💡Okay, so I have this problem where there's a rectangle ABCD with sides of lengths 12 and 14. I need to find the maximum possible area of an equilateral triangle that can fit entirely inside this rectangle. Hmm, interesting. I remember that an equilateral triangle has all sides equal and all angles equal to 60 degrees. But how do I fit the largest possible one inside a rectangle?First, I should visualize the rectangle. Let me assume that the sides of length 12 and 14 are the width and height, respectively. So, if I imagine the rectangle on a coordinate plane, one corner is at the origin (0,0), and the opposite corner is at (12,14). The other two corners are at (12,0) and (0,14). Now, I need to place an equilateral triangle inside this rectangle without any part of the triangle going outside. The goal is to maximize the area of this triangle. I know that the area of an equilateral triangle is given by the formula:[text{Area} = frac{sqrt{3}}{4} times text{side}^2]So, to maximize the area, I need to maximize the side length of the triangle. But it has to fit entirely within the rectangle. I wonder if the largest possible equilateral triangle would have its base along one side of the rectangle. Let's say I place one side of the triangle along the bottom side of the rectangle, which is 12 units long. Then, the height of the triangle would be:[text{Height} = frac{sqrt{3}}{2} times text{side}]If the side is 12, the height would be:[frac{sqrt{3}}{2} times 12 = 6sqrt{3} approx 10.39]Since the height of the rectangle is 14, which is more than 10.39, this triangle would fit vertically. So, the area would be:[frac{sqrt{3}}{4} times 12^2 = 36sqrt{3}]But wait, maybe I can fit a larger triangle if I rotate it. Sometimes, rotating a shape can allow it to fit into a space more efficiently. Let me think about how to place the triangle diagonally inside the rectangle.If I rotate the triangle so that one vertex is at a corner of the rectangle and the other two vertices touch the opposite sides, perhaps I can get a larger triangle. Let me try to model this.Let me denote the rectangle with vertices at (0,0), (12,0), (12,14), and (0,14). Suppose one vertex of the triangle is at (0,0). Let the other two vertices be at (a,0) and (b,c). Since it's an equilateral triangle, the distance from (0,0) to (a,0) should be equal to the distance from (0,0) to (b,c) and also equal to the distance from (a,0) to (b,c).So, the distance from (0,0) to (a,0) is |a|. The distance from (0,0) to (b,c) is √(b² + c²). The distance from (a,0) to (b,c) is √((b - a)² + c²). Since all sides are equal:1. |a| = √(b² + c²)2. |a| = √((b - a)² + c²)Let me square both sides to eliminate the square roots.From equation 1:a² = b² + c²From equation 2:a² = (b - a)² + c²Expanding equation 2:a² = b² - 2ab + a² + c²Simplify:a² = b² - 2ab + a² + c²Subtract a² from both sides:0 = b² - 2ab + c²But from equation 1, we know that b² + c² = a². So, substitute that into the above equation:0 = (a² - c²) - 2ab + c²Simplify:0 = a² - 2abSo, a² = 2abDivide both sides by a (assuming a ≠ 0):a = 2bSo, b = a/2Now, substitute b = a/2 into equation 1:a² = (a/2)² + c²a² = a²/4 + c²Multiply both sides by 4:4a² = a² + 4c²Subtract a²:3a² = 4c²So, c² = (3/4)a²Thus, c = (√3 / 2)aSo, the coordinates of the third vertex are (a/2, (√3 / 2)a)Now, since the triangle must fit entirely within the rectangle, the coordinates (a/2, (√3 / 2)a) must satisfy:a/2 ≤ 12 (since the x-coordinate can't exceed the width of the rectangle)and(√3 / 2)a ≤ 14 (since the y-coordinate can't exceed the height of the rectangle)So, let's solve these inequalities:1. a/2 ≤ 12 => a ≤ 242. (√3 / 2)a ≤ 14 => a ≤ (14 * 2)/√3 ≈ 28 / 1.732 ≈ 16.16So, the maximum possible a is 16.16, but since a must satisfy both conditions, the stricter condition is a ≤ 16.16. However, a cannot exceed the width of the rectangle when placed along the x-axis. Wait, but in this case, the triangle is placed with one vertex at (0,0) and another at (a,0). So, the side along the x-axis is length a, which must be ≤ 12 because the rectangle's width is 12. So, actually, a can't be more than 12.Wait, that contradicts the earlier result. Maybe I made a mistake.Hold