💡Alright, so I've got this problem here where a unit bought an experimental instrument for 32,000 yuan. They're using it continuously from the first day, and the maintenance cost on the nth day is given by the formula (n + 49)/10 yuan. The question is asking, if we want to minimize the daily average cost of using this instrument, how many days should it be used in total?Okay, let me break this down. First, the initial cost is 32,000 yuan. That's a fixed cost, right? It doesn't change based on how many days they use the instrument. Then, there are the maintenance costs, which vary each day. On day 1, the maintenance cost is (1 + 49)/10, which is 50/10 = 5 yuan. On day 2, it's (2 + 49)/10 = 51/10 = 5.1 yuan, and so on. So each day, the maintenance cost increases by 0.1 yuan.Now, the daily average cost is the total cost divided by the number of days. The total cost includes both the initial purchase and all the maintenance costs up to day n. So, I need to find the total maintenance cost over n days, add the initial cost, and then divide by n to get the average cost per day. Then, I need to find the value of n that minimizes this average cost.Let me write this out mathematically. Let’s denote the total cost as C(n) and the average cost as f(n). So,C(n) = 32,000 + sum from k=1 to n of [(k + 49)/10]And then,f(n) = C(n) / nSo, first, I need to compute the sum of the maintenance costs from day 1 to day n. The maintenance cost on day k is (k + 49)/10. So, the sum is:sum from k=1 to n of [(k + 49)/10] = (1/10) * sum from k=1 to n of (k + 49)I can split this sum into two parts:(1/10) * [sum from k=1 to n of k + sum from k=1 to n of 49]The sum of k from 1 to n is a well-known formula: n(n + 1)/2. The sum of 49 from k=1 to n is just 49n, since you're adding 49 n times.So, plugging these in:sum from k=1 to n of [(k + 49)/10] = (1/10) * [n(n + 1)/2 + 49n]Simplify this:= (1/10) * [ (n^2 + n)/2 + 49n ]= (1/10) * [ (n^2 + n + 98n)/2 ]= (1/10) * [ (n^2 + 99n)/2 ]= (n^2 + 99n)/20So, the total maintenance cost over n days is (n^2 + 99n)/20 yuan.Now, adding the initial cost of 32,000 yuan, the total cost C(n) is:C(n) = 32,000 + (n^2 + 99n)/20Therefore, the average cost per day f(n) is:f(n) = [32,000 + (n^2 + 99n)/20] / nLet me simplify this expression:First, divide each term in the numerator by n:f(n) = 32,000/n + (n^2)/20n + (99n)/20nSimplify each term:= 32,000/n + n/20 + 99/20So, f(n) = 32,000/n + n/20 + 99/20Now, to find the minimum average cost, we need to find the value of n that minimizes f(n). Since n is a positive integer, we can treat it as a continuous variable to find the minimum and then check the nearest integers if necessary.To find the minimum, we can take the derivative of f(n) with respect to n and set it equal to zero.Let’s denote f(n) as:f(n) = 32,000/n + n/20 + 99/20Differentiating f(n) with respect to n:f'(n) = -32,000/n² + 1/20Set f'(n) = 0:-32,000/n² + 1/20 = 0Solving for n:-32,000/n² + 1/20 = 0=> 1/20 = 32,000/n²=> n² = 32,000 * 20=> n² = 640,000=> n = sqrt(640,000)=> n = 800So, the critical point is at n = 800 days. Since the function f(n) is convex (the second derivative is positive), this critical point corresponds to a minimum.Therefore, the daily average cost is minimized when the instrument is used for 800 days.But just to be thorough, let me verify this by checking the average cost at n = 800, n = 799, and n = 801 to ensure that n = 800 indeed gives the minimum.First, compute f(800):f(800) = 32,000/800 + 800/20 + 99/20= 40 + 40 + 4.95= 84.95 yuanNow, f(799):f(799) = 32,000/799 + 799/20 + 99/20Compute each term:32,000 / 799 ≈ 40.05006258 yuan799 / 20 = 39.95 yuan99 / 20 = 4.95 yuanSo, f(799) ≈ 40.05006258 + 39.95 + 4.95 ≈ 84.95006258 yuanSimilarly, f(801):f(801) = 