🧠:<think>Alright, so I need to help Alex figure out the probability that a team makes a successful decision. The success is defined as at least 51% of the team members making the correct decision independently. The team size is n, each member has a probability p of making the correct decision, but groupthink reduces this probability by a factor g for each additional member. First, let's parse the problem. The effective probability that each member makes the correct decision is p - g*(n - 1). So, for each team member, their probability of being correct is adjusted based on the groupthink factor and the size of the team. That makes sense because as the team gets larger, the influence of groupthink increases, potentially reducing each member's independent decision-making ability.Now, for part 1, I need to derive the formula P(success). The success condition is that at least 51% of the team members make the correct decision. Since the team has n members, 51% of n is 0.51n. Since the number of people must be an integer, we'll need to take the ceiling of 0.51n if it's not already an integer. But maybe for the formula, we can just use 0.51n and then in calculations, we'll handle the integer part.So, the problem reduces to a binomial probability problem where each trial (team member) has a success probability of p_eff = p - g*(n - 1). We need the probability that the number of successes is at least k, where k is the smallest integer greater than or equal to 0.51n.Wait, but hold on. Is each team member's decision independent? The problem says "the effective probability that each member independently makes the correct decision in the presence of groupthink is p - g*(n - 1)". So, yes, each member's decision is independent, but their probability is reduced by g*(n - 1). So, the model is that each member has a probability p_eff = p - g*(n - 1) of making the correct decision, and these are independent events.Therefore, the number of correct decisions follows a binomial distribution with parameters n and p_eff. So, the probability of success is the sum from k = ceiling(0.51n) to n of the binomial probability mass function.So, mathematically, P(success) = sum_{k=ceil(0.51n)}^{n} C(n, k) * (p_eff)^k * (1 - p_eff)^{n - k}Where C(n, k) is the combination of n things taken k at a time.So, that's the formula. Now, for part 2, we have specific values: p = 0.7, g = 0.01, n = 10.First, let's compute p_eff. p_eff = p - g*(n - 1) = 0.7 - 0.01*(10 - 1) = 0.7 - 0.09 = 0.61.So, each team member has a 61% chance of making the correct decision independently.Now, the team size is 10, so 51% of 10 is 5.1. Since we can't have a fraction of a person, we need at least 6 correct decisions for success. So, k ranges from 6 to 10.Therefore, P(success) = sum_{k=6}^{10} C(10, k) * (0.61)^k * (0.39)^{10 - k}Now, let's compute each term:First, let's compute each combination:C(10,6) = 210C(10,7) = 120C(10,8) = 45C(10,9) = 10C(10,10) = 1Now, let's compute each term:For k=6:210 * (0.61)^6 * (0.39)^4First, compute (0.61)^6:0.61^2 = 0.37210.61^4 = (0.3721)^2 ≈ 0.13840.61^6 = 0.1384 * 0.3721 ≈ 0.0516Similarly, (0.39)^4:0.39^2 = 0.15210.39^4 = (0.1521)^2 ≈ 0.0231So, term for k=6: 210 * 0.0516 * 0.0231 ≈ 210 * 0.001191 ≈ 0.2501Wait, that seems low. Let me double-check the calculations.Wait, perhaps I should compute it more accurately.Compute (0.61)^6:0.61^1 = 0.610.61^2 = 0.61 * 0.61 = 0.37210.61^3 = 0.3721 * 0.61 ≈ 0.2269810.61^4 = 0.226981 * 0.61 ≈ 0.1383580.61^5 = 0.138358 * 0.61 ≈ 0.0844030.61^6 ≈ 0.084403 * 0.61 ≈ 0.051525Similarly, (0.39)^4:0.39^1 = 0.390.39^2 = 0.15210.39^3 = 0.1521 * 0.39 ≈ 0.0593190.39^4 ≈ 0.059319 * 0.39 ≈ 0.023134So, term for k=6: 210 * 0.051525 * 0.023134 ≈ 210 * 0.001191 ≈ 0.2501Wait, 0.051525 * 0.023134 ≈ 0.001191Then, 210 * 0.001191 ≈ 0.2501Similarly, for k=7:C(10,7) = 120(0.61)^7 = 0.051525 * 0.61 ≈ 0.031428(0.39)^3 ≈ 0.059319So, term for k=7: 120 * 0.031428 * 0.059319 ≈ 120 * 0.001868 ≈ 0.2242Wait, 0.031428 * 0.059319 ≈ 0.001868120 * 0.001868 ≈ 0.2242For k=8:C(10,8) = 45(0.61)^8 = 0.031428 * 0.61 ≈ 0.019170(0.39)^2 = 0.1521Term for k=8: 45 * 0.019170 * 0.1521 ≈ 45 * 0.002913 ≈ 0.1311For k=9:C(10,9) = 10(0.61)^9 ≈ 0.019170 * 0.61 ≈ 0.011704(0.39)^1 = 0.39Term for k=9: 10 * 0.011704 * 0.39 ≈ 10 * 0.004564 ≈ 0.04564For k=10:C(10,10) = 1(0.61)^10 ≈ 0.011704 * 0.61 ≈ 0.007140(0.39)^0 = 1Term for k=10: 1 * 0.007140 * 1 ≈ 0.007140Now, sum all these terms:k=6: ≈0.2501k=7: ≈0.2242k=8: ≈0.1311k=9: ≈0.04564k=10: ≈0.007140Total ≈0.2501 + 0.2242 = 0.47430.4743 + 0.1311 = 0.60540.6054 + 0.04564 ≈0.651040.65104 + 0.007140 ≈0.65818So, approximately 0.6582 or 65.82%.But let me verify these calculations because they seem a bit off. Maybe I should use a calculator for more precision.Alternatively, perhaps using the binomial formula with more accurate computations.Alternatively, using the binomial probability formula with exact values.Alternatively, perhaps using a calculator or software, but since I'm doing it manually, let's try to compute each term more accurately.Compute (0.61)^6:0.61^2 = 0.37210.61^4 = (0.3721)^2 = 0.13840.61^6 = 0.1384 * 0.3721 ≈ 0.0516Similarly, (0.39)^4 = (0.1521)^2 ≈ 0.0231So, term for k=6: 210 * 0.0516 * 0.0231 ≈ 210 * 0.001191 ≈ 0.2501k=7:(0.61)^7 = 0.0516 * 0.61 ≈ 0.0314(0.39)^3 = 0.0593Term: 120 * 0.0314 * 0.0593 ≈ 120 * 0.001868 ≈ 0.2242k=8:(0.61)^8 = 0.0314 * 0.61 ≈ 0.01917(0.39)^2 = 0.1521Term: 45 * 0.01917 * 0.1521 ≈ 45 * 0.002913 ≈ 0.1311k=9:(0.61)^9 ≈ 0.01917 * 0.61 ≈ 0.0117(0.39)^1 = 0.39Term: 10 * 0.0117 * 0.39 ≈ 10 * 0.004563 ≈ 0.04563k=10:(0.61)^10 ≈ 0.0117 * 0.61 ≈ 0.007137Term: 1 * 0.007137 ≈ 0.007137Adding up:0.2501 + 0.2242 = 0.47430.4743 + 0.1311 = 0.60540.6054 + 0.04563 ≈0.651030.65103 + 0.007137 ≈0.658167So, approximately 0.6582 or 65.82%.But let's check if this makes sense. With p_eff = 0.61, which is less than 0.7, but still a decent probability. For n=10, needing at least 6 correct decisions, the probability is around 65.8%.Alternatively, perhaps using the normal approximation to the binomial distribution could give a quicker estimate, but since n=10 is small, the approximation might not be very accurate. However, for the sake of verification, let's try.The expected number of successes, μ = n*p_eff = 10*0.61 = 6.1The variance, σ² = n*p_eff*(1 - p_eff) = 10*0.61*0.39 ≈ 10*0.2379 ≈ 2.379So, σ ≈ sqrt(2.379) ≈ 1.542We want P(X >=6). Since we're using the normal approximation, we can use continuity correction. So, P(X >=5.5).Z = (5.5 - μ)/σ = (5.5 - 6.1)/1.542 ≈ (-0.6)/1.542 ≈ -0.389Looking up Z=-0.389 in the standard normal table, the cumulative probability is approximately 0.3508. Therefore, P(X >=5.5) ≈ 1 - 0.3508 = 0.6492 or 64.92%, which is close to our exact calculation of 65.82%. So, that seems consistent.Therefore, the exact calculation gives approximately 65.82%, and the normal approximation gives 64.92%, which is pretty close.So, I think the exact value is around 65.8%, so 0.658 or 65.8%.But let me check if I can compute the exact value more accurately.Alternatively, perhaps using logarithms or more precise exponentiation.Alternatively, perhaps using the binomial formula with more precise calculations.Alternatively, perhaps using a calculator for each term.But since I'm doing it manually, let's try to compute each term with more precision.Compute (0.61)^6:0.61^1 = 0.610.61^2 = 0.61 * 0.61 = 0.37210.61^3 = 0.3721 * 0.61 = 0.2269810.61^4 = 0.226981 * 0.61 ≈ 0.1383580.61^5 = 0.138358 * 0.61 ≈ 0.0844030.61^6 ≈ 0.084403 * 0.61 ≈ 0.051525Similarly, (0.39)^4:0.39^1 = 0.390.39^2 = 0.15210.39^3 = 0.1521 * 0.39 ≈ 0.0593190.39^4 ≈ 0.059319 * 0.39 ≈ 0.023134So, term for k=6: 210 * 0.051525 * 0.023134First, 0.051525 * 0.023134 ≈ 0.001191Then, 210 * 0.001191 ≈ 0.25011Similarly, k=7:(0.61)^7 = 0.051525 * 0.61 ≈ 0.031428(0.39)^3 ≈ 0.059319Term: 120 * 0.031428 * 0.059319 ≈ 120 * 0.001868 ≈ 0.22416k=8:(0.61)^8 = 0.031428 * 0.61 ≈ 0.019170(0.39)^2 = 0.1521Term: 45 * 0.019170 * 0.1521 ≈ 45 * 0.002913 ≈ 0.131085k=9:(0.61)^9 = 0.019170 * 0.61 ≈ 0.011704(0.39)^1 = 0.39Term: 10 * 0.011704 * 0.39 ≈ 10 * 0.004564 ≈ 0.04564k=10:(0.61)^10 = 0.011704 * 0.61 ≈ 0.007140Term: 1 * 0.007140 ≈ 0.007140Now, summing up:0.25011 + 0.22416 = 0.474270.47427 + 0.131085 ≈ 0.6053550.605355 + 0.04564 ≈ 0.6509950.650995 + 0.007140 ≈ 0.658135So, approximately 0.6581 or 65.81%.Therefore, the probability of success is approximately 65.81%.So, to summarize:1. The formula for P(success) is the sum from k=ceil(0.51n) to n of C(n, k) * (p - g*(n - 1))^k * (1 - (p - g*(n - 1)))^{n - k}.2. For p=0.7, g=0.01, n=10, P(success) ≈ 65.81%.But let me check if I made any calculation errors. For example, in the term for k=6, 210 * 0.051525 * 0.023134.0.051525 * 0.023134 ≈ 0.001191210 * 0.001191 ≈ 0.25011Similarly, for k=7: 120 * 0.031428 * 0.059319 ≈ 120 * 0.001868 ≈ 0.22416Yes, that seems correct.k=8: 45 * 0.019170 * 0.1521 ≈ 45 * 0.002913 ≈ 0.131085k=9: 10 * 0.011704 * 0.39 ≈ 0.04564k=10: 1 * 0.007140 ≈ 0.007140Adding up: 0.25011 + 0.22416 = 0.47427+0.131085 = 0.605355+0.04564 = 0.650995+0.007140 = 0.658135Yes, that seems consistent.Alternatively, perhaps using a calculator for more precision, but I think this is sufficient.Therefore, the final answer is approximately 65.81%, which can be rounded to 65.8% or 0.658.</think>

🧠:<think>Okay, so I have this problem about a retired captain organizing a reunion and using a cipher system based on a jurisdictional map. The map is a graph with 15 precincts (nodes) and 25 communication lines (edges). He wants to use a spanning tree for encoding the invitations. The first part asks me to determine the total number of distinct spanning trees using Cayley's formula, considering the map as a complete graph on 15 precincts. Hmm, Cayley's formula is something I remember from graph theory. Let me recall... I think it's used to find the number of spanning trees in a complete graph. Cayley's formula states that the number of spanning trees in a complete graph with n nodes is n^(n-2). So, if we have 15 precincts, n would be 15. Therefore, the number of spanning trees should be 15^(15-2) = 15^13. Let me compute that. Wait, 15^13 is a huge number. Let me see if I can