💡Okay, so I have this problem where I need to find the ratio of height to radius, h over r, for a cylinder that has a fixed total surface area A, and the volume is maximized. Hmm, okay, let me try to figure this out step by step.First, I remember that the total surface area of a cylinder includes the areas of the two circular bases and the area of the side (which is like a rectangle rolled up). So, the formula for the total surface area A should be:A = 2πr² + 2πrhWhere r is the radius and h is the height. Yeah, that makes sense because the two bases each have an area of πr², so together that's 2πr², and the side area is the circumference of the base times the height, which is 2πr * h.Now, the volume V of the cylinder is given by:V = πr²hSo, we have two equations here: one for the surface area and one for the volume. The problem says that the surface area A is fixed, so I need to express the volume in terms of one variable and then find its maximum.Let me try to express h in terms of r using the surface area equation. So, starting with:A = 2πr² + 2πrhI can subtract 2πr² from both sides to get:A - 2πr² = 2πrhThen, divide both sides by 2πr to solve for h:h = (A - 2πr²) / (2πr)Okay, so now I have h expressed in terms of r. That means I can substitute this into the volume equation to have V only in terms of r.So, substituting h into V:V = πr² * [(A - 2πr²) / (2πr)]Let me simplify this expression. First, πr² divided by 2πr is (r² / (2r)) which simplifies to r/2. So, the equation becomes:V = (r/2) * (A - 2πr²)Multiplying through:V = (rA)/2 - (2πr³)/2Simplifying further:V = (A r)/2 - πr³Alright, so now the volume is expressed solely in terms of r: V(r) = (A r)/2 - πr³To find the maximum volume, I need to find the critical points of this function. That means taking the derivative of V with respect to r and setting it equal to zero.So, let's compute dV/dr:dV/dr = d/dr [(A r)/2 - πr³] = A/2 - 3πr²Set this derivative equal to zero for critical points:A/2 - 3πr² = 0Solving for r:A/2 = 3πr²Multiply both sides by 2:A = 6πr²Then, divide both sides by 6π:r² = A / (6π)Take the square root of both sides:r = sqrt(A / (6π))Okay, so that gives me the radius r that maximizes the volume. Now, I need to find the corresponding height h.Earlier, I had expressed h in terms of r:h = (A - 2πr²) / (2πr)Let me substitute r² = A / (6π) into this equation.First, compute 2πr²:2πr² = 2π * (A / (6π)) = (2πA) / (6π) = A/3So, A - 2πr² = A - A/3 = (2A)/3Now, substitute back into h:h = (2A/3) / (2πr)Simplify numerator and denominator:h = (2A/3) / (2πr) = (A/3) / (πr) = A / (3πr)But wait, I have r expressed in terms of A and π:r = sqrt(A / (6π)) = (A / (6π))^(1/2)So, let's write h in terms of A and π:h = A / (3π * sqrt(A / (6π)))Let me simplify this expression step by step.First, let's write sqrt(A / (6π)) as (A / (6π))^(1/2). So,h = A / (3π * (A / (6π))^(1/2))Let me rewrite this as:h = A / [3π * (A)^(1/2) / (6π)^(1/2))]Which is:h = A / [3π * (A)^(1/2) / (6π)^(1/2))]Simplify the denominator:3π * (A)^(1/2) / (6π)^(1/2) = 3π * sqrt(A) / sqrt(6π)So, h = A / [3π * sqrt(A) / sqrt(6π)]Divide A by this term:h = A * [sqrt(6π) / (3π sqrt(A))]Simplify A / sqrt(A) = sqrt(A):h = sqrt(A) * sqrt(6π) / (3π)Combine the square roots:h = sqrt(6π A) / (3π)Wait, let me check that again.Wait, A * sqrt(6π) / (3π sqrt(A)) = (A / sqrt(A)) * sqrt(6π) / (3π) = sqrt(A) * sqrt(6π) / (3π)Yes, that's correct.So, h = sqrt(6π A) / (3π)But sqrt(6π A) is sqrt(6π) * sqrt(A), so:h = sqrt(6π) * sqrt(A) / (3π)Simplify sqrt(6π) / (3π):sqrt(6π) / (3π) = sqrt(6π) / (3π) = sqrt(6) * sqrt(π) / (3π) = sqrt(6) / (3 sqrt(π))So, h = sqrt(A) * sqrt(6) / (3 sqrt(π)) = sqrt(6A) / (3 sqrt(π))Wait, let me see if I can express h in terms of r.Earlier, we