32,000/801 + 801/20 + 99/20Compute each term:32,000 / 801 ≈ 39.95006242 yuan801 / 20 = 40.05 yuan99 / 20 = 4.95 yuanSo, f(801) ≈ 39.95006242 + 40.05 + 4.95 ≈ 84.95006242 yuanSo, both f(799) and f(801) are approximately 84.95006258 yuan and 84.95006242 yuan, respectively, which are both slightly higher than f(800) = 84.95 yuan.This confirms that n = 800 gives the minimum average cost.Alternatively, another way to approach this problem is by using the AM-GM inequality, which states that for positive real numbers a and b, the arithmetic mean is greater than or equal to the geometric mean, with equality when a = b.Looking back at the expression for f(n):f(n) = 32,000/n + n/20 + 99/20We can consider the terms 32,000/n and n/20. Let's set a = 32,000/n and b = n/20.According to AM-GM inequality:(a + b)/2 ≥ sqrt(ab)Multiply both sides by 2:a + b ≥ 2*sqrt(ab)So,32,000/n + n/20 ≥ 2*sqrt(32,000/n * n/20) = 2*sqrt(32,000/20) = 2*sqrt(1,600) = 2*40 = 80Therefore, 32,000/n + n/20 ≥ 80Adding the constant term 99/20 to both sides:32,000/n + n/20 + 99/20 ≥ 80 + 99/20 = 80 + 4.95 = 84.95So, f(n) ≥ 84.95 yuan, with equality when 32,000/n = n/20.Solving for n:32,000/n = n/20=> n² = 32,000 * 20=> n² = 640,000=> n = 800Again, we arrive at n = 800 days as the point where the average cost is minimized.This method also confirms that the minimum average cost is 84.95 yuan, achieved when n = 800.So, both calculus and the AM-GM inequality lead us to the same conclusion: the instrument should be used for 800 days to minimize the daily average cost.Just to recap, the steps were:1. Express the total maintenance cost over n days.2. Add the initial cost to get the total cost.3. Divide by n to get the average cost function.4. Simplify the average cost function.5. Use calculus (taking the derivative and setting it to zero) or the AM-GM inequality to find the minimum.6. Verify the result by checking nearby integer values.This problem is a good example of how optimization techniques can be applied to real-world cost analysis. It also shows the power of different mathematical tools—calculus and inequalities—both leading to the same solution.Another thing to note is that the maintenance cost increases linearly with each day, which means that as the instrument ages, it becomes more expensive to maintain. However, spreading the fixed initial cost over more days reduces the average fixed cost, but the increasing maintenance costs start to have a more significant impact as n increases. The optimal point is where these two effects balance each other out, resulting in the minimum average cost.In practical terms, this means that the unit should use the instrument for 800 days to get the best cost efficiency. Using it for fewer days would mean higher average costs due to the fixed initial investment not being spread out enough, while using it for more days would increase the average costs because of the rising maintenance expenses.This kind of optimization is crucial in business and operations management, where decisions about asset utilization and maintenance schedules can significantly impact profitability and efficiency. By applying mathematical models, organizations can make data-driven decisions that minimize costs and maximize returns.In summary, through careful analysis and applying appropriate mathematical techniques, we've determined that the optimal number of days to use the instrument is 800 days.

💡First, identify the cost of the dryer, which is 490.Next, determine the cost of the washer by adding the additional amount it costs compared to the dryer: 490 + 220 = 710.Finally, calculate the total cost of the washer-dryer combination by adding the cost of the washer and the dryer: 710 + 490 = 1200.

