💡Okay, so I've got this problem here about Debby bringing soda to a potluck lunch. Let me try to understand what's going on. So, Debby brings 50 bottles of soda. That's the total number she has, right? Now, the problem says that everyone only drank 38 of the sodas. Hmm, okay, so out of the 50 she brought, 38 were consumed. Wait, so if 38 were drunk, that means those are gone, right? So, the question is asking how many she had to take back home. That would be the number of sodas that weren't drunk, right? So, I think I need to subtract the number of sodas that were drunk from the total number she brought. Let me write that down. So, total sodas brought = 50. Sodas drunk = 38. So, the sodas taken back home would be 50 minus 38. Let me do that calculation. 50 minus 38 is... let's see, 50 minus 30 is 20, and then minus 8 more would be 12. So, 12 sodas. Wait, does that make sense? If she brought 50 and 38 were drunk, then 12 are left. Yeah, that seems right. I don't think I need to do anything else here. It's a straightforward subtraction problem. I guess another way to think about it is, if she had 50 bottles and people drank 38, she's left with 12. Yeah, that makes sense. I don't think there's any trick or anything hidden here. It's just basic subtraction. So, to recap, she started with 50, subtract the 38 that were consumed, and she's left with 12 to take back home. Yeah, that seems correct. I don't see any other factors or anything else that could affect this. It's pretty straightforward. I think I'm confident with this answer. It's simple enough that I don't need to overcomplicate it. Just making sure I'm interpreting the problem correctly. She brought 50, 38 were drunk, so 12 are left. Yeah, that's it.

💡Okay, so I have this problem here: I need to find the monotonic decreasing interval of the function ( y = sqrt{x^2 + 2x - 3} ). The options are A: ( (-infty, -3] ), B: ( (-infty, -1] ), C: ( [1, +infty) ), and D: ( [-3, -1] ). Hmm, let me try to figure this out step by step.First, I remember that for a function to be defined, especially with a square root, the expression inside the square root must be non-negative. So, I should start by solving the inequality ( x^2 + 2x - 3 geq 0 ). Let me do that.To solve ( x^2 + 2x - 3 geq 0 ), I can factor the quadratic. Let's see, I need two numbers that multiply to -3 and add up to 2. Hmm, 3 and -1 work because 3 * (-1) = -3 and 3 + (-1) = 2. So, the quadratic factors as ( (x + 3)(x - 1) geq 0 ).Now, to find where this product is non-negative, I can use a sign chart. The critical points are at x = -3 and x = 1. So, I'll test intervals around these points:1. For ( x < -3 ), say x = -4: ( (-4 + 3)(-4 - 1) = (-1)(-5) = 5 ), which is positive.2. For ( -3 < x < 1 ), say x = 0: ( (0 + 3)(0 - 1) = 3*(-1) = -3 ), which is negative.3. For ( x > 1 ), say x = 2: ( (2 + 3)(2 - 1) = 5*1 = 5 ), which is positive.So, the expression ( x^2 + 2x - 3 ) is non-negative when ( x leq -3 ) or ( x geq 1 ). That means the domain of the function ( y = sqrt{x^2 + 2x - 3} ) is ( (-infty, -3] cup [1, +infty) ).Next, I need to find where this function is decreasing. Since the function is a square root of a quadratic, I can think of it as a composition of two functions: the quadratic ( t(x) = x^2 + 2x - 3 ) and the square root function ( y = sqrt{t} ).I know that the square root function ( y = sqrt{t} ) is an increasing function for ( t geq 0 ). That means if ( t(x) ) is increasing, then ( y ) will also be increasing, and if ( t(x) ) is decreasing, then ( y ) will be decreasing as well. So, the monotonicity of ( y ) depends on the monotonicity of ( t(x) ).Now, let's analyze the quadratic function ( t(x) = x^2 + 2x - 3 ). Since the coefficient of ( x^2 ) is positive, the parabola opens upwards. The vertex of this parabola is at ( x = -frac{b}{2a} ), which in this case is ( x = -frac{2}{2*1} = -1 ). So, the vertex is at x = -1.For a parabola opening upwards, the function decreases to the left of the vertex and increases to the right of the vertex. So, ( t(x) ) is decreasing on ( (-infty, -1] ) and increasing on ( [-1, +infty) ).But remember, the domain of our original function ( y ) is ( (-infty, -3] cup [1, +infty) ). So, we need to consider the intervals where ( t(x) ) is decreasing within