💡Okay, so I'm trying to figure out the average price per apple that John, Bill, and George charged when they sold apples in the morning and afternoon. The problem gives me some specific details about how each of them sold apples, and I need to calculate the average price for each boy. Let me break this down step by step.First, I know that in the morning, they sold apples at a rate of 2 apples for 5 cents. That means each apple was sold for 5/2 cents, which is 2.5 cents per apple. In the afternoon, they sold apples at a rate of 3 apples for 5 cents, so each apple was sold for 5/3 cents, approximately 1.6667 cents per apple.Now, let's look at each person individually.Starting with John: The problem states that John sold the same amount in the afternoon as he did in the morning. So, if he sold 'x' apples in the morning, he also sold 'x' apples in the afternoon. To find the average price per apple for John, I need to calculate the total revenue he made and divide it by the total number of apples he sold.In the morning, selling 'x' apples at 2.5 cents each would give him 2.5x cents. In the afternoon, selling the same 'x' apples at approximately 1.6667 cents each would give him 1.6667x cents. So, his total revenue is 2.5x + 1.6667x, which is 4.1667x cents. The total number of apples he sold is x + x = 2x. Therefore, the average price per apple for John is 4.1667x / 2x = 2.0833 cents per apple. Hmm, that seems a bit low. Let me check my calculations again.Wait, 2.5x + 1.6667x is actually 4.1667x, and dividing that by 2x gives 2.0833, which is 25/12 when expressed as a fraction. So, John's average price per apple is 25/12 cents, which is approximately 2.0833 cents.Moving on to Bill: The problem says Bill had the same revenue in the morning and afternoon. That means the amount of money he made in the morning is equal to the amount he made in the afternoon. Let's denote the number of apples he sold in the morning as 'a' and in the afternoon as 'b'. His revenue in the morning would be 2.5a cents, and in the afternoon, it would be 1.6667b cents. Since these revenues are equal, we have 2.5a = 1.6667b. To find the average price per apple for Bill, I need to find the total revenue divided by the total number of apples sold. The total revenue is 2.5a + 1.6667b, but since 2.5a = 1.6667b, the total revenue becomes 2.5a + 2.5a = 5a cents. The total number of apples sold is a + b. From the equation 2.5a = 1.6667b, I can solve for b in terms of a: b = (2.5 / 1.6667)a ≈ 1.5a. So, the total number of apples is a + 1.5a = 2.5a. Therefore, the average price per apple for Bill is 5a / 2.5a = 2 cents per apple.Lastly, George: The problem states that George sold apples in the morning and afternoon in the same ratio as 5/3 to 5/2. I need to interpret this. The ratio 5/3 to 5/2 can be simplified by dividing both parts by 5, resulting in 1/3 to 1/2. To make this ratio easier to work with, I can invert both parts to get 3:2. So, George sold apples in the morning and afternoon in a 3:2 ratio.Let's denote the number of apples he sold in the morning as 3k and in the afternoon as 2k, where 'k' is a constant. His total revenue in the morning would be 3k * 2.5 = 7.5k cents, and in the afternoon, it would be 2k * 1.6667 ≈ 3.3334k cents. The total revenue is 7.5k + 3.3334k ≈ 10.8334k cents. The total number of apples sold is 3k + 2k = 5k. Therefore, the average price per apple for George is 10.8334k / 5k ≈ 2.1667 cents per apple. Wait, that doesn't seem right because earlier calculations for Bill and John gave specific fractions. Let me re-examine.Actually, the ratio of apples sold should correspond to the inverse of the price ratio. Since the morning price is higher (2.5 cents) compared to the afternoon price (1.6667 cents), George sold more apples in the afternoon to balance the ratio. The ratio given is 5/3 to 5/2, which simplifies to 10/6 to 15/6, or 10:15, which reduces to 2:3. So, George sold apples in the morning and afternoon in a 2:3 ratio.Let me correct that. Let's denote the number of apples sold in the morning as 2k and in the afternoon as 3k. His total revenue in the morning is 2k * 2.5 = 5k cents, and in the afternoon, it's 3k * 1.6667 ≈ 5k cents. So, the total revenue is 5k + 5k = 10k cents, and the total number of apples is 2k + 3k = 5k. Therefore, the average price per apple for George is 10k / 5k = 2 cents per apple.Wait, that makes more sense. So, George's average price per apple is 2 cents.To summarize:- John's average price per apple is 25/12 cents.- Bill's average price per apple is 2 cents.- George's average price per apple is 2 cents.I think I've got it now.

