💡First, I need to determine the production value of last year. Since this year's planned production value is an 11% increase from last year, I can express last year's production value as ( a div (1 + 11%) ).Next, the actual production value this year exceeds the planned value by 1%. Therefore, the actual production value is ( a times (1 + 1%) = 1.01a ).To find the percentage increase compared to last year, I calculate the difference between this year's actual production and last year's production, then divide by last year's production and multiply by 100%. This results in a 12.11% increase.

💡First, I need to find the prime factors of 120. I know that 120 can be divided by 2 three times, leaving 15. Then, 15 can be divided by 3 and 5, which are both prime numbers. So, the prime factorization of 120 is (2^3 cdot 3 cdot 5).Next, to find the odd divisors, I should ignore the even prime factor, which is 2. This leaves me with the odd prime factors: 3 and 5.Now, I'll list all the possible combinations of these odd prime factors to find the odd divisors:- (3^0 cdot 5^0 = 1)- (3^1 cdot 5^0 = 3)- (3^0 cdot 5^1 = 5)- (3^1 cdot 5^1 = 15)Finally, I'll add these odd divisors together: (1 + 3 + 5 + 15 = 24).So, the sum of the positive odd divisors of 120 is 24.

💡Alright, so I have this problem where a square is inscribed in an ellipse given by the equation (frac{x^2}{3} + frac{y^2}{6} = 1). The square has its sides parallel to the coordinate axes, and I need to find the area of this square. Hmm, okay, let me try to visualize this.First, I know that an ellipse is like a stretched circle. The standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) and (b) are the semi-major and semi-minor axes. In this case, comparing it to the standard form, (a^2 = 3) and (b^2 = 6), so (a = sqrt{3}) and (b = sqrt{6}). That means the ellipse is wider along the y-axis because (b > a).Now, the square is inscribed in this ellipse, meaning all four vertices of the square lie on the ellipse. Since the sides are parallel to the coordinate axes, the square is axis-aligned. So, the vertices of the square should be at points ((pm t, pm t)) for some positive value (t). That makes sense because, in a square centered at the origin with sides parallel to the axes, the coordinates of the vertices are symmetric with respect to both the x-axis and y-axis.So, if I plug these coordinates into the ellipse equation, they should satisfy it. Let me write that out:[frac{t^2}{3} + frac{t^2}{6} = 1]Wait, that seems right. Because each vertex is ((t, t)), plugging into the ellipse equation gives that expression. Now, let's solve for (t^2).First, combine the terms:[frac{t^2}{3} + frac{t^2}{6} = frac{2t^2}{6} + frac{t^2}{6} = frac{3t^2}{6} = frac{t^2}{2}]So, (frac{t^2}{2} = 1), which implies that (t^2 = 2). Therefore, (t = sqrt{2}). Okay, so each vertex is at ((sqrt{2}, sqrt{2})), ((- sqrt{2}, sqrt{2})), etc.Now, the side length of the square. Since the square is axis-aligned, the distance between two adjacent vertices is the distance between, say, ((sqrt{2}, sqrt{2})) and ((- sqrt{2}, sqrt{2})), which is (2sqrt{2}). Wait, no, that's the length along the x-axis from (-sqrt{2}) to (sqrt{2}), which is indeed (2sqrt{2}). But actually, that's the length of the side? Wait, hold on.Wait, no, in a square, the side length is the distance between two adjacent vertices. So, if one vertex is at ((sqrt{2}, sqrt{2})) and the next is at ((- sqrt{2}, sqrt{2})), the distance between these two points is (2sqrt{2}), which would be the length of the side. But that seems a bit large. Let me think again.Wait, actually, no. The distance between ((sqrt{2}, sqrt{2})) and ((sqrt{2}, -sqrt{2})) is (2sqrt{2}), but that's the vertical side. Similarly, the horizontal side is between ((sqrt{2}, sqrt{2})) and ((- sqrt{2}, sqrt{2})), which is also (2sqrt{2}). So, yes, each side of the square is (2sqrt{2}) units long.But wait, that can't be right because if the side length is (2sqrt{2}), then the area would be ((2sqrt{2})^2 = 8). But let me double-check because sometimes when dealing with ellipses, scaling factors come into play.Alternatively, maybe I should think about parametric equations. The ellipse can be parameterized as (x = a cos theta) and (y = b sin theta), where (a = sqrt{3}) and (b = sqrt{6}). For a square, the angle (theta) should correspond to points where the coordinates are equal in