💡Alright, so I've got this logic problem here, and I need to figure out which proposition is true among the options given. Let me take it step by step because I want to make sure I understand everything correctly.First, let's break down the problem. There are two propositions, P and Q. Proposition P states that "All rational numbers are real numbers." Proposition Q says, "The logarithm of positive numbers is always positive." Then, I have four options: A, B, C, and D, each combining these propositions with logical operators. I need to determine which one is true.Okay, let's start by evaluating the truth values of P and Q individually because that seems like a logical starting point.Starting with proposition P: "All rational numbers are real numbers." Hmm, I remember from my math classes that rational numbers are numbers that can be expressed as a fraction of two integers, like 1/2, 3/4, or even whole numbers like 5/1. Real numbers, on the other hand, include all rational numbers and all irrational numbers. Irrational numbers can't be expressed as a simple fraction, like pi or the square root of 2. So, if all rational numbers are included in the set of real numbers, then proposition P should be true. That makes sense because I don't think there's a rational number that isn't a real number. So, P is true.Now, moving on to proposition Q: "The logarithm of positive numbers is always positive." Hmm, this one is a bit trickier. I know that logarithms can be positive or negative depending on the base and the number you're taking the logarithm of. Let me think about this. If I take the logarithm of a number greater than 1 with a base greater than 1, like log base 10 of 100, that's 2, which is positive. But what if I take the logarithm of a number between 0 and 1? For example, log base 10 of 0.1 is -1, which is negative. So, in that case, the logarithm is negative, even though 0.1 is a positive number. Therefore, the logarithm of a positive number isn't always positive—it can be negative. So, proposition Q is false.Alright, so now I know that P is true and Q is false. Let's note that down:- P: True- Q: FalseNow, let's look at the options given:A: ¬P ∨ Q B: P ∧ Q C: ¬P ∧ ¬Q D: ¬P ∨ ¬QI need to evaluate each of these propositions based on the truth values of P and Q.Starting with option A: ¬P ∨ Q. Since P is true, ¬P would be false. Q is false. So, the proposition becomes False ∨ False, which is False. Therefore, option A is false.Moving on to option B: P ∧ Q. P is true, Q is false. So, True ∧ False is False. Hence, option B is false.Next, option C: ¬P ∧ ¬Q. ¬P is false because P is true, and ¬Q is true because Q is false. So, False ∧ True is False. Therefore, option C is false.Finally, option D: ¬P ∨ ¬Q. Again, ¬P is false and ¬Q is true. So, False ∨ True is True. Therefore, option D is true.Wait, let me double-check that. So, P is true, so ¬P is false. Q is false, so ¬Q is true. So, in option D, it's saying either ¬P is true or ¬Q is true. Since ¬Q is true, the whole proposition is true. That makes sense.Just to make sure I didn't make a mistake, let me go through each step again.1. Evaluate P: All rational numbers are real numbers. Yes, that's true because rational numbers are a subset of real numbers.2. Evaluate Q: The logarithm of positive numbers is always positive. No, that's false because, for example, log(0.1) is negative.3. Therefore, ¬P is false, and ¬Q is true.4. Now, evaluate each option: - A: False ∨ False = False - B: True ∧ False = False - C: False ∧ True = False - D: False ∨ True = TrueYes, that seems consistent. So, option D is the correct one.I think I've got it. It's important to carefully evaluate each proposition and then apply the logical operators correctly. Sometimes, these can be tricky, especially with the negations and the order of operations, but breaking it down step by step helps prevent mistakes.

💡First, identify the amount of money John was given, which is 14.00, and the total cost of the gift, which is 28.00.Next, set up the ratio of the money given to the cost of the gift as 14:28.To simplify the ratio, divide both numbers by their greatest common divisor, which is 14. This results in 1:2.Therefore, the simplified ratio is 1:2.

💡Okay, so I have this hyperbola equation: (frac{x^2}{9} - frac{y^2}{m} = 1). I remember that hyperbolas have asymptotes, which are lines that the hyperbola approaches but never touches. The problem says one of the asymptotes is (y = frac{2}{3}x). I need to find the focal length of the hyperbola.First, let me recall the standard form of a hyperbola. Since the (x^2) term is positive, this is a hyperbola that opens left and right. The standard form is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a) and (b) are constants. Comparing this to the given equation, I can see that (a^2 = 9) and (b^2 = m). So, (a = 3) because the square root of 9 is 3, and (b = sqrt{m}).Now, the asymptotes of a hyperbola in this form are given by the equations (y = pm frac{b}{a}x). The problem gives one of the asymptotes as (y = frac{2}{3}x). Since the asymptotes are symmetric, the other one should be (y = -frac{2}{3}x), but I only need one to find the relationship between (b) and (a).So, setting (frac{b}{a} = frac{2}{3}), I can plug in the values I know. I know (a = 3), so substituting that in, I get:[frac{b}{3} = frac{2}{3}]Multiplying both sides by 3 gives:[b = 2]But wait, (b) is actually (sqrt{m}), right? Because (b^2 = m). So, if (b = 2), then:[sqrt{m} = 2]Squaring both sides gives:[m = 4]Okay, so now I know that (m = 4). That means the equation of the hyperbola is (frac{x^2}{9} - frac{y^2}{4} = 1).Now, I need to find the focal length. I remember that for hyperbolas, the distance from the center to each focus is (c), and the relationship between (a), (b), and (c) is given by:[c^2 = a^2 + b^2]Plugging in the known values, (a^2 = 9) and (b^2 = 4), so:[c^2 = 9 + 4 = 13]Taking the square root of both sides gives:[c = sqrt{13}]But wait, the question asks for the focal length. I think the focal length is the distance between the two foci, which would be (2c). So, if (c = sqrt{13}), then the focal length is:[2c = 2sqrt{13}]Let me double-check my steps to make sure I didn't make a mistake. I found (a = 3) correctly from (a^2 = 9). Then, using the asymptote equation, I set (frac{b}{a} = frac{2}{3}), which led me to (b = 2). Then, since (b^2 = m), I found (m = 4). Using the relationship (c^2 = a^2 + b^2), I calculated (c = sqrt{13}), and then doubled it for the focal length. Everything seems to check out.So, the focal length is (2sqrt{13}), which corresponds to option B.

