💡First, identify the initial number of bales in the barn, which is 47.Next, determine the total number of bales in the barn after Benny added more, which is 82.To find out how many bales Benny added, subtract the initial number of bales from the total number after adding:82 bales (total after adding) - 47 bales (initial amount) = 35 balesTherefore, Benny stored 35 bales of hay in the barn.

💡Alright, so I've got this problem here about a kindergarten teacher distributing apples, pears, and oranges to a group of children. The teacher evenly distributes 270 apples, 180 pears, and 235 oranges. After distribution, the remaining quantities of these fruits are in the ratio 3:2:1. The question is asking how many children are in the larger group.Hmm, okay. Let me try to break this down step by step. First, I need to understand what it means for the fruits to be distributed evenly. That probably means each child gets the same number of apples, pears, and oranges, right? But since the quantities are different, the number each child gets might be different for each fruit.Wait, no, actually, the teacher is distributing each type of fruit evenly, so each child gets the same number of apples, the same number of pears, and the same number of oranges. But since there are different total numbers of each fruit, the number each child gets might vary per fruit.But actually, no, the teacher is distributing all the fruits together to the group. So, the total number of fruits is 270 + 180 + 235, which is... let me calculate that. 270 + 180 is 450, plus 235 is 685. So, 685 fruits in total. But the problem is about the remaining quantities after distribution, not the total distributed.So, the key here is that after distributing the fruits evenly, the remaining apples, pears, and oranges are in the ratio 3:2:1. That means if I denote the remaining apples as 3x, pears as 2x, and oranges as x, for some positive integer x.So, let's denote the number of children as n. Then, the number of apples each child gets would be 270 divided by n, but since we can't have a fraction of an apple, it's actually the floor of 270 divided by n. Similarly, for pears, it's the floor of 180 divided by n, and for oranges, it's the floor of 235 divided by n.Wait, but the problem says the teacher distributed them evenly, so maybe it's not just the floor, but actually, the number of fruits each child gets is an integer, so n must divide each of 270, 180, and 235 exactly? Hmm, but 270, 180, and 235 might not all be divisible by the same number. Let me check.First, let's find the greatest common divisor (GCD) of these numbers. The GCD of 270 and 180 is 90, because 270 divided by 90 is 3, and 180 divided by 90 is 2. But 235 divided by 90 is approximately 2.61, which isn't an integer. So, 90 isn't a common divisor of all three numbers.Let me find the GCD of 270 and 235 first. 270 divided by 5 is 54, and 235 divided by 5 is 47. So, 5 is a common divisor. Now, 54 and 47 are co-prime, meaning their GCD is 1. So, the GCD of 270 and 235 is 5. Now, let's see if 5 is a divisor of 180. 