express it in terms of exponents or factorials, but I think Cayley's formula is straightforward here. So, the number is 15 to the power of 13. I don't think I need to calculate the exact numerical value unless specified, so I can leave it as 15^13. Moving on to the second part. The captain uses Kirchhoff's Matrix-Tree Theorem to find the exact spanning tree. I need to write down the Laplacian matrix of the graph and calculate the number of spanning trees using the determinant. First, what's a Laplacian matrix? If I remember correctly, the Laplacian matrix, also known as the graph Laplacian, is a matrix used in graph theory. It's defined as D - A, where D is the degree matrix and A is the adjacency matrix of the graph. Since the graph is a complete graph with 15 nodes, each node is connected to every other node. So, each node has a degree of 14. Therefore, the degree matrix D is a 15x15 diagonal matrix where each diagonal entry is 14. The adjacency matrix A for a complete graph is a 15x15 matrix where all the diagonal entries are 0 (since there are no self-loops) and all the off-diagonal entries are 1 (since every node is connected to every other node). So, the Laplacian matrix L would be D - A. That means each diagonal entry of L is 14, and each off-diagonal entry is -1. But wait, Kirchhoff's theorem says that the number of spanning trees is equal to any cofactor of the Laplacian matrix. So, I can compute the determinant of a minor of L (removing one row and one column) to get the number of spanning trees. However, calculating the determinant of a 14x14 matrix manually is going to be tedious. I wonder if there's a pattern or a formula for the determinant of such a matrix. I recall that for a complete graph with n nodes, the Laplacian matrix has a specific structure. Each diagonal entry is n-1, and each off-diagonal entry is -1. The determinant of any cofactor of this matrix is n^(n-2), which is exactly Cayley's formula. So, in this case, the determinant should be 15^13. Wait, so does that mean that regardless of which minor I take, the determinant will be 15^13? That seems consistent with what I know. So, even though the Laplacian matrix is 15x15, when I remove one row and one column, the resulting 14x14 matrix will have a determinant equal to the number of spanning trees, which is 15^13. But just to make sure, let me think about the properties of the Laplacian matrix. The Laplacian matrix is singular because the sum of each row is zero. Therefore, its determinant is zero. But when we remove a row and a column, the resulting matrix is non-singular, and its determinant gives the number of spanning trees. So, in this case, since the graph is complete, the number of spanning trees is 15^13, which matches both Cayley's formula and Kirchhoff's theorem. Therefore, for the first part, the number of spanning trees is 15^13, and for the second part, using the Laplacian matrix, the determinant of any cofactor will also give 15^13. I think I got it. The key was remembering Cayley's formula and understanding how the Laplacian matrix relates to the number of spanning trees via Kirchhoff's theorem. Since the graph is complete, the calculations simplify nicely without having to compute the determinant manually.Final Answer1. The total number of distinct spanning trees is boxed{15^{13}}.2. The number of spanning trees calculated using the Laplacian matrix determinant is also boxed{15^{13}}.</think>

🧠:<think>Alright, so I have this problem where a media publisher wants to maximize their audience engagement by allocating their 100,000 marketing budget between two platforms, A and B. The engagement rates for each platform are given by specific functions: ( E_A(x) = 5 sqrt{x} ) for Platform A and ( E_B(y) = 3 ln(y + 1) ) for Platform B. The total budget constraint is ( x + y = 100 ), where x and y are in thousands of dollars. First, I need to figure out how to maximize the total engagement, which is the sum of the engagements from both platforms. So, the total engagement function is ( E(x, y) = 5 sqrt{x} + 3 ln(y + 1) ). Since the total budget is fixed at 100,000 dollars, which is 100 in thousands, I have the constraint ( x + y = 100 ). I remember from calculus that when you have a function to maximize with a constraint, you can use the method of Lagrange multipliers or substitution. Since this seems like a two-variable optimization problem with a single constraint, substitution might be straightforward here. So, I can express y in terms of x using the constraint: ( y = 100 - x ). Then, substitute this into the total engagement function to make it a function of a single variable. Let me write that out:( E(x) = 5 sqrt{x} + 3 ln((100 - x) + 1) )Simplifying the natural log term:( E(x) = 5 sqrt{x} + 3 ln(101 - x) )Now, I need to find the value of x that maximizes E(x). To do this, I should take the derivative of E with respect to x, set it equal to zero, and solve for x. Let me compute the derivative step by step. The derivative of ( 5 sqrt{x} ) with respect to x is ( 5 * (1/(2 sqrt{x})) ), which is ( frac{5}{2 sqrt{x}} ). Next, the derivative of ( 3 ln(101 - x) ) with respect to x. Using the chain rule, the derivative of ln(u) is ( (1/u) * du/dx ). Here, u = 101 - x, so du/dx = -1. Therefore, the derivative is ( 3 * (1/(101 - x)) * (-1) ), which simplifies to ( -frac{3}{101 - x} ).Putting it all together, the derivative of E with respect to x is:( E'(x) = frac{5}{2 sqrt{x}} - frac{3}{101 - x} )To find the critical points, set E'(x) equal to zero:( frac{5}{2 sqrt{x}} - frac{3}{101 - x} = 0 )Let me solve for x. I can rearrange the equation:( frac{5}{2 sqrt{x}} = frac{3}{101 - x} )Cross-multiplying to eliminate the denominators:( 5(101 - x) = 2 sqrt{x} * 3 )Simplify both sides:Left side: ( 5 * 101 - 5x = 505 - 5x )Right side: ( 6 sqrt{x} )So, the equation becomes:( 505 - 5x = 6 sqrt{x} )Hmm, this is a bit tricky because of the square root. Maybe I can let ( t = sqrt{x} ), so that ( x = t^2 ). Let's substitute:( 505 - 5t^2 = 6t )Rewriting:( -5t^2 - 6t + 505 = 0 )Multiply both sides by -1 to make it a bit easier:( 5t^2 + 6t - 505 = 0 )Now, this is a quadratic equation in terms of t. I can use the quadratic formula to solve for t. The quadratic formula is ( t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where a = 5, b = 6, and c = -505.Calculating the discriminant first:( b^2 - 4ac = 6^2 - 4 * 5 * (-505) = 36 + 10100 = 10136 )So, the square root of 10136. Let me see, 100 squared is 10,000, so sqrt(10136) is a bit more than 100. Let me compute it:100^2 = 10,000101^2 = 10,201So, sqrt(10136) is between 100 and 101. Let's compute 100.68^2:100.68^2 = (100 + 0.68)^2 = 100^2 + 2*100*0.68 + 0.68^2 = 10,000 + 136 + 0.4624 = 10,136.4624Wait, that's actually 10,136.4624, which is very close to 10,136. So, sqrt(10136) ≈ 100.68.So, plugging back into the quadratic formula:( t = frac{-6 pm 100.68}{2 * 5} )We have two solutions:1. ( t = frac{-6 + 100.68}{10} = frac{94.68}{10} = 9.468 )2. ( t = frac{-6 - 100.68}{10} = frac{-106.68}{10} = -10.668 )Since t represents sqrt(x), it can't be negative. So, we discard the negative solution. Therefore, t ≈ 9.468.Therefore, sqrt(x) ≈ 9.468, so x ≈ (9.468)^2.Calculating 9.468 squared:9^2 = 810.468^2 ≈ 0.219Cross term: 2 * 9 * 0.468 ≈ 8.424So, total ≈ 81 + 8.424 + 0.219 ≈ 89.643Wait, that's an approximation, but let me compute it more accurately.9.468 * 9.468:First, 9 * 9 = 819 * 0.468 = 4.2120.468 * 9 = 4.2120.468 * 0.468 ≈ 0.219So, adding up:81 + 4.212 + 4.212 + 0.219 ≈ 81 + 8.424 + 0.219 ≈ 89.643So, x ≈ 89.643. Since x is in thousands of dollars, that would be approximately 89,643 allocated to Platform A.But let me check if this is correct. Let me plug x ≈ 89.643 back into the original equation to see if it satisfies the derivative equal to zero.Compute E'(x):( E'(x) = frac{5}{2 sqrt{89.643}} - frac{3}{101 - 89.643} )First, sqrt(89.643) ≈ 9.468So, 5 / (2 * 9.468) ≈ 5 / 18.936 ≈ 0.264Next, 101 - 89.643 ≈ 11.357So, 3 / 11.357 ≈ 0.264Therefore, E'(x) ≈ 0.264 - 0.264 = 0, which is correct. So, x ≈ 89.643 is indeed the critical point.Now, I should check whether this critical point is a maximum. Since we're dealing with a constrained optimization problem, and the function E(x) is likely concave, this critical point should be the maximum. But just to be thorough, I can check the second derivative or analyze the behavior.Alternatively, since the engagement functions are both increasing but with decreasing marginal returns (since the square root and log functions have decreasing derivatives), the total engagement function should be concave, meaning the critical point is indeed a maximum.Therefore, the optimal allocation is approximately x ≈ 89.643 (thousand dollars) to Platform A and y = 100 - x ≈ 10.357 (thousand dollars) to Platform B.But let me express this more precisely. Since we had t ≈ 9.468, which was sqrt(x), so x = t^2 ≈ (9.468)^2. Let me compute this more accurately.9.468 * 9.468:Compute 9 * 9 = 819 * 0.468 = 4.2120.468 * 9 = 4.2120.468 * 0.468 ≈ 0.219024Adding up:81 + 4.212 + 4.212 + 0.219024 = 81 + 8.424 + 0.219024 ≈ 89.643024So, x ≈ 89.643024, which is approximately 89.643 thousand dollars, or 89,643.Similarly, y = 100 - x ≈ 10.356976, which is approximately 10.357 thousand dollars, or 10,357.But perhaps we can express this more precisely. Let me see if I can find an exact solution.Looking back at the equation:( 505 - 5x = 6 sqrt{x} )Let me write this as:( 5x + 6 sqrt{x} - 505 = 0 )Let me let ( t = sqrt{x} ), so ( x = t^2 ). Then, the equation becomes:( 5t^2 + 6t - 505 = 0 )Which is the same quadratic as before. So, the solutions are t = [ -6 ± sqrt(36 + 10100) ] / 10 = [ -6 ± sqrt(10136) ] / 10Since sqrt(10136) is irrational, we can't express it as an exact fraction, so we have to leave it in terms of sqrt(10136). Alternatively, we can express the exact value of x as:x = ( [ -6 + sqrt(10136) ] / 10 )^2But that's a bit messy. Alternatively, we can rationalize sqrt(10136). Let me see if 10136 can be factored into squares.10136 divided by 4 is 2534. 2534 divided by 2 is 1267. 