on, if the triangle is placed with one vertex at (0,0) and another at (a,0), then a can't exceed 12 because the rectangle's width is 12. So, a ≤ 12. Then, the y-coordinate of the third vertex is (√3 / 2)a. So, to ensure that this y-coordinate is ≤ 14:(√3 / 2)a ≤ 14a ≤ (14 * 2)/√3 ≈ 28 / 1.732 ≈ 16.16But since a is already limited to 12, this condition is automatically satisfied because 12 < 16.16. So, the maximum a is 12.Therefore, the maximum side length of the triangle placed with one side along the bottom of the rectangle is 12, giving an area of 36√3.But wait, earlier I thought about rotating the triangle. Maybe if I don't place one side along the edge, I can get a larger triangle. Let me explore that.Suppose I place the triangle such that all three vertices touch the sides of the rectangle, but none are necessarily at the corners. This might allow for a larger triangle.Let me try to model this. Let me denote the rectangle with coordinates (0,0), (12,0), (12,14), (0,14). Let the three vertices of the triangle be at (x1, y1), (x2, y2), and (x3, y3). Each of these points must lie on the sides of the rectangle, meaning their x-coordinates are either 0, 12 or their y-coordinates are 0,14.But this seems complicated. Maybe there's a better way.I remember that the maximum area of a triangle inside a rectangle can sometimes be achieved by placing the triangle such that each vertex touches a different side. Maybe that's the case here.Alternatively, perhaps the largest equilateral triangle that can fit inside the rectangle is when it's inscribed such that each vertex touches a different side. Let me try to find the side length in that case.Let me assume that one vertex is on the bottom side (y=0), another on the right side (x=12), and the third on the top side (y=14). Let me denote the vertices as (a,0), (12,b), and (c,14). Since it's an equilateral triangle, all sides must be equal.So, the distance between (a,0) and (12,b) must equal the distance between (12,b) and (c,14), which must equal the distance between (c,14) and (a,0).This gives us three equations:1. √[(12 - a)² + (b - 0)²] = √[(c - 12)² + (14 - b)²]2. √[(c - 12)² + (14 - b)²] = √[(a - c)² + (0 - 14)²]3. √[(a - c)² + (0 - 14)²] = √[(12 - a)² + (b - 0)²]These equations look quite complex. Maybe I can square them to simplify.From equation 1:(12 - a)² + b² = (c - 12)² + (14 - b)²Expanding both sides:(144 - 24a + a²) + b² = (c² - 24c + 144) + (196 - 28b + b²)Simplify:144 - 24a + a² + b² = c² - 24c + 144 + 196 - 28b + b²Cancel b² from both sides:144 - 24a + a² = c² - 24c + 144 + 196 - 28bSimplify the right side:144 + 196 = 340, so:144 - 24a + a² = c² - 24c + 340 - 28bBring all terms to the left:144 - 24a + a² - c² + 24c - 340 + 28b = 0Simplify:(a² - c²) + (-24a + 24c) + (144 - 340) + 28b = 0Factor:(a - c)(a + c) -24(a - c) -196 + 28b = 0Factor out (a - c):(a - c)(a + c - 24) -196 + 28b = 0Hmm, this is getting complicated. Maybe I should try a different approach.Alternatively, perhaps using coordinate geometry and slopes. Since it's an equilateral triangle, the angles between the sides are 60 degrees. Maybe I can use the slope of the sides to find the coordinates.Let me suppose that one vertex is at (0,0), another at (x,14), and the third at (12,y). Then, the distances between these points should be equal.So, distance from (0,0) to (x,14):√(x² + 14²) = √(x² + 196)Distance from (x,14) to (12,y):√((12 - x)² + (y - 14)²)Distance from (12,y) to (0,0):√(12² + y²) = √(144 + y²)Set the first equal to the second:√(x² + 196) = √((12 - x)² + (y - 14)²)Square both sides:x² + 196 = (12 - x)² + (y - 14)²Expand (12 - x)²:144 - 24x + x²So:x² + 196 = 144 - 24x + x² + (y - 14)²Cancel x²:196 = 144 - 24x + (y - 14)²Simplify:196 - 144 = -24x + (y - 14)²52 = -24x + (y - 14)²Similarly, set the first distance equal to the third:√(x² + 196) = √(144 + y²)Square both sides:x² + 196 = 144 + y²So:x² - y² = -52Now, we have two equations:1. 52 = -24x + (y - 14)²2. x² - y² = -52Let me try to solve these equations.From equation 2:x² = y² - 52From equation 1:52 = -24x + (y - 14)²Let me expand (y - 14)²:y² - 28y + 196So, equation 1 becomes:52 = -24x + y² - 28y + 196Rearrange:-24x + y² - 28y + 196 - 52 = 0Simplify:-24x + y² - 28y + 144 = 0But from equation 2, x² = y² - 52, so x = √(y² - 52). Hmm, but this might complicate things.Alternatively, let me express x from equation 2:x² = y² - 52 => x = √(y² - 52)But plugging this into equation 1 might not be straightforward.Alternatively, let me express y² from equation 2:y² = x² + 52Plug this into equation 1:-24x + (x² + 52) - 28y + 144 = 0Simplify:-24x + x² + 52 - 28y + 144 = 0Combine constants:52 + 144 = 196So:x² -24x -28y + 196 = 0Hmm, still complicated.Wait, maybe I can express y in terms of x from equation 1.From equation 1:52 = -24x + (y - 14)²So,(y - 14)² = 52 + 24xTake square root:y - 14 = ±√(52 + 24x)So,y = 14 ± √(52 + 24x)But since y must be between 0 and 14 (because it's inside the rectangle), and the triangle is above the x-axis, y should be less than 14. So, we take the negative sign:y = 14 - √(52 + 24x)Now, plug this into equation 2:x² - y² = -52So,x² - [14 - √(52 + 24x)]² = -52Let me expand the square:[14 - √(52 + 24x)]² = 14² - 2*14*√(52 + 24x) + (√(52 + 24x))²= 196 - 28√(52 + 24x) + 52 + 24x= 196 + 52 + 24x - 28√(52 + 24x)= 248 + 24x - 28√(52 + 24x)So, equation becomes:x² - [248 + 24x - 28√(52 + 24x)] = -52Simplify:x² -248 -24x +28√(52 + 24x) = -52Bring constants to the right:x² -24x -248 +28√(52 + 24x) = -52Add 52 to both sides:x² -24x -196 +28√(52 + 24x) = 0This equation is quite complex. Maybe I can make a substitution.Let me set z = √(52 + 24x). Then, z² = 52 + 24x => x = (z² -52)/24Substitute into the equation:[(z² -52)/24]^2 -24*(z² -52)/24 -196 +28z = 0Simplify term by term:First term: [(z² -52)/24]^2 = (z² -52)^2 / 576Second term: -24*(z² -52)/24 = -(z² -52)Third term: -196Fourth term: +28zSo, the equation becomes:(z² -52)^2 / 576 - (z² -52) -196 +28z = 0Multiply all terms by 576 to eliminate the denominator:(z² -52)^2 - 576(z² -52) -196*576 +28z*576 = 0Calculate each term:First term: (z² -52)^2Second term: -576z² + 576*52Third term: -196*576Fourth term: +28*576 zSimplify:(z² -52)^2 -576z² + 576*52 -196*576 +28*576 z = 0Compute constants:576*52 = 30, let's compute 576*50=28,800 and 576*2=1,152, so total 28,800 +1,152=29,952196*576: Let's compute 200*576=115,200, subtract 4*576=2,304, so 115,200 -2,304=112,89628*576=16,128So, the equation becomes:(z² -52)^2 -576z² +29,952 -112,896 +16,128 z = 0Simplify constants:29,952 -112,896 = -82,944So:(z² -52)^2 -576z² -82,944 +16,128 z = 0Expand (z² -52)^2:z^4 -104z² +2704So, substitute:z^4 -104z² +2704 -576z² -82,944 +16,128 z = 0Combine like terms:z^4 + (-104z² -576z²) + (2704 -82,944) +16,128 z = 0Simplify:z^4 -680z² -80,240 +16,128 z = 0Rearrange:z^4 -680z² +16,128 z -80,240 = 0This is a quartic equation, which is quite difficult to solve. Maybe I made a mistake in the substitution or calculations. Alternatively, perhaps this approach is too complicated.Maybe there's a better way to approach this problem. Let me think about the maximum possible side length of an equilateral triangle that can fit inside a rectangle of 12x14.I recall that the maximum side length of an equilateral triangle that can fit inside a rectangle can be found by considering the rectangle's diagonal. The diagonal of the rectangle is √(12² +14²) = √(144 +196) = √340 ≈18.439. But an equilateral triangle with side length equal to the diagonal would have a height of (√3/2)*18.439 ≈16.16, which is larger than the rectangle's height of 14. So, that won't fit.Alternatively, maybe the maximum side length is determined by the shorter side of the rectangle. If I place the triangle with one side along the shorter side (12), the height is 6√3 ≈10.39, which fits within the height of 14. So, the area is 36√3.But earlier, I thought that maybe by rotating the triangle, I could fit a larger one. Let me check.Suppose I rotate the triangle so that one vertex is at a corner, and the other two vertices are on adjacent sides. Let me model this.Let me place one vertex at (0,0), another at (a,0), and the third at (0,b). Since it's an equilateral triangle, the distance from (0,0) to (a,0) is a, from (0,0) to (0,b) is b, and from (a,0) to (0,b) is √(a² + b²). So, setting these equal:a = b = √(a² + b²)But this implies that a = b = 0, which is not possible. So, this