the domain of ( y ).Looking at the domain ( (-infty, -3] cup [1, +infty) ), let's see how ( t(x) ) behaves in these intervals.1. On ( (-infty, -3] ): Since -3 is to the left of the vertex at x = -1, and the function ( t(x) ) is decreasing on ( (-infty, -1] ), it will also be decreasing on ( (-infty, -3] ). So, ( t(x) ) is decreasing here, which means ( y = sqrt{t(x)} ) is also decreasing here.2. On ( [1, +infty) ): Here, x is to the right of the vertex at x = -1, so ( t(x) ) is increasing on this interval. Therefore, ( y = sqrt{t(x)} ) will also be increasing on this interval.So, putting it all together, the function ( y = sqrt{x^2 + 2x - 3} ) is decreasing on ( (-infty, -3] ) and increasing on ( [1, +infty) ).Looking back at the options, A is ( (-infty, -3] ), which matches our conclusion. B is ( (-infty, -1] ), but that's not entirely correct because the domain doesn't include (-1, -3). C is ( [1, +infty) ), which is where the function is increasing, not decreasing. D is ( [-3, -1] ), but that interval isn't part of the domain of the original function.Wait, hold on. Let me double-check. The domain is ( (-infty, -3] cup [1, +infty) ). So, between -3 and 1, the function isn't defined. So, the interval [-3, -1] isn't part of the domain. Therefore, D is incorrect.So, the only interval where the function is decreasing is ( (-infty, -3] ), which is option A.But just to make sure, let me think about the derivative approach. Maybe I can take the derivative of y with respect to x and see where it's negative.So, ( y = sqrt{x^2 + 2x - 3} ). Let's compute dy/dx.Using the chain rule: dy/dx = (1/(2*sqrt(x^2 + 2x - 3))) * (2x + 2).Simplify: dy/dx = (2x + 2)/(2*sqrt(x^2 + 2x - 3)) = (x + 1)/sqrt(x^2 + 2x - 3).Now, to find where the function is decreasing, we need dy/dx < 0.So, (x + 1)/sqrt(x^2 + 2x - 3) < 0.Since sqrt(x^2 + 2x - 3) is always positive in the domain (because it's a square root), the sign of dy/dx depends on the numerator: x + 1.So, dy/dx < 0 when x + 1 < 0, which is when x < -1.But remember, the domain is ( (-infty, -3] cup [1, +infty) ). So, within the domain, where is x < -1?Looking at the domain, x can be in ( (-infty, -3] ) or ( [1, +infty) ). So, x < -1 in the domain is only ( (-infty, -3] ), because in ( [1, +infty) ), x is greater than or equal to 1, which is not less than -1.Therefore, dy/dx < 0 only on ( (-infty, -3] ), confirming that the function is decreasing there.So, both methods—the composition approach and the derivative approach—lead me to the same conclusion: the function is decreasing on ( (-infty, -3] ), which is option A.Just to make sure I didn't make a mistake, let me plug in some numbers.Take x = -4, which is in ( (-infty, -3] ). Compute y: sqrt{(-4)^2 + 2*(-4) -3} = sqrt{16 -8 -3} = sqrt{5} ≈ 2.236.Now, take x = -3: sqrt{9 + (-6) -3} = sqrt{0} = 0.So, moving from x = -4 to x = -3, y decreases from ~2.236 to 0. That's a decrease, which makes sense.Now, take x = 0, which is not in the domain, so we can't use that. Let's take x = 2, which is in [1, ∞). Compute y: sqrt{4 + 4 -3} = sqrt{5} ≈ 2.236.Take x = 3: sqrt{9 + 6 -3} = sqrt{12} ≈ 3.464.So, moving from x = 2 to x = 3, y increases from ~2.236 to ~3.464. That's an increase, so the function is indeed increasing on [1, ∞).Therefore, everything checks out. The function is decreasing on ( (-infty, -3] ), so the answer is A.**Final Answer**boxed{A}

💡To determine the highest value of 'a' such that the number 365a16 is divisible by 8, I will utilize the divisibility rule for 8. This rule states that a number is divisible by 8 if its last three digits form a number that is divisible by 8.First, I identify the last three digits of the number 365a16, which are 'a16'. My goal is to find the largest digit 'a' (ranging from 0 to 9) that makes 'a16' divisible by 8.I will test each possible value of 'a' starting from the highest (9) and move downwards:1. **For a = 9**: The last three digits become 916. Checking divisibility, 916 ÷ 8 = 114.5, which is not an integer. Therefore, 916 is not divisible by 8.2. **For a = 8**: The last three digits become 816. Checking divisibility, 816 ÷ 8 = 102, which is an integer. Therefore, 816 is divisible by 8.Since 8 is the highest digit that satisfies the condition, the highest value of 'a' is 8.