💡Okay, so I have this probability problem here: Given two real numbers x and y randomly chosen from the interval [0,2], what's the probability that x² + y² ≤ 1? Hmm, let me try to figure this out step by step.First, I know that when dealing with probabilities involving two variables like x and y, especially when they're chosen from an interval, it often relates to geometric probability. That usually involves areas, right? So, maybe I need to think about the area where x² + y² ≤ 1 and compare it to the total area from which x and y are chosen.Alright, let's break it down. The numbers x and y are both chosen from [0,2]. So, if I imagine a coordinate system where x is on the horizontal axis and y is on the vertical axis, the possible values of x and y form a square. The square would have sides of length 2 because both x and y go from 0 to 2. So, the total area of this square is 2 times 2, which is 4. Got that part.Now, the condition given is x² + y² ≤ 1. Hmm, that looks familiar. Isn't that the equation of a circle? Yeah, x² + y² = 1 is a circle with radius 1 centered at the origin (0,0). But wait, in our case, x and y are both non-negative because they're chosen from [0,2]. So, we're only looking at the part of the circle that's in the first quadrant. That would be a quarter-circle, right?So, the area where x² + y² ≤ 1 is a quarter-circle with radius 1. The area of a full circle is πr², so the area of a quarter-circle would be (πr²)/4. Since the radius r is 1, the area is (π*1²)/4, which simplifies to π/4. Okay, so the area we're interested in is π/4.But wait, hold on. The quarter-circle is entirely within our square [0,2] x [0,2], right? Because the radius is 1, so it doesn't go beyond 1 in either the x or y direction. So, the area π/4 is entirely contained within the square of area 4. That means the probability we're looking for is the ratio of the area of the quarter-circle to the area of the square.So, probability P = (Area of quarter-circle) / (Area of square) = (π/4) / 4. Let me compute that. Dividing π/4 by 4 is the same as multiplying π/4 by 1/4, which gives π/16. So, the probability is π/16.Wait, let me make sure I didn't make a mistake. The total area is 4, correct. The quarter-circle area is π/4, correct. So, π/4 divided by 4 is indeed π/16. Yeah, that seems right.Is there another way to think about this? Maybe using integration? Let's see. The probability can be calculated as the double integral over the region where x² + y² ≤ 1, with x and y both ranging from 0 to 2. So, the integral would be ∫ (from x=0 to 1) ∫ (from y=0 to sqrt(1 - x²)) dy dx. Because beyond x=1, y would have to be imaginary, which isn't in our interval.Calculating that integral: The inner integral with respect to y is just sqrt(1 - x²). So, the integral becomes ∫ (from x=0 to 1) sqrt(1 - x²) dx. I remember that the integral of sqrt(1 - x²) dx from 0 to 1 is π/4. That's because it's the area of a quarter-circle, which we already established. So, that confirms the area is π/4.Therefore, the probability is π/4 divided by the total area 4, which is π/16. So, yeah, that seems consistent.I guess another way to think about it is that since both x and y are chosen uniformly from [0,2], the joint probability distribution is uniform over the square [0,2] x [0,2]. So, the probability is just the area of the region where x² + y² ≤ 1 divided by the area of the square.Just to visualize it, the square is from (0,0) to (2,2), and the quarter-circle is from (0,0) to (1,1). So, the quarter-circle is entirely within the square, and it's a much smaller area. That makes sense why the probability is π/16, which is approximately 0.196, or about 19.6%. That seems reasonable because the quarter-circle isn't taking up too much of the square.Wait, just to double-check, what's π/16 numerically? π is approximately 3.1416, so 3.1416 divided by 16 is roughly 0.19635. So, about 19.635%. That seems plausible because the area of the quarter-circle is about 0.7854, and the total area is 4, so 0.7854 / 4 is approximately 0.19635. Yep, that matches.Is there any chance I messed up the limits of integration? Let me think. If I set up the double integral for the region x² + y² ≤ 1 with x and y from 0 to 2, I have to make sure that I'm only integrating over the part where x and y are less than or equal to 1, right? Because beyond x=1 or y=1, the equation x² + y² ≤ 1 wouldn't hold.So, actually, the limits for x would be from 0 to 1, and for each x, y would go from 0 to sqrt(1 - x²). That makes sense. If I tried to integrate beyond x=1, y would have to be imaginary, which isn't in our interval [0,2], so we can't include that. So, the integral is correctly set up from 0 to 1 for x.Another thing to consider: is the quarter-circle entirely within the square? Yes, because the square goes up to 2, and the quarter-circle only goes up to 1 in both x and y. So, there's no overlap beyond that. So, the area π/4 is entirely within the square, so we don't have to worry about parts of the circle being outside the square or anything like that.So, putting it all together, the probability is π/16. I think that's the answer.**Final Answer**The probability is boxed{dfrac{pi}{16}}.

