🧠:<think>First, I need to calculate the total annual energy output from all 10 power plants. Each plant produces 500 MW, and they operate 24 hours a day. Converting megawatts to megawatt-hours per year involves multiplying by 24 hours and 365 days.Next, I'll determine the current energy loss due to the 5% transmission loss. This is done by taking 5% of the total annual energy output.Then, I'll calculate the energy loss if the transmission loss is reduced to 2%. This involves finding 2% of the total annual energy output.To find the additional energy saved annually by reducing the loss from 5% to 2%, I'll subtract the new energy loss from the current energy loss.For the break-even analysis, I know that each 1% reduction in transmission loss saves 1 million annually. Reducing the loss by 3% (from 5% to 2%) will result in 3 million in annual savings.Finally, I'll determine how many years it will take for the 3 million annual savings to cover the initial 20 million investment. This is done by dividing the investment by the annual savings.</think>

🧠:<think>Alright, so I have this problem about cognitive load modeling for developers using graph theory and information theory. Hmm, let me try to break it down step by step.Starting with part 1: We have a directed graph G where each vertex represents a concept or tool, and each directed edge represents a dependency or usage relationship. For each tool T_i, E_i is the set of edges that are direct dependencies needed to understand T_i. The cognitive load L_i is the sum of the weights of these edges. The weight of each edge e is given by w(e) = -log(p(e)), where p(e) is the probability of the dependency e being used correctly.Okay, so for each tool T_i, I need to find L_i in terms of p(e). Since L_i is the sum of the weights of edges in E_i, and each weight is -log(p(e)), then L_i should just be the sum of -log(p(e)) for all edges e in E_i.Let me write that out:L_i = Σ_{e ∈ E_i} w(e) = Σ_{e ∈ E_i} (-log(p(e)))So, that's straightforward. It's just the sum of the negative logarithms of the probabilities for each dependency edge of tool T_i.Now, moving on to part 2: The overall cognitive load L is the sum of all individual cognitive loads L_i. It's given that L follows a normal distribution N(μ, σ²). We need to find the mean μ and variance σ² based on the individual L_i, assuming they are independent and identically distributed (i.i.d) random variables.Alright, so if each L_i is i.i.d, then the mean of the sum is the sum of the means, and the variance of the sum is the sum of the variances. Since they are identically distributed, each L_i has the same mean and variance.Let me denote the number of tools as n. So, L = L_1 + L_2 + ... + L_n.Then, the mean μ of L is:μ = E[L] = E[L_1 + L_2 + ... + L_n] = E[L_1] + E[L_2] + ... + E[L_n] = n * E[L_i]Similarly, the variance σ² is:σ² = Var(L) = Var(L_1 + L_2 + ... + L_n) = Var(L_1) + Var(L_2) + ... + Var(L_n) = n * Var(L_i)But wait, the problem doesn't specify the number of tools, so maybe it's just in terms of each L_i? Or perhaps it's considering the sum over all tools, but without knowing n, we can't give a numerical value. Hmm.Wait, actually, the problem says "the overall cognitive load L for a developer is modeled as the sum of the cognitive loads for all tools". So, if there are n tools, then L = Σ_{i=1}^n L_i.Therefore, the mean μ is n times the mean of a single L_i, and the variance σ² is n times the variance of a single L_i.But since the problem doesn't specify n, maybe we can express μ and σ² in terms of the individual means and variances. Alternatively, if each L_i is identical, then μ = n * μ_i and σ² = n * σ_i², where μ_i is the mean of L_i and σ_i² is the variance of L_i.But hold on, the problem says "derive the expressions for the mean μ and the variance σ² of the total cognitive load based on the individual cognitive loads L_i". So, probably, if we have n tools, then:μ = Σ_{i=1}^n E[L_i] = n * E[L_i] (since they are identical)σ² = Σ_{i=1}^n Var(L_i) = n * Var(L_i)But without knowing n, we can't