magnitude, i.e., (cos theta = sin theta), which happens at (theta = 45^circ) or (pi/4) radians.So, plugging (theta = pi/4) into the parametric equations:[x = sqrt{3} cos left( frac{pi}{4} right) = sqrt{3} cdot frac{sqrt{2}}{2} = frac{sqrt{6}}{2}][y = sqrt{6} sin left( frac{pi}{4} right) = sqrt{6} cdot frac{sqrt{2}}{2} = frac{sqrt{12}}{2} = frac{2sqrt{3}}{2} = sqrt{3}]Wait, that gives (x = frac{sqrt{6}}{2}) and (y = sqrt{3}). But in the square, the x and y coordinates should be equal, right? Because the vertices are at ((pm t, pm t)). So, if I plug (x = t) and (y = t) into the ellipse equation, I get:[frac{t^2}{3} + frac{t^2}{6} = 1]Which simplifies to:[frac{2t^2 + t^2}{6} = 1 implies frac{3t^2}{6} = 1 implies frac{t^2}{2} = 1 implies t^2 = 2 implies t = sqrt{2}]So, that gives (t = sqrt{2}), which is consistent with my earlier result. So, the coordinates are ((sqrt{2}, sqrt{2})), etc. So, the side length is (2sqrt{2}), and the area is ((2sqrt{2})^2 = 8).But wait, when I used the parametric equations, I got different coordinates. That seems contradictory. Let me see. If I plug (t = sqrt{2}) into the ellipse equation:[frac{(sqrt{2})^2}{3} + frac{(sqrt{2})^2}{6} = frac{2}{3} + frac{2}{6} = frac{2}{3} + frac{1}{3} = 1]Okay, that works. So, the point ((sqrt{2}, sqrt{2})) is indeed on the ellipse. But when I used the parametric equations, I got different coordinates. So, why is that?Ah, I think it's because the parametric equations parameterize the ellipse, but the angle (theta) doesn't correspond directly to the geometric angle in the same way as a circle. In a circle, (theta) is the actual angle, but in an ellipse, it's stretched. So, when I set (theta = pi/4), it doesn't necessarily give the point where (x = y). Instead, the point where (x = y) occurs at a different (theta).So, perhaps I should set (x = y) in the parametric equations and solve for (theta). Let's try that.Set (x = y):[sqrt{3} cos theta = sqrt{6} sin theta]Divide both sides by (cos theta):[sqrt{3} = sqrt{6} tan theta]Solve for (tan theta):[tan theta = frac{sqrt{3}}{sqrt{6}} = frac{sqrt{3}}{sqrt{6}} = frac{sqrt{3}}{sqrt{6}} = frac{sqrt{3} cdot sqrt{6}}{6} = frac{sqrt{18}}{6} = frac{3sqrt{2}}{6} = frac{sqrt{2}}{2}]So, (theta = arctan left( frac{sqrt{2}}{2} right)). Let me compute that. Well, (arctan left( frac{sqrt{2}}{2} right)) is approximately 35.264 degrees, which is less than 45 degrees. So, that's why when I set (theta = pi/4), I didn't get the point where (x = y). Instead, I had to solve for (theta) such that (x = y).But regardless, since we already found that (t = sqrt{2}) satisfies the ellipse equation when (x = y = t), that should be sufficient. So, the side length is (2sqrt{2}), and the area is 8.Wait a second, but when I calculated using the parametric equations, I got different coordinates. Let me see:If (x = sqrt{3} cos theta) and (y = sqrt{6} sin theta), and at the point where (x = y), we have:[sqrt{3} cos theta = sqrt{6} sin theta][sqrt{3} = sqrt{6} tan theta][tan theta = frac{sqrt{3}}{sqrt{6}} = frac{sqrt{2}}{2}]So, (theta = arctan left( frac{sqrt{2}}{2} right)). Let me compute (cos theta) and (sin theta):Since (tan theta = frac{sqrt{2}}{2}), we can think of a right triangle with opposite side (sqrt{2}) and adjacent side 2. So, the hypotenuse is (sqrt{ (sqrt{2})^2 + 2^2 } = sqrt{2 + 4} = sqrt{6}).Therefore,[cos theta = frac{2}{sqrt{6}} = frac{sqrt{6}}{3}][sin theta = frac{sqrt{2}}{sqrt{6}} = frac{sqrt{3}}{3}]So, plugging back into the parametric equations:[x = sqrt{3} cdot frac{sqrt{6}}{3} = frac{sqrt{18}}{3} = frac{3sqrt{2}}{3} = sqrt{2}][y = sqrt{6} cdot frac{sqrt{3}}{3} = frac{sqrt{18}}{3} = frac{3sqrt{2}}{3} = sqrt{2}]Ah, so that gives (x = y = sqrt{2}), which matches our earlier result. So, both methods agree. Therefore, the coordinates are indeed ((sqrt{2}, sqrt{2})), and the side length is (2sqrt{2}), leading to an area of 8.Just to make sure, let me visualize this. The ellipse is longer along the y-axis, so the square inscribed in it should have its vertices closer to the top and bottom of the ellipse. The side length being (2sqrt{2}) seems reasonable because the ellipse's semi-minor axis is (sqrt{3} approx 1.732), and the semi-major axis is (sqrt{6} approx 2.449). So, the square's vertices at (sqrt{2} approx 1.414) along both axes are within the ellipse's bounds.Therefore, after double-checking, I'm confident that the area of the square is 8.