💡Alright, so I have this problem where I need to find the maximum and minimum values of the expression (2x + y), given that ((x, y)) are related by the equation (sqrt{x - 1} + sqrt{y - 4} = 2). Hmm, okay. Let me try to figure this out step by step.First, I know that expressions involving square roots often require some substitution to simplify them. Maybe I can let (a = sqrt{x - 1}) and (b = sqrt{y - 4}). That way, the equation becomes (a + b = 2). That seems manageable.So, substituting back, (x = a^2 + 1) and (y = b^2 + 4). Now, I need to express (2x + y) in terms of (a) and (b). Let's do that:(2x + y = 2(a^2 + 1) + (b^2 + 4)).Simplifying that, I get:(2x + y = 2a^2 + 2 + b^2 + 4 = 2a^2 + b^2 + 6).But since (a + b = 2), I can express (b) in terms of (a): (b = 2 - a). Substituting that into the equation:(2x + y = 2a^2 + (2 - a)^2 + 6).Let me expand ((2 - a)^2):((2 - a)^2 = 4 - 4a + a^2).So now, substituting back:(2x + y = 2a^2 + 4 - 4a + a^2 + 6 = 3a^2 - 4a + 10).Okay, so now I have (2x + y) expressed as a quadratic function in terms of (a): (f(a) = 3a^2 - 4a + 10). I need to find the maximum and minimum values of this function given the constraints on (a).Wait, what are the constraints on (a)? Since (a = sqrt{x - 1}) and (b = sqrt{y - 4}), both (a) and (b) must be non-negative. Also, since (a + b = 2), (a) can range from 0 to 2. So, (a in [0, 2]).Alright, so I need to find the maximum and minimum of (f(a) = 3a^2 - 4a + 10) on the interval ([0, 2]).To find extrema, I can take the derivative of (f(a)) with respect to (a) and set it equal to zero to find critical points.Calculating the derivative:(f'(a) = 6a - 4).Setting this equal to zero:(6a - 4 = 0)(6a = 4)(a = frac{4}{6} = frac{2}{3}).So, there's a critical point at (a = frac{2}{3}). Now, I need to evaluate (f(a)) at the endpoints of the interval and at this critical point to determine where the maximum and minimum occur.First, at (a = 0):(f(0) = 3(0)^2 - 4(0) + 10 = 10).At (a = 2):(f(2) = 3(2)^2 - 4(2) + 10 = 12 - 8 + 10 = 14).At (a = frac{2}{3}):(fleft(frac{2}{3}right) = 3left(frac{2}{3}right)^2 - 4left(frac{2}{3}right) + 10).Calculating each term:(3left(frac{4}{9}right) = frac{12}{9} = frac{4}{3}),(-4left(frac{2}{3}right) = -frac{8}{3}),Adding these together with 10:(frac{4}{3} - frac{8}{3} + 10 = -frac{4}{3} + 10 = frac{26}{3} approx 8.6667).So, comparing the values:- At (a = 0), (f(a) = 10).- At (a = frac{2}{3}), (f(a) = frac{26}{3} approx 8.6667).- At (a = 2), (f(a) = 14).Therefore, the minimum value of (2x + y) is (frac{26}{3}) and the maximum value is 14.Now, I need to find the corresponding ((x, y)) points where these extrema occur.Starting with the maximum at (a = 2):Since (a = 2), (b = 2 - a = 0).So,(x = a^2 + 1 = 2^2 + 1 = 4 + 1 = 5),(y = b^2 + 4 = 0^2 + 4 = 0 + 4 = 4).Thus, the maximum occurs at ((5, 4)).For the minimum at (a = frac{2}{3}):(b = 2 - frac{2}{3} = frac{4}{3}).Calculating (x) and (y):(x = left(frac{2}{3}right)^2 + 1 = frac{4}{9} + 1 = frac{13}{9}),(y = left(frac{4}{3}right)^2 + 4 = frac{16}{9} + 4 = frac{16}{9} + frac{36}{9} = frac{52}{9}).So, the minimum occurs at (left(frac{13}{9}, frac{52}{9}right)).Let me just double-check my calculations to make sure I didn't make any mistakes.For the maximum:- (a = 2), (b = 0),- (x = 2^2 + 1 = 5),- (y = 0^2 + 4 = 4),- (2x + y = 2*5 + 4 = 10 + 4 = 14). That checks out.For the minimum:- (a = frac{2}{3}),- (b = frac{4}{3}),- (x = left(frac{2}{3}right)^2 + 1 = frac{4}{9} + 1 = frac{13}{9}),- (y = left(frac{4}{3}right)^2 + 4 = frac{16}{9} + 4 = frac{52}{9}),- (2x + y = 2*frac{13}{9} + frac{52}{9} = frac{26}{9} + frac{52}{9} = frac{78}{9} = frac{26}{3}). That also checks out.I think I've got it right. So, the maximum value of (2x + y) is 14 at the point ((5, 4)), and the minimum value is (frac{26}{3}) at the point (left(frac{13}{9}, frac{52}{9}right)).**Final Answer**The maximum value is (boxed{14}) at ((5, 4)) and the minimum value is (boxed{dfrac{26}{3}}) at (left(dfrac{13}{9}, dfrac{52}{9}right)).