💡Okay, so I have this problem where there's a rectangle EFGH with sides of lengths 8 and 15. I need to find the maximum possible area of an equilateral triangle that can fit entirely inside this rectangle. Hmm, interesting. Let me think about how to approach this.First, I know that an equilateral triangle has all sides equal and all angles equal to 60 degrees. The area of an equilateral triangle is given by the formula (√3/4) * side². So, to maximize the area, I need to maximize the side length of the triangle while ensuring that the entire triangle stays within the rectangle.Now, the rectangle has sides 8 and 15. I guess the maximum possible side of the triangle can't exceed the longer side of the rectangle, which is 15, but I'm not sure if that's the case. Maybe the triangle can be rotated in such a way that it fits diagonally inside the rectangle, allowing for a larger side length.Wait, if I place the triangle inside the rectangle, the triangle's vertices have to lie on the sides or corners of the rectangle. So, perhaps one vertex is at a corner, and the other two are somewhere on the adjacent sides. That might allow the triangle to be larger than just the side of the rectangle.Let me visualize this. Imagine the rectangle with length 15 and width 8. If I place one vertex of the triangle at the bottom-left corner, then the other two vertices can be somewhere on the top and right sides of the rectangle. Since the triangle is equilateral, the sides from the corner vertex should make 60-degree angles with each other.Maybe I can model this using coordinates. Let's place the rectangle in a coordinate system with the bottom-left corner at (0,0), the bottom side along the x-axis from (0,0) to (15,0), the left side along the y-axis from (0,0) to (0,8), and the top-right corner at (15,8).So, if one vertex is at (0,0), the other two vertices, let's call them P and Q, must lie somewhere on the sides of the rectangle. Let's say P is on the bottom side, so its coordinates are (x, 0), and Q is on the left side, so its coordinates are (0, y). But wait, if the triangle is equilateral, the distance from (0,0) to P and from (0,0) to Q should be equal, and the distance from P to Q should also be equal to those.So, the distance from (0,0) to P is x, and the distance from (0,0) to Q is y. Therefore, x = y. Also, the distance between P and Q should be equal to x as well. The distance between (x,0) and (0,y) is √(x² + y²). Since x = y, this becomes √(2x²) = x√2. But for the triangle to be equilateral, this distance should be equal to x, so x√2 = x, which implies √2 = 1, which is not true. Hmm, that doesn't make sense. Maybe my initial assumption is wrong.Perhaps P and Q aren't on the bottom and left sides but on other sides. Maybe P is on the top side and Q is on the right side. Let me try that. So, P would be at (x,8) and Q would be at (15,y). Then, the distances from (0,0) to P and Q should be equal, and the distance from P to Q should also be equal.Distance from (0,0) to P is √(x² + 8²) = √(x² + 64). Distance from (0,0) to Q is √(15² + y²) = √(225 + y²). Setting these equal: √(x² + 64) = √(225 + y²). Squaring both sides: x² + 64 = 225 + y². So, x² - y² = 161.Now, the distance between P and Q should be equal to these distances as well. The distance between (x,8) and (15,y) is √((15 - x)² + (y - 8)²). This should equal √(x² + 64). So, √((15 - x)² + (y - 8)²) = √(x² + 64). Squaring both sides: (15 - x)² + (y - 8)² = x² + 64.Expanding (15 - x)²: 225 - 30x + x². Expanding (y - 8)²: y² - 16y + 64. So, putting it all together: 225 - 30x + x² + y² - 16y + 64 = x² + 64. Simplify: 225 - 30x + y² - 16y + 64 = 64. So, 225 - 30x + y² - 16y = 0.But from earlier, we have x² - y² = 161, so y² = x² - 161. Substitute this into the equation: 225 - 30x + (x² - 161) - 16y = 0. Simplify: 225 - 30x + x² - 161 - 16y = 0. So, x² - 30x + 64 - 16y = 0.Hmm, this is getting complicated. Maybe there's a better way to approach this. Perhaps using trigonometry since it's an equilateral triangle, which has 60-degree angles.If I place one vertex at (0,0), then the other two vertices can be found by moving along the sides of the rectangle at 60-degree angles. So, from (0,0), moving along the x-axis for some distance, and then moving at a 60-degree angle to reach the other vertex.Wait, maybe I can model this using vectors. Let me denote the two sides from (0,0) as vectors. Let vector A be along the x-axis with length 'a', and vector B be at a 60-degree angle from vector A with length 'b'. Since it's an equilateral triangle, a = b, and the angle between them is 60 degrees.So, vector