1267 is a prime number? Let me check: 1267 divided by 7 is 181, which is prime. So, 10136 = 4 * 2534 = 4 * 2 * 1267 = 8 * 1267. Since 1267 is 7 * 181, which are both primes. So, sqrt(10136) = sqrt(8 * 7 * 181) = 2 * sqrt(2 * 7 * 181) = 2 * sqrt(2534). So, it doesn't simplify further. Therefore, we can leave the exact value as:x = ( ( -6 + sqrt(10136) ) / 10 )^2But for practical purposes, we can use the approximate decimal value, which is about 89.643.Therefore, the optimal allocation is approximately 89,643 to Platform A and 10,357 to Platform B.Now, moving on to part 2: calculating the maximum possible engagement.We need to compute E(x, y) at x ≈ 89.643 and y ≈ 10.357.First, compute E_A(x) = 5 * sqrt(x). So, sqrt(89.643) ≈ 9.468, so 5 * 9.468 ≈ 47.34.Next, compute E_B(y) = 3 * ln(y + 1). So, y + 1 = 10.357 + 1 = 11.357. ln(11.357) is approximately... Let me recall that ln(10) ≈ 2.3026, ln(11) ≈ 2.3979, ln(12) ≈ 2.4849. So, 11.357 is between 11 and 12. Let me compute ln(11.357).Using a calculator approximation, ln(11.357) ≈ 2.430.Therefore, E_B(y) ≈ 3 * 2.430 ≈ 7.29.Adding both engagements: 47.34 + 7.29 ≈ 54.63.But let me compute this more accurately.First, sqrt(89.643) is exactly 9.468, as we saw earlier. So, 5 * 9.468 = 47.34.Now, for ln(11.357):We can use the Taylor series expansion around ln(11) to approximate ln(11.357). Let me recall that ln(a + h) ≈ ln(a) + h/a - (h^2)/(2a^2) + ... for small h.Here, a = 11, h = 0.357.So, ln(11.357) ≈ ln(11) + 0.357/11 - (0.357)^2/(2 * 11^2)Compute each term:ln(11) ≈ 2.39790.357 / 11 ≈ 0.03245(0.357)^2 ≈ 0.1274490.127449 / (2 * 121) ≈ 0.127449 / 242 ≈ 0.000526So, ln(11.357) ≈ 2.3979 + 0.03245 - 0.000526 ≈ 2.4298Therefore, ln(11.357) ≈ 2.4298So, E_B(y) = 3 * 2.4298 ≈ 7.2894Therefore, total engagement E ≈ 47.34 + 7.2894 ≈ 54.6294So, approximately 54.63 interactions per thousand dollars.But let me verify this with a calculator for more precision.Alternatively, using a calculator, ln(11.357) is approximately 2.430, so 3 * 2.430 = 7.29.Adding to 47.34 gives 54.63.Therefore, the maximum engagement is approximately 54.63.But perhaps I can compute it more precisely.Alternatively, let's use more accurate values.Compute sqrt(89.643):We know that 9.468^2 = 89.643, so sqrt(89.643) = 9.468 exactly.So, E_A(x) = 5 * 9.468 = 47.34.Now, for E_B(y):y = 100 - x = 10.357So, y + 1 = 11.357Compute ln(11.357):Using a calculator, ln(11.357) ≈ 2.4301Therefore, E_B(y) = 3 * 2.4301 ≈ 7.2903Total engagement: 47.34 + 7.2903 ≈ 54.6303So, approximately 54.63.But to be precise, let's carry out the calculation with more decimal places.First, let's compute sqrt(89.643):We have x = 89.643, so sqrt(89.643) = 9.468 exactly, as per earlier.So, E_A(x) = 5 * 9.468 = 47.34.Now, for E_B(y):y = 100 - 89.643 = 10.357y + 1 = 11.357Compute ln(11.357):Using a calculator, ln(11.357) ≈ 2.4301Therefore, E_B(y) = 3 * 2.4301 ≈ 7.2903Total engagement: 47.34 + 7.2903 ≈ 54.6303So, approximately 54.63.But perhaps we can express this more precisely. Let me see if I can compute ln(11.357) more accurately.Using the Taylor series expansion around a point where I know the ln value. Let's take a = 11.357 is close to e^2.43, since e^2.43 ≈ 11.357.But perhaps a better approach is to use the natural logarithm properties.Alternatively, use the fact that ln(11.357) can be expressed as ln(11 + 0.357). Let me use the expansion around a = 11.So, f(x) = ln(x), f(a + h) ≈ f(a) + h*f’(a) + (h^2/2)*f''(a) + ...Where a = 11, h = 0.357f(a) = ln(11) ≈ 2.397895272798f’(a) = 1/11 ≈ 0.0909090909091f''(a) = -1/(11)^2 ≈ -0.00826446280992So, f(a + h) ≈ 2.397895272798 + 0.357*(0.0909090909091) + (0.357^2)/2*(-0.00826446280992)Compute each term:First term: 2.397895272798Second term: 0.357 * 0.0909090909091 ≈ 0.0324545454545Third term: (0.357^2)/2 * (-0.00826446280992)Compute 0.357^2 = 0.127449Divide by 2: 0.0637245Multiply by -0.00826446280992: ≈ -0.000526So, total approximation:2.397895272798 + 0.0324545454545 - 0.000526 ≈ 2.42982381825So, ln(11.357) ≈ 2.42982381825Therefore, E_B(y) = 3 * 2.42982381825 ≈ 7.28947145475Adding to E_A(x):47.34 + 7.28947145475 ≈ 54.62947145475So, approximately 54.6295.Rounding to four decimal places, that's 54.6295.But perhaps we can carry this further. Let me compute ln(11.357) with more precision.Alternatively, use a calculator for a more accurate value.Using a calculator, ln(11.357) ≈ 2.4301So, 3 * 2.4301 ≈ 7.2903Thus, total engagement ≈ 47.34 + 7.2903 ≈ 54.6303So, approximately 54.63.Therefore, the maximum possible engagement is approximately 54.63 interactions per thousand dollars.But let me check if there's a more precise way to compute this.Alternatively, since we have the exact value of x, which is ( ( -6 + sqrt(10136) ) / 10 )^2, we can express the total engagement in terms of exact expressions, but that might be complicated.Alternatively, we can use more precise decimal approximations.Given that sqrt(10136) ≈ 100.677 (since 100.677^2 ≈ 10136.0009), so sqrt(10136) ≈ 100.677.Therefore, t = ( -6 + 100.677 ) / 10 ≈ 94.677 / 10 ≈ 9.4677So, t ≈ 9.4677, so x = t^2 ≈ 9.4677^2 ≈ 89.643Therefore, sqrt(x) ≈ 9.4677, so E_A(x) = 5 * 9.4677 ≈ 47.3385Now, y = 100 - x ≈ 