configuration doesn't work.Alternatively, maybe the triangle is placed such that one vertex is at (0,0), another at (12, y), and the third at (x,14). Let me try this.So, the distances:From (0,0) to (12,y): √(12² + y²) = √(144 + y²)From (12,y) to (x,14): √((x -12)² + (14 - y)²)From (x,14) to (0,0): √(x² +14²) = √(x² +196)Set all equal:√(144 + y²) = √((x -12)² + (14 - y)²) = √(x² +196)This is similar to the earlier approach, leading to complex equations. Maybe I can use some geometric insights instead.I remember that the largest equilateral triangle that can fit inside a rectangle can sometimes be found by considering the rectangle's aspect ratio. The rectangle here is 12x14, which is not a square, so the triangle might be placed in a way that utilizes both dimensions.Alternatively, perhaps the maximum area is indeed 36√3, as calculated earlier, since placing the triangle with a side along the shorter side of the rectangle allows it to fit within the height. But I'm not entirely sure if a larger triangle can be placed by rotating it.Wait, let me think about the height of the triangle. If I place the triangle with a side along the longer side of the rectangle (14), the height would be (√3/2)*14 ≈12.124, which is less than the width of 12. Wait, no, 12.124 is greater than 12, so it wouldn't fit. So, placing the triangle with a side along the longer side would cause the height to exceed the width of the rectangle, which is 12. Therefore, that triangle wouldn't fit.So, placing the triangle with a side along the shorter side (12) gives a height of ≈10.39, which fits within the height of 14. Therefore, the area is 36√3.But wait, maybe if I rotate the triangle so that it's not aligned with the sides, I can fit a larger triangle. Let me consider the maximum possible side length when the triangle is rotated.The maximum side length of an equilateral triangle that can fit inside a rectangle can be found by considering the rectangle's width and height in relation to the triangle's dimensions. The formula for the maximum side length S is given by:S = min( width / cos(30°), height / sin(30°) )But wait, let me think about this. If the triangle is rotated such that one side is at an angle θ with respect to the rectangle's sides, then the projections onto the width and height must be less than or equal to the rectangle's dimensions.For an equilateral triangle, the angle between the base and the height is 90°, but when rotated, the projections change.Alternatively, perhaps the maximum side length is determined by the rectangle's diagonal, but as I thought earlier, the height would be too large.Wait, maybe I can use the following approach. The maximum side length S of an equilateral triangle that can fit inside a rectangle of width W and height H is given by:S = min( W / cos(θ), H / sin(θ) )where θ is the angle between the base of the triangle and the width of the rectangle. To maximize S, we need to find θ such that W / cos(θ) = H / sin(θ). This occurs when tan(θ) = H / W.So, tan(θ) = 14 / 12 = 7/6 ≈1.1667, so θ ≈49.4 degrees.Then, S = W / cos(θ) = 12 / cos(49.4°)Compute cos(49.4°):cos(49.4°) ≈0.65So, S ≈12 /0.65 ≈18.46But wait, the height of the triangle would be (√3/2)*S ≈1.732/2 *18.46 ≈16.16, which is larger than the rectangle's height of 14. So, this doesn't work.Alternatively, maybe the maximum S is when the height of the triangle equals the rectangle's height. So, (√3/2)S =14 => S=28/√3≈16.16. Then, the width required would be S*cos(θ). But θ is such that tan(θ)=H/W=14/12=7/6. So, cos(θ)=6/√(6²+7²)=6/√85≈0.666. So, width required=16.16*0.666≈10.77, which is less than 12. So, this would fit.Therefore, the maximum side length is 28/√3≈16.16, and the area is (√3/4)*(28/√3)^2.Compute that:(√3/4)*(784/3)= (√3/4)*(784/3)= (784√3)/12≈65.33√3≈113.14But wait, the rectangle's width is 12, and the width required is≈10.77, which is less than 12, so it fits. Therefore, the area is≈65.33√3.But earlier, when placing the triangle with a side along the shorter side, the area was 36√3≈62.35, which is less than 65.33√3. So, this seems better.But wait, is this correct? Because when I place the triangle rotated, the side length is larger, but