A is (a, 0). Vector B can be represented as (b*cos(60°), b*sin(60°)) = (b*(1/2), b*(√3/2)). Since a = b, let's just call it 's'. So, vector A is (s, 0), and vector B is (s/2, (s√3)/2).The third vertex of the triangle would be at the sum of vectors A and B: (s + s/2, 0 + (s√3)/2) = (3s/2, (s√3)/2). But this point has to lie within the rectangle, so its x-coordinate must be ≤ 15 and y-coordinate must be ≤ 8.So, 3s/2 ≤ 15 and (s√3)/2 ≤ 8.From the first inequality: 3s/2 ≤ 15 ⇒ s ≤ 10.From the second inequality: (s√3)/2 ≤ 8 ⇒ s ≤ (16)/√3 ≈ 9.24.So, s must be ≤ 9.24 to satisfy both conditions. Therefore, the maximum possible side length is approximately 9.24, which is 16/√3.Wait, 16/√3 is approximately 9.24, right? So, s = 16/√3.Then, the area of the triangle would be (√3/4) * s² = (√3/4) * (256/3) = (256√3)/12 = (64√3)/3 ≈ 36.94.But wait, is this the maximum possible? Because maybe if we rotate the triangle differently, we can get a larger area.Alternatively, perhaps placing the triangle such that one side is along the diagonal of the rectangle. The diagonal of the rectangle is √(15² + 8²) = √(225 + 64) = √289 = 17. So, the diagonal is 17 units long. If we can fit an equilateral triangle with side length 17, but that's impossible because the rectangle itself is only 15x8, so the triangle can't have a side longer than the diagonal.Wait, but an equilateral triangle with side length 17 would have a height of (√3/2)*17 ≈ 14.73, which is larger than the rectangle's height of 8. So, that won't fit.Alternatively, maybe the maximum side is when the triangle is inscribed such that each vertex touches a different side of the rectangle. That might allow for a larger triangle.Let me think about that. If each vertex is on a different side, then the triangle is kind of stretched across the rectangle. Maybe I can set up equations for that.Let me denote the rectangle with vertices at (0,0), (15,0), (15,8), and (0,8). Let the triangle have vertices at (a,0), (15,b), and (c,8), where a, b, c are between 0 and 15, 0 and 8, etc.Since it's an equilateral triangle, all sides must be equal. So, the distance between (a,0) and (15,b) must equal the distance between (15,b) and (c,8), and also equal the distance between (c,8) and (a,0).This seems complex, but maybe I can set up equations.Distance between (a,0) and (15,b): √[(15 - a)² + (b - 0)²] = √[(15 - a)² + b²]Distance between (15,b) and (c,8): √[(c - 15)² + (8 - b)²]Distance between (c,8) and (a,0): √[(a - c)² + (0 - 8)²] = √[(a - c)² + 64]So, setting these equal:√[(15 - a)² + b²] = √[(c - 15)² + (8 - b)²] = √[(a - c)² + 64]This is a system of equations with variables a, b, c. It might be difficult to solve, but maybe I can find a relationship.Alternatively, perhaps using rotation. If I place one vertex at (0,0), another at (s,0), and the third somewhere in the plane, rotated 60 degrees. But that third vertex has to lie within the rectangle.Wait, earlier I tried placing one vertex at (0,0), another at (s,0), and the third at (s/2, (s√3)/2). For this point to be within the rectangle, (s√3)/2 ≤ 8 ⇒ s ≤ 16/√3 ≈ 9.24, as before.But maybe if I rotate the triangle differently, such that the third vertex is on a different side, I can get a larger s.Alternatively, perhaps placing the triangle so that one vertex is at (0,0), another at (15, y), and the third at (x,8). Then, ensuring all sides are equal.Let me try that. So, vertices at (0,0), (15,y), and (x,8). Then, the distances between these points must be equal.Distance from (0,0) to (15,y): √(225 + y²)Distance from (15,y) to (x,8): √[(x - 15)² + (8 - y)²]Distance from (x,8) to (0,0): √(x² + 64)So, setting these equal:√(225 + y²) = √[(x - 15)² + (8 - y)²] = √(x² + 64)First, set √(225 + y²) = √(x² + 64). Squaring both sides: 225 + y² = x² + 64 ⇒ x² - y² = 161.Next, set √(225 + y²) = √[(x - 15)² + (8 - y)²]. Squaring both sides: 225 + y² = (x - 15)² + (8 - y)².Expanding the right side: (x² - 30x + 225) + (64 - 16y + y²) = x² - 30x + 225 + 64 - 16y + y² = x² - 30x + y² - 16y + 289.So, 225 + y² = x² - 30x + y² - 16y + 289.Simplify: 225 = x² - 30x - 16y + 289 ⇒ x² - 30x - 16y = -64.But from earlier, x² - y² = 161 ⇒ y² = x² - 161.So, substitute y² into the equation: x² - 30x - 16y = -64.But we still have y in there. Maybe express y in terms of x from y² = x² - 161. So, y = √(x² - 161). But that might complicate things.Alternatively, maybe express y from the second equation. From x² - 30x - 16y = -64, we can write y = (x² - 30x + 64)/16.Then, substitute this into y² = x² - 161:[(x² - 30x + 64)/16]^2 = x² - 161.This will lead to a quartic equation, which might be messy, but let's try.Let me denote z = x² - 30x + 64. Then, (z/16)^2 = x² - 161 ⇒ z² = 256(x² - 161).But z = x² - 30x + 64, so:(x² - 30x + 64)^2 = 256x² - 256*161.Expanding the left side:(x² - 30x + 64)^2 = x^4 - 60x^3 + (900 + 128)x² - (2*30*64)x + 64²Wait, let me compute it step by step.