💡Okay, so I need to find the equation of the normal line to the curve y = 8 times the fourth root of x minus 70 at the point where the abscissa (which is the x-coordinate) is 16. Alright, let me break this down step by step.First, I remember that the normal line to a curve at a given point is perpendicular to the tangent line at that point. So, to find the normal line, I first need to find the slope of the tangent line at that point, and then take the negative reciprocal of that slope for the normal line.Alright, let's start by finding the derivative of the function y = 8∛x - 70. Wait, hold on, the original problem says the fourth root, right? Because it's written as y = 8√[1]{x} - 70. Hmm, actually, that notation is a bit confusing. The fourth root is usually written as ∛x, but here it's written with a 1 inside the root symbol. Maybe that's a typo? Or maybe it's supposed to be the first root, which is just x. Wait, that doesn't make sense. Let me double-check.Oh, maybe it's supposed to be the fourth root. Because in some notations, the index is written inside the root symbol. So, if it's √[4]{x}, that would be the fourth root. So, perhaps the original problem is y = 8 times the fourth root of x minus 70. Yeah, that makes more sense because if it were the first root, it's just x, and the problem would be too simple. So, I think it's safe to assume that it's the fourth root.So, y = 8x^(1/4) - 70. Now, to find the derivative, I'll use the power rule. The derivative of x^n is n*x^(n-1). So, the derivative of 8x^(1/4) is 8*(1/4)*x^(1/4 - 1) = 2x^(-3/4). The derivative of -70 is 0. So, the derivative y' is 2x^(-3/4), which can also be written as 2/(x^(3/4)).Okay, now I need to evaluate this derivative at x = 16 to find the slope of the tangent line at that point. So, plugging in x = 16:y' = 2/(16^(3/4)). Let's compute 16^(3/4). First, 16^(1/4) is 2 because 2^4 = 16. Then, 16^(3/4) is (16^(1/4))^3 = 2^3 = 8. So, y' at x = 16 is 2/8 = 1/4.So, the slope of the tangent line at x = 16 is 1/4. Therefore, the slope of the normal line, which is the negative reciprocal, would be -4.Now, I need to find the point on the curve where x = 16. So, plugging x = 16 into the original equation:y = 8*(16)^(1/4) - 70. Again, 16^(1/4) is 2, so y = 8*2 - 70 = 16 - 70 = -54.So, the point is (16, -54). Now, with the slope of the normal line being -4 and passing through the point (16, -54), I can write the equation of the normal line using the point-slope form: y - y1 = m(x - x1).Plugging in the values: y - (-54) = -4(x - 16). Simplifying this, y + 54 = -4x + 64. Subtracting 54 from both sides gives y = -4x + 10.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting from the derivative: y' = 2x^(-3/4). At x = 16, that's 2/(16^(3/4)) = 2/8 = 1/4. Correct. The slope of the normal is -4. The point is (16, -54). Using point-slope: y + 54 = -4(x - 16). Expanding: y + 54 = -4x + 64. Subtracting 54: y = -4x + 10. Yeah, that seems right.I think I've got it. The equation of the normal line is y = -4x + 10.