found that r = sqrt(A / (6π)). Let me express h in terms of r.We had h = A / (3πr)But A = 6πr², so substitute that in:h = (6πr²) / (3πr) = (6πr²) / (3πr) = 2rWait, that's much simpler! So, h = 2r.Therefore, h / r = 2r / r = 2.Wait, so that's different from what I got earlier. Hmm, maybe I made a mistake in the earlier steps when I tried to express h in terms of A and π.Let me go back.We had h = A / (3πr)But since A = 6πr², substituting that in:h = (6πr²) / (3πr) = 2rYes, that's correct. So, h = 2r, which means h / r = 2.Wait, so why did I get a different result earlier when I tried to express h in terms of A and π? Maybe I messed up the algebra.Let me try that again.We had h = A / (3πr)But r = sqrt(A / (6π)), so plug that into h:h = A / (3π * sqrt(A / (6π)))Let me compute the denominator:3π * sqrt(A / (6π)) = 3π * sqrt(A) / sqrt(6π) = 3π * sqrt(A) / (sqrt(6) sqrt(π)) ) = 3π sqrt(A) / (sqrt(6) sqrt(π)) )Simplify sqrt(π) in the denominator:3π / sqrt(π) = 3 sqrt(π)So, denominator becomes 3 sqrt(π) * sqrt(A) / sqrt(6)Wait, no, let me do it step by step.Denominator: 3π * sqrt(A / (6π)) = 3π * sqrt(A) / sqrt(6π) = 3π * sqrt(A) / (sqrt(6) sqrt(π)) )So, 3π / sqrt(π) = 3 sqrt(π)Therefore, denominator is 3 sqrt(π) * sqrt(A) / sqrt(6)So, h = A / [3 sqrt(π) * sqrt(A) / sqrt(6)] = A * sqrt(6) / (3 sqrt(π) sqrt(A)) )Simplify A / sqrt(A) = sqrt(A):h = sqrt(A) * sqrt(6) / (3 sqrt(π)) = sqrt(6A) / (3 sqrt(π))But earlier, we saw that h = 2r, and r = sqrt(A / (6π)), so h = 2 sqrt(A / (6π)) = 2 sqrt(A) / sqrt(6π) = 2 sqrt(A) / (sqrt(6) sqrt(π)) ) = (2 / sqrt(6)) sqrt(A) / sqrt(π) = sqrt(6)/3 * sqrt(A) / sqrt(π)Wait, because 2 / sqrt(6) = sqrt(6)/3, since sqrt(6)/3 = (sqrt(6)/sqrt(6)) * sqrt(6)/3 = 1 * sqrt(6)/3.Yes, so h = sqrt(6)/3 * sqrt(A)/sqrt(π) = sqrt(6A) / (3 sqrt(π))Which matches the earlier result. So, both methods give the same h.But when I express h in terms of r, it's simpler: h = 2r, so h/r = 2.Therefore, the ratio h/r is 2.Wait, but in the initial problem, the user had a different thought process where they ended up with h/r = 1. But that seems incorrect because when I do it step by step, I get h/r = 2.Let me double-check.We have:From surface area: A = 2πr² + 2πrhExpress h: h = (A - 2πr²)/(2πr)Volume: V = πr²h = πr² * (A - 2πr²)/(2πr) = (r(A - 2πr²))/2Take derivative: dV/dr = (A - 4πr²)/2Set to zero: A - 4πr² = 0 => r² = A/(4π) => r = sqrt(A/(4π))Wait, hold on, in the initial problem, the user had:dV/dr = (A - 4πr²)/2 = 0 => A = 4πr² => r = sqrt(A/(4π))But in my calculation, I had:From dV/dr = A/2 - 3πr² = 0 => A = 6πr² => r = sqrt(A/(6π))Wait, so which one is correct?Wait, let's go back.In the initial problem, the user had:V = (r(A - 2πr²))/2Then, derivative dV/dr = (A - 4πr²)/2Set to zero: A - 4πr² = 0 => r² = A/(4π)But in my calculation, I had:V = (A r)/2 - πr³Then, dV/dr = A/2 - 3πr² = 0 => A = 6πr² => r² = A/(6π)Wait, so which is correct?Wait, let's compute V correctly.From surface area: A = 2πr² + 2πrh => h = (A - 2πr²)/(2πr)Then, V = πr²h = πr² * (A - 2πr²)/(2πr) = (r(A - 2πr²))/2So, V = (A r - 2πr³)/2Therefore, V = (A r)/2 - πr³So, dV/dr = A/2 - 3πr²Set to zero: A/2 - 3πr² = 0 => A = 6πr² => r² = A/(6π) => r = sqrt(A/(6π))So, my calculation is correct, and the initial user's calculation was incorrect because they had dV/dr = (A - 4πr²)/2, which would imply that V = (A r - 2πr³)/2, but actually, the derivative of (A r)/2 is A/2, and the derivative of -πr³ is -3πr², so dV/dr = A/2 - 3πr².Therefore, the correct critical point is at r = sqrt(A/(6π)), leading to h = 2r, so h/r = 2.Therefore, the initial user's answer of h/r = 1 was incorrect, and the correct answer is h/r = 2.So, to summarize, the ratio h/r when the volume is maximized for a given surface area A is 2.

💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. First, it says that {a_n} is an arithmetic sequence with a non-zero common difference. That means each term increases by a fixed amount, right? So, if the first term is a_1 and the common difference is d, then a_2 = a_1 + d, a_3 = a_1 + 2d, and so on. Got that.Next, it says that a_2 is the geometric mean of a_1 and a_4. Hmm, geometric mean. I remember that the geometric mean of two numbers x and y is sqrt(xy). So, does that mean a_2 = sqrt(a_1 * a_4)? Let me write that down:a_2 = sqrt(a_1 * a_4)But since it's an arithmetic sequence, I can express a_2 and a_4 in terms of a_1 and d. a_2 = a_1 + da_4 = a_1 + 3dSo substituting these into the equation:a_1 + d = sqrt(a_1 * (a_1 + 3d))Hmm, that looks a bit complicated. Maybe I can square both sides to get rid of the square root:(a_1 + d)^2 = a_1 * (a_1 + 3d)Let me expand the left side:(a_1)^2 + 2a_1d + d^2 = (a_1)^2 + 3a_1dNow, subtract (a_1)^2 from both sides:2a_1d + d^2 = 3a_1dSubtract 2a_1d from both sides:d^2 = a_1dSince d is non-zero, I can divide both sides by d:d = a_1Oh, interesting! So the common difference d is equal to the first term a_1. That simplifies things.Now, the second condition is that a_5 + a_6 = 11. Let me express a_5 and a_6 in terms of a_1 and d:a_5 = a_1 + 4da_6 = a_1 + 5dSo, adding them together:a_5 + a_6 = (a_1 + 4d) + (a_1 + 5d) = 2a_1 + 9d = 11But earlier, I found that d = a_1. So, substituting d with a_1:2a_1 + 9a_1 = 1111a_1 = 11a_1 = 1Therefore, d = a_1 = 1. So, the arithmetic sequence has a first term of 1 and a common difference of 1. That means the sequence is just 1, 2, 3, 4, 5, 6, and so on.So, the general formula for a_n is:a_n = a_1 + (n - 1)d = 1 + (n - 1)*1 = nAlright, that seems straightforward. Let me just double-check my steps to make sure I didn't make a mistake.1. Expressed a_2 and a_4 in terms of a_1 and d.2. Set up the equation for the geometric mean.3. Squared both sides and simplified to find that d = a_1.4. Used the second condition to solve for a_1, which turned out to be 1.5. Concluded that the sequence is just the natural numbers.Seems solid. I think part (I) is done.Now, moving on to part (II). It says to choose one of the following conditions to find the sum of the first n terms of the sequence {b_n}. The options are:① b_n = a_n * 2^{a_n}② b_n = (a_{2n})^2 / [(2a_n - 1)(2a_n + 1)]I need to choose one of these. Since I have to write a detailed thought process, I'll consider both, but in the end, I'll choose one to solve. Maybe I'll try both and see which one seems easier.Starting with condition ①: b_n = a_n * 2^{a_n}Since I found that a_n = n, this simplifies to:b_n = n * 2^nSo, I need to find the sum S_n = sum_{k=1}^n k * 2^kHmm, I remember that there's a formula for this kind of sum. It's a standard sum involving arithmetic-geometric series. Let me recall how to approach this.One method is to consider the sum S = sum_{k=1}^n k * r^k. There's a formula for this, which is:S = r(1 - (n + 1)r^n + n r^{n + 1}) / (1 - r)^2But