(x² - 30x + 64)(x² - 30x + 64) =x²*x² + x²*(-30x) + x²*64 + (-30x)*x² + (-30x)*(-30x) + (-30x)*64 + 64*x² + 64*(-30x) + 64*64Simplify term by term:x^4 - 30x^3 + 64x² - 30x^3 + 900x² - 1920x + 64x² - 1920x + 4096Combine like terms:x^4 + (-30x^3 - 30x^3) + (64x² + 900x² + 64x²) + (-1920x - 1920x) + 4096So:x^4 - 60x^3 + (64 + 900 + 64)x² - 3840x + 4096Calculate coefficients:64 + 900 + 64 = 1028So:x^4 - 60x^3 + 1028x² - 3840x + 4096Set equal to 256x² - 256*161:x^4 - 60x^3 + 1028x² - 3840x + 4096 = 256x² - 41216Bring all terms to left side:x^4 - 60x^3 + 1028x² - 3840x + 4096 - 256x² + 41216 = 0Simplify:x^4 - 60x^3 + (1028 - 256)x² - 3840x + (4096 + 41216) = 0Calculate:1028 - 256 = 7724096 + 41216 = 45312So:x^4 - 60x^3 + 772x² - 3840x + 45312 = 0This is a quartic equation, which is quite complex. Maybe I can factor it or find rational roots.Using Rational Root Theorem, possible roots are factors of 45312 divided by factors of 1 (leading coefficient). So possible integer roots are ±1, ±2, ..., up to ±45312. That's a lot, but maybe trying small integers.Let me test x=12:12^4 - 60*12^3 + 772*12² - 3840*12 + 45312= 20736 - 60*1728 + 772*144 - 46080 + 45312Calculate each term:2073660*1728 = 103680772*144 = 111, let's compute 700*144=100,800 and 72*144=10,368, so total 111,168-46080+45312So:20736 - 103680 + 111168 - 46080 + 45312Compute step by step:20736 - 103680 = -82944-82944 + 111168 = 28,22428,224 - 46080 = -17,856-17,856 + 45312 = 27,456 ≠ 0Not zero.Try x=16:16^4 - 60*16^3 + 772*16² - 3840*16 + 45312= 65536 - 60*4096 + 772*256 - 61440 + 45312Calculate each term:6553660*4096=245,760772*256=197, let's compute 700*256=179,200 and 72*256=18,432, so total 197,632-61440+45312So:65536 - 245,760 + 197,632 - 61,440 + 45,312Step by step:65536 - 245,760 = -180,224-180,224 + 197,632 = 17,40817,408 - 61,440 = -44,032-44,032 + 45,312 = 1,280 ≠ 0Not zero.Try x=24:24^4 - 60*24^3 + 772*24² - 3840*24 + 45312= 331,776 - 60*13,824 + 772*576 - 92,160 + 45,312Calculate each term:331,77660*13,824=829,440772*576=444, let's compute 700*576=403,200 and 72*576=41,472, so total 444,672-92,160+45,312So:331,776 - 829,440 + 444,672 - 92,160 + 45,312Step by step:331,776 - 829,440 = -497,664-497,664 + 444,672 = -52,992-52,992 - 92,160 = -145,152-145,152 + 45,312 = -99,840 ≠ 0Not zero.This is getting tedious. Maybe there's a better approach.Alternatively, maybe using calculus to maximize the area. Let me consider the side length s as a function of x, and then find its maximum.Earlier, I considered placing one vertex at (0,0), another at (s,0), and the third at (s/2, (s√3)/2). For this point to lie within the rectangle, (s√3)/2 ≤ 8 ⇒ s ≤ 16/√3 ≈ 9.24.But maybe if I rotate the triangle so that the third vertex is on the top side, I can get a larger s.Wait, if I rotate the triangle such that the third vertex is on the top side, then the y-coordinate is 8. So, let's model that.Let me denote the triangle with vertices at (0,0), (a,0), and (b,8). The triangle is equilateral, so all sides are equal.Distance from (0,0) to (a,0) is a.Distance from (a,0) to (b,8) is √[(b - a)² + 64].Distance from (b,8) to (0,0) is √(b² + 64).Since it's equilateral, a = √[(b - a)² + 64] = √(b² + 64).So, set a = √(b² + 64). Squaring both sides: a² = b² + 64 ⇒ b² = a² - 64.Also, a = √[(b - a)² + 64]. Squaring both sides: a² = (b - a)² + 64.Expanding (b - a)²: b² - 2ab + a². So:a² = b² - 2ab + a² + 64 ⇒ 0 = b² - 2ab + 64.But from earlier, b² = a² - 64. Substitute:0 = (a² - 64) - 2ab + 64 ⇒ 0 = a² - 2ab.So, a² = 2ab ⇒ a = 2b (since a ≠ 0).But from b² = a² - 64, substitute a = 2b:b² = (4b²) - 64 ⇒ 3b² = 64 ⇒ b² = 64/3 ⇒ b = 8/√3 ≈ 4.618.Then, a = 2b = 16/√3 ≈ 9.237.So, the side length s = a ≈ 9.237, which is the same as before.So, the maximum side length is 16/√3, and the area is (√3/4)*(16/√3)^2 = (√3/4)*(256/3) = (256√3)/12 = (64√3)/3 ≈ 36.94.But wait, earlier I thought the area was 64√3, but that was a mistake. The correct calculation gives (64√3)/3.Wait, let me double-check:(√3/4) * (16/√3)^2 = (√3/4) * (256/3) = (256√3)/12 = (64√3)/3.Yes, that's correct.But is this the maximum possible? Because earlier I considered the triangle with vertices at (0,0), (16/√3, 0), and (8/√3, 8). Let me check if this triangle fits within the rectangle.The x-coordinate of the third vertex is 8/√3 ≈ 4.618, which is less than 15, so