💡Okay, so I have this problem here: there's a rectangular prism with dimensions 10 inches by 5 inches by 12 inches, and I need to find the surface area of a cube that has the same volume as this prism. Hmm, let's break this down step by step.First, I remember that the volume of a rectangular prism is calculated by multiplying its length, width, and height. So, for this prism, the volume should be 10 inches times 5 inches times 12 inches. Let me write that out:Volume = length × width × height = 10 × 5 × 12.Okay, let's do the math. 10 times 5 is 50, and then 50 times 12 is... 600. So, the volume of the rectangular prism is 600 cubic inches.Now, the problem says that there's a cube with the same volume. So, the volume of the cube is also 600 cubic inches. I need to find the surface area of this cube. But first, I think I need to find the length of one edge of the cube because once I have that, I can figure out the surface area.I recall that the volume of a cube is given by edge length cubed, which is s³, where s is the length of one edge. So, if the volume is 600, then:s³ = 600.To find s, I need to take the cube root of 600. Hmm, cube roots can be tricky. I know that 8 cubed is 512 and 9 cubed is 729, so the cube root of 600 should be somewhere between 8 and 9. Maybe around 8.4 or something? I could use a calculator to get a more precise value, but since this is a math problem, maybe I can leave it in terms of the cube root for now.But wait, the question asks for the surface area, not the edge length. Okay, so let's remember that the surface area of a cube is calculated by 6 times the area of one face. Since all edges are equal, each face is a square with area s². So, the surface area formula is:Surface Area = 6 × s².Since I know that s³ = 600, I can express s as the cube root of 600, which is ∛600. So, s = ∛600.Now, plugging that into the surface area formula:Surface Area = 6 × (∛600)².Hmm, that looks a bit complicated. Maybe I can simplify it or find a numerical value. Let's see, if s = ∛600, then s² = (∛600)². I wonder if there's a way to write this differently. I know that exponents can be tricky, but maybe I can express it as 600 raised to the power of 2/3.So, s² = (600)^(2/3).Therefore, the surface area becomes:Surface Area = 6 × (600)^(2/3).Now, I need to calculate this. I think I can use logarithms or a calculator to find the numerical value, but let's see if there's a simpler way. Maybe I can factor 600 to make it easier.600 can be factored into prime factors: 600 = 2^3 × 3 × 5^2. Let me check that:2^3 is 8, 8 × 3 is 24, and 24 × 5^2 is 24 × 25, which is 600. Yes, that's correct.So, 600 = 2^3 × 3 × 5^2.Now, let's plug that back into the expression for s²:s² = (2^3 × 3 × 5^2)^(2/3).Using exponent rules, when you raise a product to a power, you raise each factor to that power:s² = (2^3)^(2/3) × 3^(2/3) × (5^2)^(2/3).Simplify each term:(2^3)^(2/3) = 2^(3 × 2/3) = 2^2 = 4.(5^2)^(2/3) = 5^(2 × 2/3) = 5^(4/3).3^(2/3) remains as it is.So, s² = 4 × 3^(2/3) × 5^(4/3).Hmm, that still looks a bit complicated. Maybe I can combine the terms with exponents:3^(2/3) × 5^(4/3) = (3^2 × 5^4)^(1/3) = (9 × 625)^(1/3) = (5625)^(1/3).Wait, is that right? Let me check:3^(2/3) × 5^(4/3) = (3^2 × 5^4)^(1/3) because when you multiply terms with the same exponent, you can combine the bases under that exponent.So, 3^2 is 9, and 5^4 is 625. Multiplying those gives 9 × 625 = 5625.So, (5625)^(1/3) is the cube root of 5625.Hmm, I know that 17^3 is 4913 and 18^3 is 5832, so the cube root of 5625 should be between 17 and 18. Maybe around 17.7 or something.But I'm not sure if this is helpful. Maybe I should just calculate the numerical value step by step.Let's go back to s = ∛600.I can approximate ∛600:Since 8^3 = 512 and 9^3 = 729, and 