💡To determine the horizontal shift needed to transform the graph of ( y = sin(2x) ) into the graph of ( y = sin(2x + frac{pi}{3}) ), we analyze the function ( y = sin(2x + frac{pi}{3}) ).1. **Identify the phase shift**: The general form of a sine function is ( y = sin(Bx + C) ). The phase shift is given by ( -frac{C}{B} ).2. **Apply the phase shift formula**: For ( y = sin(2x + frac{pi}{3}) ), ( B = 2 ) and ( C = frac{pi}{3} ). Thus, the phase shift is ( -frac{pi/3}{2} = -frac{pi}{6} ).3. **Interpret the phase shift**: A negative phase shift indicates a shift to the left. Therefore, shifting the graph of ( y = sin(2x) ) to the left by ( frac{pi}{6} ) units will result in the graph of ( y = sin(2x + frac{pi}{3}) ).**Conclusion**: The correct answer is (boxed{text{B}}).

💡First, I need to determine the volume of the aquarium by multiplying its length, width, and height. The dimensions are given in different units, so I'll convert them all to meters for consistency.Next, I'll calculate the volume using the formula for the volume of a rectangular parallelepiped.After finding the volume, I'll convert the filling rate from liters per minute to cubic meters per second to ensure the units are consistent.Finally, I'll divide the total volume of the aquarium by the filling rate to determine the time required to fill the aquarium completely.

💡First, calculate the total length of rope needed by multiplying the number of stories by the height of each story.Next, determine the total length of the ropes purchased by multiplying the number of pieces by the length of each piece.Then, find the difference between the total length purchased and the required length to determine the amount of rope lost during lashing.Finally, calculate the percentage of rope lost by dividing the lost length by the total purchased length and multiplying by 100%.

💡Okay, so I have this problem where I need to find the maximal value of the expression:[frac{1}{a^2 - 4a + 9} + frac{1}{b^2 - 4b + 9} + frac{1}{c^2 - 4c + 9}]given that ( a, b, c ) are nonnegative real numbers and ( a + b + c = 1 ).Hmm, let me start by understanding each term in the expression. Each term is of the form ( frac{1}{x^2 - 4x + 9} ). Maybe I can simplify this or find a way to bound it.First, let me look at the denominator: ( x^2 - 4x + 9 ). I can complete the square here to make it easier to analyze.Completing the square for ( x^2 - 4x + 9 ):[x^2 - 4x + 9 = (x^2 - 4x + 4) + 5 = (x - 2)^2 + 5]So, each denominator can be written as ( (x - 2)^2 + 5 ). That means each term in the expression is:[frac{1}{(x - 2)^2 + 5}]Since ( (x - 2)^2 ) is always non-negative, the smallest value the denominator can take is 5, which happens when ( x = 2 ). Therefore, the maximum value of each term ( frac{1}{(x - 2)^2 + 5} ) is ( frac{1}{5} ).But wait, ( a, b, c ) are nonnegative and their sum is 1. So, each of ( a, b, c ) is between 0 and 1. That means ( x ) in each term is between 0 and 1. So, ( x ) can't be 2, which is outside the range. Therefore, the minimum value of the denominator ( (x - 2)^2 + 5 ) in the given range occurs at ( x = 1 ), since 1 is the closest