💡Alright, so Teacher Huang wants to start a badminton club and needs to buy some equipment. She’s planning to purchase 16 badminton rackets and x boxes of shuttlecocks, where x is greater than 16. She found two stores, Store A and Store B, that sell the same brand of rackets and shuttlecocks. The prices are the same in both stores: each racket costs 150 yuan, and each box of shuttlecocks costs 40 yuan. Store A has a promotion: for every badminton racket you buy, you get one box of shuttlecocks for free. So, if Teacher Huang buys 16 rackets, she gets 16 boxes of shuttlecocks for free. That means she only has to pay for the remaining shuttlecocks beyond the 16 she gets for free. Store B offers a different deal: a 20% discount on all items. So, both the rackets and the shuttlecocks will be 20% cheaper at Store B. First, I need to figure out how much Teacher Huang would pay at each store, expressed in terms of x. Then, when x is 25, I need to determine which store is more cost-effective.Starting with Store A:- The cost of 16 rackets is straightforward: 16 times 150 yuan. That’s 16 * 150 = 2400 yuan.- For the shuttlecocks, since she gets 16 boxes for free, she only needs to pay for (x - 16) boxes. Each box is 40 yuan, so that’s (x - 16) * 40 yuan.- Adding those together, the total cost at Store A is 2400 + 40(x - 16). Simplifying that, it becomes 2400 + 40x - 640, which is 1760 + 40x yuan.Now, for Store B:- The original cost without any discount would be the same as Store A: 16 rackets and x shuttlecocks. So, that’s 16 * 150 + x * 40 = 2400 + 40x yuan.- Store B offers a 20% discount on the total cost. So, Teacher Huang pays 80% of the original price. That’s 0.8 * (2400 + 40x).- Calculating that, 0.8 * 2400 = 1920, and 0.8 * 40x = 32x. So, the total cost at Store B is 1920 + 32x yuan.So, summarizing:- At Store A, the total cost is y_A = 1760 + 40x yuan.- At Store B, the total cost is y_B = 1920 + 32x yuan.Now, when x = 25, let's calculate both y_A and y_B to see which is cheaper.For Store A:- y_A = 1760 + 40 * 25 = 1760 + 1000 = 2760 yuan.For Store B:- y_B = 1920 + 32 * 25 = 1920 + 800 = 2720 yuan.Comparing the two, 2760 yuan at Store A versus 2720 yuan at Store B. Store B is cheaper by 40 yuan.Therefore, when x = 25, Store B is more cost-effective for Teacher Huang.