💡Okay, so I have this geometry problem here with four propositions about lines and planes. Let me try to figure out which ones are correct. I remember some basic properties from my geometry class, but I need to think carefully about each one.First, let's restate the problem to make sure I understand it. We have two non-coincident lines, a and b, and three pairwise non-coincident planes, α, β, and γ. There are four propositions given, and I need to determine which ones are correct.Let's go through each proposition one by one.**Proposition ①:** If a is perpendicular to α and a is perpendicular to β, then α is parallel to β.Hmm, okay. So if a single line is perpendicular to two different planes, does that mean the planes are parallel? I think so. Because if a line is perpendicular to a plane, it means it's perpendicular to every line in that plane. So if the same line is perpendicular to both α and β, then their normals are in the same direction, right? That should mean the planes are parallel. Yeah, I think this one is correct.**Proposition ②:** If α is perpendicular to γ and β is perpendicular to γ, then α is parallel to β.Wait, this is similar to the first one, but with planes instead of lines. So if two planes are both perpendicular to a third plane, does that make them parallel? Hmm, I'm not sure. Let me visualize this. Imagine γ is like the floor, and α and β are walls. If both walls are perpendicular to the floor, does that mean they are parallel? No, they could intersect along a line, like two walls meeting at a corner. So they don't have to be parallel. So this proposition is not necessarily true. It might be incorrect.**Proposition ③:** If α is parallel to β, a is contained in α, and b is contained in β, then a is parallel to b.Okay, so if two planes are parallel, and each has a line in them, does that make the lines parallel? I don't think so. The lines could be going in different directions. For example, imagine two parallel sheets of paper. If I draw a horizontal line on one and a vertical line on the other, they aren't parallel. They might even be skew lines if they're not in the same space. So this proposition doesn't hold. It's incorrect.**Proposition ④:** If α is parallel to β, α intersects γ at a, and β intersects γ at b, then a is parallel to b.Alright, so two parallel planes intersecting another plane γ. The lines of intersection are a and b. Are these lines parallel? I think they should be. Because if two parallel planes are cut by a third plane, the lines of intersection should be parallel. It's like cutting two parallel sheets of paper with a knife; the cuts should be parallel lines. Yeah, I think this one is correct.So, summarizing my thoughts:- Proposition ①: Correct.- Proposition ②: Incorrect.- Proposition ③: Incorrect.- Proposition ④: Correct.Therefore, the correct propositions are ① and ④.

💡Okay, so I have this quadratic equation: (x^2 + kx + 8 = 0). The problem says that the roots of this equation differ by (sqrt{73}), and I need to find the greatest possible value of (k). Hmm, let me think about how to approach this.First, I remember that for a quadratic equation (ax^2 + bx + c = 0), the roots can be found using the quadratic formula: (frac{-b pm sqrt{b^2 - 4ac}}{2a}). In this case, (a = 1), (b = k), and (c = 8). So, plugging these into the formula, the roots would be:[frac{-k pm sqrt{k^2 - 4 cdot 1 cdot 8}}{2 cdot 1} = frac{-k pm sqrt{k^2 - 32}}{2}]Okay, so the two roots are (frac{-k + sqrt{k^2 - 32}}{2}) and (frac{-k - sqrt{k^2 - 32}}{2}). The problem states that the difference between these two roots is (sqrt{73}). Let me calculate the difference between the two roots.Subtracting the smaller root from the larger one:[left( frac{-k + sqrt{k^2 - 32}}{2} right) - left( frac{-k - sqrt{k^2 - 32}}{2} right)]Simplifying this expression:[frac{-k + sqrt{k^2 - 32} + k + sqrt{k^2 - 32}}{2} = frac{2sqrt{k^2 - 32}}{2} = sqrt{k^2 - 32}]So, the difference between the roots is (sqrt{k^2 - 32}). According to the problem, this difference is equal to (sqrt{73}). Therefore, I can set up the equation:[sqrt{k^2 - 32} = sqrt{73}]To solve for (k), I'll square both sides of the equation to eliminate the square roots:[(sqrt{k^2 - 32})^2 = (sqrt{73})^2 k^2 - 32 = 73]Now, solving for (k^2):[k^2 = 73 + 32 k^2 = 105]Taking the square root of both sides gives:[k = pm sqrt{105}]So, (k) can be either (sqrt{105}) or (-sqrt{105}). The problem asks for the greatest possible value of (k), so between these two, (sqrt{105}) is the larger value.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the quadratic formula, I correctly identified the roots. Then, I subtracted the smaller root from the larger one, which gave me (sqrt{k^2 - 32}). Setting that equal to (sqrt{73}) and squaring both sides seems right. Solving for (k^2) gives 105, and taking the square root gives (pm sqrt{105}). So, yes, the greatest value is (sqrt{105}).I think that makes sense. The quadratic equation will have real roots because the discriminant (k^2 - 32) is positive when (k^2 = 105), which is definitely positive. So, everything checks out. I don't see any errors in my reasoning.**Final Answer**The greatest possible value of (k) is (boxed{sqrt{105}}).