compute exact values, but we can express μ and σ² in terms of the individual means and variances. However, since they are i.i.d, the total mean is n times the individual mean, and total variance is n times the individual variance.But wait, in the problem statement, it's mentioned that L follows a normal distribution N(μ, σ²). So, if each L_i is normal, then the sum is also normal with parameters as above.But hold on, actually, each L_i is a sum of logs, which might not necessarily be normal, but the problem states that the overall L is normal. So, maybe the Central Limit Theorem is in play here, but the problem says L is normal, so we can take it as given.So, to sum up, the mean μ is the sum of the means of each L_i, which, since they are identical, is n times the mean of a single L_i. Similarly, the variance σ² is n times the variance of a single L_i.But since the problem doesn't specify n, perhaps we can express μ and σ² in terms of the individual L_i's mean and variance. Alternatively, if we consider that each L_i is a random variable, then the total mean is the sum of individual means, and the total variance is the sum of individual variances.But since they are i.i.d, we can write:μ = Σ E[L_i] = n * E[L_i]σ² = Σ Var(L_i) = n * Var(L_i)But without knowing n, we can't proceed further. Wait, maybe the problem is considering that each L_i is a term in the sum, and since they are i.i.d, the total mean is n * μ_i and variance is n * σ_i².But the problem doesn't specify n, so perhaps we can just express μ and σ² in terms of the individual means and variances. Alternatively, maybe it's considering that each L_i is a term, and since they are i.i.d, the total mean is the sum of individual means, and variance is the sum of individual variances.But without knowing how many tools there are, we can't give a numerical expression. Hmm.Wait, perhaps the problem is just asking for the expressions in terms of the individual L_i's. So, if L = Σ L_i, then μ = Σ E[L_i] and σ² = Σ Var(L_i). Since they are i.i.d, μ = n * E[L_i] and σ² = n * Var(L_i). But since n isn't given, maybe we can leave it as μ = Σ E[L_i] and σ² = Σ Var(L_i).But the problem says "derive the expressions for the mean μ and the variance σ² of the total cognitive load based on the individual cognitive loads L_i". So, probably, it's expecting expressions in terms of the individual means and variances, not necessarily involving n.Wait, but if L is the sum of all L_i, then μ is the sum of all individual means, and σ² is the sum of all individual variances. Since the L_i are i.i.d, each has the same mean and variance, so μ = n * μ_i and σ² = n * σ_i². But without knowing n, we can't write it in terms of n. Hmm.Alternatively, maybe the problem is considering that each L_i is a term, and since they are i.i.d, the total mean is the sum of individual means, and variance is the sum of individual variances. So, if there are k tools, then μ = k * E[L_i] and σ² = k * Var(L_i). But since k isn't specified, perhaps the answer is just μ = Σ E[L_i] and σ² = Σ Var(L_i).But the problem says "based on the individual cognitive loads L_i", so maybe it's expecting expressions in terms of the individual L_i's. So, if L = Σ L_i, then μ = Σ E[L_i] and σ² = Σ Var(L_i). Since the L_i are independent, the variance adds up.But since the L_i are i.i.d, E[L_i] is the same for all i, say μ_i, and Var(L_i) is the same, say σ_i². Then, μ = n * μ_i and σ² = n * σ_i². But without knowing n, we can't write it in terms of n. So, perhaps the answer is just μ = Σ E[L_i] and σ² = Σ Var(L_i).Alternatively, maybe the problem is considering that each L_i is a term, and since they are i.i.d, the total mean is the sum of individual means, and variance is the sum of individual variances. So, if there are k tools, then μ = k * E[L_i] and σ² = k * Var(L_i). But since k isn't specified, perhaps the answer is just μ = Σ E[L_i] and σ² = Σ Var(L_i).Wait, but the problem says "derive the expressions for the mean μ and the variance σ² of the total cognitive load based on the individual