does the entire triangle fit within the rectangle?Wait, let me double-check. If the side length is 28/√3≈16.16, then the height is 14, which matches the rectangle's height. The width required is S*cos(θ)=16.16*(6/√85)=16.16*(6/9.2195)=16.16*0.65≈10.5, which is less than 12. So, yes, it fits.Therefore, the maximum area is (√3/4)*(28/√3)^2= (√3/4)*(784/3)= (784√3)/12= (196√3)/3≈113.14But wait, the rectangle's area is 12*14=168, so the triangle's area is about 113, which is less than half of the rectangle's area. That seems plausible.But earlier, I thought that placing the triangle with a side along the shorter side gives an area of 36√3≈62.35, which is much less. So, clearly, the rotated triangle has a larger area.But wait, is there a way to fit an even larger triangle? Maybe by placing it differently.Alternatively, perhaps the maximum area is indeed 36√3, but I'm not sure. Let me check some references or similar problems.Wait, I recall that the maximum area of an equilateral triangle inside a rectangle can be found by considering the rectangle's dimensions and the triangle's orientation. The formula I used earlier seems correct, but let me verify.If I set the height of the triangle equal to the rectangle's height, then the side length is 2H/√3=28/√3≈16.16, and the width required is S*cos(θ)=16.16*(6/√85)=10.5, which is less than 12. So, it fits.Therefore, the area is (√3/4)*(28/√3)^2= (784√3)/12=196√3/3≈113.14.But wait, let me compute 196√3/3:196/3≈65.33, so 65.33√3≈65.33*1.732≈113.14.Yes, that's correct.But earlier, I thought that placing the triangle with a side along the shorter side gives 36√3≈62.35, which is less. So, the rotated triangle is larger.But wait, is this the maximum? Or can we fit an even larger triangle by placing it differently?Alternatively, perhaps the maximum area is when the triangle is inscribed such that each vertex touches a different side of the rectangle. Let me try to compute that.Let me denote the rectangle with width W=12 and height H=14. Let the triangle have vertices on each side: one on the bottom side (y=0), one on the right side (x=12), and one on the top side (y=14). Let me denote the vertices as (a,0), (12,b), and (c,14).Since it's an equilateral triangle, all sides must be equal. So, the distance between (a,0) and (12,b) must equal the distance between (12,b) and (c,14), which must equal the distance between (c,14) and (a,0).This gives us three equations:1. √[(12 - a)^2 + (b - 0)^2] = √[(c - 12)^2 + (14 - b)^2]2. √[(c - 12)^2 + (14 - b)^2] = √[(a - c)^2 + (0 - 14)^2]3. √[(a - c)^2 + (0 - 14)^2] = √[(12 - a)^2 + (b - 0)^2]These equations are symmetric, so solving them might give the side length.Let me square the first equation:(12 - a)^2 + b^2 = (c - 12)^2 + (14 - b)^2Expand both sides:144 -24a +a² +b² = c² -24c +144 +196 -28b +b²Simplify:144 -24a +a² +b² = c² -24c +340 -28b +b²Cancel b²:144 -24a +a² = c² -24c +340 -28bRearrange:a² -24a +144 = c² -24c +340 -28bSimilarly, square the second equation:(c - 12)^2 + (14 - b)^2 = (a - c)^2 + (0 - 14)^2Expand:c² -24c +144 +196 -28b +b² = a² -2ac +c² +196Simplify:c² -24c +340 -28b +b² = a² -2ac +c² +196Cancel c²:-24c +340 -28b +b² = a² -2ac +196Rearrange:a² -2ac +196 = -24c +340 -28b +b²Now, we have two equations:1. a² -24a +144 = c² -24c +340 -28b2. a² -2ac +196 = -24c +340 -28b +b²Let me subtract equation 1 from equation 2:(a² -2ac +196) - (a² -24a +144) = (-24c +340 -28b +b²) - (c² -24c +340 -28b)Simplify left side:a² -2ac +196 -a² +24a -144 = -2ac +24a +52Right side:-24c +340 -28b +b² -c² +24c -340 +28b = b² -c²So, we have:-2ac +24a +52 = b² -c²Let me rearrange:b² = -2ac +24a +52 +c²Now, from equation 1:a² -24a +144 = c² -24c +340 -28bLet me express b from this equation:a² -24a +144 -c² +24c -340 = -28bSimplify:a² -24a -c² +24c -196 = -28bSo,b = (c² -24c +a² -24a +196)/28Now, substitute this into the equation b² = -2ac +24a +52 +c²This will lead to a very complex equation. Maybe I can assume some symmetry or make an intelligent guess.Alternatively, perhaps the maximum area is indeed 36√3, as calculated earlier, and the rotated triangle approach is incorrect because it doesn't account for the entire triangle fitting within the