that's fine. The y-coordinate is 8, which is exactly the top of the rectangle. So, yes, this triangle fits.But is there a way to get a larger triangle by placing the vertices differently?Maybe if I place the triangle such that one vertex is at a corner, and the other two are on adjacent sides, but not necessarily at the midpoints.Wait, let's consider placing one vertex at (0,0), another at (x,0), and the third at (0,y). Then, the triangle is right-angled, not equilateral. So, that's not helpful.Alternatively, placing one vertex at (0,0), another at (x,8), and the third at (15,y). Then, ensuring all sides are equal.This seems similar to what I tried earlier, leading to the quartic equation, which was complicated.Alternatively, maybe using the fact that the maximum area occurs when the triangle is inscribed such that each vertex touches a different side of the rectangle. But I'm not sure how to model that.Wait, perhaps using the concept of affine transformations. The maximum area of an equilateral triangle inscribed in a rectangle can be found by considering the rectangle's aspect ratio.But I'm not sure about that. Maybe looking for similar problems or known results.Wait, I recall that the maximum area of an equilateral triangle that can fit inside a rectangle of sides a and b is min(a, b√3) * min(b, a√3) / 2. But I'm not sure if that's correct.Alternatively, perhaps the maximum area is achieved when the triangle is rotated such that its base is along the longer side of the rectangle, and its height is limited by the shorter side.Given the rectangle is 15x8, the longer side is 15. The height of the equilateral triangle is (√3/2)*s. To fit within the rectangle, (√3/2)*s ≤ 8 ⇒ s ≤ 16/√3 ≈ 9.24, as before.So, the maximum side length is 16/√3, and the area is (√3/4)*(16/√3)^2 = (64√3)/3.Therefore, the maximum area is (64√3)/3.Wait, but earlier I thought the area was 64√3, but that was a mistake. The correct area is (64√3)/3.But let me confirm this with another approach.Suppose I place the triangle with one vertex at (0,0), another at (16/√3, 0), and the third at (8/√3, 8). Then, the distances between these points should be equal.Distance from (0,0) to (16/√3, 0) is 16/√3.Distance from (16/√3, 0) to (8/√3, 8): √[(8/√3 - 16/√3)^2 + (8 - 0)^2] = √[(-8/√3)^2 + 64] = √[(64/3) + 64] = √[(64 + 192)/3] = √[256/3] = 16/√3.Distance from (8/√3, 8) to (0,0): √[(8/√3)^2 + 8^2] = √[(64/3) + 64] = √[256/3] = 16/√3.So, all sides are equal, confirming it's an equilateral triangle.Therefore, the area is (√3/4)*(16/√3)^2 = (√3/4)*(256/3) = (64√3)/3.So, the maximum area is (64√3)/3.But wait, earlier I thought the area was 64√3, but that was incorrect. The correct area is (64√3)/3.However, I recall that sometimes the maximum area is achieved when the triangle is inscribed such that each vertex touches a different side, but in this case, the triangle I found only touches three sides: two at the bottom and one at the top. Maybe there's a way to have a larger triangle by touching all four sides, but since it's a triangle, it can only touch three sides.Alternatively, maybe placing the triangle such that one vertex is at a corner, and the other two are on the opposite sides, allowing for a larger side length.Wait, let me consider placing one vertex at (0,0), another at (15,y), and the third at (x,8). Then, ensuring all sides are equal.This is similar to what I tried earlier, leading to the quartic equation. Maybe instead of solving the quartic, I can use calculus to maximize the area.Let me denote the side length as s. Then, the area is (√3/4)s². To maximize the area, I need to maximize s.From the earlier setup, we have:From (0,0) to (15,y): s = √(225 + y²)From (15,y) to (x,8): s = √[(x - 15)² + (8 - y)²]From (x,8) to (0,0): s = √(x² + 64)So, we have:√(225 + y²) = √[(x - 15)² + (8 - y)²] = √(x² + 64)Let me square the first and third equations:225 + y² = x² + 64 ⇒ x² - y² = 161.Now, square the first and second equations:225 + y² = (x - 15)² + (8 - y)².Expanding the right side:(x² - 30x + 225) + (64 - 16y + y²) = x² - 30x + y² - 16y + 289.So, 225 + y² = x² - 30x + y² - 16y + 289 ⇒ 225 = x² - 30x - 16y + 289 ⇒ x² - 30x - 16y = -64.From x² - y² = 161, we can write y² = x² - 161 ⇒ y = √(x² - 161).Substitute y into x² - 30x - 16y = -64:x² - 30x - 16√(x² - 161) = -64.This is a complicated equation, but maybe I can solve for x numerically.Let me denote f(x) = x² - 30x - 16√(x² - 161) + 64.We need to find x such that f(x) = 0.Let me try x=16:f(16) = 256 - 480 - 16√(256 - 161) + 64 = 256 - 480 - 16√95 + 64.Calculate:256 - 480 = -224-224 + 64 = -160√95 ≈ 9.74716*9.747 ≈ 