600 is closer to 512 than to 729, so ∛600 is closer to 8 than to 9.Let me try 8.4:8.4^3 = 8.4 × 8.4 × 8.4.First, 8 × 8 = 64, 8 × 0.4 = 3.2, 0.4 × 8 = 3.2, and 0.4 × 0.4 = 0.16.So, 8.4 × 8.4 = (8 + 0.4)^2 = 64 + 2×8×0.4 + 0.16 = 64 + 6.4 + 0.16 = 70.56.Now, 70.56 × 8.4:Let's break it down:70 × 8.4 = 588.0.56 × 8.4 = 4.704.So, total is 588 + 4.704 = 592.704.Hmm, that's close to 600 but still a bit less. Let's try 8.43:8.43^3.First, 8.43 × 8.43:Let's calculate 8 × 8 = 64.8 × 0.43 = 3.44.0.43 × 8 = 3.44.0.43 × 0.43 = 0.1849.So, (8 + 0.43)^2 = 64 + 3.44 + 3.44 + 0.1849 = 64 + 6.88 + 0.1849 = 71.0649.Now, multiply 71.0649 × 8.43:Let's do 70 × 8.43 = 590.1.1.0649 × 8.43 ≈ 1 × 8.43 = 8.43, and 0.0649 × 8.43 ≈ 0.547.So, total ≈ 590.1 + 8.43 + 0.547 ≈ 599.077.That's very close to 600. So, 8.43^3 ≈ 599.077, which is just a bit less than 600.Let's try 8.44:8.44^3.First, 8.44 × 8.44:8 × 8 = 64.8 × 0.44 = 3.52.0.44 × 8 = 3.52.0.44 × 0.44 = 0.1936.So, (8 + 0.44)^2 = 64 + 3.52 + 3.52 + 0.1936 = 64 + 7.04 + 0.1936 = 71.2336.Now, multiply 71.2336 × 8.44:70 × 8.44 = 590.8.1.2336 × 8.44 ≈ 1 × 8.44 = 8.44, and 0.2336 × 8.44 ≈ 1.976.So, total ≈ 590.8 + 8.44 + 1.976 ≈ 601.216.Okay, so 8.44^3 ≈ 601.216, which is just a bit more than 600.So, the cube root of 600 is between 8.43 and 8.44. Let's say approximately 8.434.So, s ≈ 8.434 inches.Now, to find the surface area, which is 6 × s².First, let's calculate s²:s² = (8.434)^2.8 × 8 = 64.8 × 0.434 = 3.472.0.434 × 8 = 3.472.0.434 × 0.434 ≈ 0.188.So, (8 + 0.434)^2 = 64 + 3.472 + 3.472 + 0.188 ≈ 64 + 6.944 + 0.188 ≈ 71.132.So, s² ≈ 71.132.Now, multiply by 6:Surface Area ≈ 6 × 71.132 ≈ 426.792.So, approximately 426.79 square inches.But let me double-check my calculations to make sure I didn't make any mistakes.First, volume of the prism: 10 × 5 × 12 = 600. That seems correct.Cube volume: s³ = 600, so s = ∛600 ≈ 8.434. That seems reasonable.Surface area: 6 × s² ≈ 6 × 71.132 ≈ 426.79. That seems consistent.Alternatively, I could use a calculator for more precision, but since this is a math problem, maybe I can express it in terms of exponents or radicals.Wait, earlier I had:Surface Area = 6 × (600)^(2/3).And I factored 600 as 2^3 × 3 × 5^2.So, (600)^(2/3) = (2^3 × 3 × 5^2)^(2/3) = 2^(3 × 2/3) × 3^(2/3) × 5^(4/3) = 2^2 × 3^(2/3) × 5^(4/3) = 4 × 3^(2/3) × 5^(4/3).So, Surface Area = 6 × 4 × 3^(2/3) × 5^(4/3) = 24 × 3^(2/3) × 5^(4/3).Hmm, that's another way to write it, but it's still not a simple numerical value. So, I think the approximate value of 426.79 square inches is acceptable.Wait, but let me check if I can simplify 3^(2/3) × 5^(4/3).3^(2/3) × 5^(4/3) = (3^2 × 5^4)^(1/3) = (9 × 625)^(1/3) = 5625^(1/3).And 5625^(1/3) is the cube root of 5625, which we approximated earlier as around 17.7.So, 24 × 17.7 ≈ 24 × 17 + 24 × 0.7 = 408 + 16.8 = 424.8.Hmm, that's slightly less than our previous approximation of 426.79. Maybe my approximation of 5625^(1/3) as 17.7 was a bit low.Wait, earlier when I calculated s = ∛600 ≈ 8.434, and then s² ≈ 71.132, leading to Surface Area ≈ 426.79.But using the other method, I got 24 × 5625^(1/3) ≈ 24 × 17.7 ≈ 424.8.There's a slight discrepancy here, probably due to rounding errors in the approximations.To get a more accurate value, maybe I should use a calculator for ∛600 and then compute s² and the surface area.But since I don't have a calculator here, I'll stick with the first method where I approximated s ≈ 8.434, leading to Surface Area ≈ 426.79.So, rounding to two decimal places, it's approximately 426.79 square inches.Alternatively, if I want to express it as a fraction or a more exact form, but I think for the purposes of this problem, a decimal approximation is sufficient.Therefore, the surface area of the cube is approximately 426.79 square inches.