cognitive loads L_i". So, I think it's expecting the general expressions, not in terms of n. So, if L = Σ L_i, then μ = Σ E[L_i] and σ² = Σ Var(L_i). Since the L_i are independent, the variance is the sum of variances.But since the L_i are i.i.d, each has the same mean and variance, so μ = n * E[L_i] and σ² = n * Var(L_i). But without knowing n, we can't write it in terms of n. Hmm.Wait, maybe the problem is just asking for the general formula, regardless of the number of tools. So, for the mean, it's the sum of the means of each L_i, and for the variance, it's the sum of the variances of each L_i. Since they are independent, the covariance terms are zero.So, in symbols:μ = Σ_{i=1}^n E[L_i]σ² = Σ_{i=1}^n Var(L_i)But since the L_i are i.i.d, this simplifies to:μ = n * E[L_i]σ² = n * Var(L_i)But since n isn't given, perhaps we can just express it in terms of the sum.Alternatively, maybe the problem is considering that each L_i is a term, and since they are i.i.d, the total mean is the sum of individual means, and variance is the sum of individual variances. So, if there are k tools, then μ = k * E[L_i] and σ² = k * Var(L_i). But since k isn't specified, perhaps the answer is just μ = Σ E[L_i] and σ² = Σ Var(L_i).I think that's the way to go. So, the mean is the sum of the expected values of each L_i, and the variance is the sum of the variances of each L_i. Since they are independent, the covariance terms are zero.So, putting it all together:For part 1, L_i is the sum of -log(p(e)) for each edge e in E_i.For part 2, μ is the sum of E[L_i] and σ² is the sum of Var(L_i). Since the L_i are i.i.d, μ = n * E[L_i] and σ² = n * Var(L_i), but without knowing n, we can express it as the sum.Wait, but the problem says "derive the expressions for the mean μ and the variance σ² of the total cognitive load based on the individual cognitive loads L_i". So, maybe it's expecting the general formula, not necessarily in terms of n.So, final answers:1. L_i = Σ_{e ∈ E_i} (-log(p(e)))2. μ = Σ E[L_i], σ² = Σ Var(L_i)But since the L_i are i.i.d, we can write μ = n * E[L_i] and σ² = n * Var(L_i), but without knowing n, perhaps it's better to leave it as the sum.Alternatively, if we consider that each L_i is a term, and since they are i.i.d, the total mean is the sum of individual means, and variance is the sum of individual variances. So, if there are k tools, then μ = k * E[L_i] and σ² = k * Var(L_i). But since k isn't specified, perhaps the answer is just μ = Σ E[L_i] and σ² = Σ Var(L_i).I think that's the way to go.</think>

🧠:<think>Okay, so I need to solve these two sub-problems related to the functions S(t) and C(t). Let me start by understanding the given information.First, S(t) is given by the function S(t) = A cos(Bt) + D. The constants are A = 5, B = π/6, and D = 10. So plugging those in, S(t) = 5 cos(πt/6) + 10. That makes sense—it's a cosine function with amplitude 5, period related to π/6, and a vertical shift of 10. So the satisfaction level oscillates between 5 and 15, right? Because cosine varies between -1 and 1, so 5*(-1) +10 = 5 and 5*(1) +10 = 15.Then, the number of complaints C(t) is inversely proportional to S(t), so C(t) = k / S(t). They gave k = 100, so C(t) = 100 / (5 cos(πt/6) + 10). That simplifies to C(t) = 100 / (5 cos(πt/6) + 10). Maybe I can simplify that further. Let me factor out a 5 from the denominator: 5(cos(πt/6) + 2). So C(t) = 100 / [5(cos(πt/6) + 2)] = 20 / (cos(πt/6) + 2). That might be easier to work with.Alright, moving on to the first sub-problem: Determine the critical points of C(t) over the interval [0, 12] months. I need to find where the derivative of C(t) is zero or undefined, and then determine if those points are local maxima, minima, or saddle points.So, to find critical points, I need to compute C'(t) and set it equal to zero. Let me write down C(t):C(t) = 20 / (cos(πt/6) + 2)Let me denote the denominator as D(t) = cos(πt/6) + 2. So C(t) = 20 / D(t). Then, the derivative C'(t) is -20 * D'(t) / [D(t)]².First, compute D'(t):D(t) = cos(πt/6) + 2D'(t) = -sin(πt/6) * (π/6)So