155.95So, f(16) ≈ -160 - 155.95 ≈ -315.95 ≠ 0.Try x=17:f(17) = 289 - 510 - 16√(289 - 161) + 64 = 289 - 510 - 16√128 + 64.Calculate:289 - 510 = -221-221 + 64 = -157√128 ≈ 11.31416*11.314 ≈ 181.02So, f(17) ≈ -157 - 181.02 ≈ -338.02 ≠ 0.Wait, this is going further negative. Maybe try a smaller x.Try x=14:f(14) = 196 - 420 - 16√(196 - 161) + 64 = 196 - 420 - 16√35 + 64.Calculate:196 - 420 = -224-224 + 64 = -160√35 ≈ 5.91616*5.916 ≈ 94.656So, f(14) ≈ -160 - 94.656 ≈ -254.656 ≠ 0.Hmm, still negative. Maybe try x=13:f(13) = 169 - 390 - 16√(169 - 161) + 64 = 169 - 390 - 16√8 + 64.Calculate:169 - 390 = -221-221 + 64 = -157√8 ≈ 2.82816*2.828 ≈ 45.25So, f(13) ≈ -157 - 45.25 ≈ -202.25 ≠ 0.Still negative. Maybe x=12:f(12) = 144 - 360 - 16√(144 - 161) + 64.Wait, √(144 - 161) is √(-17), which is imaginary. So, x must be ≥ √161 ≈ 12.69.So, x must be greater than approximately 12.69.Let me try x=13:As before, f(13) ≈ -202.25.x=14: f(14) ≈ -254.656.Wait, but as x increases, f(x) becomes more negative. So, maybe there's no solution in this setup, meaning that the maximum area is indeed achieved when the triangle is placed with one vertex at (0,0), another at (16/√3,0), and the third at (8/√3,8), giving an area of (64√3)/3.Therefore, the maximum area is (64√3)/3.But wait, let me check if this is indeed the maximum. Suppose I rotate the triangle such that it's not aligned with the sides, but somehow fits better.Alternatively, maybe the maximum area is achieved when the triangle is inscribed such that each vertex touches a different side of the rectangle. Let me try that.Let the rectangle have vertices at (0,0), (15,0), (15,8), (0,8). Let the triangle have vertices at (a,0), (15,b), and (c,8). All sides must be equal.So, distance from (a,0) to (15,b): √[(15 - a)² + b²]Distance from (15,b) to (c,8): √[(c - 15)² + (8 - b)²]Distance from (c,8) to (a,0): √[(a - c)² + 64]Set these equal:√[(15 - a)² + b²] = √[(c - 15)² + (8 - b)²] = √[(a - c)² + 64]This is a system of equations. Let me square the first two:(15 - a)² + b² = (c - 15)² + (8 - b)²Expanding:225 - 30a + a² + b² = c² - 30c + 225 + 64 - 16b + b²Simplify:225 - 30a + a² = c² - 30c + 289 - 16bRearrange:a² - 30a - c² + 30c + 16b = 64.Now, set the first and third distances equal:√[(15 - a)² + b²] = √[(a - c)² + 64]Squaring both sides:(15 - a)² + b² = (a - c)² + 64Expanding:225 - 30a + a² + b² = a² - 2ac + c² + 64Simplify:225 - 30a + b² = -2ac + c² + 64Rearrange:-30a + b² + 2ac - c² = -161.Now, we have two equations:1) a² - 30a - c² + 30c + 16b = 642) -30a + b² + 2ac - c² = -161This system is still complex, but maybe we can find a relationship.From equation 1: a² - 30a - c² + 30c + 16b = 64From equation 2: -30a + b² + 2ac - c² = -161Let me subtract equation 2 from equation 1:(a² - 30a - c² + 30c + 16b) - (-30a + b² + 2ac - c²) = 64 - (-161)Simplify:a² - 30a - c² + 30c + 16b + 30a - b² - 2ac + c² = 225Simplify terms:a² + 30c + 16b - b² - 2ac = 225This is still complicated. Maybe express b from one equation and substitute.From equation 1: 16b = 64 - a² + 30a + c² - 30cSo, b = (64 - a² + 30a + c² - 30c)/16Substitute this into equation 2:-30a + [(64 - a² + 30a + c² - 30c)/16]^2 + 2ac - c² = -161This is getting too messy. Maybe this approach isn't the best.Given the time I've spent and the complexity of the equations, I think the initial approach where the triangle is placed with one vertex at (0,0), another at (16/√3,0), and the third at (8/√3,8) giving an area of (64√3)/3 is the correct maximum area.Therefore, the maximum possible area of such a triangle is (64√3)/3.But wait, I just realized that 16/√3 is approximately 9.24, which is less than 15, so it fits on the bottom side. The third vertex is at (8/√3,8), which is approximately (4.618,8), which is within the rectangle.Therefore, this configuration is valid, and the area is indeed (64√3)/3.However, I recall that sometimes the maximum area can be larger if the triangle is rotated differently. Let me check if there's a way to have a larger triangle by rotating it such that the base is not along the bottom side but somewhere else.Suppose I rotate the triangle so that one vertex is at (0,0), another at (x,y), and the third at (u,v), all within the rectangle. The challenge is to maximize the side length s while keeping all points within the rectangle.This seems too vague. Maybe using parametric equations or optimization techniques.Alternatively, perhaps the maximum area is indeed (64√3)/3, as found earlier, and that's the answer.Therefore, after considering various configurations and verifying the calculations, I conclude that the maximum area is (64√3)/3.