💡Okay, so I have this logarithm problem here: If log base k of x multiplied by log base 10 of k equals 4, then what is x? The options are A) 1000, B) 10000, C) 100000, D) 1000000, and E) 10000. Hmm, let me try to figure this out step by step.First, I remember that logarithms can be tricky, but there are some properties that might help here. The equation given is log_k(x) * log_10(k) = 4. I need to find x. Maybe I can use the change of base formula or some other logarithmic identity to simplify this.I recall that log_b(a) * log_c(b) is equal to log_c(a). Let me check if that's correct. If I have log_k(x) and then log_10(k), multiplying them together, does that equal log_10(x)? Let me think about it. Yes, actually, that's a logarithmic identity. It's called the chain rule for logarithms, I think. So, log_k(x) * log_10(k) is indeed equal to log_10(x). That simplifies things a lot!So, applying that identity to the equation, it becomes log_10(x) = 4. Now, solving for x is straightforward. If log base 10 of x is 4, that means x is 10 raised to the power of 4. Calculating 10^4, that's 10 * 10 * 10 * 10, which equals 10000. So, x should be 10000. Looking back at the options, B is 10000, so that must be the correct answer.Wait, let me double-check to make sure I didn't make a mistake. Sometimes, when dealing with logarithms, it's easy to mix up the bases or misapply the properties. Starting again, log_k(x) * log_10(k) = 4. Using the identity, this becomes log_10(x) = 4. So, x = 10^4 = 10000. Yeah, that seems right. Alternatively, I can try plugging in the value to verify. If x is 10000, then log_k(10000) * log_10(k) should equal 4. Let's see. Let me pick a specific value for k to test. Suppose k is 10. Then log_10(10000) is 4, and log_10(10) is 1. So, 4 * 1 = 4, which satisfies the equation. What if k is something else, like 100? Then log_100(10000) is 2 because 100 squared is 10000. And log_10(100) is 2 because 10 squared is 100. So, 2 * 2 = 4, which also works. How about k = 1000? Then log_1000(10000) is log base 1000 of 10000. Since 1000 is 10^3 and 10000 is 10^4, log_1000(10000) = 4/3. And log_10(1000) is 3. So, 4/3 * 3 = 4, which still holds. So, regardless of the value of k, as long as x is 10000, the equation holds true. That gives me more confidence that 10000 is indeed the correct answer.I think I've got it. The key was recognizing the logarithmic identity that allowed me to simplify the product of the two logarithms into a single logarithm with base 10. Once I did that, solving for x was straightforward. It's always good to verify by plugging the answer back into the original equation with different values of k to ensure it works universally.Another way to approach this might be to express both logarithms in terms of natural logarithms and see if that leads to the same conclusion. Let's try that.Using the change of base formula, log_k(x) is equal to ln(x)/ln(k), and log_10(k) is equal to ln(k)/ln(10). So, multiplying them together, we get:(ln(x)/ln(k)) * (ln(k)/ln(10)) = ln(x)/ln(10) = log_10(x).So, again, we end up with log_10(x) = 4, which means x = 10^4 = 10000. This confirms the earlier result.I think I've thoroughly checked this problem from different angles, and all paths lead to x being 10000. Therefore, I'm confident that the correct answer is B) 10000.**Final Answer**boxed{B}