💡Alright, so I'm trying to figure out which of the four statements about the Cassini lemniscate are correct. I remember that a Cassini lemniscate is defined as the set of points where the product of the distances from two fixed points (foci) is a constant. The foci are given as F₁(-c, 0) and F₂(c, 0), and the product |PF₁|·|PF₂| equals a².Let me start by writing down the equation for a point P(x, y) on the curve. The distance from P to F₁ is √[(x + c)² + y²], and the distance from P to F₂ is √[(x - c)² + y²]. So, the product of these distances is:√[(x + c)² + y²] · √[(x - c)² + y²] = a²If I square both sides to eliminate the square roots, I get:[(x + c)² + y²] · [(x - c)² + y²] = a⁴Expanding this might help me analyze the properties. Let me try expanding the left side:First, expand (x + c)² and (x - c)²:(x + c)² = x² + 2xc + c²(x - c)² = x² - 2xc + c²So, the product becomes:(x² + 2xc + c² + y²)(x² - 2xc + c² + y²)This looks like the product of (A + B)(A - B) where A = x² + c² + y² and B = 2xc. So, it simplifies to:A² - B² = (x² + c² + y²)² - (2xc)²Expanding this further:(x² + c² + y²)² = x⁴ + 2x²c² + c⁴ + 2x²y² + 2c²y² + y⁴(2xc)² = 4x²c²So subtracting:(x² + c² + y²)² - (2xc)² = x⁴ + 2x²c² + c⁴ + 2x²y² + 2c²y² + y⁴ - 4x²c²Simplify:x⁴ - 2x²c² + c⁴ + 2x²y² + 2c²y² + y⁴Hmm, that seems complicated. Maybe there's a better way to analyze the properties without expanding everything.Let me go back to the original equation:√[(x + c)² + y²] · √[(x - c)² + y²] = a²I can consider the symmetry of this equation. If I replace x with -x, the equation remains the same because (x + c) becomes (-x + c) and (x - c) becomes (-x - c), but squaring them would make it the same as before. Similarly, replacing y with -y also doesn't change the equation. So, the curve is symmetric with respect to both the x-axis and y-axis, which means it's axisymmetric. Also, replacing both x and y with their negatives doesn't change the equation, so it's centrally symmetric. Therefore, statement (1) seems correct.Next, statement (2): If a = c, then the curve passes through the origin. Let's test the origin (0,0). Plugging into the equation:√[(0 + c)² + 0²] · √[(0 - c)² + 0²] = √[c²] · √[c²] = c · c = c²If a = c, then a² = c², so the equation holds. Therefore, the origin is on the curve when a = c. So, statement (2) is correct.Statement (3): If 0 < a < c, then the curve does not exist. Let me think about the minimum value of |PF₁|·|PF₂|. For any point P, the product |PF₁|·|PF₂| is minimized when P is on the line segment joining F₁ and F₂. The minimum occurs at the midpoint, which is the origin. Wait, no, actually, the minimum might not necessarily be at the origin. Let me recall that for two points, the product |PF₁|·|PF₂| is minimized when P is at the midpoint, which is the origin in this case.But wait, if P is at the origin, then |PF₁| = |PF₂| = c, so the product is c². If a < c, then a² < c², which would mean that the product |PF₁|·|PF₂| is less than c², but we just saw that the minimum product is c². Therefore, if a² < c², there are no points P such that |PF₁|·|PF₂| = a². So, the curve does not exist. Therefore, statement (3) is correct.Statement (4): If 0 < c < a, then a² - c² ≤ x² + y² ≤ a² + c². Let me see. I need to find bounds on x² + y² for points on the curve. Let me denote r² = x² + y². Then, I can express the product |PF₁|·|PF₂| in terms of r and x.Wait, |PF₁|·|PF₂| can be expressed using the formula for the product of distances from two points. I recall that |PF₁|·|PF₂| = (r² + c²) - (2xc)² / (r² + c²). Hmm, maybe that's too complicated.Alternatively, I can use the identity that |PF₁|·|PF₂| = r² - c² when P is on the lemniscate. Wait, no, that's not quite right. Let me think again.Actually, for a Cassini lemniscate, the equation can be rewritten in polar coordinates. Let me try that. Let me switch to polar coordinates where x = r cosθ and y = r sinθ. Then, the distance from P to F₁ is √[(r cosθ + c)² + (r sinθ)²] = √[r² + 2rc cosθ + c²]. Similarly, the distance to F₂ is √[r² - 2rc cosθ + c²].So, the product is:√[r² + 2rc cosθ + c²] · √[r² - 2rc cosθ + c²] = a²Squaring both sides:(r² + 2rc cosθ + c²)(r² - 2rc cosθ + c²) = a⁴This simplifies to:(r² + c²)² - (2rc cosθ)² = a⁴Expanding:r⁴ + 2r²c² + c⁴ - 4r²c² cos²θ = a⁴Hmm, this seems complicated, but maybe I can find bounds on r².Let me consider the maximum and minimum values of cos²θ. Since cos²θ ranges from 0 to 1, the term -4r²c² cos²θ ranges from -4r²c² to 0.So, the equation becomes:r⁴ + 2r²c² + c⁴ - 4r²c² cos²θ = a⁴Let me rearrange:r⁴ + 2r²c² + c⁴ - a⁴ = 4r²c² cos²θSince cos²θ is between 0 and 1, the right side is between 0 and 4r²c².Therefore:0 ≤ r⁴ + 2r²c² + c⁴ - a⁴ ≤ 4r²c²Let me solve the inequalities separately.First inequality:r⁴ + 2r²c² + c⁴ - a⁴ ≥ 0Second inequality:r⁴ + 2r²c² + c⁴ - a⁴ ≤ 4r²c²Simplify the second inequality:r⁴ + 2r²c² + c⁴ - a⁴ - 4r²c² ≤ 0r⁴ - 2r²c² + c⁴ - a⁴ ≤ 0This can be written as:(r² - c²)² - a⁴ ≤ 0So,(r² - c²)² ≤ a⁴Taking square roots:|r² - c²| ≤ a²Which implies:- a² ≤ r² - c² ≤ a²Adding c²:c² - a² ≤ r² ≤ c² + a²But since r² is always non-negative, the lower bound is max(0, c² - a²). However, in the case where 0 < c < a, c² - a² is negative, so the lower bound is 0. But wait, the original inequality was |r² - c²| ≤ a², which implies:c² - a² ≤ r² ≤ c² + a²But since r² is non-negative, the effective bounds are:max(0, c² - a²) ≤ r² ≤ c² + a²But in the case where 0 < c < a, c² - a² is negative, so the lower bound is 0. However, the statement (4) says a² - c² ≤ x² + y² ≤ a² + c². Wait, that's different from what I have here.Wait, I have c² - a² ≤ r² ≤ c² + a², but since r² is non-negative, the lower bound is 0 when c < a. However, the statement says a² - c² ≤ r². But a² - c² is positive when a > c, so the lower bound should be a² - c², not 0.Hmm, maybe I made a mistake in interpreting the inequality. Let me go back.From the second inequality:(r² - c²)² ≤ a⁴Taking square roots:|r² - c²| ≤ a²Which means:- a² ≤ r² - c² ≤ a²Adding c²:c² - a² ≤ r² ≤ c² + a²But since r² is non-negative, the lower bound is max(0, c² - a²). However, when 0 < c < a, c² - a² is negative, so the lower bound is 0. But the statement says a² - c² ≤ r², which is positive because a > c. So, is the lower bound a² - c² or 0?Wait, maybe I need to consider the first inequality as well. The first inequality was:r⁴ + 2r²c² + c⁴ - a⁴ ≥ 0Let me denote this as:(r² + c²)² - a⁴ ≥ 0Which implies:(r² + c²)² ≥ a⁴Taking square roots:r² + c² ≥ a²So,r² ≥ a² - c²Since r² is non-negative, this gives:r² ≥ max(0, a² - c²)But when 0 < c < a, a² - c² is positive, so r² ≥ a² - c²Therefore, combining both inequalities:a² - c² ≤ r² ≤ a² + c²Which matches statement (4). So, statement (4) is correct.Wait, but earlier I thought the lower bound was max(0, c² - a²), but considering the first inequality, it's actually r² ≥ a² - c². So, when 0 < c < a, a² - c² is positive, so r² must be at least a² - c². Therefore, statement (4) is correct.So, all four statements seem to be correct. But let me double-check statement (3). If 0 < a < c, then the curve does not exist. As I thought earlier, the minimum value of |PF₁|·|PF₂| is c², so if a² < c², there are no points P such that |PF₁|·|PF₂| = a². Therefore, the curve does not exist, so statement (3) is correct.Wait, but I'm a bit confused about statement (4). When 0 < c < a, the bounds are a² - c² ≤ r² ≤ a² + c². But when c > a, what happens? Well, in that case, a² - c² would be negative, so the lower bound would be 0. But statement (4) only applies when 0 < c < a, so in that case, a² - c² is positive, and the bounds make sense.Okay, I think I've convinced myself that all four statements are correct. But let me just visualize the Cassini lemniscate to make sure. When a = c, it passes through the origin, which makes sense. When a > c, it forms a lemniscate, and when a < c, it doesn't exist. The bounds on r² also make sense when a > c. So, yeah, all four statements seem correct.