rectangle.Wait, let me think again. If I place the triangle with a side along the shorter side (12), the height is 6√3≈10.39, which fits within the height of 14. So, the area is 36√3.But if I rotate the triangle so that its base is not along the side, but instead, it's placed diagonally, perhaps I can fit a larger triangle. However, when I tried that earlier, the calculations became too complex.Alternatively, maybe the maximum area is indeed 36√3, and the rotated triangle approach is not possible because the triangle would exceed the rectangle's boundaries.Wait, let me try to visualize this. If I rotate the triangle so that one vertex is at (0,0), another at (12, y), and the third at (x,14), then the triangle's height would be from (0,0) to (x,14), which is 14 units. But the height of the triangle is (√3/2)*S, so S=28/√3≈16.16. But the width required is S*cos(θ)=16.16*(6/√85)=10.5, which is less than 12. So, the triangle would fit.But wait, the vertex at (x,14) must lie on the top side of the rectangle, so x must be between 0 and12. Similarly, the vertex at (12,y) must lie on the right side, so y must be between0 and14.But when I calculated earlier, the side length was≈16.16, which seems larger than the rectangle's diagonal≈18.439. Wait, no, 16.16 is less than 18.439, so it's possible.But I'm getting confused. Let me try to compute the coordinates.If the side length is 28/√3≈16.16, then the height is14, as desired. The base of the triangle would be along the line from (0,0) to (x,14), which has length16.16. Wait, no, the base is not along that line. The base is one side of the triangle, which is placed diagonally.Wait, perhaps I'm overcomplicating this. Let me try to find the coordinates.Let me place one vertex at (0,0), another at (a,0), and the third at (b,c). Since it's an equilateral triangle, the distance between (0,0) and (a,0) is a, between (a,0) and (b,c) is √((b -a)^2 +c^2), and between (b,c) and (0,0) is √(b² +c²). Setting these equal:a = √((b -a)^2 +c^2) = √(b² +c²)From the first equality:a = √(b² +c²)From the second equality:a = √((b -a)^2 +c^2)Square both sides:a² = (b -a)^2 +c²Expand:a² = b² -2ab +a² +c²Simplify:0 = b² -2ab +c²But from the first equality, a² = b² +c², so substitute:0 = (a² -c²) -2ab +c²Simplify:0 = a² -2abSo, a² =2ab => a=2b (since a≠0)Thus, b=a/2From a² =b² +c²:a² = (a²/4) +c² => c²= (3/4)a² => c=(√3/2)aSo, the third vertex is at (a/2, (√3/2)a)Now, since the triangle must fit within the rectangle, the coordinates must satisfy:a/2 ≤12 => a ≤24and(√3/2)a ≤14 => a ≤28/√3≈16.16So, the maximum a is≈16.16, which gives c≈14.But wait, a=16.16, so b=8.08, and c=14.So, the vertices are at (0,0), (16.16,0), and (8.08,14). But wait, the point (16.16,0) is outside the rectangle, since the rectangle's width is12. So, this is not possible.Ah, here's the mistake. Earlier, I assumed that the triangle could be placed with a vertex at (a,0) where a>12, but the rectangle's width is only12. Therefore, a cannot exceed12. So, the maximum a is12, which gives b=6, c=6√3≈10.39.Thus, the vertices are at (0,0), (12,0), and (6,10.39). This triangle fits within the rectangle, as the y-coordinate is≈10.39<14.Therefore, the maximum side length is12, and the area is36√3.So, my initial approach was correct. The maximum area is36√3.But wait, earlier I thought that by rotating the triangle, I could get a larger area, but that led to a vertex outside the rectangle. So, it seems that placing the triangle with a side along the shorter side of the rectangle gives the maximum area.Therefore, the maximum area is36√3.

💡First, I need to compute ( f(3) ). Since 3 is greater than or equal to 0, I use the first case of the piecewise function:[ f(3) = -3^3 = -27. ]Next, I compute ( f(-27) ). Since -27 is less than 0, I use the second case:[ f(-27) = -27 + 9 = -18. ]Then, I compute ( f(-18) ). Again, -18 is less than 0, so I use the second case:[ f(-18) = -18 + 9 = -9. ]Next, I compute ( f(-9) ). Since -9 is less than 0, I use the second case:[ f(-9) = -9 + 9 = 0. ]Finally, I compute ( f(0) ). Since 0 is greater than or equal to 0, I use the first case:[ f(0) = -0^3 = 0. ]Therefore, the value of ( f(f(f(f(f(3))))) ) is (boxed{0}).