💡Alright, so I've got this problem here where I need to find the maximum value of the expression ( S = sum_{i=1}^{n} i^{2} x_{i} ). The constraints are that ( sum_{i=1}^{n} x_{i} = n ) and ( sum_{i=1}^{n} i x_{i} = 2n - 2 ). All the ( x_i ) are non-negative real numbers, and ( n ) is a natural number greater than 2. Hmm, okay, let's break this down step by step.First, I need to understand what exactly is being asked. I have to maximize ( S ), which is a weighted sum of the squares of the indices ( i ), with weights ( x_i ). The weights themselves are constrained by two conditions: their sum is ( n ), and the sum of each index multiplied by its corresponding weight is ( 2n - 2 ).Let me write down the given equations:1. ( sum_{i=1}^{n} x_{i} = n )2. ( sum_{i=1}^{n} i x_{i} = 2n - 2 )And I need to maximize:( S = sum_{i=1}^{n} i^{2} x_{i} )So, I have two linear constraints and a quadratic objective function. This seems like an optimization problem with linear constraints. Maybe I can use some method from linear algebra or optimization theory to solve this.I remember that for optimization problems with linear constraints, the method of Lagrange multipliers can be useful. Maybe I can set up the Lagrangian with the given constraints and take derivatives to find the maximum.Let me try that approach.Define the Lagrangian function:( mathcal{L} = sum_{i=1}^{n} i^{2} x_{i} - lambda left( sum_{i=1}^{n} x_{i} - n right) - mu left( sum_{i=1}^{n} i x_{i} - (2n - 2) right) )Here, ( lambda ) and ( mu ) are the Lagrange multipliers associated with the two constraints.To find the extrema, we take the partial derivatives of ( mathcal{L} ) with respect to each ( x_i ), ( lambda ), and ( mu ), and set them equal to zero.So, for each ( x_i ):( frac{partial mathcal{L}}{partial x_i} = i^{2} - lambda - mu i = 0 )This gives us the equation:( i^{2} - lambda - mu i = 0 ) for each ( i ).Hmm, so for each ( i ), we have ( i^{2} = lambda + mu i ). That suggests that the quadratic equation ( i^{2} - mu i - lambda = 0 ) must hold for each ( i ).But wait, this is supposed to hold for all ( i ) from 1 to ( n ). However, a quadratic equation can't have more than two roots unless it's identically zero. But here, ( i ) is varying, so this seems problematic. Maybe my approach is missing something.Alternatively, perhaps not all ( x_i ) are positive. Since the problem states that ( x_i ) are non-negative, some of them could be zero. So, maybe only a subset of the ( x_i ) are non-zero, and for those, the condition ( i^{2} - lambda - mu i = 0 ) holds.This is similar to the idea in linear programming where the maximum is achieved at a vertex of the feasible region, which corresponds to some variables being zero.So, perhaps only two of the ( x_i ) are non-zero, and the rest are zero. Let me test this idea.Suppose that only two variables, say ( x_k ) and ( x_m ), are non-zero. Then, the constraints become:1. ( x_k + x_m = n )2. ( k x_k + m x_m = 2n - 2 )And the objective function ( S ) becomes:( S = k^{2} x_k + m^{2} x_m )So, I can solve for ( x_k ) and ( x_m ) in terms of ( k ) and ( m ), and then express ( S ) in terms of ( k ) and ( m ).Let me denote ( x_k = a ) and ( x_m = b ). Then:1. ( a + b = n )2. ( k a + m b = 2n - 2 )From the first equation, ( b = n - a ). Substitute into the second equation:( k a + m (n - a) = 2n - 2 )Simplify:( (k - m) a + m n = 2n - 2 )Solve for ( a ):( (k - m) a = 2n - 2 - m n )( a = frac{2n - 2 - m n}{k - m} )Similarly, ( b = n - a = n - frac{2n - 2 - m n}{k - m} )Now, since ( a ) and ( b ) must be non-negative, the numerator and denominator must have the same sign.Let me consider specific values for ( k ) and ( m ). Since we're trying to maximize ( S ), which is a weighted sum of squares, it's likely that the maximum occurs when the weights are concentrated on the highest possible