💡To determine the minimum number of 5 cubic cm cubes needed to construct the box, I will start by calculating the volume of the box.The box has a length of 10 cm, a width of 13 cm, and a height of 5 cm. Using the formula for the volume of a rectangular prism:[text{Volume of the box} = text{length} times text{width} times text{height}][text{Volume of the box} = 10 , text{cm} times 13 , text{cm} times 5 , text{cm} = 650 , text{cubic cm}]Next, I will calculate how many 5 cubic cm cubes are required to fill this volume:[text{Number of cubes} = frac{text{Volume of the box}}{text{Volume of one cube}} = frac{650 , text{cubic cm}}{5 , text{cubic cm}} = 130]Therefore, the minimum number of 5 cubic cm cubes needed to build the box is 130.

💡Okay, so I have this problem about triangle ABC. It says that 2 times the sine squared of (A+B)/2 minus cosine of 2C equals 1, and the radius of the circumcircle R is 2. I need to find angle C and then the maximum area of triangle ABC. Hmm, let me try to break this down step by step.First, part (1): Find angle C. The equation given is 2 sin²[(A+B)/2] - cos(2C) = 1. I remember that in any triangle, the sum of angles A, B, and C is π radians (or 180 degrees). So, A + B + C = π. That means A + B = π - C. Maybe I can substitute that into the equation.Let me rewrite the equation:2 sin²[(A+B)/2] - cos(2C) = 1Since A + B = π - C, then (A+B)/2 = (π - C)/2. So, sin[(A+B)/2] is sin[(π - C)/2]. I also recall that sin[(π - C)/2] is equal to cos(C/2). Because sin(π/2 - x) = cos x. So, sin[(π - C)/2] = cos(C/2). Therefore, sin²[(A+B)/2] = cos²(C/2).So, substituting back into the equation:2 cos²(C/2) - cos(2C) = 1Hmm, okay. Now, I can use some trigonometric identities here. I know that cos²(x) can be written as (1 + cos(2x))/2. So, cos²(C/2) = (1 + cos C)/2. Let me substitute that in:2 * [(1 + cos C)/2] - cos(2C) = 1Simplify this:(1 + cos C) - cos(2C) = 1So, 1 + cos C - cos(2C) = 1Subtract 1 from both sides:cos C - cos(2C) = 0Hmm, that simplifies to cos C = cos(2C). Now, I need to solve this equation for C. Let me recall that cos(2C) can be written in terms of cos C. There are a few identities, like cos(2C) = 2 cos² C - 1 or cos(2C) = 1 - 2 sin² C. Maybe I can use one of these to rewrite the equation.Let me use cos(2C) = 2 cos² C - 1. Then, the equation becomes:cos C = 2 cos² C - 1Let me rearrange this:2 cos² C - cos C - 1 = 0So, this is a quadratic equation in terms of cos C. Let me denote x = cos C for simplicity:2x² - x - 1 = 0Now, solving this quadratic equation. The quadratic formula is x = [ -b ± sqrt(b² - 4ac) ] / (2a). Here, a = 2, b = -1, c = -1.So, x = [1 ± sqrt(1 + 8)] / 4 = [1 ± 3] / 4So, two solutions:x = (1 + 3)/4 = 4/4 = 1x = (1 - 3)/4 = (-2)/4 = -1/2So, cos C = 1 or cos C = -1/2. Now, in a triangle, angle C is between 0 and π radians. So, cos C = 1 would mean C = 0, which is impossible in a triangle. So, we discard that solution.Therefore, cos C = -1/2. So, what angle between 0 