💡First, I need to determine how many students have cats. Since three-fifths of the 25 students have cats, I calculate:[frac{3}{5} times 25 = 15 text{ students}]Next, I calculate the number of students with dogs. 20% of the students have dogs, so:[20% times 25 = frac{20}{100} times 25 = 5 text{ students}]The problem states that 3 students have elephants.Now, I add up all the students with pets:[15 , (text{cats}) + 5 , (text{dogs}) + 3 , (text{elephants}) = 23 text{ students}]Finally, to find out how many students have no pets, I subtract the number of students with pets from the total number of students:[25 - 23 = 2 text{ students}]

💡First, I need to determine the breadth of the floor. Given that the area of the floor is 529/3 square meters and the length is 23 meters, I can use the formula for the area of a rectangle, which is length multiplied by breadth.Next, I'll set up the equation:529/3 = 23 × breadthSolving for breadth:breadth = (529/3) / 23breadth = 529 / 69breadth = 7.6667 metersNow, I need to find the percentage by which the length exceeds the breadth. The difference between the length and breadth is:23 - 7.6667 = 15.3333 metersTo find the percentage increase:percentage = (15.3333 / 7.6667) × 100percentage = 200%Therefore, the length of the floor is 200% more than its breadth.

💡Okay, so I have this problem where I need to arrange the letters a, a, b, b, c, c into three rows and two columns. The requirements are that each row has different letters and each column also has different letters. The question is asking for the total number of different arrangements, and the options are A: 12, B: 18, C: 24, D: 36.Hmm, let me try to visualize this. I have three rows and two columns, so it's like a 3x2 grid. Each row must have two different letters, and each column must have three different letters. Since we have two of each letter (a, b, c), we need to make sure that each letter appears exactly twice in the grid, but not in the same row or column more than once.First, maybe I should think about how to arrange the letters in the first row. Since each row needs different letters, the first row can be any combination of two different letters. The letters we have are a, b, c, so the possible combinations for the first row are: a & b, a & c, or b & c.Let me pick one combination and see how it affects the rest of the grid. Suppose the first row is a & b. Then, the second row can't have a or b in the same column as the first row. So, in the first column, which has an a, the second and third rows must have b or c. But wait, the second row can't have a in the first column, so it has to be either b or c. Similarly, in the second column, which has a b, the second and third rows can't have b, so they have to be a or c.But hold on, we have two of each letter, so we need to make sure that each letter is used exactly twice. If the first row is a & b, then we have one a and one b used. That leaves one a, one b, and two c's remaining.So, in the second row, the first column can't be a, so it has to be b or c. But we already have one b used in the first row, so if we put b in the second row, first column, that would use up the second b. Then, the second column in the second row can't be b, so it has to be a or c. But we have one a and two c's left. If we put a in the second column, second row, that would use up the second a. Then, the third row would have to be c in both columns, but that's not allowed because each row must have different letters. So, that doesn't work.Alternatively, if the second row, first column is c instead of b, then we have c in the first column, second row. Then, the second column in the second row can't be b, so it has to be a or c. If we put a in the second column, second row, that uses up the second a. Then, the third row would have to be b in the first column and c in the second column. That works because we have one b and one c left. So, that gives us a valid arrangement.So, one possible arrangement is:Row 1: a bRow 2: c aRow 3: b cBut wait, let me check the columns. First column: a, c, b – all different. Second column: b, a, c – all different. Yes, that works.Now, let me see if there are other possibilities. If I choose a different combination for the first row, say a & c. Then, similar logic applies. The first row is a & c, so we have one a and one c used. Remaining letters: one a, one c, and two b's.Second row, first column can't be a, so it has to be b or c. If we put b in the first column, second row, then the second column can't be c, so it has to be a or b. If we put a in the second column, second row, that uses up the second a. Then, the third row would have to be c in the first column and b in the second column. That works.So, another arrangement:Row 1: a cRow 2: b aRow 3: c bChecking columns: first column: a, b, c – all different. Second column: c, a, b – all different. Good.Similarly, if the first row is b & c, we can follow the same process.First row: b cRemaining letters: one b, one c, two a's.Second row, first column can't be b, so it has to be a or c. If we put a in the first column, second row, then the second column can't be c, so it has to be a or b. If we put a in the