in this case, r = 2, so plugging in r = 2:S = 2(1 - (n + 1)2^n + n 2^{n + 1}) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1So,S = 2(1 - (n + 1)2^n + n 2^{n + 1})Let me expand this:= 2 * 1 - 2*(n + 1)2^n + 2*n*2^{n + 1}Simplify each term:= 2 - (n + 1)2^{n + 1} + n*2^{n + 2}Wait, that seems a bit messy. Maybe I made a mistake in applying the formula. Let me check the formula again.Wait, actually, I think the formula is:sum_{k=0}^n k r^k = r(1 - (n + 1)r^n + n r^{n + 1}) / (1 - r)^2But since our sum starts at k=1, it's the same as sum_{k=0}^n k r^k minus the term at k=0, which is 0. So, the formula still applies.But let me try a different approach. Maybe using generating functions or recursion.Let me denote S = sum_{k=1}^n k 2^kI can write S = 2 + 2*2^2 + 3*2^3 + ... + n*2^nMultiply both sides by 2:2S = 2*2 + 2*2*2^2 + 3*2*2^3 + ... + n*2*2^n= 2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n + 1}Now, subtract the original S from this:2S - S = S = (2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n + 1}) - (2 + 2*2^2 + 3*2^3 + ... + n*2^n)Let me align the terms:= [2^2 - 2] + [2*2^3 - 2*2^2] + [3*2^4 - 3*2^3] + ... + [n*2^{n + 1} - n*2^n]Factor out the common terms:= (2^2 - 2) + 2*(2^3 - 2^2) + 3*(2^4 - 2^3) + ... + n*(2^{n + 1} - 2^n)Notice that 2^{k + 1} - 2^k = 2^k(2 - 1) = 2^kSo, each term becomes:= (4 - 2) + 2*(8 - 4) + 3*(16 - 8) + ... + n*(2^{n + 1} - 2^n)= 2 + 2*4 + 3*8 + ... + n*2^nWait, that doesn't seem to help much. Maybe I need to adjust my approach.Alternatively, let's consider the sum S = sum_{k=1}^n k 2^kLet me write S as:S = 2 + 2*2^2 + 3*2^3 + ... + n*2^nNow, consider 2S:2S = 2*2 + 2*2*2^2 + 3*2*2^3 + ... + n*2*2^n= 2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n + 1}Subtract S from 2S:2S - S = S = (2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n + 1}) - (2 + 2*2^2 + 3*2^3 + ... + n*2^n)Let me write this out term by term:= (2^2 - 2) + (2*2^3 - 2*2^2) + (3*2^4 - 3*2^3) + ... + (n*2^{n + 1} - n*2^n)Factor out the common terms:= (4 - 2) + 2*(8 - 4) + 3*(16 - 8) + ... + n*(2^{n + 1} - 2^n)Simplify each term:= 2 + 2*4 + 3*8 + ... + n*2^nWait, this seems to be going in circles. Maybe I need to express this differently.Let me consider the difference between terms:Looking at the expression:S = sum_{k=1}^n k 2^kI can write this as:S = 2 + 2*2^2 + 3*2^3 + ... + n*2^nNow, let's consider S - 2S:Wait, no, that's not helpful. Maybe I can write S in terms of another sum.Let me recall that sum_{k=0}^n 2^k = 2^{n + 1} - 1Also, sum_{k=0}^n k 2^k = 2(2^n(n - 1) + 1)Wait, maybe I can use differentiation. Let me think.Consider the sum G = sum_{k=0}^n 2^k = 2^{n + 1} - 1If I differentiate G with respect to 2, I get sum_{k=0}^n k 2^{k - 1} = (n 2^{n + 1} - 2^{n + 1} + 2) / (2 - 1)^2Wait, that might be overcomplicating. Maybe I should look up the formula.Wait, I think the formula for sum_{k=1}^n k r^k is r(1 - (n + 1) r^n + n r^{n + 1}) / (1 - r)^2Let me plug in r = 2:sum_{k=1}^n k 2^k = 2(1 - (n + 1)2^n + n 2^{n + 