10.357So, y + 1 = 11.357Compute ln(11.357):Using a calculator, ln(11.357) ≈ 2.4301Therefore, E_B(y) = 3 * 2.4301 ≈ 7.2903Adding together: 47.3385 + 7.2903 ≈ 54.6288So, approximately 54.6288, which rounds to 54.63.Therefore, the maximum engagement is approximately 54.63.But let me check if I can get a more precise value for ln(11.357).Using a calculator, ln(11.357) ≈ 2.430101So, 3 * 2.430101 ≈ 7.290303Adding to 47.3385:47.3385 + 7.290303 ≈ 54.628803So, approximately 54.6288, which is about 54.63.Therefore, the maximum engagement is approximately 54.63.But perhaps we can express this as a fraction or a more precise decimal.Alternatively, we can note that the exact value is 5*sqrt(x) + 3*ln(y + 1) where x ≈ 89.643 and y ≈ 10.357, but without exact expressions, we have to rely on decimal approximations.Therefore, the optimal allocation is approximately x ≈ 89.643 (thousand dollars) to Platform A and y ≈ 10.357 (thousand dollars) to Platform B, resulting in a maximum engagement of approximately 54.63 interactions per thousand dollars.But let me check if I made any errors in the calculations.Wait, when I computed the quadratic equation, I had:5t^2 + 6t - 505 = 0Solutions:t = [ -6 ± sqrt(36 + 10100) ] / 10 = [ -6 ± sqrt(10136) ] / 10Which is correct.Then, sqrt(10136) ≈ 100.677, so t ≈ ( -6 + 100.677 ) / 10 ≈ 94.677 / 10 ≈ 9.4677So, x ≈ (9.4677)^2 ≈ 89.643Yes, that's correct.Then, y ≈ 100 - 89.643 ≈ 10.357Then, E_A(x) = 5*sqrt(x) ≈ 5*9.4677 ≈ 47.3385E_B(y) = 3*ln(11.357) ≈ 3*2.4301 ≈ 7.2903Total ≈ 54.6288So, all steps seem correct.Therefore, the optimal allocation is approximately 89,643 to Platform A and 10,357 to Platform B, yielding a maximum engagement of approximately 54.63.But let me check if the initial substitution was correct.We had E(x, y) = 5 sqrt(x) + 3 ln(y + 1)Constraint: x + y = 100So, y = 100 - xThus, E(x) = 5 sqrt(x) + 3 ln(101 - x)Yes, that's correct.Then, derivative:E’(x) = (5)/(2 sqrt(x)) - 3/(101 - x)Set to zero:5/(2 sqrt(x)) = 3/(101 - x)Cross-multiplied:5(101 - x) = 6 sqrt(x)Which leads to 505 - 5x = 6 sqrt(x)Yes, that's correct.Then, substitution t = sqrt(x), leading to 5t^2 + 6t - 505 = 0Solutions t = [ -6 ± sqrt(36 + 10100) ] / 10 = [ -6 ± sqrt(10136) ] / 10Positive solution t ≈ 9.4677, so x ≈ 89.643Yes, all correct.Therefore, the final answers are:1. x ≈ 89.643 thousand dollars to Platform A, y ≈ 10.357 thousand dollars to Platform B.2. Maximum engagement ≈ 54.63.But let me express these in boxed form as per the instructions.For part 1, the optimal allocation is x ≈ 89.643 and y ≈ 10.357. Since the problem mentions that x and y are in thousands of dollars, we can present them as such.For part 2, the maximum engagement is approximately 54.63.But perhaps we can express these with more decimal places or as exact expressions, but given the context, decimal approximations are sufficient.Alternatively, we can express the exact value of x as:x = ( ( -6 + sqrt(10136) ) / 10 )^2But that's quite complex, so decimal approximations are preferable.Therefore, the final answers are:1. Allocate approximately 89,643 to Platform A and 10,357 to Platform B.2. The maximum engagement is approximately 54.63 interactions per thousand dollars.But let me check if the problem expects the answers in thousands or in actual dollars. The problem states that x and y are in thousands of dollars, so the answers should be in thousands.Therefore, x ≈ 89.643 (thousand dollars), y ≈ 10.357 (thousand dollars), and engagement ≈ 54.63.So, to present them as:1. x ≈ 89.643, y ≈ 10.3572. Maximum engagement ≈ 54.63But perhaps we can round to two decimal places for clarity.x ≈ 89.64, y ≈ 10.36, engagement ≈ 54.63Alternatively, since the problem might expect more precise answers, perhaps we can carry out the calculations to more decimal places.But given the context, two decimal places are probably sufficient.Therefore, the optimal allocation is approximately x = 89.64 thousand dollars to Platform A and y = 10.36 thousand dollars to Platform B, resulting in a maximum engagement of approximately 54.63.So, summarizing:1. x ≈ 89.64, y ≈ 10.362. Maximum engagement ≈ 54.63But let me verify the engagement calculation once more.With x = 89.64, sqrt(x) ≈ 9.468, so E_A = 5 * 9.468 ≈ 47.34y = 10.36, y + 1 = 11.36, ln(11.36) ≈ 2.430, so E_B = 3 * 2.430 ≈ 7.29Total ≈ 47.34 + 7.29 ≈ 54.63Yes, correct.Therefore, the final answers are:1. Allocate approximately 89.64 thousand dollars to Platform A and 10.36 thousand dollars to Platform B.2. The maximum engagement is approximately 54.63.But to express these precisely, perhaps we can write them as:1. x = (sqrt(10136) - 6)/10 squared, but that's complicated. Alternatively, just present the decimal approximations.Therefore, the optimal allocation is x ≈ 89.64 and y ≈ 10.36, with maximum engagement ≈ 54.63.I think that's thorough enough.</think>