and π has a cosine of -1/2? That would be 2π/3 radians, or 120 degrees.So, angle C is 2π/3 radians. That answers part (1). Okay, moving on to part (2): Find the maximum value of the area of triangle ABC.We know that the radius R of the circumcircle is 2. The area of a triangle can be given by several formulas. One formula is (1/2)ab sin C, where a and b are sides, and C is the included angle. Another formula is (abc)/(4R), where a, b, c are the sides, and R is the circumradius. Alternatively, using the sine formula: a = 2R sin A, b = 2R sin B, c = 2R sin C.Since we know R = 2, maybe we can express sides a, b, c in terms of angles A, B, C. Let me recall the Law of Sines: a/sin A = b/sin B = c/sin C = 2R = 4. So, a = 4 sin A, b = 4 sin B, c = 4 sin C.Given that angle C is 2π/3, so sin C = sin(2π/3) = sqrt(3)/2. Therefore, c = 4*(sqrt(3)/2) = 2 sqrt(3).So, side c is 2 sqrt(3). Now, to find the area, I can use the formula (1/2)ab sin C. So, if I can express a and b in terms of angles A and B, which are related because A + B = π - C = π - 2π/3 = π/3.So, A + B = π/3. Let me denote angle A as x, so angle B is π/3 - x. Then, sides a and b can be written as:a = 4 sin xb = 4 sin(π/3 - x)So, the area S is:S = (1/2) * a * b * sin CSubstituting the values:S = (1/2) * 4 sin x * 4 sin(π/3 - x) * (sqrt(3)/2)Simplify this:First, 1/2 * 4 * 4 = (1/2)*16 = 8Then, 8 * sin x * sin(π/3 - x) * (sqrt(3)/2) = 8 * (sqrt(3)/2) * sin x * sin(π/3 - x) = 4 sqrt(3) * sin x * sin(π/3 - x)So, S = 4 sqrt(3) * sin x * sin(π/3 - x)Now, I need to find the maximum value of S with respect to x, where x is between 0 and π/3 (since A and B are both positive angles adding up to π/3).Let me denote f(x) = sin x * sin(π/3 - x). I need to find the maximum of f(x) in the interval (0, π/3). Then, multiply by 4 sqrt(3) to get the maximum area.So, let's compute f(x):f(x) = sin x * sin(π/3 - x)I can use the identity sin A sin B = [cos(A - B) - cos(A + B)] / 2.So, f(x) = [cos(x - (π/3 - x)) - cos(x + (π/3 - x))]/2Simplify the arguments:x - (π/3 - x) = 2x - π/3x + (π/3 - x) = π/3So, f(x) = [cos(2x - π/3) - cos(π/3)] / 2We know that cos(π/3) = 1/2, so:f(x) = [cos(2x - π/3) - 1/2] / 2 = [cos(2x - π/3)]/2 - 1/4So, f(x) = (1/2) cos(2x - π/3) - 1/4Now, to find the maximum of f(x), we need to find the maximum of cos(2x - π/3). The maximum value of cosine is 1, so the maximum of f(x) is (1/2)(1) - 1/4 = 1/2 - 1/4 = 1/4.Therefore, the maximum value of f(x) is 1/4, achieved when cos(2x - π/3) = 1, which occurs when 2x - π/3 = 2π k, where k is integer. Since x is between 0 and π/3, let's solve for x:2x - π/3 = 0 => 2x = π/3 => x = π/6So, when x = π/6, f(x) is maximized. Therefore, the maximum area S is:S_max = 4 sqrt(3) * (1/4) = sqrt(3)Wait, that seems too straightforward. Let me verify.Alternatively, maybe I can use calculus to find the maximum. Let's take f(x) = sin x sin(π/3 - x). Let me compute its derivative.f'(x) = cos x sin(π/3 - x) + sin x (-cos(π/3 - x)) = cos x