💡Okay, so I'm trying to figure out this problem where Juan has three rectangular solids, each with a volume of 128. He wants to stack them to make the tallest tower possible, and we need to find the minimum exposed surface area, not counting the base. Hmm, okay, let's break this down step by step.First, I need to figure out the dimensions of each solid. Each solid has a volume of 128, which is length × width × height. For each solid, we're given two face areas. I remember that the area of a face is length × width, so if I have two areas, I can find the third dimension using the volume.Let me start with the first solid. It has two faces with areas 4 and 32. Let's call the dimensions of this solid x, y, and z. So, if one face is 4, that could be x × y = 4, and another face is y × z = 32. Since the volume is 128, we have x × y × z = 128. If x × y is 4, then z must be 128 / 4 = 32. Then, since y × z is 32, and z is 32, then y must be 1. Then, x is 4 / y, which is 4 / 1 = 4. So, the dimensions of the first solid are 4 × 1 × 32.Okay, moving on to the second solid. It has two faces with areas 64 and 16. Let's call its dimensions a, b, and c. So, a × b = 64 and b × c = 16. The volume is 128, so a × b × c = 128. From a × b = 64, we get c = 128 / 64 = 2. Then, from b × c = 16, and c = 2, we get b = 16 / 2 = 8. Then, a = 64 / b = 64 / 8 = 8. So, the dimensions of the second solid are 8 × 8 × 2.Now, the third solid has two faces with areas 8 and 32. Let's call its dimensions p, q, and r. So, p × q = 8 and q × r = 32. The volume is 128, so p × q × r = 128. From p × q = 8, we get r = 128 / 8 = 16. Then, from q × r = 32, and r = 16, we get q = 32 / 16 = 2. Then, p = 8 / q = 8 / 2 = 4. So, the dimensions of the third solid are 4 × 2 × 16.Alright, so now I have the dimensions of all three solids:1. 4 × 1 × 322. 8 × 8 × 23. 4 × 2 × 16Now, Juan wants to stack them to make the tallest tower. So, I need to decide the orientation of each solid to maximize the height. The height of the tower will be the sum of the heights of each solid when stacked. But since we can rotate the solids, the height of each solid can be any of its dimensions.But wait, actually, when stacking, the height of each solid in the stack will be one of its dimensions, and the base will be another dimension. So, to maximize the total height, I should choose the tallest possible dimension for each solid.Looking at the first solid, 4 × 1 × 32. The tallest dimension is 32. The second solid, 8 × 8 × 2, the tallest dimension is 8. The third solid, 4 × 2 × 16, the tallest dimension is 16. So, if I stack them with their tallest sides up, the total height will be 32 + 8 + 16 = 56. Hmm, that seems pretty tall.But wait, I need to make sure that the bases can fit on top of each other. The base of the top solid must fit within the top face of the lower solid. So, the dimensions of the base of the upper solid must be less than or equal to the dimensions of the top face of the lower solid.So, I need to arrange the solids in such a way that each upper solid's base dimensions are less than or equal to the lower solid's top dimensions. To minimize the exposed surface area, I should try to have as much overlap as possible, meaning the upper solid's base should be as large as possible to cover the lower solid's top face, reducing the exposed area.Let me think about the order of stacking. If I put the tallest solid at the bottom, that might give me a larger base, which could help in covering more area when stacking the next solid on top. Alternatively, putting a shorter solid at the bottom might allow for a more compact stack, but I need to maximize the height.Wait, actually, the height is fixed by the sum of the heights of each solid, regardless of the order. So, regardless of the order, the total height will be 32 + 8 + 16 = 56. So, the order doesn't affect the total height, but it does affect the exposed surface area.Therefore, to minimize the exposed surface area, I need to arrange the solids in such a way that the areas of the sides are minimized. That is, I want the sides of the stack to have as small an area as possible.So, I need to consider the dimensions of each solid and how they can be oriented to minimize the side areas when stacked.Let me list the dimensions again:1. Solid A: 4 × 1 × 322. Solid B: 8 × 8 × 23. Solid C: 4 × 2 × 16I need to decide for each solid which dimension will be the height, which will be the depth, and which will be the width. The goal is to have the sides (depth × height and width × height) as small as possible.But since we're stacking them, the depth and width of each solid (except the bottom one) must be less than or equal to the depth and width of the solid below it.Wait, actually, the base of the upper solid must fit within the top face of the lower solid. So, if the top face of the lower solid is, say, 8 × 8, then the base of the upper solid must be ≤ 8 × 8. Similarly, if the top face is 4 × 1, the base of the upper solid must be ≤ 4 × 1.So, to minimize the exposed surface area, I want the upper solid's base to be as large as possible to cover more of the lower solid's top face, thereby reducing the exposed area.Therefore, I should arrange the solids in decreasing order of their base areas. That is, the solid with the largest base area should