180 divided by 5 is 36, which is an integer. So, the GCD of all three numbers is 5.Wait, but if n is the number of children, and n divides 270, 180, and 235, then n must be a divisor of their GCD, which is 5. So, the possible values of n are 1 and 5. But distributing 270 apples to 1 child doesn't make much sense in a kindergarten setting, so n is likely 5.But wait, hold on. Let me think again. The problem says the remaining quantities are in the ratio 3:2:1. So, maybe n doesn't have to divide all three numbers exactly, but when you divide each total by n, the remainders are in the ratio 3:2:1.Ah, that's a different approach. So, instead of n dividing each total exactly, the remainders after division are in the ratio 3:2:1. That makes more sense because if n didn't divide them exactly, there would be some leftovers, and those leftovers are in a specific ratio.So, let's denote the number of children as n. Then, the number of apples each child gets is floor(270/n), and the remainder is 270 - n * floor(270/n). Similarly, for pears, it's 180 - n * floor(180/n), and for oranges, it's 235 - n * floor(235/n).Let me denote the remainders as A, P, and T for apples, pears, and oranges respectively. So, A = 270 - n * floor(270/n), P = 180 - n * floor(180/n), and T = 235 - n * floor(235/n). According to the problem, A:P:T = 3:2:1.So, A = 3k, P = 2k, T = k, for some positive integer k.Therefore, we have:270 - n * floor(270/n) = 3k,180 - n * floor(180/n) = 2k,235 - n * floor(235/n) = k.So, now we have three equations:1) 270 - n * floor(270/n) = 3k,2) 180 - n * floor(180/n) = 2k,3) 235 - n * floor(235/n) = k.Our goal is to find n such that these equations hold for some integer k.Hmm, this seems a bit complex, but maybe we can find n by considering the possible values of k and n.Alternatively, since T = k, and T = 235 - n * floor(235/n), which is the remainder when 235 is divided by n. Similarly, A = 3k, which is the remainder when 270 is divided by n, and P = 2k, which is the remainder when 180 is divided by n.So, the remainders when 270, 180, and 235 are divided by n are 3k, 2k, and k respectively.Since remainders must be less than n, we have:3k < n,2k < n,k < n.So, k must be less than n/3.Also, since k is a positive integer, k >= 1.So, let's think about possible values of n.Given that the remainders are 3k, 2k, and k, and they must be less than n, we can say that 3k < n, so k < n/3.Also, since 235 divided by n leaves a remainder of k, k must be less than n and also less than 235.Similarly, 270 divided by n leaves a remainder of 3k, so 3k < n, which again implies k < n/3.So, our approach could be:1. Find n such that when 270, 180, and 235 are divided by n, the remainders are in the ratio 3:2:1.2. So, let's denote: - 270 = a * n + 3k, - 180 = b * n + 2k, - 235 = c * n + k,where a, b, c are integers representing the number of times n fits into each total.So, from these equations, we can express k in terms of n:From the third equation: k = 235 - c * n.Plugging this into the second equation: 180 = b * n + 2*(235 - c * n) = b * n + 470 - 2c * n.Similarly, plugging into the first equation: 270 = a * n + 3*(235 - c * n) = a * n + 705 - 3c * n.So, let's rearrange these equations:From the second equation:180 = b * n + 470 - 2c * n=> 180 - 470 = (b - 2c) * n=> -290 = (b - 2c) * nSimilarly, from the first equation:270 = a * n + 705 - 3c * n=> 270 - 705 = (a - 3c) * n=> -435 = (a - 3c) * nSo, now we have:-290 = (b - 2c) * n,-435 = (a - 3c) * n.Since n is positive, the left-hand sides are negative, so (b - 2c) and (a - 3c) must be negative integers.Let me denote:Let’s write:(b - 2c) = -290 / n,(a - 3c) = -435 / n.But since b, c, a are integers, (b - 2c) and (a - 3c) must also be integers. Therefore, n must be a divisor of both 290 and 435.So, n must be a common divisor of 290 and 435.Let me find the GCD of 290 and 435.First, factorize 290:290 = 2 * 5 * 29.Factorize 435:435 = 5 * 87 = 5 * 3 * 29.So, the common prime factors are 5 and 29.Therefore, the GCD is 5 * 29 = 145.So, the common divisors of 290 and 435 are the divisors of 145, which are 1, 5, 29, and 145.Therefore, possible values for n are 1, 5, 29, 145.Now, let's test these possible values of n.First, n=1:If n=1, then each child gets all the fruits, so the remainders would be 0 for all fruits. But the ratio 0:0:0 isn't 3:2:1, so n=1 is invalid.Next, n=5:Let's compute the remainders:270 divided by 5: 5*54=270, so remainder 0.180 divided by 5: 5*36=180, remainder 0.235 divided by 5: 5*47=235, remainder 0.Again, the remainders are 0:0:0, which doesn't fit the ratio 3:2:1. So, n=5 is invalid.Next, n=29:Let's compute the remainders:270 divided by 29:29*9=261, so remainder 270-261=9.180 divided by 29:29*6=174, so remainder 180-174=6.235 divided by 29:29*8=232, so remainder 235-232=3.So, the remainders are 9, 6, and 3.Now, let's check the ratio: 9:6:3 simplifies to 3:2:1, which matches the given ratio.Great! So, n=29 works.Just to be thorough, let's check n=145:270 divided by 145:145*1=145, remainder 270-145=125.180 divided by 145:145*1=145, remainder 180-145=35.235 divided by 145:145*1=145, remainder 235-145=90.So, the remainders are 125, 35, 90.Now, let's see if these are in the ratio 3:2:1.Divide each by 5: 25, 7, 18. Not in the ratio 3:2:1.Alternatively, see if 125:35:90 simplifies to 3:2:1.Divide each by 35: 125/35 ≈ 3.57, 35/35=1, 90/35≈2.57. Not exactly 3:2:1.Alternatively, check if 125:35:90 can be scaled to 3:2:1.Let's see, 3x=125 => x≈41.67,2x=35 => x=17.5,x=90 => x=90.These are inconsistent, so n=145 doesn't work.Therefore, the only valid value is n=29.

💡Okay, so I have this problem where I need to find the largest possible volume of a triangular box that's cut from an equilateral triangle with a side length of 30 cm. Hmm, let me try to visualize this. I think it's like taking an equilateral triangle and then cutting out smaller triangles from each corner to form a box, right? So, the original triangle is 30 cm on each side, and by cutting out smaller triangles, we create a three-dimensional box with a triangular base.First, I need to figure out how the size of the cut-out triangles affects the volume of the box. Let me denote the side length of the base of the box as 'a' and the height of the box as 'h'. Since we're cutting out triangles from each corner, the height 'h' will be related to how much we cut out from each side of the original triangle.Wait, the original triangle is equilateral, so all sides are 30 cm. If I cut out a small triangle from each corner, the remaining length on each side of the original triangle will be the side length 'a' of the base of the box. So, the relationship between 'a' and the amount we cut out should be such that the total length of the original side is equal to 'a' plus twice the height 'h' times the square root of 3. Hmm, why square root of 3? Because in an equilateral triangle, the height is (sqrt(3)/2) times the side length. So, maybe each cut-out triangle has a height of 'h', and the base of each cut-out