sin(π/3 - x) - sin x cos(π/3 - x)Using the sine subtraction formula: sin(A - B) = sin A cos B - cos A sin B. So, sin(π/3 - 2x) = sin π/3 cos 2x - cos π/3 sin 2x. Wait, maybe that's not directly applicable here.Wait, f'(x) can be written as sin(π/3 - x - x) = sin(π/3 - 2x). Because, using the identity sin(A - B) = sin A cos B - cos A sin B. So, if I factor out, f'(x) = sin(π/3 - 2x). Let me check:sin(π/3 - 2x) = sin π/3 cos 2x - cos π/3 sin 2xBut f'(x) is cos x sin(π/3 - x) - sin x cos(π/3 - x). Let me compute this:Let me denote A = π/3 - x, then f'(x) = cos x sin A - sin x cos A = sin(A - x) = sin(π/3 - x - x) = sin(π/3 - 2x). Yes, that's correct.So, f'(x) = sin(π/3 - 2x). To find critical points, set f'(x) = 0:sin(π/3 - 2x) = 0Which implies π/3 - 2x = nπ, where n is integer.So, 2x = π/3 - nπx = π/6 - (nπ)/2Now, x must be between 0 and π/3. Let's find n such that x is in this interval.For n = 0: x = π/6 ≈ 0.523, which is within (0, π/3 ≈ 1.047). So, x = π/6 is a critical point.For n = 1: x = π/6 - π/2 = -π/3, which is negative, so discard.For n = -1: x = π/6 + π/2 = 2π/3, which is greater than π/3, so discard.So, the only critical point in (0, π/3) is x = π/6.Now, let's test the endpoints and the critical point.At x = 0: f(0) = sin 0 * sin(π/3 - 0) = 0 * sin π/3 = 0At x = π/3: f(π/3) = sin(π/3) * sin(π/3 - π/3) = sin(π/3) * sin 0 = 0At x = π/6: f(π/6) = sin(π/6) * sin(π/3 - π/6) = sin(π/6) * sin(π/6) = (1/2)*(1/2) = 1/4So, indeed, the maximum occurs at x = π/6, and f(x) = 1/4. Therefore, the maximum area is 4 sqrt(3) * 1/4 = sqrt(3).Alternatively, maybe I can think about the triangle in another way. Since angle C is fixed at 2π/3, the area is maximized when sides a and b are such that their product is maximized, given the constraint from the Law of Cosines.Wait, let me recall that in a triangle with fixed angle C and fixed side c, the area is maximized when the triangle is isoceles with a = b. Is that the case here?Wait, in this problem, we don't have side c fixed, but we have R fixed. Hmm, but we can express sides a and b in terms of angles A and B, which sum to π/3.Wait, but earlier, I used the Law of Sines to express a and b in terms of angles, and then expressed the area in terms of x, found the maximum. So, that seems consistent.Alternatively, maybe using the formula for area in terms of sides and angles, and using the Law of Cosines.We have c = 2 sqrt(3), as found earlier. Then, by the Law of Cosines:c² = a² + b² - 2ab cos CSo, (2 sqrt(3))² = a² + b² - 2ab cos(2π/3)12 = a² + b² - 2ab (-1/2) [since cos(2π/3) = -1/2]So, 12 = a² + b² + abSo, we have a² + b² + ab = 12We need to maximize the area, which is (1/2)ab sin C = (1/2)ab*(sqrt(3)/2) = (sqrt(3)/4)abSo, to maximize the area, we need to maximize ab, given that a² + b² + ab = 12.So, let me denote ab = k. Then, a² + b² = 12 - kBut, we know that (a + b)² = a² + 2ab + b² = (12 - k) + 2k = 12 + kBut I don't know if that helps directly. Alternatively, using