be at the bottom, then the next largest, and so on.Let's calculate the base areas for each solid when oriented to have their maximum height:- Solid A: Height = 32, so the base area is 4 × 1 = 4- Solid B: Height = 8, so the base area is 8 × 8 = 64- Solid C: Height = 16, so the base area is 4 × 2 = 8So, arranging them in decreasing order of base area: Solid B (64), Solid C (8), Solid A (4). So, the order should be Solid B at the bottom, Solid C in the middle, and Solid A on top.Wait, but Solid B has a base area of 64, which is much larger than the others. If I put Solid B at the bottom, then Solid C on top of it, which has a base area of 8, which is much smaller than 64, so there will be a lot of exposed area on Solid B. Then, Solid A on top of Solid C, which has a base area of 4, which is even smaller, so even more exposed area.Alternatively, maybe arranging them differently could result in less exposed area.Let me think. If I put Solid A at the bottom, which has a base area of 4, then Solid C on top of it, which has a base area of 8, which is larger than 4, so that won't fit. So, that's not possible.If I put Solid C at the bottom, which has a base area of 8, then Solid B on top of it, which has a base area of 64, which is larger than 8, so that won't fit either.So, the only possible order is Solid B at the bottom, then Solid C, then Solid A on top. Because Solid B has the largest base area, and the others have smaller base areas, so they can fit on top.So, the order is Solid B (8 × 8 × 2) at the bottom, Solid C (4 × 2 × 16) in the middle, and Solid A (4 × 1 × 32) on top.Now, let's calculate the exposed surface area.First, the base of the entire tower is the base of Solid B, which is 8 × 8. Since the base is not exposed, we don't count that.Now, let's consider each solid:1. Solid B (bottom): It has dimensions 8 × 8 × 2. When placed on the base, its height is 2. The sides of Solid B are: - Front and back: 8 × 2 each, so total 2 × (8 × 2) = 32 - Left and right: 8 × 2 each, so total 2 × (8 × 2) = 32 - Top: 8 × 8, but it's covered by Solid C, so not exposed - Bottom: 8 × 8, not exposed So, total exposed area from Solid B: 32 + 32 = 642. Solid C (middle): It has dimensions 4 × 2 × 16. When placed on top of Solid B, its height is 16. The base of Solid C is 4 × 2, which fits on the 8 × 8 top face of Solid B. The sides of Solid C are: - Front and back: 4 × 16 each, so total 2 × (4 × 16) = 128 - Left and right: 2 × 16 each, so total 2 × (2 × 16) = 64 - Top: 4 × 2, which is covered by Solid A, so not exposed - Bottom: 4 × 2, which is attached to Solid B, so not exposed So, total exposed area from Solid C: 128 + 64 = 1923. Solid A (top): It has dimensions 4 × 1 × 32. When placed on top of Solid C, its height is 32. The base of Solid A is 4 × 1, which fits on the 4 × 2 top face of Solid C. The sides of Solid A are: - Front and back: 4 × 32 each, so total 2 × (4 × 32) = 256 - Left and right: 1 × 32 each, so total 2 × (1 × 32) = 64 - Top: 4 × 1, which is exposed - Bottom: 4 × 1, which is attached to Solid C, so not exposed So, total exposed area from Solid A: 256 + 64 + (4 × 1) = 256 + 64 + 4 = 324Wait, but the top face of Solid A is exposed, so we need to add that. So, 4 × 1 = 4.Now, let's add up all the exposed areas:- Solid B: 64- Solid C: 192- Solid A: 324Total exposed surface area: 64 + 192 + 324 = 580But wait, I think I might have missed something. When stacking, the top face of Solid B is covered by Solid C, and the top face of Solid C is covered by Solid A, so those areas are not exposed. However, the sides of each solid are exposed, but we also need to consider the areas where the upper solid doesn't cover the lower solid's top face.For example, Solid C is placed on top of Solid B. Solid B's top face is 8 × 8, but Solid C's base is only 4 × 2. So, the area of Solid B's top face that is not covered by Solid C is 8 × 8 - 4 × 2 = 64 - 8 = 56. This area is exposed.Similarly, Solid A is placed on top of Solid C. Solid C's top face is 4 × 2, and Solid A's base is 4 × 1. So, the area of Solid C's top face not covered by Solid A is 4 × 2 - 4 × 1 = 8 - 4 = 4. This area is exposed.And finally, the top face of Solid A is fully exposed, which is 4 × 1 = 4.So, we need to add these exposed areas to our total.So, let's recalculate:- Solid B: 64 (sides) + 56 (uncovered top) = 120- Solid C: 192 (sides) + 4 (uncovered top) = 196- Solid A: 324 (sides and top)Total exposed surface area: 120 + 196 + 324 = 640Wait, that seems higher than before. Maybe I'm double-counting something.Alternatively, perhaps I should calculate the total surface area of all solids and then subtract the areas that are covered when stacking.Total surface area of each solid:- Solid A: 2(4×1 + 1×32 + 4×32) = 2(4 + 32 + 128) = 2(164) = 328- Solid B: 2(8×8 + 8×2 + 8×2) = 2(64 + 16 + 16) = 2(96) = 192- Solid C: 2(4×2 + 2×16 + 4×16) = 2(8 + 32 + 64) = 2(104) = 208Total surface area: 328 + 192 + 208 = 728Now, when stacking, the areas where the solids are glued together are not exposed. Specifically, the top face of Solid B is glued to the bottom face of Solid C, and the top face of Solid C is glued to the bottom face of Solid A.So, the areas to subtract are:- Top face of Solid B: 8 × 8 = 64- Top face of Solid C: 4 × 2 = 8But wait, the bottom face of Solid C is 4 × 2, which