.7072 | cos(π/2) = 03 | cos(3π/4) ≈ -0.7074 | cos(π) = -15 | cos(5π/4) ≈ -0.7076 | cos(3π/2) = 07 | cos(7π/4) ≈ 0.7078 | cos(2π) = 19 | cos(9π/4) ≈ 0.70710 | cos(5π/2) = 011 | cos(11π/4) ≈ -0.70712 | cos(3π) = -1Wait, let me verify these:For x=1: cos(π/4) ≈ 0.707x=2: cos(π/2)=0x=3: cos(3π/4)= -√2/2 ≈ -0.707x=4: cos(π)= -1x=5: cos(5π/4)= -√2/2 ≈ -0.707x=6: cos(3π/2)=0x=7: cos(7π/4)=√2/2 ≈ 0.707x=8: cos(2π)=1x=9: cos(9π/4)=cos(π/4)=√2/2 ≈0.707x=10: cos(5π/2)=0x=11: cos(11π/4)=cos(3π/4)= -√2/2 ≈-0.707x=12: cos(3π)= -1Okay, that seems correct.Now, let's compute R(x)^2 for each x from 1 to 12.Starting with x=1:First term: 480.62 + 3000*0.5 = 480.62 + 1500 = 1980.62Second term: 483.37 + 2000*0.707 ≈ 483.37 + 1414 ≈ 1897.37R(x)^2 ≈ (1980.62)^2 + (1897.37)^2 ≈ 3,923,000 + 3,600,000 ≈ 7,523,000x=2:First term: 480.62 + 3000*0.866 ≈ 480.62 + 2598 ≈ 3078.62Second term: 483.37 + 2000*0 ≈ 483.37R(x)^2 ≈ (3078.62)^2 + (483.37)^2 ≈ 9,475,000 + 233,000 ≈ 9,708,000x=3:First term: 480.62 + 3000*1 = 480.62 + 3000 = 3480.62Second term: 483.37 + 2000*(-0.707) ≈ 483.37 - 1414 ≈ -930.63R(x)^2 ≈ (3480.62)^2 + (-930.63)^2 ≈ 12,115,000 + 866,000 ≈ 12,981,000x=4:First term: 480.62 + 3000*0.866 ≈ 480.62 + 2598 ≈ 3078.62Second term: 483.37 + 2000*(-1) ≈ 483.37 - 2000 ≈ -1516.63R(x)^2 ≈ (3078.62)^2 + (-1516.63)^2 ≈ 9,475,000 + 2,300,000 ≈ 11,775,000x=5:First term: 480.62 + 3000*0.5 = 480.62 + 1500 = 1980.62Second term: 483.37 + 2000*(-0.707) ≈ 483.37 - 1414 ≈ -930.63R(x)^2 ≈ (1980.62)^2 + (-930.63)^2 ≈ 3,923,000 + 866,000 ≈ 4,789,000x=6:First term: 480.62 + 3000*0 = 480.62Second term: 483.37 + 2000*0 = 483.37R(x)^2 ≈ (480.62)^2 + (483.37)^2 ≈ 231,000 + 233,000 ≈ 464,000x=7:First term: 480.62 + 3000*(-0.5) = 480.62 - 1500 ≈ -1019.38Second term: 483.37 + 2000*0.707 ≈ 483.37 + 1414 ≈ 1897.37R(x)^2 ≈ (-1019.38)^2 + (1897.37)^2 ≈ 1,039,000 + 3,600,000 ≈ 4,639,000x=8:First term: 480.62 + 3000*(-0.866) ≈ 480.62 - 2598 ≈ -2117.38Second term: 483.37 + 2000*1 ≈ 483.37 + 2000 ≈ 2483.37R(x)^2 ≈ (-2117.38)^2 + (2483.37)^2 ≈ 4,483,000 + 6,169,000 ≈ 10,652,000x=9:First term: 480.62 + 3000*(-1) = 480.62 - 3000 ≈ -2519.38Second term: 483.37 + 2000*0.707 ≈ 483.37 + 1414 ≈ 1897.37R(x)^2 ≈ (-2519.38)^2 + (1897.37)^2 ≈ 6,347,000 + 3,600,000 ≈ 9,947,000x=10:First term: 480.62 + 3000*(-0.866) ≈ 480.62 - 2598 ≈ -2117.38Second term: 483.37 + 2000*(-0.707) ≈ 483.37 - 1414 ≈ -930.63R(x)^2 ≈ (-2117.38)^2 + (-930.63)^2 ≈ 4,483,000 + 866,000 ≈ 5,349,000x=11:First term: 480.62 + 3000*(-0.5) = 480.62 - 1500 ≈ -1019.38Second term: 483.37 + 2000*(-0.707) ≈ 483.37 - 1414 ≈ -930.63R(x)^2 ≈ (-1019.38)^2 + (-930.63)^2 ≈ 1,039,000 + 866,000 ≈ 1,905,000x=12:First term: 480.62 + 3000*0 = 480.62Second term: 483.37 + 2000*(-1) ≈ 483.37 - 2000 ≈ -1516.63R(x)^2 ≈ (480.62)^2 + (-1516.63)^2 ≈ 231,000 + 2,300,000 ≈ 2,531,000Now, let's list all R(x)^2 values:x | R(x)^21 | ~7,523,0002 | ~9,708,0003 | ~12,981,0004 | ~11,775,0005 | ~4,789,0006 | ~464,0007 | ~4,639,0008 | ~10,652,0009 | ~9,947,00010 | ~5,349,00011 | ~1,905,00012 | ~2,531,000Looking at these, the smallest R(x)^2 is at x=6, which is approximately 464,000.Wait, let me check x=6:First term: 480.62 + 0 = 480.62Second term: 483.37 + 0 = 483.37So, R(x)^2 = (480.62)^2 + (483.37)^2 ≈ 231,000 + 233,000 ≈ 464,000Yes, that's the smallest.But wait, let me check x=11:R(x)^2 ≈ 1,905,000, which is bigger than x=6.x=12: ~2,531,000x=5: ~4,789,000x=7: ~4,639,000So, x=6 is indeed the smallest.Therefore, the month that minimizes the risk function is x=6, which is June.Interpreting this, the organization should focus on June as the month where the risk is minimized, meaning that the deviations from the average donations for both strategies are the smallest. This implies that June is a relatively stable month in terms of donations, with both strategies performing close to their average performances, thus minimizing the risk of large fluctuations.Alternatively, since x=6 is June, it's the midpoint of the year, and perhaps the strategies have stabilized by then, leading to less variability.So, summarizing:1. The expected total donation amount per month is ( h(x) = 4600 + 1800 sinleft(frac{pi}{6}xright) + 800 cosleft(frac{pi}{4}xright) ).2. The month that minimizes the risk function is June (x=6), indicating that this is the most stable month for donations, balancing both strategies effectively.</think>