to 2.Let me calculate the denominator at ( x = 1 ):[(1 - 2)^2 + 5 = 1 + 5 = 6]So, the maximum value of each term ( frac{1}{(x - 2)^2 + 5} ) when ( x ) is in [0,1] is ( frac{1}{6} ), occurring at ( x = 1 ).Similarly, at ( x = 0 ):[(0 - 2)^2 + 5 = 4 + 5 = 9]So, the term is ( frac{1}{9} ) when ( x = 0 ).Therefore, each term in the expression is between ( frac{1}{9} ) and ( frac{1}{6} ).Now, since we have three such terms, the maximum value of the entire expression would occur when each term is as large as possible. But since ( a + b + c = 1 ), we can't have all three variables equal to 1. In fact, only one of them can be 1, and the others have to be 0.Let me test this case. Let’s set ( a = 1 ), ( b = 0 ), ( c = 0 ).Then, the expression becomes:[frac{1}{1^2 - 4(1) + 9} + frac{1}{0^2 - 4(0) + 9} + frac{1}{0^2 - 4(0) + 9} = frac{1}{1 - 4 + 9} + frac{1}{0 - 0 + 9} + frac{1}{0 - 0 + 9} = frac{1}{6} + frac{1}{9} + frac{1}{9}]Calculating this:[frac{1}{6} + frac{1}{9} + frac{1}{9} = frac{3}{18} + frac{2}{18} + frac{2}{18} = frac{7}{18}]So, the expression equals ( frac{7}{18} ) when ( a = 1 ), ( b = c = 0 ).Is this the maximum? Let me check another case where two variables are 0.5 and the third is 0, but wait, ( a + b + c = 1 ), so if two variables are 0.5, the third would have to be 0. But 0.5 + 0.5 + 0 = 1.Let me compute the expression for ( a = 0.5 ), ( b = 0.5 ), ( c = 0 ):[frac{1}{(0.5)^2 - 4(0.5) + 9} + frac{1}{(0.5)^2 - 4(0.5) + 9} + frac{1}{0^2 - 4(0) + 9}]Calculating each term:First term:[(0.5)^2 - 4(0.5) + 9 = 0.25 - 2 + 9 = 7.25]So, ( frac{1}{7.25} approx 0.1379 )Second term is the same as the first: ( approx 0.1379 )Third term:[0^2 - 4(0) + 9 = 9]So, ( frac{1}{9} approx 0.1111 )Adding them up:[0.1379 + 0.1379 + 0.1111 approx 0.3869]Now, ( frac{7}{18} approx 0.3889 ), which is slightly larger than 0.3869. So, this case gives a slightly smaller value.What if I set ( a = 1 ), ( b = 0 ), ( c = 0 ), as before, giving ( frac{7}{18} approx 0.3889 ).Another case: Let’s try ( a = 0.8 ), ( b = 0.2 ), ( c = 0 ).Compute each term:First term:[0.8^2 - 4(0.8) + 9 = 0.64 - 3.2 + 9 = 6.44]So, ( frac{1}{6.44} approx 0.1553 )Second term:[0.2^2 - 4(0.2) + 9 = 0.04 - 0.8 + 9 = 8.24]So, ( frac{1}{8.24} approx 0.1214 )Third term:[0^2 - 4(0) + 9 = 9]So, ( frac{1}{9} approx 0.1111 )Adding them up:[0.1553 + 0.1214 + 0.1111 approx 0.3878]Again, this is slightly less than ( frac{7}{18} approx 0.3889 ).Hmm, so far, the maximum seems to occur when one variable is 1 and the others are 0.Let me try another case where one variable is 0.9, another is 0.1, and the third is 0.Compute each term:First term:[0.9^2 - 4(0.9) + 9 = 0.81 - 3.6 + 9 = 6.21]So, ( frac{1}{6.21} approx 0.1610 )Second term:[0.1^2 - 4(0.1) + 9 = 0.01 - 0.4 + 9 = 8.61]So, ( frac{1}{8.61} approx 0.1162 )Third term:[0^2 - 4(0) + 9 = 9]So, ( frac{1}{9} approx 0.1111 )Adding them up:[0.1610 + 0.1162 + 0.1111 approx 0.3883]Still, this is slightly less than ( frac{7}{18} approx 0.3889 ).Wait, so it seems that as I increase one variable towards 1 and decrease the others towards 0, the expression approaches ( frac{7}{18} ).Let me try ( a = 1 ), ( b = 0 ), ( c = 0 ):As before, the expression is ( frac{1}{6} + frac{1}{9} + frac{1}{9} = frac{7}{18} approx 0.3889 ).Is this the maximum? Let me see if I can find a case where the expression is larger.Suppose I set ( a = 1 ), ( b = 0 ), ( c = 0 ), as before, giving ( frac{7}{18} ).Alternatively, suppose I set ( a = 0.6 ), ( b = 0.4 ), ( c = 0 ).Compute each term:First term:[0.6^2 - 4(0.6) + 9 = 0.36 - 2.4 + 9 = 6.96]So, ( frac{1}{6.96} approx 0.1437 )Second term:[0.4^2 - 4(0.4) + 9 = 0.16 - 1.6 + 9 = 7.56]So, ( frac{1}{7.56} approx 0.1323 )Third term:[0^2 - 4(0) + 9 = 9]So, ( frac{1}{9} approx 0.1111 )Adding them up:[0.1437 + 0.1323 + 0.1111 approx 0.3871]Again, this is less than ( frac{7}{18} ).Wait a minute, maybe I should try to see if the function ( f(x) = frac{1}{x^2 - 4x + 9} ) is convex or concave in the interval [0,1]. If it's convex, then by Jensen's inequality, the maximum would occur at the endpoints, which are 0 and 1.Let me compute the second derivative of ( f(x) ) to check for convexity.First, ( f(x) = frac{1}{x^2 - 4x + 9} ).First derivative:[f'(x) = frac{ - (2x - 4) }{(x^2 - 4x + 9)^2}]Second derivative:Let me compute it step by step.Let ( u = x^2 - 4x + 9 ), so ( f(x) = frac{1}{u} ).First derivative: ( f'(x) = - frac{u'}{u^2} ), where ( u' = 2x - 4 ).Second derivative:Using the quotient rule:[f''(x) = frac{ - [ (2x - 4)' cdot u^2 - u' cdot 2u cdot u' ] }{u^4}]Wait, that might be a bit messy. Alternatively, using the chain rule:( f'(x) = - (2x - 4) u^{-2} )So, ( f''(x) = - [ 2 cdot u^{-2} + (2x - 4)(-2) u^{-3} cdot u' ] )Wait, perhaps it's better to use the product rule.Let me write ( f'(x) = - (2x - 4) cdot u^{-2} ).Then, ( f''(x) = - [ 2 cdot u^{-2} + (2x - 4) cdot (-2) u^{-3} cdot u' ] )Compute ( u' = 2x - 4 ).So,[f''(x) = - [ 2 u^{-2} - 2(2x - 4)^2 u^{-3} ]]Factor out ( 2 u^{-3} ):[f''(x) = -2 u^{-3} [ u - (2x - 4)^2 ]]Now, ( u = x^2 - 4x + 9 ).So,[f''(x) = -2 (x^2 - 4x + 9)^{-3} [ (x^2 - 4x + 9) - (2x - 4)^2 ]]Compute ( (2x - 4)^2 = 4x^2 - 16x + 16 ).So,[(x^2 - 4x + 9) - (4x^2 - 16x + 16) = x^2 - 4x + 9 - 4x^2 + 16x - 16 = -3x^2 + 12x - 7]Therefore,[f''(x) = -2 (x^2 - 4x + 9)^{-3} ( -3x^2 + 12x - 7 )]Simplify:[f''(x) = 2 (x^2 - 4x + 9)^{-3} (3x^2 - 12x + 7)]Now, let's analyze the sign of ( f''(x) ) in the interval [0,1].The term ( (x^2 - 4x + 9)^{-3} ) is always positive because ( x^2 - 4x + 9 = (x - 2)^2 + 5 ) is always positive.So, the sign of ( f''(x) ) depends on ( 3x^2 - 12x + 7 ).Let me find the roots of ( 3x^2 - 12x + 7 = 0 ):Using quadratic formula:[x = frac{12 pm sqrt{144 - 84}}{6} = frac{12 pm sqrt{60}}{6} = frac{12 pm 2sqrt{15}}{6} = 2 pm frac{sqrt{15}}{3}]Approximately, ( sqrt{15} approx 3.87298 ), so:[x approx 2 pm 1.29099]So, the roots are approximately ( x approx 3.29099 ) and ( x approx 0.70901 ).Therefore, in the interval [0,1], the quadratic ( 3x^2 - 12x + 7 ) is positive when ( x < 0.70901 ) and negative when ( x > 0.70901 ).Since 0.70901 is approximately 0.71, which is within [0,1], we can say that:- For ( x in [0, 0.71) ), ( f''(x) > 0 ) (convex)- For ( x in (0.71, 1] ), ( f''(x) < 0 ) (concave)Therefore, the function ( f(x) ) is convex on [0, 