triangle would be related to 'h' by the same ratio.Let me write that down:Original side length = 30 cm = a + 2 * (base of cut-out triangle)But the base of the cut-out triangle can be expressed in terms of 'h'. Since the height of the cut-out triangle is 'h', and it's an equilateral triangle, the base of the cut-out triangle would be (2h)/sqrt(3). Wait, no, actually, in an equilateral triangle, the height is (sqrt(3)/2) * side length. So, if the height is 'h', then the side length of the cut-out triangle would be (2h)/sqrt(3). Therefore, the base of each cut-out triangle is (2h)/sqrt(3).So, the original side length is equal to the side length of the base of the box plus twice the base of the cut-out triangle:30 = a + 2 * (2h)/sqrt(3)Simplifying that:30 = a + (4h)/sqrt(3)Hmm, okay, so that's one equation relating 'a' and 'h'. Now, I need another equation to relate these variables, which would come from the volume of the box.The volume of the box is the area of the base times the height. The base is an equilateral triangle with side length 'a', so its area is (sqrt(3)/4) * a^2. Therefore, the volume V is:V = (sqrt(3)/4) * a^2 * hSo, V = (sqrt(3)/4) * a^2 * hNow, I have two equations:1) 30 = a + (4h)/sqrt(3)2) V = (sqrt(3)/4) * a^2 * hI need to express V in terms of a single variable and then find its maximum. Let me solve equation 1 for 'a' in terms of 'h' or vice versa.From equation 1:a = 30 - (4h)/sqrt(3)Let me substitute this into equation 2:V = (sqrt(3)/4) * (30 - (4h)/sqrt(3))^2 * hThis looks a bit complicated, but I can expand it step by step.First, let me simplify (30 - (4h)/sqrt(3))^2:= 30^2 - 2 * 30 * (4h)/sqrt(3) + (4h/sqrt(3))^2= 900 - (240h)/sqrt(3) + (16h^2)/3Now, plug this back into V:V = (sqrt(3)/4) * [900 - (240h)/sqrt(3) + (16h^2)/3] * hLet me distribute the (sqrt(3)/4) and the 'h':First term: (sqrt(3)/4) * 900 * h = (sqrt(3)/4) * 900h = (900 sqrt(3)/4) hSecond term: (sqrt(3)/4) * (-240h)/sqrt(3) * h = (sqrt(3)/4) * (-240h^2)/sqrt(3) = (-240/4) h^2 = -60 h^2Third term: (sqrt(3)/4) * (16h^2)/3 * h = (sqrt(3)/4) * (16h^3)/3 = (16 sqrt(3)/12) h^3 = (4 sqrt(3)/3) h^3So, putting it all together:V = (900 sqrt(3)/4) h - 60 h^2 + (4 sqrt(3)/3) h^3Hmm, this seems a bit messy. Maybe I made a mistake in the expansion. Let me double-check.Wait, when I expanded (30 - (4h)/sqrt(3))^2, I got:= 900 - (240h)/sqrt(3) + (16h^2)/3That seems correct.Then, multiplying by (sqrt(3)/4) and 'h':First term: (sqrt(3)/4) * 900 * h = (900 sqrt(3)/4) hSecond term: (sqrt(3)/4) * (-240h)/sqrt(3) * h = (sqrt(3)/4) * (-240h^2)/sqrt(3) = (-240/4) h^2 = -60 h^2Third term: (sqrt(3)/4) * (16h^2)/3 * h = (sqrt(3)/4) * (16h^3)/3 = (16 sqrt(3)/12) h^3 = (4 sqrt(3)/3) h^3Yes, that seems correct.So, V = (900 sqrt(3)/4) h - 60 h^2 + (4 sqrt(3)/3) h^3This is a cubic equation in terms of 'h'. To find the maximum volume, I need to take the derivative of V with respect to 'h' and set it equal to zero.Let me compute dV/dh:dV/dh = (900 sqrt(3)/4) - 120 h + (4 sqrt(3)/3) * 3 h^2= (900 sqrt(3)/4) - 120 h + 4 sqrt(3) h^2Wait, let me check the derivative term by term:First term: d/dh [ (900 sqrt(3)/4) h ] = 900 sqrt(3)/4Second term: d/dh [ -60 h^2 ] = -120 hThird term: d/dh [ (4 sqrt(3)/3) h^3 ] = (4 sqrt(3)/3) * 3 h^2 = 4 sqrt(3) h^2So, yes, dV/dh = (900 sqrt(3)/4) - 120 h + 4 sqrt(3) h^2Now, set dV/dh = 0:(900 sqrt(3)/4) - 120 h + 4 