D'(t) = - (π/6) sin(πt/6)Therefore, C'(t) = -20 * [ - (π/6) sin(πt/6) ] / [cos(πt/6) + 2]^2Simplify that:C'(t) = (20 * π / 6) * sin(πt/6) / [cos(πt/6) + 2]^2Simplify 20π/6 to 10π/3:C'(t) = (10π/3) * sin(πt/6) / [cos(πt/6) + 2]^2So, critical points occur where C'(t) = 0 or where C'(t) is undefined.Looking at the denominator [cos(πt/6) + 2]^2, since cos(πt/6) ranges between -1 and 1, so cos(πt/6) + 2 ranges between 1 and 3. So the denominator is always positive, never zero. Therefore, C(t) is differentiable everywhere in [0,12], so no undefined points.Therefore, critical points occur only where C'(t) = 0. So set numerator equal to zero:(10π/3) * sin(πt/6) = 0Since 10π/3 is non-zero, sin(πt/6) = 0.Solutions to sin(πt/6) = 0 are when πt/6 = nπ, where n is integer. So t/6 = n, so t = 6n.In the interval [0,12], n can be 0,1,2. So t = 0,6,12.So critical points at t = 0,6,12.Wait, but t=0 and t=12 are endpoints of the interval. So when considering critical points, sometimes endpoints are not considered unless specified. But the problem says "over the interval [0,12]", so I think we include endpoints as critical points? Or do we only consider interior points? Hmm.Wait, critical points are points in the domain where derivative is zero or undefined. Since t=0 and t=12 are endpoints, but the derivative is defined there (as the denominator is non-zero), so they are critical points.But when determining if they are maxima or minima, we have to consider the behavior around them.But let's see. So critical points at t=0,6,12.Now, to determine whether each critical point is a local maximum, minimum, or saddle point, we can use the second derivative test or analyze the sign changes of the first derivative.But since this is a continuous function on a closed interval, we can also evaluate the function at these points and see.Alternatively, since we have a function C(t) which is 20 / (cos(πt/6) + 2), let's analyze its behavior.Note that cos(πt/6) has a period of 12 months, right? Because the period of cos(Bt) is 2π / B, so here B = π/6, so period is 2π / (π/6) = 12. So over [0,12], it completes one full cycle.So, the function S(t) = 5 cos(πt/6) +10 oscillates between 5 and 15, as I thought earlier.Therefore, C(t) = 20 / (cos(πt/6) + 2) will oscillate between 20 / (1 + 2) = 20/3 ≈6.666 and 20 / ( -1 + 2 ) = 20/1 =20. So C(t) varies between approximately 6.666 and 20.So when cos(πt/6) is at its maximum (1), C(t) is at its minimum (20/3), and when cos(πt/6) is at its minimum (-1), C(t) is at its maximum (20).But wait, cos(πt/6) is 1 when πt/6 = 0, 2π, 4π,... So t=0,12,24,... So in [0,12], t=0 and t=12 are points where cos(πt/6)=1. Similarly, cos(πt/6)=-1 when πt/6=π, 3π, 5π,... So t=6, 18, etc. So in [0,12], t=6 is where cos(πt/6)=-1.Therefore, t=0 and t=12 are points where C(t) is minimized, and t=6 is where C(t) is maximized.But wait, let's check:At t=0: cos(0)=1, so C(0)=20/(1+2)=20/3≈6.666At t=6: cos(π*6/6)=cos(π)= -1, so C(6)=20/(-1 +2)=20/1=20At t=12: cos(π*12/6)=cos(2π)=1, so C(12)=20/3≈6.666So, yes, t=0 and t=12 are minima, and t=6 is a maximum.But wait, are t=0 and t=12 considered local minima? Since they are endpoints, sometimes they are considered as global minima or maxima, but for critical points, they are still critical points.But in terms of being local maxima or minima, since they are endpoints, they can only be local if the function doesn't go beyond that in the neighborhood. But since t=0 is the left endpoint, and t=12 is the right endpoint, they can be considered local minima if the function increases after t=0 and decreases before t=12.But let's check the behavior around t=0 and t=12.Looking at C'(t):C'(t) = (10π/3) * sin(πt/6) / [cos(πt/6) + 2]^2At t=0: sin(0)=0, so derivative is zero.Just to the right of t=0, t=0+, sin(πt/6) is positive (since πt/6 is small positive), so C'(t) is positive. So function is increasing at t=0. Therefore, t=0 is a local minimum.Similarly, at t=12: sin(π*12/6)=sin(2π)=0, so derivative is zero.Just to the left of t=12, t=12-, sin(πt/6)=sin(2π - ε)= -sin(ε)≈-ε, which is negative. So C'(t) is negative just before t=12, meaning function is decreasing as it approaches t=12. Therefore, t=12 is also a local minimum.At t=6: sin(π*6/6)=sin(π)=0, so derivative is zero.Looking at the derivative around t=6:Just before t=6, say t=6-ε, sin(π*(6-ε)/6)=sin(π - πε/6)=sin(π - x)=sin(x)≈sin(πε/6)≈πε/6, which is positive. So C'(t) is positive before t=6.Just after t=6, t=6+ε, sin(π*(6+ε)/6)=sin(π + πε/6)= -sin(πε/6)≈-πε/6, which is negative. So C'(t) changes from positive to negative at t=6. Therefore, t=6 is a local maximum.So, summarizing:Critical points at t=0,6,12.t=0: local minimumt=6: local maximumt=12: local minimumSo that's the first sub-problem.Now, moving on to the second sub-problem: Find the total number of complaints filed over the 12-month period by evaluating the integral ∫₀¹² C(t) dt.We have C(t) = 20 / (cos(πt/6) + 2). So the integral is ∫₀¹² [20 / (cos(πt/6) + 2)] dt.This integral might be a bit tricky, but let's see if we can compute it.First, let me make a substitution to simplify the integral.Let me set u = πt/6. Then, du/dt = π/6, so dt = (6/π) du.When t=0, u=0; when t=12, u=π*12/6=2π.So the integral becomes:∫₀²π [20 / (cos(u) + 2)] * (6/π) duSimplify constants:20*(6/π) = 120/πSo integral is (120/π) ∫₀²π [1 / (cos u + 2)] duSo now, we need to compute ∫ [1 / (cos u + 2)] du from 0 to 2π.I recall that integrals of the form ∫ 1/(a + b cos u) du can be evaluated using the substitution t = tan(u/2), which is the Weierstrass substitution.Let me try that.Let me set t = tan(u/2). Then, cos u = (1 - t²)/(1 + t²), and du = 2 dt / (1 + t²).So substituting into the integral:∫ [1 / ( (1 - t²)/(1 + t²) + 2 ) ] * [2 dt / (1 + t²) ]Simplify the denominator:(1 - t²)/(1 + t²) + 2 = [1 - t² + 2(1 + t²)] / (1 + t²) = [1 - t² + 2 + 2t²] / (1 + t²) = (3 + t²) / (1 + t²)So the integral becomes:∫ [1 / ( (3 + t²)/(1 + t²) ) ] * [2 dt / (1 + t²) ] = ∫ [ (1 + t²)/(3 + t²) ] * [2 / (1 + t²) ] dt = ∫ [2 / (3 + t²) ] dtThat simplifies nicely to 2 ∫ 1/(t² + 3) dt.The integral of 1/(t² + a²) dt is (1/a) arctan(t/a) + C. So here, a = sqrt(3).Therefore, ∫ [1/(t² + 3)] dt = (1/√3) arctan(t/√3) + C.So putting it all together:2 ∫ [1/(t² + 3)] dt = 2*(1/√3) arctan(t/√3) + C = (2/√3) arctan(t/√3) + C.Now, revert back to u:t = tan(u/2), so arctan(t/√3) = arctan( tan(u/2) / √3 )But let's evaluate the definite integral from u=0 to u=2π.So when u=0, t = tan(0) = 0.When u=2π, t = tan(π) = 0.Wait, that's a problem because tan(u/2) at u=2π is tan(π) = 0, same as u=0. So the substitution causes the limits to coincide, which complicates things.Alternatively, maybe I can exploit the periodicity of the integrand.The integrand 1/(cos u + 2) has period 2π, so integrating over 0 to 2π is the same as integrating over any interval of length 2π. But perhaps we can compute it over 0 to π and double it, but let's see.Alternatively, maybe use symmetry.Wait, let me think again. The substitution t = tan(u/2) maps u from 0 to 2π to t from 0 to ∞ and back to 0, which complicates the limits. Maybe a better substitution is needed.Alternatively, perhaps use the identity:∫₀^{2π} 1/(a + b cos u) du = 2π / sqrt(a² - b²) when a > |b|In our case, a = 2, b = 1, so a² - b² = 4 -1 =3, which is positive. So the integral is 2π / sqrt(3).Therefore, ∫₀²π [1 / (cos u + 2)] du = 2π / sqrt(3)So going back, our integral was (120/π) * (2π / sqrt(3)) = (120/π)*(2π)/sqrt(3) = 240 / sqrt(3)Simplify 240 / sqrt(3): Multiply numerator and denominator by sqrt(3):240 sqrt(3) / 3 = 80 sqrt(3)So the total number of complaints over 12 months is 80 sqrt(3).Wait, let me verify that.Wait, the integral ∫₀²π [1/(cos u + 2)] du = 2π / sqrt(a² - b²) where a=2, b=1, so sqrt(4 -1)=sqrt(3). So yes, 2π / sqrt(3).Multiply by (120/π):(120/π)*(2π / sqrt(3)) = (120 * 2) / sqrt(3) = 240 / sqrt(3) = 80 sqrt(3) after rationalizing.Yes, that seems correct.So the total number of complaints is 80√3.Let me just recap:We had C(t) = 20 / (cos(πt/6) + 2). The integral over 0 to 12 is transformed via substitution u=πt/6, leading to an integral