the AM-GM inequality.We have a² + b² + ab = 12We can write this as a² + ab + b² = 12Note that a² + ab + b² is equal to (a + b)² - abWait, (a + b)² = a² + 2ab + b², so a² + ab + b² = (a + b)² - abSo, (a + b)² - ab = 12But I don't know if that helps.Alternatively, perhaps express in terms of variables. Let me set a = b for maximum product, as in symmetric cases often give extrema.If a = b, then:a² + a² + a² = 3a² = 12 => a² = 4 => a = 2So, a = b = 2. Then, ab = 4. So, the area is (sqrt(3)/4)*4 = sqrt(3). So, that's the same result as before.But is this indeed the maximum? Let me check with another approach.Let me consider a² + b² + ab = 12. Let me treat this as a quadratic in terms of a, given b.So, a² + ab + (b² - 12) = 0For real solutions, discriminant must be non-negative:b² - 4*(1)*(b² - 12) ≥ 0b² - 4b² + 48 ≥ 0-3b² + 48 ≥ 0 => 3b² ≤ 48 => b² ≤ 16 => |b| ≤ 4Since b is positive, b ≤ 4.Similarly, same for a.But this might not directly help.Alternatively, perhaps using Lagrange multipliers. Let me set f(a, b) = ab, subject to the constraint g(a, b) = a² + b² + ab - 12 = 0.The gradient of f is (b, a), and the gradient of g is (2a + b, 2b + a). Setting gradient f = λ gradient g:b = λ(2a + b)a = λ(2b + a)From the first equation: b = λ(2a + b) => b = 2λ a + λ b => b - λ b = 2λ a => b(1 - λ) = 2λ a => a = [b(1 - λ)] / (2λ)From the second equation: a = λ(2b + a) => a = 2λ b + λ a => a - λ a = 2λ b => a(1 - λ) = 2λ b => b = [a(1 - λ)] / (2λ)So, from both equations, a = [b(1 - λ)] / (2λ) and b = [a(1 - λ)] / (2λ). Let me substitute b from the second equation into the first equation:a = [ ( [a(1 - λ)] / (2λ) ) (1 - λ) ] / (2λ )Simplify:a = [ a (1 - λ)^2 ] / (4λ² )Multiply both sides by 4λ²:4λ² a = a (1 - λ)^2Assuming a ≠ 0 (since a is a side length, it must be positive), we can divide both sides by a:4λ² = (1 - λ)^2Take square roots? Or expand:4λ² = 1 - 2λ + λ²Bring all terms to left:4λ² - λ² + 2λ - 1 = 0 => 3λ² + 2λ - 1 = 0Solve quadratic equation:λ = [ -2 ± sqrt(4 + 12) ] / (2*3) = [ -2 ± sqrt(16) ] / 6 = [ -2 ± 4 ] / 6So, two solutions:λ = ( -2 + 4 ) / 6 = 2/6 = 1/3λ = ( -2 - 4 ) / 6 = -6/6 = -1So, λ = 1/3 or λ = -1If λ = -1, then from the first equation:b = (-1)(2a + b) => b = -2a - b => 2b = -2a => b = -aBut a and b are positive lengths, so b = -a is impossible. So, discard λ = -1.Thus, λ = 1/3.Now, from the first equation:a = [b(1 - 1/3)] / (2*(1/3)) = [b*(2/3)] / (2/3) = bSo, a = b.Therefore, the maximum occurs when a = b. So, a = b.Thus, plugging back into the constraint:a² + a² + a² = 3a² = 12 => a² = 4 => a = 2So, a = b = 2, as before.Therefore, the maximum area is (sqrt(3)/4)*ab = (sqrt(3)/4)*4 = sqrt(3). So, same result.Therefore, regardless of the method, the maximum area is sqrt(3).So, summarizing:(1) Angle C is 2π/3 radians.(2) The maximum area of triangle ABC is sqrt(3).**Final Answer**(1) The measure of angle ( C ) is (boxed{dfrac{2pi}{3}}).(2) The maximum area of ( triangle ABC ) is (boxed{sqrt{3}}).