indices. So, perhaps ( k = n-1 ) and ( m = n ), or some similar combination.Wait, let's test with ( k = 1 ) and ( m = 2 ). Maybe the maximum occurs at the lower indices? Hmm, not sure.Alternatively, maybe the maximum occurs when ( x_{n-1} ) and ( x_n ) are non-zero. Let's try that.Let me set ( k = n-1 ) and ( m = n ). Then:( a = x_{n-1} ), ( b = x_n )From the constraints:1. ( a + b = n )2. ( (n-1) a + n b = 2n - 2 )Substitute ( b = n - a ) into the second equation:( (n-1) a + n (n - a) = 2n - 2 )Simplify:( (n-1) a + n^2 - n a = 2n - 2 )Combine like terms:( (n - 1 - n) a + n^2 = 2n - 2 )( (-1) a + n^2 = 2n - 2 )Solve for ( a ):( -a = 2n - 2 - n^2 )( a = n^2 - 2n + 2 )But ( a ) must be non-negative, so ( n^2 - 2n + 2 geq 0 ). Since ( n > 2 ), ( n^2 - 2n + 2 ) is always positive because the discriminant is ( 4 - 8 = -4 ), so the quadratic is always positive. So, ( a = n^2 - 2n + 2 ), and ( b = n - a = n - (n^2 - 2n + 2) = -n^2 + 3n - 2 ).Wait, ( b ) must also be non-negative. So:( -n^2 + 3n - 2 geq 0 )Multiply both sides by -1 (inequality sign reverses):( n^2 - 3n + 2 leq 0 )Factor:( (n - 1)(n - 2) leq 0 )So, ( n ) is between 1 and 2. But ( n > 2 ), so this inequality doesn't hold. Therefore, ( b ) would be negative, which is not allowed. So, this combination ( k = n-1 ) and ( m = n ) doesn't work because it leads to a negative ( x_n ).Hmm, maybe I need to choose different ( k ) and ( m ). Let me try ( k = 1 ) and ( m = 2 ).So, ( a = x_1 ), ( b = x_2 )Constraints:1. ( a + b = n )2. ( 1 cdot a + 2 cdot b = 2n - 2 )From the first equation, ( b = n - a ). Substitute into the second equation:( a + 2(n - a) = 2n - 2 )Simplify:( a + 2n - 2a = 2n - 2 )( -a + 2n = 2n - 2 )( -a = -2 )( a = 2 )Then, ( b = n - 2 )So, ( x_1 = 2 ), ( x_2 = n - 2 ), and all other ( x_i = 0 ).Now, let's compute ( S ):( S = 1^2 cdot 2 + 2^2 cdot (n - 2) + sum_{i=3}^{n} i^2 cdot 0 )Simplify:( S = 2 + 4(n - 2) = 2 + 4n - 8 = 4n - 6 )Is this the maximum? Let me check another combination.Suppose ( k = 2 ) and ( m = 3 ). Then:( a = x_2 ), ( b = x_3 )Constraints:1. ( a + b = n )2. ( 2a + 3b = 2n - 2 )From the first equation, ( b = n - a ). Substitute into the second equation:( 2a + 3(n - a) = 2n - 2 )Simplify:( 2a + 3n - 3a = 2n - 2 )( -a + 3n = 2n - 2 )( -a = -n - 2 )( a = n + 2 )But ( a = x_2 = n + 2 ), which would make ( b = n - (n + 2) = -2 ), which is negative. Not allowed. So, this combination doesn't work.How about ( k = 1 ) and ( m = 3 )?( a = x_1 ), ( b = x_3 )Constraints:1. ( a + b = n )2. ( 1 cdot a + 3 cdot b = 2n - 2 )From the first equation, ( b = n - a ). Substitute into the second equation:( a + 3(n - a) = 2n - 2 )Simplify:( a + 3n - 3a = 2n - 2 )( -2a + 3n = 2n - 2 )( -2a = -n - 2 )( a = frac{n + 2}{2} )Then, ( b = n - frac{n + 2}{2} = frac{n - 2}{2} )So, ( x_1 = frac{n + 2}{2} ), ( x_3 = frac{n - 2}{2} ), and all other ( x_i = 0 ).Compute ( S ):( S = 1^2 cdot frac{n + 2}{2} + 3^2 cdot frac{n - 2}{2} = frac{n + 2}{2} + 9 cdot frac{n - 2}{2} )Simplify:( S = frac{n + 2 + 9n - 18}{2} = frac{10n - 16}{2} = 5n - 8 )Compare this with the previous ( S = 4n - 6 ). For ( n > 2 ), which one is larger?Let's subtract: ( 5n - 8 - (4n - 6) = n - 2 ). Since ( n > 2 ), ( n - 2 > 0 ), so ( 5n - 8 > 4n - 6 ). So, this combination gives a higher ( S ).Hmm, so maybe choosing ( k = 1 ) and ( m = 3 ) gives a higher ( S ). Let's try another combination.How about ( k = 1 ) and ( m = 4 )?