💡First, I need to simplify each square root separately. The first square root is √(5·3), which simplifies to √15. The second square root is √(3³·5³). I can rewrite 3³ as 3²·3 and 5³ as 5²·5, so the second square root becomes √(3²·3·5²·5).Next, I can take the square roots of the squared terms. The square root of 3² is 3, and the square root of 5² is 5. This gives me 3·5·√(3·5), which simplifies to 15·√15.Now, I multiply the simplified forms of both square roots: √15·15·√15. Multiplying √15 by √15 gives 15, so the expression becomes 15·15, which equals 225.Therefore, the simplified expression is 225.

💡Okay, so I have this quadratic equation: (x^2 - 4x + 2 = 0). The roots are (x_1) and (x_2). I need to find the value of the expression (x_1^2 - 4x_1 + 2x_1x_2). Hmm, let me think about how to approach this.First, I remember that for a quadratic equation (ax^2 + bx + c = 0), the sum of the roots (x_1 + x_2) is equal to (-b/a), and the product of the roots (x_1x_2) is equal to (c/a). In this case, the equation is (x^2 - 4x + 2 = 0), so (a = 1), (b = -4), and (c = 2). That means the sum of the roots (x_1 + x_2 = -(-4)/1 = 4), and the product (x_1x_2 = 2/1 = 2). Okay, so I have that (x_1 + x_2 = 4) and (x_1x_2 = 2).Now, looking at the expression I need to evaluate: (x_1^2 - 4x_1 + 2x_1x_2). I notice that (x_1^2 - 4x_1) looks similar to the original quadratic equation. In fact, since (x_1) is a root of the equation (x^2 - 4x + 2 = 0), plugging (x_1) into the equation gives (x_1^2 - 4x_1 + 2 = 0). So, if I rearrange that, I get (x_1^2 - 4x_1 = -2). That's useful because it simplifies part of the expression I need to find.So, substituting (x_1^2 - 4x_1) with (-2) in the original expression, I have:[x_1^2 - 4x_1 + 2x_1x_2 = (-2) + 2x_1x_2]Now, I already know that (x_1x_2 = 2), so substituting that in:[-2 + 2 times 2 = -2 + 4 = 2]So, the value of the expression (x_1^2 - 4x_1 + 2x_1x_2) is 2.Wait, let me double-check that. I know (x_1^2 - 4x_1 = -2) because (x_1) satisfies the equation. Then, adding (2x_1x_2) which is (2 times 2 = 4). So, (-2 + 4 = 2). Yeah, that seems right.Alternatively, I could try plugging in the actual roots to verify. Let me find the roots using the quadratic formula. The roots are:[x = frac{4 pm sqrt{(-4)^2 - 4 times 1 times 2}}{2 times 1} = frac{4 pm sqrt{16 - 8}}{2} = frac{4 pm sqrt{8}}{2} = frac{4 pm 2sqrt{2}}{2} = 2 pm sqrt{2}]So, the roots are (2 + sqrt{2}) and (2 - sqrt{2}). Let's take (x_1 = 2 + sqrt{2}) and (x_2 = 2 - sqrt{2}).Now, compute (x_1^2 - 4x_1 + 2x_1x_2):First, (x_1^2 = (2 + sqrt{2})^2 = 4 + 4sqrt{2} + 2 = 6 + 4sqrt{2}).Then, (-4x_1 = -4(2 + sqrt{2}) = -8 - 4sqrt{2}).Adding these together: (6 + 4sqrt{2} - 8 - 4sqrt{2} = -2).Now, (2x_1x_2 = 2 times (2 + sqrt{2})(2 - sqrt{2}) = 2 times (4 - 2) = 2 times 2 = 4).Adding that to the previous result: (-2 + 4 = 2). Yep, same answer. So, that confirms it.I think I'm confident now that the value is indeed 2.