1}) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1So,= 2(1 - (n + 1)2^n + n 2^{n + 1})Expand:= 2 - 2(n + 1)2^n + 2n 2^{n + 1}Simplify each term:= 2 - (n + 1)2^{n + 1} + n 2^{n + 2}Wait, that seems too complicated. Maybe I made a mistake in the formula.Wait, let me double-check the formula. The standard formula for sum_{k=1}^n k r^k is:r(1 - (n + 1) r^n + n r^{n + 1}) / (1 - r)^2Yes, that's correct. So plugging in r = 2:= 2(1 - (n + 1)2^n + n 2^{n + 1}) / (1 - 2)^2Denominator is 1, so:= 2(1 - (n + 1)2^n + n 2^{n + 1})Let me factor out 2^n:= 2[1 + 2^n(- (n + 1) + 2n)]= 2[1 + 2^n(n - 1)]So,= 2 + 2^{n + 1}(n - 1)Therefore, the sum S_n = (n - 1)2^{n + 1} + 2Let me test this with small n to see if it works.For n = 1:b_1 = 1*2^1 = 2Sum S_1 = 2Using formula: (1 - 1)2^{2} + 2 = 0 + 2 = 2. Correct.For n = 2:b_1 = 2, b_2 = 2*2^2 = 8Sum S_2 = 2 + 8 = 10Using formula: (2 - 1)2^{3} + 2 = 1*8 + 2 = 10. Correct.For n = 3:b_3 = 3*2^3 = 24Sum S_3 = 2 + 8 + 24 = 34Using formula: (3 - 1)2^{4} + 2 = 2*16 + 2 = 34. Correct.Okay, seems to work. So, the sum is S_n = (n - 1)2^{n + 1} + 2Alright, that seems solid. So, if I choose condition ①, the sum is S_n = (n - 1)2^{n + 1} + 2Now, let me consider condition ② just to see if it's easier or not.Condition ②: b_n = (a_{2n})^2 / [(2a_n - 1)(2a_n + 1)]Since a_n = n, this becomes:b_n = (2n)^2 / [(2n - 1)(2n + 1)] = 4n^2 / [(2n - 1)(2n + 1)]Simplify denominator: (2n - 1)(2n + 1) = 4n^2 - 1So,b_n = 4n^2 / (4n^2 - 1) = [4n^2 - 1 + 1] / (4n^2 - 1) = 1 + 1/(4n^2 - 1)So,b_n = 1 + 1/(4n^2 - 1)Now, 4n^2 - 1 can be factored as (2n - 1)(2n + 1), so:b_n = 1 + 1/[(2n - 1)(2n + 1)]This looks like it can be expressed using partial fractions. Let me try that.Express 1/[(2n - 1)(2n + 1)] as A/(2n - 1) + B/(2n + 1)Multiply both sides by (2n - 1)(2n + 1):1 = A(2n + 1) + B(2n - 1)Let me solve for A and B.Let me set up equations by choosing suitable n values.Let n = 1/2:1 = A(2*(1/2) + 1) + B(2*(1/2) - 1)= A(1 + 1) + B(1 - 1)= 2A + 0So, 2A = 1 => A = 1/2Similarly, let n = -1/2:1 = A(2*(-1/2) + 1) + B(2*(-1/2) - 1)= A(-1 + 1) + B(-1 - 1)= 0 + B(-2)So, -2B = 1 => B = -1/2Therefore,1/[(2n - 1)(2n + 1)] = (1/2)/(2n - 1) - (1/2)/(2n + 1)So,b_n = 1 + (1/2)/(2n - 1) - (1/2)/(2n + 1)Therefore, the sum S_n = sum_{k=1}^n b_k = sum_{k=1}^n [1 + (1/2)/(2k - 1) - (1/2)/(2k + 1)]This can be split into three separate sums:S_n = sum_{k=1}^n 1 + (1/2) sum_{k=1}^n 1/(2k - 1) - (1/2) sum_{k=1}^n 1/(2k + 1)Simplify each sum:First sum: sum_{k=1}^n 1 = nSecond sum: (1/2) sum_{k=1}^n 1/(2k - 1)Third sum: -(1/2) sum_{k=1}^n 1/(2k + 1)Let me write the second and third sums together:(1/2)[sum_{k=1}^n 1/(2k - 1) - sum_{k=1}^n 1/(2k + 1)]Notice that the second sum can be reindexed. Let me set m = k + 1 in the third sum:sum_{k=1}^n 1/(2k + 1) = sum_{m=2}^{n + 1} 1/(2m - 1)So, the difference becomes:sum_{k=1}^n 1/(2k - 1) - sum_{m=2}^{n + 1} 1/(2m - 1) = 1/(2*1 - 1) - 1/(2(n + 1) - 1) = 1/1 - 1/(2n + 1) = 1 - 1/(2n + 1)Therefore, the second and third sums together:(1/2)[1 - 1/(2n + 1)] = (1/2) - 1/(2(2n + 1)) = (1/2) - 1/(4n + 2)So, putting it all together:S_n = n + (1/2) - 1/(4n + 2)Simplify:= n + 1/2 - 1/(4n + 2)To combine the constants, let me write 1/2 as (2n + 1)/(4n + 2):= n + (2n + 1)/(4n + 2) - 1/(4n + 2)= n + (2n + 1 - 1)/(4n + 2)= n + (2n)/(4n + 2)= n + (2n)/(2(2n + 1))= n + n/(2n + 1)To combine these terms, let me write n as (n(2n + 1))/(2n + 1):= [n(2n + 1) + n]/(2n + 1)= [2n^2 + n + n]/(2n + 1)= [2n^2 + 2n]/(2n + 1)Factor numerator:= 2n(n + 1)/(2n + 1)So, S_n = (2n(n + 1))/(2n + 1)Let me test this with small n.For n = 1:b_1 = (2*1)^2 / [(2*1 - 1)(2*1 + 1)] = 4 / (1*3) = 4/3Sum S_1 = 4/3Using formula: (2*1*2)/(2*1 + 1) = 4/3. Correct.For n = 2:b_1 = 4/3, b_2 = (4)^2 / (3*5) = 16/15Sum S_2 = 4/3 + 16/15 = (20/15 + 16/15) = 36/15 = 12/5Using formula: (2*2*3)/(5) = 12/5. Correct.For n = 3:b_3 = (6)^2 / (5*7) = 36/35Sum S_3 = 4/3 + 16/15 + 36/35Convert to common denominator 105:= 140/105 + 112/105 + 108/105 = (140 + 112 + 108)/105 = 360/105 = 24/7Using formula: (2*3*4)/(7) = 24/7. Correct.Great, so the formula works.So, if I choose condition ②, the sum is S_n = (2n(n + 1))/(2n + 1)Comparing both conditions, condition ① leads to a sum involving 2^{n + 1}, which is exponential, while condition ② leads to a rational function. Both are interesting, but since the problem allows choosing one, I think I'll go with condition ① because it's more straightforward in terms of steps, even though it involves a bit more complex algebra.Wait, actually, condition ② was easier in terms of telescoping series, which made the sum collapse nicely. But condition ① required using the arithmetic-geometric series formula, which is a bit more involved.But since I already solved both, I can present both solutions. However, the note says that if multiple conditions are chosen and answered separately, the first answer will be scored. So, maybe I should just choose one. But since I have to write a detailed thought process, I'll present both.But in the end, I think condition ① is more challenging and thus more interesting, so I'll go with that.So, to recap:Part (I): a_n = nPart (II): Choosing condition ①, sum S_n = (n - 1)2^{n + 1} + 2Alternatively, choosing condition ②, sum S_n = (2n(n + 1))/(2n + 1)But since the problem says to choose one, I'll present both solutions but note that only the first will be scored. However, for thoroughness, I'll include both.But in the final answer, I'll present both as separate solutions.

💡First, calculate the time taken for Route X:- Distance: 8 miles- Speed: 40 mph- Time: 8 / 40 = 0.2 hours = 12 minutesNext, calculate the time taken for Route Y:- Total Distance: 6 miles - Non-construction zone: 5 miles - Construction zone: 1 mile- Speeds: - Non-construction zone: 50 mph - Construction zone: 10 mph- Time for non-construction zone: 5 / 50 = 0.1 hours = 6 minutes- Time for construction zone: 1 / 10 = 0.1 hours = 6 minutes- Total time for Route Y: 6 + 6 = 12 minutesFinally, compare the times:- Difference: 12 minutes - 12 minutes = 0 minutesThus, Route Y is not faster than Route X.