0.71) and concave on (0.71, 1].This means that Jensen's inequality can't be directly applied over the entire interval [0,1], but we can consider the behavior in each subinterval.However, since the function changes concavity, it complicates the use of Jensen's inequality. Maybe instead, I can consider the extremal values.Given that ( f(x) ) is convex on [0, 0.71) and concave on (0.71, 1], the maximum of the sum might occur at the endpoints, i.e., when variables are at 0 or 1.But let's think about it: if one variable is 1, the others are 0, which gives us the expression value ( frac{7}{18} ).If I try to set two variables to 0.71 and the third to ( 1 - 2 times 0.71 = 1 - 1.42 = -0.42 ), which is negative, but variables can't be negative. So, that's not feasible.Alternatively, maybe setting two variables to 0.71 isn't possible because their sum would exceed 1.Wait, 0.71 + 0.71 = 1.42 > 1, so we can't have two variables at 0.71.Therefore, perhaps the maximum occurs when one variable is as large as possible, i.e., 1, and the others are 0.Alternatively, maybe distributing the values around 0.71 could give a higher sum, but since the function is concave beyond 0.71, it might not.Wait, let me test ( a = 0.7 ), ( b = 0.3 ), ( c = 0 ).Compute each term:First term:[0.7^2 - 4(0.7) + 9 = 0.49 - 2.8 + 9 = 6.69]So, ( frac{1}{6.69} approx 0.1495 )Second term:[0.3^2 - 4(0.3) + 9 = 0.09 - 1.2 + 9 = 7.89]So, ( frac{1}{7.89} approx 0.1267 )Third term:[0^2 - 4(0) + 9 = 9]So, ( frac{1}{9} approx 0.1111 )Adding them up:[0.1495 + 0.1267 + 0.1111 approx 0.3873]Still less than ( frac{7}{18} approx 0.3889 ).Wait, maybe if I set ( a = 0.75 ), ( b = 0.25 ), ( c = 0 ).Compute each term:First term:[0.75^2 - 4(0.75) + 9 = 0.5625 - 3 + 9 = 6.5625]So, ( frac{1}{6.5625} approx 0.1525 )Second term:[0.25^2 - 4(0.25) + 9 = 0.0625 - 1 + 9 = 8.0625]So, ( frac{1}{8.0625} approx 0.1240 )Third term:[0^2 - 4(0) + 9 = 9]So, ( frac{1}{9} approx 0.1111 )Adding them up:[0.1525 + 0.1240 + 0.1111 approx 0.3876]Still less than ( frac{7}{18} ).Hmm, it seems that as I increase one variable towards 1, the sum approaches ( frac{7}{18} ).Let me try ( a = 0.95 ), ( b = 0.05 ), ( c = 0 ).First term:[0.95^2 - 4(0.95) + 9 = 0.9025 - 3.8 + 9 = 6.1025]So, ( frac{1}{6.1025} approx 0.1639 )Second term:[0.05^2 - 4(0.05) + 9 = 0.0025 - 0.2 + 9 = 8.8025]So, ( frac{1}{8.8025} approx 0.1136 )Third term:[0^2 - 4(0) + 9 = 9]So, ( frac{1}{9} approx 0.1111 )Adding them up:[0.1639 + 0.1136 + 0.1111 approx 0.3886]This is very close to ( frac{7}{18} approx 0.3889 ).So, as ( a ) approaches 1, the sum approaches ( frac{7}{18} ).Therefore, it seems that the maximum value of the expression is indeed ( frac{7}{18} ), achieved when one variable is 1 and the others are 0.To confirm, let me check another case where ( a = 1 ), ( b = 0 ), ( c = 0 ):[frac{1}{1^2 - 4(1) + 9} + frac{1}{0^2 - 4(0) + 9} + frac{1}{0^2 - 4(0) + 9} = frac{1}{6} + frac{1}{9} + frac{1}{9} = frac{3}{18} + frac{2}{18} + frac{2}{18} = frac{7}{18}]Yes, that's correct.Therefore, the maximal value of the given expression is ( frac{7}{18} ), achieved when one of the variables is 1 and the others are 0.