is glued to the top face of Solid B, which is 8 × 8. So, only 4 × 2 of Solid B's top face is covered, leaving 8 × 8 - 4 × 2 = 64 - 8 = 56 exposed.Similarly, the bottom face of Solid A is 4 × 1, which is glued to the top face of Solid C, which is 4 × 2. So, only 4 × 1 of Solid C's top face is covered, leaving 4 × 2 - 4 × 1 = 8 - 4 = 4 exposed.Therefore, the total area subtracted is:- The area of Solid C's bottom face: 4 × 2 = 8- The area of Solid A's bottom face: 4 × 1 = 4So, total subtracted area: 8 + 4 = 12But wait, actually, when stacking, the top face of the lower solid and the bottom face of the upper solid are both covered, so we should subtract both areas. So, for each glued interface, we subtract twice the area: once for the top face of the lower solid and once for the bottom face of the upper solid.So, for Solid B and Solid C:- Top face of Solid B: 8 × 8 = 64- Bottom face of Solid C: 4 × 2 = 8But only 4 × 2 of Solid B's top face is covered, so the area subtracted is 8 (from Solid C's bottom) and 8 (from Solid B's top). Wait, no, actually, the area subtracted is the area of the glued face, which is 4 × 2 = 8. So, we subtract 8 from Solid B's top face and 8 from Solid C's bottom face.Similarly, for Solid C and Solid A:- Top face of Solid C: 4 × 2 = 8- Bottom face of Solid A: 4 × 1 = 4Only 4 × 1 of Solid C's top face is covered, so the area subtracted is 4 (from Solid A's bottom) and 4 (from Solid C's top).Therefore, total subtracted area:- From Solid B: 8 (covered top face)- From Solid C: 8 (covered bottom face) + 4 (covered top face)- From Solid A: 4 (covered bottom face)So, total subtracted area: 8 + 8 + 4 + 4 = 24Wait, but actually, each glued interface subtracts twice the area: once from the lower solid's top face and once from the upper solid's bottom face. So, for Solid B and Solid C, the glued area is 4 × 2 = 8, so we subtract 8 from Solid B's top face and 8 from Solid C's bottom face. Similarly, for Solid C and Solid A, the glued area is 4 × 1 = 4, so we subtract 4 from Solid C's top face and 4 from Solid A's bottom face.Therefore, total subtracted area:- From Solid B: 8- From Solid C: 8 (bottom) + 4 (top) = 12- From Solid A: 4Total subtracted: 8 + 12 + 4 = 24So, total exposed surface area is total surface area minus subtracted area:728 - 24 = 704But wait, that doesn't seem right because earlier, when I calculated by adding up the sides and the exposed top faces, I got 640. There's a discrepancy here.I think the confusion arises because when calculating the total surface area and subtracting the glued areas, we also need to consider that the base of the entire tower is not exposed. The base is the bottom face of Solid B, which is 8 × 8 = 64. So, we need to subtract that as well.So, total surface area: 728Subtract:- Glued areas: 24- Base area: 64Total exposed surface area: 728 - 24 - 64 = 640Okay, that matches my earlier calculation. So, the total exposed surface area is 640.But wait, is this the minimum? Maybe there's a way to arrange the solids differently to get a lower exposed surface area.Let me consider another stacking order. What if I put Solid C at the bottom, Solid B in the middle, and Solid A on top? Let's see if that's possible.Solid C has dimensions 4 × 2 × 16. If I put it at the bottom, its height would be 16. Then, Solid B has dimensions 8 × 8 × 2. If I put it on top of Solid C, the base of Solid B must fit within the top face of Solid C. The top face of Solid C is 4 × 2, but Solid B's base is 8 × 8, which is much larger. So, it won't fit. Therefore, this order is not possible.What if I put Solid A at the bottom? Solid A has dimensions 4 × 1 × 32. Its base is 4 × 1. Then, Solid C has a base of 4 × 2, which is larger than 4 × 1, so it won't fit. Similarly, Solid B has a base of 8 × 8, which is way larger. So, Solid A can't be at the bottom.Therefore, the only possible order is Solid B at the bottom, Solid C in the middle, and Solid A on top.But let's double-check the exposed areas.Solid B (bottom):- Sides: 8 × 2 (front and back) × 2 = 32- Sides: 8 × 2 (left and right) × 2 = 32- Top: 8 × 8, but 4 × 2 is covered by Solid C, so exposed area: 8 × 8 - 4 × 2 = 64 - 8 = 56- Total exposed: 32 + 32 + 56 = 120Solid C (middle):- Sides: 4 × 16 (front and back) × 2 = 128- Sides: 2 × 16 (left and right) × 2 = 64- Top: 4 × 2, but 4 × 1 is covered by Solid A, so exposed area: 4 × 2 - 4 × 1 = 8 - 4 = 4- Total exposed: 128 + 64 + 4 = 196Solid A (top):- Sides: 4 × 32 (front and back) × 2 = 256- Sides: 1 × 32 (left and right) × 2 = 64- Top: 4 × 1 = 4- Total exposed: 256 + 64 + 4 = 324Adding them up: 120 + 196 + 324 = 640So, the total exposed surface area is indeed 640.But wait, is there a way to orient the solids differently to reduce this?For example, maybe rotating Solid C so that its larger face is on the sides, reducing the exposed area.Let me think. Solid C has dimensions 4 × 2 × 16. If I rotate it so that the 4 × 16 face is the side, then the height would still be 16, but the depth and width would be 2 and 4. Wait, but that's the same as before.Alternatively, if I rotate Solid C so that the 2 × 16 face is the side, then the depth and width would be 4 and 2, which is the same as before.So, rotating Solid C doesn't seem to help.What about Solid A? It has dimensions 4 × 1 × 32. If I rotate it so that