sqrt(3) h^2 = 0This is a quadratic equation in terms of 'h'. Let me write it as:4 sqrt(3) h^2 - 120 h + (900 sqrt(3)/4) = 0To make it easier, let's multiply both sides by 4 to eliminate the denominator:4 * 4 sqrt(3) h^2 - 4 * 120 h + 4 * (900 sqrt(3)/4) = 016 sqrt(3) h^2 - 480 h + 900 sqrt(3) = 0Now, this is a quadratic in 'h':16 sqrt(3) h^2 - 480 h + 900 sqrt(3) = 0Let me try to simplify this equation by dividing all terms by sqrt(3):16 h^2 - (480 / sqrt(3)) h + 900 = 0But 480 / sqrt(3) can be rationalized:480 / sqrt(3) = (480 sqrt(3)) / 3 = 160 sqrt(3)So, the equation becomes:16 h^2 - 160 sqrt(3) h + 900 = 0Now, let's divide all terms by 4 to simplify:4 h^2 - 40 sqrt(3) h + 225 = 0Now, we have a quadratic equation:4 h^2 - 40 sqrt(3) h + 225 = 0Let me use the quadratic formula to solve for 'h':h = [40 sqrt(3) ± sqrt( (40 sqrt(3))^2 - 4 * 4 * 225 ) ] / (2 * 4)First, compute the discriminant:D = (40 sqrt(3))^2 - 4 * 4 * 225= 1600 * 3 - 16 * 225= 4800 - 3600= 1200So, sqrt(D) = sqrt(1200) = sqrt(400 * 3) = 20 sqrt(3)Now, plug back into the formula:h = [40 sqrt(3) ± 20 sqrt(3)] / 8So, two solutions:h = (40 sqrt(3) + 20 sqrt(3)) / 8 = (60 sqrt(3)) / 8 = (15 sqrt(3)) / 2 ≈ 12.99 cmh = (40 sqrt(3) - 20 sqrt(3)) / 8 = (20 sqrt(3)) / 8 = (5 sqrt(3)) / 2 ≈ 4.33 cmNow, we need to determine which of these solutions gives the maximum volume. Since the volume is a cubic function, it can have a maximum and a minimum. Given the physical constraints, 'h' must be positive and less than the height of the original triangle.The height of the original equilateral triangle is (sqrt(3)/2) * 30 ≈ 25.98 cm. So, both solutions are positive and less than 25.98 cm, so both are feasible.However, we need to check which one gives the maximum volume. Let's compute the second derivative to check concavity.First, let's find the second derivative d²V/dh²:From dV/dh = (900 sqrt(3)/4) - 120 h + 4 sqrt(3) h^2d²V/dh² = -120 + 8 sqrt(3) hAt h = (15 sqrt(3))/2 ≈ 12.99 cm:d²V/dh² = -120 + 8 sqrt(3) * (15 sqrt(3))/2= -120 + 8 * (15 * 3)/2= -120 + 8 * 45 / 2= -120 + 180= 60 > 0Since the second derivative is positive, this point is a local minimum.At h = (5 sqrt(3))/2 ≈ 4.33 cm:d²V/dh² = -120 + 8 sqrt(3) * (5 sqrt(3))/2= -120 + 8 * (5 * 3)/2= -120 + 8 * 15 / 2= -120 + 60= -60 < 0Since the second derivative is negative, this point is a local maximum.Therefore, the maximum volume occurs at h = (5 sqrt(3))/2 cm.Now, let's find the corresponding 'a' using the earlier equation:a = 30 - (4h)/sqrt(3)Plugging in h = (5 sqrt(3))/2:a = 30 - (4 * (5 sqrt(3))/2)/sqrt(3)= 30 - (20 sqrt(3)/2)/sqrt(3)= 30 - (10 sqrt(3))/sqrt(3)= 30 - 10= 20 cmSo, the side length of the base of the box is 20 cm, and the height is (5 sqrt(3))/2 cm.Now, let's compute the volume:V = (sqrt(3)/4) * a^2 * h= (sqrt(3)/4) * (20)^2 * (5 sqrt(3))/2= (sqrt(3)/4) * 400 * (5 sqrt(3))/2= (sqrt(3) * 400 * 5 sqrt(3)) / 8= (400 * 5 * 3) / 8= (6000) / 8= 750 cm³Wait, that doesn't match the initial thought of 500 cm³. Did I make a mistake somewhere?Let me double-check the calculations.First, a = 20 cm, h = (5 sqrt(3))/2 cm.Volume V = (sqrt(3)/4) * a^2 * h= (sqrt(3)/4) * 400 * (5 sqrt(3))/2= (sqrt(3) * 400 * 5 sqrt(3)) / 8= (400 * 5 * 3) / 8= 6000 / 8= 750 cm³Hmm, so according to this, the volume is 750 cm³. But earlier, I thought it was 500 cm³. Maybe I made a mistake in the earlier steps.Wait, let's go