💡Alright, so I've got this problem here with five real numbers: a, b, c, d, and e. They satisfy two conditions: their sum is 8, and the sum of their squares is 16. The question is asking for the maximum possible value of e. The options are A: 2, B: 16/5, C: 3, and D: 2/5. Hmm, okay.Let me write down the given equations to make it clearer:1. a + b + c + d + e = 82. a² + b² + c² + d² + e² = 16I need to find the maximum value of e. So, maybe I can express the other variables in terms of e and then use some inequality to find the maximum.From the first equation, if I subtract e from both sides, I get:a + b + c + d = 8 - eSimilarly, from the second equation, subtracting e² gives:a² + b² + c² + d² = 16 - e²Now, I have four variables a, b, c, d with their sum and sum of squares known. I remember there's something called the Cauchy-Schwarz inequality which relates sums and sums of squares. Maybe that can help here.The Cauchy-Schwarz inequality in its basic form is:(a₁² + a₂² + ... + aₙ²)(b₁² + b₂² + ... + bₙ²) ≥ (a₁b₁ + a₂b₂ + ... + aₙbₙ)²But in this case, I think I can use a version where all the b's are 1. So, for four variables, it would be:(a² + b² + c² + d²)(1² + 1² + 1² + 1²) ≥ (a + b + c + d)²Plugging in the values I have:(16 - e²)(4) ≥ (8 - e)²Let me write that out:4*(16 - e²) ≥ (8 - e)²Now, let's expand both sides.Left side: 4*(16 - e²) = 64 - 4e²Right side: (8 - e)² = 64 - 16e + e²So, putting it all together:64 - 4e² ≥ 64 - 16e + e²Hmm, let's subtract 64 from both sides to simplify:-4e² ≥ -16e + e²Now, let's bring all terms to one side:-4e² - (-16e + e²) ≥ 0Wait, that's not the best way. Maybe I should move everything to the left side:-4e² - (-16e + e²) ≥ 0Which simplifies to:-4e² + 16e - e² ≥ 0Combine like terms:-5e² + 16e ≥ 0Hmm, that's a quadratic inequality. Let me rewrite it:-5e² + 16e ≥ 0I can factor out a negative sign to make it easier:5e² - 16e ≤ 0So, 5e² - 16e ≤ 0Factor out an e:e(5e - 16) ≤ 0Okay, so this inequality holds when the product is less than or equal to zero. That happens when one factor is non-positive and the other is non-negative.So, the critical points are e = 0 and e = 16/5.Let me test intervals:1. e < 0: Let's pick e = -1. Then, e is negative, and 5e - 16 is also negative. Negative times negative is positive, which is not ≤ 0. So, this interval doesn't satisfy.2. 0 < e < 16/5: Let's pick e = 1. Then, e is positive, and 5e - 16 is 5 - 16 = -11, which is negative. Positive times negative is negative, which satisfies the inequality.3. e > 16/5: Let's pick e = 4. Then, e is positive, and 5e - 16 is 20 - 16 = 4, which is positive. Positive times positive is positive, which doesn't satisfy the inequality.So, the solution is 0 ≤ e ≤ 16/5.But wait, the question is asking for the maximum value of e, so the maximum is 16/5.Let me just verify if this is achievable. If e = 16/5, then a + b + c + d = 8 - 16/5 = (40/5 - 16/5) = 24/5.And a² + b² + c² + d² = 16 - (256/25) = (400/25 - 256/25) = 144/25.Now, if a, b, c, d are all equal, then each would be (24/5)/4 = 6/5.So, a = b = c = d = 6/5.Then, a² = (6/5)² = 36/25. So, four of them would be 4*(36/25) = 144/25, which matches the sum of squares.Therefore, e can indeed be 16/5 when a, b, c, d are each 6/5.So, the maximum value of e is 16/5, which is option B.**Final Answer**boxed{B}

💡Okay, so I have this problem where a car salesperson sold a total of 300 cars. There's a table that lists the percentage of each type of car sold: Mercedes, Nissan, Ford, and Chevrolet. The question is asking how many BMWs were sold, assuming that the remaining cars after accounting for the ones listed in the table were BMWs.Alright, let's break this down step by step. First, I need to understand what the table is telling me. It lists four types of cars and the percentage of total cars sold that each type represents. Mercedes accounts for 20%, Nissan for 25%, Ford for 10%, and Chevrolet for 18%. So, these percentages add up to a certain total, and whatever is left over from 100% must be the percentage of BMWs sold.So, my first task is to add up these percentages to see what portion of the total cars are accounted for by these four brands. Let me do that:Mercedes: 20%Nissan: 25%Ford: 10%Chevrolet: 18%Adding them together: 20 + 25 + 10 + 18. Let's see, 20 plus 25 is 45, plus 10 is 55, plus 18 is 73. So, altogether, these four brands account for 73% of the total cars sold.Now, since the total percentage of all cars sold should be 100%, the remaining percentage must be BMWs. So, to find out what percentage