over 0 to 2π of [20/(cos u + 2)]*(6/π) du. Simplify constants to get (120/π) ∫₀²π [1/(cos u + 2)] du. Using the standard integral formula, we find that integral is 2π / sqrt(3). Multiply by (120/π) to get 240 / sqrt(3) = 80 sqrt(3).So, the total number of complaints is 80√3.Therefore, the answers are:1. Critical points at t=0 (local minimum), t=6 (local maximum), and t=12 (local minimum).2. Total complaints: 80√3.Final Answer1. The critical points are at ( t = 0 ) (local minimum), ( t = 6 ) (local maximum), and ( t = 12 ) (local minimum).2. The total number of complaints filed over the 12-month period is (boxed{80sqrt{3}}).</think>

🧠:<think>Okay, so I have this problem about an elliptical park where a community fair is being organized. The park has a major axis of 200 meters and a minor axis of 120 meters. The volunteer wants to set up booths along the perimeter and in the interior. There are two parts to the problem: first, calculating the perimeter using Ramanujan's approximation formula, and second, figuring out the total area available for other activities after placing the maximum number of booths.Let me start with the first part. I need to calculate the perimeter of the ellipse. I remember that ellipses can be a bit tricky because they don't have a simple formula like circles. Ramanujan provided an approximation formula for the perimeter, which is given here as:[P approx pi left( 3(a + b) - sqrt{(3a + b)(a + 3b)} right)]Where ( a ) is the semi-major axis and ( b ) is the semi-minor axis. So, first, I need to find the semi-major and semi-minor axes from the given major and minor axes.The major axis is 200 meters, so the semi-major axis ( a ) is half of that, which is 100 meters. Similarly, the minor axis is 120 meters, so the semi-minor axis ( b ) is 60 meters. Got that down.Now, plugging these into Ramanujan's formula. Let me write that out step by step.First, calculate ( 3(a + b) ). That would be 3 times (100 + 60). So, 3*(160) = 480.Next, compute the square root term: ( sqrt{(3a + b)(a + 3b)} ). Let me compute each part inside the square root.Compute ( 3a + b ): 3*100 + 60 = 300 + 60 = 360.Compute ( a + 3b ): 100 + 3*60 = 100 + 180 = 280.Multiply these two results: 360 * 280. Hmm, let me calculate that. 360*280. Well, 360*200 is 72,000, and 360*80 is 28,800. So, adding those together: 72,000 + 28,800 = 100,800.So, the square root term is ( sqrt{100,800} ). Let me compute that. Hmm, 100,800 is 1008 * 100, so sqrt(1008*100) = sqrt(1008)*10. What's sqrt(1008)? Let me see.I know that 31^2 is 961 and 32^2 is 1024. So sqrt(1008) is between 31 and 32. Let's see, 31.7^2 is 31.7*31.7. 30*30=900, 30*1.7=51, 1.7*30=51, 1.7*1.7=2.89. So, adding up: 900 + 51 + 51 + 2.89 = 1004.89. Hmm, that's 31.7^2=1004.89. But we have 1008, which is 3.11 more. So, maybe 31.7 + some decimal.Alternatively, maybe it's easier to approximate sqrt(1008) as approximately 31.75, since 31.75^2 is (31 + 0.75)^2 = 31^2 + 2*31*0.75 + 0.75^2 = 961 + 46.5 + 0.5625 = 1008.0625. Oh, that's pretty close. So sqrt(1008) ≈ 31.75. Therefore, sqrt(100,800) = 31.75 * 10 = 317.5.So, going back to the formula, we have:[P approx pi left( 480 - 317.5 right) = pi (162.5)]Calculating that, 162.5 * π. Since π is approximately 3.1416, so 162.5 * 3.1416. Let me compute that.First, 160 * 3.1416 = 502.656.Then, 2.5 * 3.1416 = 7.854.Adding them together: 502.656 + 7.854 = 510.51 meters.So, the approximate perimeter is 510.51 meters.Wait, let me double-check my calculations because sometimes when approximating square roots, errors can creep in. So, I approximated sqrt(1008) as 31.75, which gave me sqrt(100,800) as 317.5. Then, 3(a + b) was 480, so 480 - 317.5 is 162.5, multiplied by π gives approximately 510.51 meters. That seems reasonable.Alternatively, maybe I can use a calculator for a more precise value, but since I don't have one, I think 510.51 is a good approximation.So, moving on to the second part. The volunteer wants to place booths along the perimeter. Each booth takes up 5 meters of perimeter space and requires an additional 10 