💡Alright, so I'm trying to figure out how to arrange the counties on the Big Island, which has an odd number of counties. I know that each county has a road along one of its diagonals, and these roads need to form a closed path without passing through any point more than once. First, I'll think about the Small Island example with 6 counties. It's arranged in a way that the roads form a closed loop. I'm guessing that the arrangement alternates the direction of the diagonals in some pattern to achieve this. Maybe it's like a checkerboard pattern where the diagonals alternate between going from top-left to bottom-right and top-right to bottom-left.Now, for the Big Island, which has an odd number of counties, this might complicate things because odd numbers can sometimes lead to issues with alternating patterns. I'm wondering if it's possible to have a closed path with an odd number of counties or if there's a specific way to arrange them to make it work.Let me try to visualize a small odd number first, like 3 counties. If I arrange them in a straight line, each with a diagonal road, can I form a closed path? Probably not, because the roads would either all go in the same direction or alternate in a way that doesn't loop back.What about 5 counties? Maybe arranging them in a cross shape? If I have one county in the center and four surrounding it, perhaps the roads can connect in a way that forms a loop. But I'm not sure if that would work without overlapping or leaving gaps.Going back to the example with 6 counties, I notice that the roads form a continuous loop by alternating the direction of the diagonals. Maybe for an odd number, I need to adjust the pattern slightly. Perhaps adding an extra county in a way that still allows the loop to close.I recall that in graph theory, a closed path that visits each edge exactly once is called an Eulerian circuit. For an Eulerian circuit to exist, every vertex must have an even degree. Translating this to the counties, each intersection point (vertex) must have an even number of roads (edges) meeting there. If I think of each county as a vertex, then each county has two roads (diagonals) meeting at its corners. Wait, no, each county has one road, which is a diagonal. So maybe each corner is a vertex where multiple roads meet. This is getting a bit confusing. Maybe I need to think differently. If each county has a diagonal road, then each road connects two corners of the county. For the entire island, the roads need to form a single closed loop without crossing over itself.I think the key is to ensure that the arrangement of diagonals allows the loop to continue seamlessly from one county to the next. For an even number of counties, this seems straightforward by alternating the diagonals. But with an odd number, there might be a point where the pattern breaks, making it impossible to close the loop without overlapping.Is there a way to adjust the pattern to accommodate an odd number? Maybe by having one county with a different diagonal to "reset" the pattern. But I'm not sure if that would work without creating an overlap or a dead end.I wonder if there's a specific number of counties that makes this possible. The problem mentions that the Big Island has an odd number, but it doesn't specify which one. Maybe the smallest odd number greater than 6 that allows such an arrangement is 9. If I try arranging 9 counties in a 3x3 grid, perhaps I can alternate the diagonals in a way that forms a closed loop. Starting from the top-left county, I can have a diagonal going from top-left to bottom-right. Then, the next county to the right would have a diagonal from top-right to bottom-left, and so on, alternating each time. But when I get to the end of the row, I need to continue the loop to the next row. This might require adjusting the direction of the diagonals in the next row to maintain the closed path. It might look like a checkerboard pattern where the diagonals alternate in both rows and columns.After setting up the 3x3 grid with alternating diagonals, I can trace the path to see if it forms a closed loop without crossing over itself. Starting from the top-left, moving along the diagonal to the bottom-right of the first county, then continuing to the next county's diagonal, and so on. If done correctly, the path should loop back to the starting point after traversing all 9 counties.This seems promising. The alternation of diagonals in both directions allows the path to weave through the grid and close back on itself. Even though it's an odd number of counties, the alternating pattern compensates for it by ensuring the path doesn't get stuck or overlap.So, based on this reasoning, I think the Big Island can be arranged with 9 counties, forming a 3x3 grid with alternating diagonal roads that create a closed loop.

💡Alright, let's tackle this problem step by step. I've got four options here, labeled A through D, each presenting a different mathematical calculation. My goal is to figure out which one of these is correct. I'll go through each option one by one, think through the math, and see where it leads me.Starting with option A: (sqrt{3})^2 = 9. Hmm, okay. So, the square of the square root of 3. I remember that when you square a square root, they essentially cancel each other out. So, (sqrt{3})^2 should just equal 3, right? Because the square and the square root are inverse operations. But here, it's saying it equals 9. That doesn't seem right. 3 squared is 9, but the square of the square root of 3 should just be 3. So, I'm thinking option A is incorrect.Moving on to option B: sqrt{(-2)^2} = -2. Alright, let's break this down. First, I have (-2)^2. Squaring a negative number should give a positive result because a negative times a negative is a positive. So, (-2)^2 is 4. Now, taking the square root of 4, which is 2. But the option says it equals -2. Wait a minute, the square root of a positive number is always positive, right? So, sqrt{4} is 2, not -2. Therefore, option B is also incorrect.Next up is option C: sqrt{3} times sqrt{2} = 6. Okay, multiplying square roots. I recall that the product of square roots is the square root of the product. So, sqrt{3} times sqrt{2} should be sqrt{3 times 2}, which is sqrt{6}. But here, it's saying the result is 6. That's not correct because sqrt{6} is approximately 2.45, not 6. So, option C is incorrect as well.Finally, option D: sqrt{8} div sqrt{2} = 2. Let's see. Dividing square roots. I think the rule here is similar to multiplication; the quotient of square roots is the square root of the quotient. So, sqrt{8} div sqrt{2} should be sqrt{8 div 2}, which is sqrt{4}. And the square root of 4 is 2. That matches what's given in option D. So, this one seems correct.Wait, let me double-check to make sure I didn't make any mistakes. For option A, squaring the square root of 3 definitely gives 3, not 9. For option B, the square root of 4 is 2, not -2. For option C, multiplying sqrt{3} and sqrt{2} gives sqrt{6}, not 6. And for option D, dividing sqrt{8} by sqrt{2} simplifies to sqrt{4}, which is 2. Yep, that all checks out. So, option D is the correct calculation.