🧠:<think>Okay, so I have this differential equation to solve: dP/dt = k1 P + k2 sin(ω t). Hmm, it's a first-order linear ordinary differential equation. I remember that these can be solved using an integrating factor. Let me try to recall the steps.First, the standard form of a linear ODE is dy/dt + P(t) y = Q(t). So, I need to rewrite the given equation in that form. Let me see, dP/dt - k1 P = k2 sin(ω t). Yeah, that looks right. So here, P(t) is -k1 and Q(t) is k2 sin(ω t).The integrating factor, μ(t), is usually e^(∫P(t) dt). So in this case, it should be e^(∫-k1 dt) which is e^(-k1 t). Okay, got that.Now, multiply both sides of the equation by the integrating factor. So, e^(-k1 t) dP/dt - k1 e^(-k1 t) P = k2 e^(-k1 t) sin(ω t). The left side should now be the derivative of (P * integrating factor). So, d/dt [P e^(-k1 t)] = k2 e^(-k1 t) sin(ω t).Next, I need to integrate both sides with respect to t. So, ∫ d/dt [P e^(-k1 t)] dt = ∫ k2 e^(-k1 t) sin(ω t) dt. That simplifies to P e^(-k1 t) = k2 ∫ e^(-k1 t) sin(ω t) dt + C, where C is the constant of integration.Now, the integral on the right side is a bit tricky. I think I need to use integration by parts or maybe a standard integral formula. Let me recall: ∫ e^(at) sin(bt) dt. There's a formula for that. I think it's e^(at)/(a² + b²) [a sin(bt) - b cos(bt)] + C. Let me verify that by differentiating.Let me set u = e^(at) sin(bt). Then, du/dt = a e^(at) sin(bt) + b e^(at) cos(bt). If I multiply by the integrating factor, which in this case is e^(-k1 t), but wait, maybe I should just proceed with the integral.So, in our case, a is -k1 and b is ω. So, the integral becomes:∫ e^(-k1 t) sin(ω t) dt = e^(-k1 t)/[(-k1)^2 + ω^2] [ -k1 sin(ω t) - ω cos(ω t) ] + CWait, let me compute that step by step. Let me denote the integral as I.I = ∫ e^(-k1 t) sin(ω t) dtLet me use integration by parts. Let u = sin(ω t), dv = e^(-k1 t) dt. Then, du = ω cos(ω t) dt, v = -1/k1 e^(-k1 t).So, I = uv - ∫ v du = -sin(ω t)/k1 e^(-k1 t) + (ω/k1) ∫ e^(-k1 t) cos(ω t) dt.Now, let me compute the remaining integral, let's call it J = ∫ e^(-k1 t) cos(ω t) dt.Again, integration by parts. Let u = cos(ω t), dv = e^(-k1 t) dt. Then, du = -ω sin(ω t) dt, v = -1/k1 e^(-k1 t).So, J = uv - ∫ v du = -cos(ω t)/k1 e^(-k1 t) - (ω/k1) ∫ e^(-k1 t) sin(ω t) dt.Wait, but the integral on the right is our original I. So, J = -cos(ω t)/k1 e^(-k1 t) - (ω/k1) I.Now, plug J back into the expression for I:I = -sin(ω t)/k1 e^(-k1 t) + (ω/k1) [ -cos(ω t)/k1 e^(-k1 t) - (ω/k1) I ]Simplify this:I = -sin(ω t)/k1 e^(-k1 t) - (ω cos(ω t))/k1² e^(-k1 t) - (ω²/k1²) INow, bring the (ω²/k1²) I term to the left:I + (ω²/k1²) I = -sin(ω t)/k1 e^(-k1 t) - (ω cos(ω t))/k1² e^(-k1 t)Factor I:I [1 + (ω²/k1²)] = -e^(-k1 t) [ sin(ω t)/k1 + ω cos(ω t)/k1² ]Multiply both sides by k1²:I [k1² + ω²] = -e^(-k1 t) [ k1 sin(ω t) + ω cos(ω t) ]Therefore,I = -e^(-k1 t) [ k1 sin(ω t) + ω cos(ω t) ] / (k1² + ω²) + CSo, going back to our original equation:P e^(-k1 t) = k2 I + CSubstitute I:P e^(-k1 t) = k2 [ -e^(-k1 t) (k1 sin(ω t) + ω cos(ω t)) / (k1² + ω²) ] + CMultiply both sides by e^(k1 t):P(t) = -k2 (k1 sin(ω t) + ω cos(ω t)) / (k1² + ω²) + C e^(k1 t)Now, apply the initial condition P(0) = P0.At t = 0:P0 = -k2 (k1 sin(0) + ω cos(0)) / (k1² + ω²) + C e^(0)Simplify:P0 = -k2 (0 + ω * 1) / (k1² + ω²) + CSo,C = P0 + k2 ω / (k1² + ω²)Therefore, the general solution is:P(t) = -k2 (k1 sin(ω t) + ω cos(ω t)) / (k1² + ω²) + [ P0 + k2 ω / (k1² + ω²) ] e^(k1 t)Hmm, let me double-check the signs. When I multiplied both sides by e^(k1 t), the negative sign from I should carry over. So, the expression is correct.Alternatively, sometimes the homogeneous solution is written with a positive exponential, so maybe I can write it as:P(t) = [ P0 + (k2 ω)/(k1² + ω²) ] e^(k1 t) - [ k2 (k1 sin(ω t) + ω cos(ω t)) ] / (k1² + ω²)Yes, that seems correct.So, that's the general solution.Now, moving on to part 2. We have specific values: k1 = 0.05 per year, k2 = 1000 dollars/year, ω = π radians/year, and P0 = 10,000 dollars. We need to find P(t) after 10 years.First, let me plug these values into the general solution.Compute the constants:k1 = 0.05k2 = 1000ω = πSo, compute the denominator: k1² + ω² = (0.05)^2 + (π)^2 ≈ 0.0025 + 9.8696 ≈ 9.8721Compute the terms:First term: [ P0 + (k2 ω)/(k1² + ω²) ] e^(k1 t)Compute (k2 ω)/(k1² + ω²): (1000 * π)/9.8721 ≈ (3141.59265)/9.8721 ≈ 317.78So, P0 + that term: 10,000 + 317.78 ≈ 10,317.78Multiply by e^(0.05 * 10): e^0.5 ≈ 1.64872So, first term: 10,317.78 * 1.64872 ≈ Let's compute that.10,317.78 * 1.64872 ≈ 10,317.78 * 1.6 = 16,508.45 and 10,317.78 * 0.04872 ≈ 503.54, so total ≈ 16,508.45 + 503.54 ≈ 17,011.99Second term: [ k2 (k1 sin(ω t) + ω cos(ω t)) ] / (k1² + ω²)Compute numerator: 1000 [0.05 sin(π * 10) + π cos(π * 10)]Compute sin(π * 10): sin(10π) = 0, since sin(nπ) = 0 for integer n.Compute cos(π * 10): cos(10π) = 1, since cos(nπ) = (-1)^n. For even n, it's 1.So, numerator: 1000 [0 + π * 1] = 1000 π ≈ 3141.59Denominator: 9.8721So, second term: 3141.59 / 9.8721 ≈ 317.78Therefore, P(t) = first term - second term ≈ 17,011.99 - 317.78 ≈ 16,694.21Wait, that seems a bit low. Let me double-check the calculations.Wait, the first term was 10,317.78 * e^0.5 ≈ 10,317.78 * 1.64872 ≈ Let me compute this more accurately.10,000 * 1.64872 = 16,487.2317.78 * 1.64872 ≈ 317.78 * 1.6 = 508.45, and 317.78 * 0.04872 ≈ 15.53, so total ≈ 508.45 + 15.53 ≈ 523.98So, total first term ≈ 16,487.2 + 523.98 ≈ 17,011.18Second term: 3141.59 / 9.8721 ≈ Let me compute 3141.59 ÷ 9.8721.Divide 3141.59 by 9.8721:9.8721 * 317 ≈ 9.8721 * 300 = 2961.63, 9.8721 * 17 ≈ 167.8257, so total ≈ 2961.63 + 167.8257 ≈ 3129.4557Difference: 3141.59 - 3129.4557 ≈ 12.1343So, 12.1343 / 9.8721 ≈ 1.229So, total second term ≈ 317 + 1.229 ≈ 318.229Therefore, P(t) ≈ 17,011.18 - 318.229 ≈ 16,692.95So, approximately 16,693.Wait, but let me check the expression again. The general solution was:P(t) = [ P0 + (k2 ω)/(k1² + ω²) ] e^(k1 t) - [ k2 (k1 sin(ω t) + ω cos(ω t)) ] / (k1² + ω²)So, plugging in t = 10:First term: [10,000 + (1000 * π)/(0.05² + π²)] e^(0.05*10)Second term: [1000 (0.05 sin(10π) + π cos(10π)) ] / (0.05² + π²)As computed, sin(10π)=0, cos(10π)=1.So, second term numerator: 1000*(0 + π*1)=1000π≈3141.59Denominator: 0.0025 + π²≈9.8721So, second term≈3141.59/9.8721≈317.78First term: [10,000 + 317.78] e^0.5≈10,317.78 *1.64872≈17,011.18So, P(t)=17,011.18 -317.78≈16,693.40So, approximately 16,693.40 after 10 years.Wait, but let me check if I made a mistake in the sign. The general solution was:P(t) = [ P0 + (k2 ω)/(k1² + ω²) ] e^(k1 t) - [ k2 (k1 sin(ω t) + ω cos(ω t)) ] / (k1² + ω²)So, yes, the second term is subtracted. So, the calculation seems correct.Alternatively, maybe I should compute it more precisely.Compute e^0.5: e^0.5 ≈ 1.6487212707Compute 10,317.78 * 1.6487212707:Let me compute 10,000 * 1.6487212707 = 16,487.212707317.78 * 1.6487212707:First, 300 * 1.6487212707 = 494.616381217.78 * 1.6487212707 ≈ Let's compute 17 *1.6487212707≈28.0282616, and 0.78*1.6487212707≈1.286283So, total ≈28.0282616 +1.286283≈29.3145446So, total 317.78 *1.6487212707≈494.6163812 +29.3145446≈523.9309258So, total first term≈16,487.212707 +523.9309258≈17,011.14363Second term: 3141.59265 /9.8721≈3141.59265 /9.8721≈317.78So, P(t)=17,011.14363 -317.78≈16,693.36363So, approximately 16,693.36Rounding to the nearest dollar, it's approximately 16,693.Wait, but let me check if I made a mistake in the initial integration. Let me go back to the integral:I = ∫ e^(-k1 t) sin(ω t) dtI used integration by parts twice and arrived at:I = -e^(-k1 t) [k1 sin(ω t) + ω cos(ω t)] / (k1² + ω²) + CBut let me verify this by differentiating:Let me compute d/dt [ -e^(-k1 t) (k1 sin(ω t) + ω cos(ω t)) / (k1² + ω²) ]Using product rule:First, derivative of -e^(-k1 t) is k1 e^(-k1 t)Multiply by (k1 sin(ω t) + ω cos(ω t)) / (k1² + ω²)Plus (-e^(-k1 t)) times derivative of (k1 sin(ω t) + ω cos(ω t)) / (k1² + ω²)Derivative of (k1 sin(ω t) + ω cos(ω t)) is k1 ω cos(ω t) - ω² sin(ω t)So, putting it all together:k1 e^(-k1 t) (k1 sin(ω t) + ω cos(ω t)) / (k1² + ω²) - e^(-k1 t) (k1 ω cos(ω t) - ω² sin(ω t)) / (k1² + ω²)Factor out e^(-k1 t) / (k1² + ω²):[ k1 (k1 sin(ω t) + ω cos(ω t)) - (k1 ω cos(ω t) - ω² sin(ω t)) ] e^(-k1 t) / (k1² + ω²)Simplify inside the brackets:k1² sin(ω t) + k1 ω cos(ω t) - k1 ω cos(ω t) + ω² sin(ω t) = (k1² + ω²) sin(ω t)So, the derivative is (k1² + ω²) sin(ω t) e^(-k1 t) / (k1² + ω²) = sin(ω t) e^(-k1 t)Which matches the integrand. So, the integral was correct.Therefore, the solution is correct.So, after 10 years, the price is approximately 16,693.Wait, but let me check if I made a mistake in the sign of the second term. The general solution was:P(t) = [ P0 + (k2 ω)/(k1² + ω²) ] e^(k1 t) - [ k2 (k1 sin(ω t) + ω cos(ω t)) ] / (k1² + ω²)So, yes, the second term is subtracted. So, the calculation is correct.Alternatively, maybe I should consider that the particular solution might have a different sign. Let me think.Wait, when we solved the ODE, we had:dP/dt - k1 P = k2 sin(ω t)So, the integrating factor was e^(-k1 t), and after multiplying, we had:d/dt [P e^(-k1 t)] = k2 e^(-k1 t) sin(ω t)Then, integrating both sides:P e^(-k1 t) = k2 ∫ e^(-k1 t) sin(ω t) dt + CWhich led to:P(t) = [C + k2 ∫ e^(-k1 t) sin(ω t) dt ] e^(k1 t)But when we computed the integral, we had:∫ e^(-k1 t) sin(ω t) dt = -e^(-k1 t) [k1 sin(ω t) + ω cos(ω t)] / (k1² + ω²) + CSo, substituting back:P(t) = [ C + k2 ( -e^(-k1 t) [k1 sin(ω t) + ω cos(ω t)] / (k1² + ω²) ) ] e^(k1 t)= C e^(k1 t) - k2 [k1 sin(ω t) + ω cos(ω t)] / (k1² + ω²)So, yes, the second term is subtracted. So, the solution is correct.Therefore, after 10 years, the price is approximately 16,693.Wait, but let me check the calculation again because sometimes constants can be tricky.Compute [ P0 + (k2 ω)/(k1² + ω²) ] e^(k1 t):P0 = 10,000k2 ω = 1000 * π ≈ 3141.59265k1² + ω² ≈ 0.0025 + 9.8696 ≈ 9.8721So, (k2 ω)/(k1² + ω²) ≈ 3141.59265 /9.8721≈317.78So, P0 + that ≈10,000 +317.78≈10,317.78Multiply by e^(0.05*10)=e^0.5≈1.6487210,317.78 *1.64872≈17,011.14Second term: [k2 (k1 sin(ω t) + ω cos(ω t)) ] / (k1² + ω²)At t=10, sin(10π)=0, cos(10π)=1So, numerator:1000*(0.05*0 + π*1)=1000π≈3141.59Denominator:9.8721So, second term≈3141.59 /9.8721≈317.78Thus, P(t)=17,011.14 -317.78≈16,693.36So, approximately 16,693.36, which we can round to 16,693.I think that's the correct answer.</think>