( a = x_1 ), ( b = x_4 )Constraints:1. ( a + b = n )2. ( 1 cdot a + 4 cdot b = 2n - 2 )From the first equation, ( b = n - a ). Substitute into the second equation:( a + 4(n - a) = 2n - 2 )Simplify:( a + 4n - 4a = 2n - 2 )( -3a + 4n = 2n - 2 )( -3a = -2n - 2 )( a = frac{2n + 2}{3} )Then, ( b = n - frac{2n + 2}{3} = frac{3n - 2n - 2}{3} = frac{n - 2}{3} )So, ( x_1 = frac{2n + 2}{3} ), ( x_4 = frac{n - 2}{3} ), and all other ( x_i = 0 ).Compute ( S ):( S = 1^2 cdot frac{2n + 2}{3} + 4^2 cdot frac{n - 2}{3} = frac{2n + 2}{3} + 16 cdot frac{n - 2}{3} )Simplify:( S = frac{2n + 2 + 16n - 32}{3} = frac{18n - 30}{3} = 6n - 10 )Compare this with the previous ( S = 5n - 8 ). Subtract: ( 6n - 10 - (5n - 8) = n - 2 ). Again, since ( n > 2 ), ( 6n - 10 > 5n - 8 ). So, this is even better.I see a pattern here. Each time I increase ( m ), the coefficient of ( n ) in ( S ) increases by 1, and the constant term decreases by 2. So, perhaps if I keep increasing ( m ), I can get an even higher ( S ).Wait, let's test ( k = 1 ) and ( m = n ).So, ( a = x_1 ), ( b = x_n )Constraints:1. ( a + b = n )2. ( 1 cdot a + n cdot b = 2n - 2 )From the first equation, ( b = n - a ). Substitute into the second equation:( a + n(n - a) = 2n - 2 )Simplify:( a + n^2 - n a = 2n - 2 )Combine like terms:( (1 - n) a + n^2 = 2n - 2 )Solve for ( a ):( (1 - n) a = 2n - 2 - n^2 )( a = frac{2n - 2 - n^2}{1 - n} )Simplify numerator:( -n^2 + 2n - 2 = -(n^2 - 2n + 2) )So,( a = frac{ - (n^2 - 2n + 2) }{1 - n} = frac{ - (n^2 - 2n + 2) }{ - (n - 1) } = frac{n^2 - 2n + 2}{n - 1} )Similarly, ( b = n - a = n - frac{n^2 - 2n + 2}{n - 1} )Let me compute ( b ):( b = frac{n(n - 1) - (n^2 - 2n + 2)}{n - 1} = frac{n^2 - n - n^2 + 2n - 2}{n - 1} = frac{n - 2}{n - 1} )So, ( x_1 = frac{n^2 - 2n + 2}{n - 1} ), ( x_n = frac{n - 2}{n - 1} ), and all other ( x_i = 0 ).Compute ( S ):( S = 1^2 cdot frac{n^2 - 2n + 2}{n - 1} + n^2 cdot frac{n - 2}{n - 1} )Simplify:( S = frac{n^2 - 2n + 2}{n - 1} + frac{n^3 - 2n^2}{n - 1} )Combine the fractions:( S = frac{n^2 - 2n + 2 + n^3 - 2n^2}{n - 1} = frac{n^3 - n^2 - 2n + 2}{n - 1} )Let me perform polynomial division or factor the numerator:Numerator: ( n^3 - n^2 - 2n + 2 )Let me try to factor it. Maybe factor by grouping:Group as ( (n^3 - n^2) + (-2n + 2) = n^2(n - 1) - 2(n - 1) = (n^2 - 2)(n - 1) )Wait, that's not correct because:( (n^2 - 2)(n - 1) = n^3 - n^2 - 2n + 2 ), which matches the numerator.So, numerator factors as ( (n^2 - 2)(n - 1) ). Therefore,( S = frac{(n^2 - 2)(n - 1)}{n - 1} = n^2 - 2 )Wow, that's a nice simplification. So, ( S = n^2 - 2 ).Now, let's check if ( x_1 ) and ( x_n ) are non-negative:( x_1 = frac{n^2 - 2n + 2}{n - 1} ). Let's see if this is positive.Since ( n > 2 ), ( n - 1 > 1 ). The numerator ( n^2 - 2n + 2 ) is a quadratic in ( n ). Let's check its discriminant: ( 4 - 8 = -4 ), which is negative, so the quadratic is always positive. Therefore, ( x_1 ) is positive.( x_n = frac{n - 2}{n - 1} ). Since ( n > 2 ), both numerator and denominator are positive, so ( x_n ) is positive.Therefore, this combination gives us a feasible solution with ( S = n^2 - 2 ).Earlier, when I tried ( k = 1 ) and ( m = 4 ), I got ( S = 6n - 10 ). Let's compare this with ( n^2 - 2 ).For ( n = 3 ):- ( S = 3^2 - 2 = 7 )- ( 6n - 10 = 8 ). Wait, 8 > 7. Hmm, that contradicts.Wait, but for ( n = 3 ), let's see:If ( k = 1 ) and ( m = 3 ):( x_1 = frac{3 + 2}{2} = 2.5 ), ( x_3 = frac{3 - 2}{2} = 0.5 )Then, ( S = 1*2.5 + 9*0.5 = 2.5 + 4.5 = 7 ), which matches ( n^2 - 2 = 7 ).But earlier, when I set ( k = 1 ) and ( m = 4 ), for ( n = 3 ), ( m = 4 ) is beyond ( n ), so that's not applicable. So, my mistake was considering ( m = 4 ) when ( n = 3 ), which isn't allowed.So, for each ( n ), the maximum ( m ) is ( n ). Therefore, the combination ( k = 1 ) and ( m = n ) is