second column, second row, that uses up the second a. Then, the third row would have to be b in the first column and c in the second column. That works.So, another arrangement:Row 1: b cRow 2: a aWait, no, that's not allowed because the second row would have two a's, which violates the rule that each row must have different letters. Oops, that's a mistake.Let me correct that. If the first row is b c, and the second row, first column is a, then the second column can't be c, so it has to be a or b. If we put b in the second column, second row, that uses up the second b. Then, the third row would have to be a in the first column and c in the second column. That works.So, arrangement:Row 1: b cRow 2: a bRow 3: a cChecking columns: first column: b, a, a – wait, that's two a's in the first column, which is not allowed. Oops, that's another mistake.I need to make sure that each column also has different letters. So, in the first column, after putting b in the first row, the second and third rows can't be b. So, if I put a in the second row, first column, then the third row, first column has to be c. Similarly, in the second column, after putting c in the first row, the second and third rows can't be c. So, if I put b in the second row, second column, then the third row, second column has to be a.So, correct arrangement:Row 1: b cRow 2: a bRow 3: c aChecking columns: first column: b, a, c – all different. Second column: c, b, a – all different. Yes, that works.So, from this, it seems that for each initial choice of the first row, there are two possible arrangements. Since there are three possible choices for the first row (a & b, a & c, b & c), and for each, there are two arrangements, that would give 3 * 2 = 6 arrangements.But wait, I think I might be missing something. Let me try to count differently.Another approach is to consider the permutations of the letters in the grid while satisfying the constraints.Since we have three rows and two columns, and each letter must appear exactly twice, with no repeats in any row or column.This is similar to arranging the letters in a 3x2 grid such that each row and column contains distinct letters.This is equivalent to finding the number of 3x2 Latin rectangles using the letters a, b, c, with each letter appearing exactly twice.A Latin rectangle is an m x n array filled with n different symbols, each occurring exactly once in each row and column.But in this case, we have three symbols (a, b, c) and a 3x2 grid, so it's a Latin rectangle of order 3x2.The number of Latin rectangles of order 3x2 is known to be 12.Wait, but I'm not sure about that. Let me think.Alternatively, we can calculate it step by step.First, choose the first row. There are 3 choices: a & b, a & c, or b & c.For each choice of the first row, we need to arrange the remaining letters in the second and third rows such that each column has distinct letters.Let's take the first row as a & b.Then, the remaining letters are a, b, c, c.We need to arrange these in the second and third rows, two letters each, such that:- In the first column, we have a in the first row, so the second and third rows must be b or c.- In the second column, we have b in the first row, so the second and third rows must be a or c.But we have two c's left, so we need to place them in the remaining positions.Let me try to fill in the second row.Second row, first column: can be b or c.Case 1: Second row, first column is b.Then, second row, second column can't be b, so it has to be a or c. But we have only one a left, so if we put a in the second column, second row, then the third row would have to be c in the first column and c in the second column, which is not allowed because each row must have different letters.Alternatively, if second row, second column is c, then we have:Row 1: a bRow 2: b cThen, the remaining letters are a and c.Third row, first column can't be a or b, so it has to be c.Third row, second column can't be c, so it has to be a.So, Row 3: c aThis works.Case 2: Second row, first column is c.Then, second row, second column can't be b, so it has to be a or c. If we put a in the second column, second row, then:Row 1: a bRow 2: c aRemaining letters: b and c.Third row, first column can't be a or c, so it has to be b.Third row, second column can't be a, so it has to be c.So, Row 3: b cThis also works.Therefore, for the first row being a & b, there are two possible arrangements.Similarly, if the first row is a & c, we can follow the same logic and find two arrangements.And if the first row is b & c, we can also find two arrangements.So, total arrangements: 3 (choices for first row) * 2 (arrangements for each) = 6.Wait, but earlier I thought it might be 12. Maybe I'm missing something.Alternatively, perhaps I need to consider that the first row can be arranged in different orders. For example, a & b can be arranged as a b or b a.So, for each combination, there are two permutations.Therefore, total first row choices: 3 combinations * 2 permutations = 6.Then, for each of these 6, we have two arrangements as above, leading to 6 * 2 = 12.Yes, that makes sense.So, the total number of arrangements is 12.Therefore, the answer is A: 12.