square meters of interior space. We need to find the maximum number of booths that can be placed and then calculate the remaining area for other activities.First, let's find the maximum number of booths. Since each booth takes up 5 meters of perimeter, the number of booths is the total perimeter divided by 5.We have the perimeter as approximately 510.51 meters. So, 510.51 / 5 ≈ 102.102. Since we can't have a fraction of a booth, we take the integer part, which is 102 booths.Wait, but let me think again. Is 510.51 divided by 5 exactly 102.102? Yes, because 102*5=510, and 0.102*5=0.51, so 510 + 0.51=510.51. So, yes, 102.102 booths, but since we can't have 0.102 of a booth, we have to round down to 102 booths.So, maximum number of booths is 102.Each booth requires 10 square meters of interior space. So, total interior space used by booths is 102 * 10 = 1020 square meters.Now, we need to find the total area of the ellipse and subtract the area used by the booths to find the remaining area for other activities.First, let's compute the area of the ellipse. The formula for the area of an ellipse is:[A = pi a b]Where ( a ) is the semi-major axis and ( b ) is the semi-minor axis. We already have ( a = 100 ) meters and ( b = 60 ) meters.So, plugging in the values:[A = pi * 100 * 60 = 6000 pi]Calculating that, 6000 * π ≈ 6000 * 3.1416 ≈ 18,849.6 square meters.So, the total area of the ellipse is approximately 18,849.6 square meters.Now, subtracting the area used by the booths: 18,849.6 - 1020 = 17,829.6 square meters.Therefore, the total area available for other activities is approximately 17,829.6 square meters.Wait, let me just make sure I didn't make any calculation errors. So, area of ellipse: 100*60=6000, times π is about 18,849.6. Then, 102 booths * 10 square meters each is 1020. Subtracting that gives 17,829.6. That seems correct.But just to double-check, maybe I should compute 6000π more precisely. Let's see, π is approximately 3.1415926535. So, 6000 * 3.1415926535.Calculating 6000 * 3 = 18,000.6000 * 0.1415926535 ≈ 6000 * 0.1415926535.0.1415926535 * 6000: 0.1*6000=600, 0.04*6000=240, 0.0015926535*6000≈9.555921.Adding those together: 600 + 240 = 840, plus 9.555921 ≈ 849.555921.So, total area is 18,000 + 849.555921 ≈ 18,849.555921 square meters, which is approximately 18,849.56 square meters. So, my earlier approximation was pretty accurate.Subtracting 1020 gives 18,849.56 - 1020 = 17,829.56 square meters.So, rounding to a reasonable decimal place, maybe two decimal places, it's 17,829.56 square meters.But since the problem didn't specify the required precision, I think it's safe to present it as approximately 17,829.6 square meters.Wait, but let me think again about the perimeter calculation. I approximated sqrt(1008) as 31.75, which gave me sqrt(100,800) as 317.5. But let me verify that sqrt(1008) is indeed approximately 31.75.Calculating 31.75^2: 31^2=961, 0.75^2=0.5625, and the cross term is 2*31*0.75=46.5. So, 961 + 46.5 + 0.5625 = 1008.0625. So, yes, 31.75^2=1008.0625, which is very close to 1008. So, my approximation was correct.Therefore, the perimeter calculation is accurate.Another thing to consider: when calculating the number of booths, I divided the perimeter by 5 and got 102.102, which I rounded down to 102. Is that the correct approach? Because you can't have a fraction of a booth, so yes, you have to take the floor of that number. So, 102 booths is correct.Therefore, the area used by the booths is 102*10=1020 square meters, and the remaining area is 18,849.56 - 1020 = 17,829.56 square meters.So, summarizing:1. The perimeter is approximately 510.51 meters.2. The maximum number of booths is 102, and the remaining area is approximately 17,829.56 square meters.I think that's it. I don't see any mistakes in my reasoning. I double-checked the critical steps, so I'm confident in the answers.Final Answer1. The total perimeter of the ellipse is approximately boxed{510.51} meters.2. The total area available for other activities is approximately boxed{17829.56} square meters.</think>