💡Okay, so I have this problem where I need to find positive integers x and y such that x is not equal to y, and the sum of their reciprocals is 1/15. Then, I need to find the smallest possible value for x + y. Hmm, let me think about how to approach this.First, the equation given is 1/x + 1/y = 1/15. I remember that when dealing with equations involving reciprocals, it's often helpful to combine them into a single fraction. So, let me try that.Combining the fractions on the left side, I get (y + x) / (xy) = 1/15. So, that simplifies to (x + y) / (xy) = 1/15. Now, if I cross-multiply, I can get rid of the fractions. That would give me 15(x + y) = xy.Okay, so now I have 15(x + y) = xy. Let me rearrange this equation to make it a bit more manageable. If I move all terms to one side, I get xy - 15x - 15y = 0. Hmm, this looks a bit like a quadratic equation, but it's in two variables. Maybe I can factor it somehow.I recall that sometimes adding a constant term can help factor such equations. Let me see if that works here. If I add 225 to both sides, I get xy - 15x - 15y + 225 = 225. Now, let me check if the left side factors nicely. It looks like it might factor into (x - 15)(y - 15) because when I expand that, I get xy -15x -15y + 225, which matches the left side. So, (x - 15)(y - 15) = 225.Great, so now I have the product of (x - 15) and (y - 15) equal to 225. Since x and y are positive integers, both (x - 15) and (y - 15) must also be positive integers. So, I need to find pairs of positive integers that multiply to 225. These are the factor pairs of 225.Let me list the factor pairs of 225. Starting from the smallest:1 and 225,3 and 75,5 and 45,9 and 25,15 and 15.Wait, but the problem states that x ≠ y. So, if (x - 15) and (y - 15) are both 15, then x and y would both be 30, which is not allowed. So, I have to exclude the pair (15, 15).Now, I need to consider the other factor pairs. For each pair, I can find the corresponding x and y by adding 15 back to each factor. Then, I can calculate x + y for each pair and find the smallest sum.Let's go through each factor pair:1. (1, 225): So, x - 15 = 1 and y - 15 = 225. Therefore, x = 16 and y = 240. The sum x + y = 16 + 240 = 256.2. (3, 75): x - 15 = 3 and y - 15 = 75. So, x = 18 and y = 90. The sum x + y = 18 + 90 = 108.3. (5, 45): x - 15 = 5 and y - 15 = 45. Thus, x = 20 and y = 60. The sum x + y = 20 + 60 = 80.4. (9, 25): x - 15 = 9 and y - 15 = 25. Hence, x = 24 and y = 40. The sum x + y = 24 + 40 = 64.5. (15, 15): As mentioned earlier, this would give x = 30 and y = 30, but since x ≠ y, we can't use this pair.So, from the valid factor pairs, the sums are 256, 108, 80, and 64. The smallest sum is 64.Wait a second, let me double-check to make sure I didn't miss any factor pairs. 225 is 15 squared, so its factors are all the numbers that multiply together to give 225. The factors are 1, 3, 5, 9, 15, 25, 45, 75, 225. So, the factor pairs I listed are indeed all the possible ones. Therefore, 64 is the smallest possible sum.Just to be thorough, let me plug x = 24 and y = 40 back into the original equation to verify.1/24 + 1/40. Let's compute this:The least common denominator of 24 and 40 is 120. So, 1/24 is 5/120 and 1/40 is 3/120. Adding them together gives 8/120, which simplifies to 1/15. Perfect, that matches the given equation.So, I'm confident that the smallest possible value for x + y is 64.**Final Answer**The smallest possible value for ( x + y ) is boxed{64}.