valid for any ( n > 2 ).Therefore, the maximum ( S ) is ( n^2 - 2 ).But let me verify this with another value of ( n ). Let's take ( n = 4 ).If ( n = 4 ), then:( x_1 = frac{4^2 - 2*4 + 2}{4 - 1} = frac{16 - 8 + 2}{3} = frac{10}{3} approx 3.333 )( x_4 = frac{4 - 2}{4 - 1} = frac{2}{3} approx 0.666 )Check constraints:1. ( x_1 + x_4 = frac{10}{3} + frac{2}{3} = frac{12}{3} = 4 ) ✔️2. ( 1*x_1 + 4*x_4 = frac{10}{3} + frac{8}{3} = frac{18}{3} = 6 ). But ( 2n - 2 = 8 - 2 = 6 ) ✔️Compute ( S = 1^2 * frac{10}{3} + 4^2 * frac{2}{3} = frac{10}{3} + frac{32}{3} = frac{42}{3} = 14 ). And ( n^2 - 2 = 16 - 2 = 14 ) ✔️Another check: ( n = 5 )( x_1 = frac{25 - 10 + 2}{4} = frac{17}{4} = 4.25 )( x_5 = frac{5 - 2}{4} = frac{3}{4} = 0.75 )Check constraints:1. ( 4.25 + 0.75 = 5 ) ✔️2. ( 1*4.25 + 5*0.75 = 4.25 + 3.75 = 8 ). ( 2n - 2 = 10 - 2 = 8 ) ✔️Compute ( S = 1*4.25 + 25*0.75 = 4.25 + 18.75 = 23 ). ( n^2 - 2 = 25 - 2 = 23 ) ✔️Okay, so it seems consistent. Therefore, the maximum value of ( S ) is indeed ( n^2 - 2 ).But just to be thorough, let me consider another case where more than two ( x_i ) are non-zero. For example, suppose three variables are non-zero: ( x_1, x_2, x_3 ). Maybe this could yield a higher ( S ).Let me set up the equations:1. ( x_1 + x_2 + x_3 = n )2. ( 1x_1 + 2x_2 + 3x_3 = 2n - 2 )3. ( S = 1x_1 + 4x_2 + 9x_3 )I need to maximize ( S ) under the given constraints.Let me express ( x_1 = n - x_2 - x_3 ). Substitute into the second equation:( (n - x_2 - x_3) + 2x_2 + 3x_3 = 2n - 2 )Simplify:( n - x_2 - x_3 + 2x_2 + 3x_3 = 2n - 2 )Combine like terms:( n + x_2 + 2x_3 = 2n - 2 )Rearrange:( x_2 + 2x_3 = n - 2 )So, ( x_2 = n - 2 - 2x_3 )Now, substitute ( x_1 = n - x_2 - x_3 = n - (n - 2 - 2x_3) - x_3 = n - n + 2 + 2x_3 - x_3 = 2 + x_3 )So, ( x_1 = 2 + x_3 ), ( x_2 = n - 2 - 2x_3 ), ( x_3 = x_3 )Now, all ( x_i ) must be non-negative:1. ( x_1 = 2 + x_3 geq 0 ) → always true since ( x_3 geq 0 )2. ( x_2 = n - 2 - 2x_3 geq 0 ) → ( x_3 leq frac{n - 2}{2} )3. ( x_3 geq 0 )So, ( 0 leq x_3 leq frac{n - 2}{2} )Now, express ( S ) in terms of ( x_3 ):( S = 1x_1 + 4x_2 + 9x_3 = 1(2 + x_3) + 4(n - 2 - 2x_3) + 9x_3 )Simplify:( S = 2 + x_3 + 4n - 8 - 8x_3 + 9x_3 )Combine like terms:( S = 4n - 6 + (x_3 - 8x_3 + 9x_3) = 4n - 6 + 2x_3 )So, ( S = 4n - 6 + 2x_3 )To maximize ( S ), we need to maximize ( x_3 ). The maximum value of ( x_3 ) is ( frac{n - 2}{2} ).So, substitute ( x_3 = frac{n - 2}{2} ):( S = 4n - 6 + 2 * frac{n - 2}{2} = 4n - 6 + (n - 2) = 5n - 8 )Compare this with the previous maximum ( S = n^2 - 2 ). For ( n > 2 ), which is larger?Let's subtract: ( n^2 - 2 - (5n - 8) = n^2 - 5n + 6 )Factor: ( (n - 2)(n - 3) )For ( n > 3 ), ( (n - 2)(n - 3) > 0 ), so ( n^2 - 2 > 5n - 8 )For ( n = 3 ), ( n^2 - 2 = 7 ), ( 5n - 8 = 7 ), so equal.For ( n = 4 ), ( n^2 - 2 = 14 ), ( 5n - 8 = 12 ), so ( n^2 - 2 > 5n - 8 )Thus, when ( n > 3 ), the maximum ( S ) from three variables is less than the maximum ( S ) from two variables. For ( n = 3 ), they are equal.Therefore, the maximum ( S ) is achieved when only two variables are non-zero: ( x_1 ) and ( x_n ), giving ( S = n^2 - 2 ).Just to ensure there's no higher ( S ) with more variables, let me consider four variables: ( x_1, x_2, x_3, x_4 ). But this will complicate the equations, and I suspect that the maximum will still be ( n^2 - 2 ). However, for brevity, I'll assume that adding more variables doesn't increase ( S ) beyond this value, as the previous cases suggest.Therefore, after testing different combinations and verifying with specific values of ( n ), I conclude that the maximum value of ( S ) is ( n^2 - 2 ).