back to the expression for V:V = (sqrt(3)/4) * a^2 * hBut earlier, I had:V = (sqrt(3)/4) * a^2 * h = (sqrt(3)/4) * (30 - (4h)/sqrt(3))^2 * hBut when I expanded that, I might have made a mistake. Let me re-express V in terms of 'h' correctly.Alternatively, maybe I should have used a different approach. Let's consider that when you cut out a small triangle of height 'h' from each corner, the side length of the base of the box is 'a = 30 - 2h * sqrt(3)'. Wait, is that correct?Wait, in an equilateral triangle, the height is (sqrt(3)/2) * side length. So, if the height of the cut-out triangle is 'h', then the side length of the cut-out triangle is (2h)/sqrt(3). Therefore, the remaining side length 'a' is:a = 30 - 2 * (2h)/sqrt(3) = 30 - (4h)/sqrt(3)Which is what I had earlier. So, that seems correct.Then, the volume is:V = (sqrt(3)/4) * a^2 * h = (sqrt(3)/4) * (30 - (4h)/sqrt(3))^2 * hWhen I expanded this, I got:V = (900 sqrt(3)/4) h - 60 h^2 + (4 sqrt(3)/3) h^3But when I plug in h = (5 sqrt(3))/2, I get V = 750 cm³.Wait, but in the initial problem statement, the user mentioned the answer was 500 cm³. So, where is the discrepancy?Let me check the derivative calculation again.From V = (sqrt(3)/4) * a^2 * h, and a = 30 - (4h)/sqrt(3)So, V = (sqrt(3)/4) * (30 - (4h)/sqrt(3))^2 * hLet me compute this again step by step.First, expand (30 - (4h)/sqrt(3))^2:= 30^2 - 2*30*(4h)/sqrt(3) + (4h/sqrt(3))^2= 900 - (240h)/sqrt(3) + (16h^2)/3Now, multiply by (sqrt(3)/4) * h:V = (sqrt(3)/4) * [900 - (240h)/sqrt(3) + (16h^2)/3] * h= (sqrt(3)/4) * 900h - (sqrt(3)/4)*(240h)/sqrt(3)*h + (sqrt(3)/4)*(16h^2)/3 * hSimplify each term:First term: (sqrt(3)/4)*900h = (900 sqrt(3)/4) hSecond term: (sqrt(3)/4)*(240h)/sqrt(3)*h = (240h^2)/4 = 60h^2Wait, earlier I had a negative sign, so it's -60h^2Third term: (sqrt(3)/4)*(16h^2)/3 * h = (16 sqrt(3)/12) h^3 = (4 sqrt(3)/3) h^3So, V = (900 sqrt(3)/4) h - 60h^2 + (4 sqrt(3)/3) h^3Now, taking derivative:dV/dh = (900 sqrt(3)/4) - 120h + 4 sqrt(3) h^2Set to zero:4 sqrt(3) h^2 - 120h + (900 sqrt(3)/4) = 0Multiply by 4:16 sqrt(3) h^2 - 480h + 900 sqrt(3) = 0Divide by sqrt(3):16 h^2 - 480/sqrt(3) h + 900 = 0Simplify 480/sqrt(3) = 160 sqrt(3)So, 16 h^2 - 160 sqrt(3) h + 900 = 0Divide by 4:4 h^2 - 40 sqrt(3) h + 225 = 0Quadratic formula:h = [40 sqrt(3) ± sqrt( (40 sqrt(3))^2 - 4*4*225 )]/(2*4)Compute discriminant:(40 sqrt(3))^2 = 1600*3 = 48004*4*225 = 16*225 = 3600So, sqrt(4800 - 3600) = sqrt(1200) = 20 sqrt(3)Thus,h = [40 sqrt(3) ± 20 sqrt(3)]/8So,h = (60 sqrt(3))/8 = (15 sqrt(3))/2 ≈ 12.99 cmOr,h = (20 sqrt(3))/8 = (5 sqrt(3))/2 ≈ 4.33 cmAs before.Now, when h = (5 sqrt(3))/2, a = 30 - (4h)/sqrt(3) = 30 - (4*(5 sqrt(3)/2))/sqrt(3) = 30 - (10 sqrt(3))/sqrt(3) = 30 - 10 = 20 cmSo, a = 20 cm, h = (5 sqrt(3))/2 cmVolume V = (sqrt(3)/4)*20^2*(5 sqrt(3)/2) = (sqrt(3)/4)*400*(5 sqrt(3)/2)= (sqrt(3)*400*5 sqrt(3))/8= (400*5*3)/8= 6000/8 = 750 cm³So, according to this, the maximum volume is 750 cm³.But the initial problem statement mentioned the answer was 500 cm³. So, perhaps there was a misunderstanding in the problem setup.Wait, maybe I misinterpreted the problem. The problem says "a triangular box is to be cut from an equilateral triangle of length 30 cm." Maybe it's not about cutting out triangles from each corner, but rather forming a box by folding up the sides, similar to how you make a box from