the 4 × 32 face is the side, then the depth and width would be 1 and 4, which is the same as before. Alternatively, rotating it so that the 1 × 32 face is the side, but that would make the depth and width 4 and 1, which is the same.So, no help there either.What about Solid B? It has dimensions 8 × 8 × 2. If I rotate it so that the 8 × 2 face is the side, then the depth and width would be 8 and 8, which is the same as before. Alternatively, rotating it so that the 8 × 8 face is the side, but that would make the height 2, which is already the case.So, it seems that no matter how I rotate the solids, the exposed surface area remains the same.Wait, but maybe I can arrange the solids in a different order to minimize the exposed areas.Earlier, I thought that the only possible order is Solid B at the bottom, Solid C in the middle, and Solid A on top. But let's see if there's another way.If I put Solid C at the bottom, its base is 4 × 2, and its height is 16. Then, Solid B has a base of 8 × 8, which is larger than 4 × 2, so it won't fit. Similarly, Solid A has a base of 4 × 1, which is smaller than 4 × 2, so it could fit on top of Solid C.Wait, so if I put Solid C at the bottom (4 × 2 × 16), then Solid A on top of it (4 × 1 × 32). But Solid A's base is 4 × 1, which fits on Solid C's top face of 4 × 2. Then, on top of Solid A, I can't put Solid B because Solid B's base is 8 × 8, which is too large. So, the total height would be 16 + 32 = 48, which is less than the previous total height of 56. But we want the tallest tower, so this order is worse in terms of height, even though it might have a different exposed surface area.Similarly, if I put Solid A at the bottom (4 × 1 × 32), then Solid C on top of it (4 × 2 × 16), which fits because Solid C's base is 4 × 2, which is larger than Solid A's top face of 4 × 1. Wait, no, the base of the upper solid must fit within the top face of the lower solid. So, Solid C's base is 4 × 2, which is larger than Solid A's top face of 4 × 1. So, it won't fit. Therefore, this order is not possible.So, the only way to get the maximum height of 56 is to have Solid B at the bottom, Solid C in the middle, and Solid A on top.Therefore, the minimum exposed surface area is 640.Wait, but earlier when I calculated using the total surface area minus the glued areas and the base, I got 640, which matches the detailed calculation.But I'm a bit unsure because sometimes when stacking, the sides can overlap in a way that reduces the exposed area further. Maybe I'm missing something.Let me think about the sides again. When stacking, the sides of the upper solid can cover parts of the sides of the lower solid, potentially reducing the exposed area.For example, Solid C is placed on top of Solid B. Solid C's sides are 4 × 16 and 2 × 16. Solid B's sides are 8 × 2 and 8 × 2. So, when Solid C is placed on top, its sides are smaller than Solid B's sides, so the sides of Solid B are still fully exposed, and the sides of Solid C are also fully exposed. Therefore, there's no overlap in the sides that would reduce the exposed area.Similarly, Solid A is placed on top of Solid C. Solid A's sides are 4 × 32 and 1 × 32, which are larger than Solid C's sides of 4 × 16 and 2 × 16. So, Solid A's sides are fully exposed, and Solid C's sides are also fully exposed.Therefore, there's no overlap in the sides that would reduce the exposed area. So, the total exposed surface area remains 640.Wait, but I think I might have made a mistake in calculating the sides. Let me double-check.Solid B (bottom):- Front and back: 8 × 2 each, so 2 × (8 × 2) = 32- Left and right: 8 × 2 each, so 2 × (8 × 2) = 32- Top: 8 × 8, but 4 × 2 is covered, so exposed area: 64 - 8 = 56- Total: 32 + 32 + 56 = 120Solid C (middle):- Front and back: 4 × 16 each, so 2 × (4 × 16) = 128- Left and right: 2 × 16 each, so 2 × (2 × 16) = 64- Top: 4 × 2, but 4 × 1 is covered, so exposed area: 8 - 4 = 4- Total: 128 + 64 + 4 = 196Solid A (top):- Front and back: 4 × 32 each, so 2 × (4 × 32) = 256- Left and right: 1 × 32 each, so 2 × (1 × 32) = 64- Top: 4 × 1 = 4- Total: 256 + 64 + 4 = 324Adding them up: 120 + 196 + 324 = 640Yes, that seems correct.But wait, I just realized that when calculating the sides of Solid A, I included the top face, which is correct, but I also included the sides. However, the sides of Solid A are 4 × 32 and 1 × 32, which are indeed fully exposed because they are larger than the sides of Solid C.Therefore, the total exposed surface area is indeed 640.But I'm still a bit unsure because sometimes in these problems, there's a way to orient the solids to have more overlap and thus reduce the exposed area. Maybe I should consider different orientations for the solids.Let me try to see if I can orient Solid C differently to reduce the exposed area.Solid C has dimensions 4 × 2 × 16. If I rotate it so that the 2 × 16 face is the side, then the depth and width would be 4 and 2, which is the same as before. Alternatively, if I rotate it so that the 4 × 2 face is the side, then the depth and width would be 16 and 2, but that would make the height 4, which is shorter than 16, so the total height would be less, which we don't want.Similarly, for Solid A, rotating it to have a different face as the side doesn't help because the sides would still be 4 × 32 and 1 × 