of cars are BMWs, I subtract the sum of the other percentages from 100%.100% - 73% = 27%. So, BMWs make up 27% of the total cars sold.Now, I need to find out how many cars that is. The total number of cars sold is 300. To find 27% of 300, I can use the formula:Number of BMWs = (Percentage of BMWs / 100) * Total number of carsPlugging in the numbers:Number of BMWs = (27 / 100) * 300Let me calculate that. 27 divided by 100 is 0.27. Then, 0.27 multiplied by 300. Hmm, 0.27 times 300. I can think of it as 0.27 times 300 is the same as 27 times 3, because 0.27 times 100 is 27, and 300 is three times 100. So, 27 times 3 is 81.Therefore, the number of BMWs sold is 81.Wait, let me double-check that to make sure I didn't make a mistake. So, 27% of 300 cars is 81. If I add up all the cars:Mercedes: 20% of 300 is 60 carsNissan: 25% of 300 is 75 carsFord: 10% of 300 is 30 carsChevrolet: 18% of 300 is 54 carsBMW: 27% of 300 is 81 carsNow, let's add these numbers together to see if they total 300:60 + 75 = 135135 + 30 = 165165 + 54 = 219219 + 81 = 300Yes, that adds up correctly. So, the calculations seem to be accurate. Therefore, the number of BMWs sold is indeed 81.

💡First, I need to convert the binary number (11010_{(2)}) to a decimal number. I'll do this by expanding each digit multiplied by its corresponding power of 2.Next, I'll use the division by 8 method to convert the resulting decimal number to an octal number. This involves dividing the decimal number by 8 and keeping track of the remainders.Finally, I'll compile the remainders from the last to the first to obtain the octal representation of the original binary number.

💡Okay, so I have this function g(x) that's defined in two parts. It's a piecewise function, right? For x greater than 2, it's just x plus 4, and for x less than or equal to 2, it's 3x plus some constant b. The question is asking me what value of b will make this function continuous. Hmm, continuity is important because if a function isn't continuous at a point, it means there's a jump or a break there, which might not be desirable depending on the context.Alright, so I remember that for a function to be continuous at a point, specifically at x = 2 in this case, the limit of the function as x approaches 2 from the left side should be equal to the limit as x approaches 2 from the right side, and both should equal the function's value at x = 2. That makes sense because if the left and right limits aren't the same, there's a jump discontinuity, and if the function value at that point doesn't match the limits, it's a removable discontinuity.So, let me break this down. First, I need to find the limit as x approaches 2 from the right side. Since for x greater than 2, the function is x + 4, I can plug in x = 2 into that expression. That would be 2 + 4, which is 6. So, the right-hand limit is 6.Next, I need to find the limit as x approaches 2 from the left side. For x less than or equal to 2, the function is 3x + b. So, plugging in x = 2 into this expression gives me 3 times 2 plus b, which is 6 + b. So, the left-hand limit is 6 + b.Now, for the function to be continuous at x = 2, these two limits must be equal. That means 6 (from the right side) must equal 6 + b (from the left side). So, setting them equal: 6 = 6 + b. If I subtract 6 from both sides, I get 0 = b. So, b must be 0.Wait, let me double-check that. If b is 0, then for x less than or equal to 2, the function is 3x. So, at x = 2, that would be 6, and the other part of the function at x = 2 is 6 as well. So, yes, both sides meet at 6 when b is 0. That seems correct.I think I got it. So, the value of b that makes the function continuous is 0. It makes sense because without that, the left side would be 6 + b, which would be higher or lower than 6 depending on the value of b, causing a discontinuity. By setting b to 0, both expressions meet smoothly at x = 2, ensuring continuity.Just to visualize it, if I imagine the graph of g(x), for x > 2, it's a straight line with a slope of 1 starting from (2,6) going upwards. For x ≤ 2, it's another straight line with a steeper slope of 3. If b were anything other than 0, the left side would either be above or below the right side at x = 2, creating a jump. But with b = 0, both lines meet exactly at the point (2,6), making the function continuous there.So, yeah, I'm pretty confident now that b has to be 0 to ensure that the function doesn't have a discontinuity at x = 2. It all checks out when I plug in the values and think about the graph.