32, which are fully exposed.So, it seems that no matter how I rotate the solids, the exposed surface area remains the same.Therefore, the minimum possible exposed surface area of the tallest tower Juan can construct is 640.Wait, but earlier when I calculated using the total surface area minus the glued areas and the base, I got 640, which matches the detailed calculation. So, I think that's the correct answer.But just to be thorough, let me consider if there's any other way to stack the solids to get a lower exposed surface area.Suppose I put Solid B at the bottom, Solid A in the middle, and Solid C on top. Let's see if that's possible.Solid B has a base of 8 × 8. Solid A has a base of 4 × 1, which fits on Solid B's top face. Then, Solid C has a base of 4 × 2, which needs to fit on Solid A's top face, which is 4 × 1. But 4 × 2 is larger than 4 × 1, so it won't fit. Therefore, this order is not possible.Alternatively, Solid B at the bottom, Solid C in the middle, and Solid A on top, which is the same as before.Another idea: Maybe if I rotate Solid B so that its height is 8, but that's already the case. Its base is 8 × 8, which is fine.Wait, what if I rotate Solid B so that its height is 2, making its base 8 × 8. Then, Solid C can be placed on top of it with its base 4 × 2, which fits on 8 × 8. Then, Solid A can be placed on top of Solid C with its base 4 × 1, which fits on 4 × 2. So, the total height would be 2 (Solid B) + 16 (Solid C) + 32 (Solid A) = 50, which is less than 56. So, not the tallest.Therefore, to get the tallest tower, we need to have Solid B's height as 8, Solid C's height as 16, and Solid A's height as 32, totaling 56.Thus, the minimum exposed surface area is 640.Wait, but I just realized that when I calculated the total surface area and subtracted the glued areas and the base, I got 640, which matches the detailed calculation. So, I think that's correct.Therefore, the minimum possible exposed surface area of the tallest tower Juan can construct is 640.

💡Okay, so I have this function f(x) = x times e to the power of x, right? And I need to figure out whether x=1 is a local maximum or minimum, and similarly for x=-1. Hmm, let me think about how to approach this.First, I remember that to find local maxima or minima, I need to look at the critical points of the function. Critical points occur where the derivative is zero or undefined. Since f(x) is a product of x and e^x, which are both smooth functions, the derivative should exist everywhere, so I just need to find where the derivative is zero.Alright, let me compute the derivative of f(x). Using the product rule, which says that the derivative of u*v is u'v + uv', where u is x and v is e^x. So, the derivative f'(x) should be the derivative of x times e^x plus x times the derivative of e^x. The derivative of x is 1, and the derivative of e^x is e^x. So putting that together, f'(x) = 1 * e^x + x * e^x. That simplifies to e^x (1 + x). So, f'(x) = e^x (x + 1). Now, to find the critical points, I set this equal to zero. So, e^x (x + 1) = 0. But wait, e^x is never zero for any real x, right? Because the exponential function is always positive. So, the only way this product is zero is if x + 1 = 0, which means x = -1. So, the only critical point is at x = -1.Now, I need to determine whether this critical point is a local maximum or a local minimum. I think I can use the second derivative test for this. Let me compute the second derivative of f(x).First, I have f'(x) = e^x (x + 1). To find f''(x), I'll differentiate f'(x). Again, using the product rule: the derivative of e^x is e^x, and the derivative of (x + 1) is 1. So, f''(x) = e^x (x + 1) + e^x * 1. Simplifying that, f''(x) = e^x (x + 1 + 1) = e^x (x + 2).Now, evaluate the second derivative at the critical point x = -1. Plugging in, f''(-1) = e^{-1} (-1 + 2) = e^{-1} (1) = 1/e. Since e is approximately 2.718, 1/e is about 0.368, which is positive. I remember that if the second derivative at a critical point is positive, then the function is concave up there, which means it's a local minimum. If it's negative, it's concave down, meaning a local maximum. So, since f''(-1) is positive, x = -1 is a local minimum. That would make option D correct.Wait, but what about x = 1? The options mention x = 1 as well. Did I miss something? Let me check. The critical point was only at x = -1, right? Because f'(x) = 0 only when x = -1. So, x = 1 isn't a critical point. Therefore, it can't be a local maximum or minimum. So, options A and B are incorrect.Just to be thorough, maybe I should analyze the behavior of the function around x = -1 and x = 1. For x < -1, say x = -2, f'(x) = e^{-2} (-2 + 1) = e^{-2} (-1), which is negative. So, the function is decreasing there. For x between -1 and 1, say x = 0, f'(x) = e^{0} (0 + 1) = 1, which is positive. So, the function is increasing from x = -1 onwards. What about x > 1, say x = 2? f'(x) = e^{2} (2 + 1) = 3e^{2}, which is positive. So, the function is increasing beyond x = -1, including at x = 1. Since the function is increasing at x = 1, it's not a local maximum or minimum. It's just a point where the function is increasing.So, putting it all together, the only critical point is at x = -1, and it's a local minimum. Therefore, the correct answer is D.**Final Answer**boxed{D}