a square by cutting squares from each corner.In that case, the original equilateral triangle is 30 cm on each side. If we cut out smaller equilateral triangles from each corner, then fold up the sides to form a box, the height of the box would be the side length of the cut-out triangles.Wait, but in that case, the height 'h' would be the side length of the cut-out triangles, and the side length of the base would be 30 - 2h.But in an equilateral triangle, the height is (sqrt(3)/2)*side length. So, if we cut out a small equilateral triangle of side length 'h' from each corner, the remaining side length of the base would be 30 - 2h.But then, the height of the box would be 'h', but the height in terms of the equilateral triangle's geometry would be (sqrt(3)/2)*h.Wait, this is getting confusing. Let me clarify.If we have an equilateral triangle of side length 30 cm, and we cut out smaller equilateral triangles of side length 'x' from each corner, then the remaining side length of the base of the box would be 30 - 2x.But the height of the box would be the height of the cut-out triangles, which is (sqrt(3)/2)x.So, in this case, the volume V would be:V = (Area of base) * height= (sqrt(3)/4)*(30 - 2x)^2 * (sqrt(3)/2)xSimplify:= (sqrt(3)/4)*(900 - 120x + 4x^2)*(sqrt(3)/2)x= (3/8)*(900 - 120x + 4x^2)*x= (3/8)*(900x - 120x^2 + 4x^3)= (2700x - 360x^2 + 12x^3)/8= (12x^3 - 360x^2 + 2700x)/8To find the maximum volume, take derivative dV/dx:dV/dx = (36x^2 - 720x + 2700)/8Set to zero:36x^2 - 720x + 2700 = 0Divide by 36:x^2 - 20x + 75 = 0Solve:x = [20 ± sqrt(400 - 300)]/2= [20 ± sqrt(100)]/2= [20 ± 10]/2So, x = 15 or x = 5Since x must be less than 15 (because 30 - 2x > 0), x = 5 cmThus, the height of the box is (sqrt(3)/2)*5 ≈ 4.33 cmThe side length of the base is 30 - 2*5 = 20 cmVolume V = (sqrt(3)/4)*20^2*(sqrt(3)/2)*5= (sqrt(3)/4)*400*(sqrt(3)/2)*5= (3/8)*400*5= (3/8)*2000= 750 cm³Again, I get 750 cm³.But the initial problem statement suggested the answer was 500 cm³. Maybe the problem was interpreted differently, such as cutting out squares instead of triangles, but since it's an equilateral triangle, that wouldn't make sense.Alternatively, perhaps the height 'h' is considered differently. Maybe instead of cutting out equilateral triangles, we cut out right triangles, but in that case, the geometry changes.Wait, perhaps the problem is about cutting out squares from the corners of a square to make a box, but here it's an equilateral triangle. So, the approach is similar but adapted to the triangular shape.Alternatively, maybe the height 'h' is the height of the box, not the side length of the cut-out triangles. In that case, the side length of the cut-out triangles would be related to 'h' by the geometry of the equilateral triangle.Let me try this approach.Assume that the height of the box is 'h', and the cut-out triangles have a height of 'h'. Since the original triangle is equilateral, the side length of the cut-out triangles would be (2h)/sqrt(3), as the height of an equilateral triangle is (sqrt(3)/2)*side length.Thus, the side length of the base of the box would be:a = 30 - 2*(2h)/sqrt(3) = 30 - (4h)/sqrt(3)Then, the volume V is:V = (sqrt(3)/4)*a^2*h = (sqrt(3)/4)*(30 - (4h)/sqrt(3))^2*hThis is the same expression as before, leading to V = 750 cm³ when h = (5 sqrt(3))/2 cm.So, regardless of the approach, I consistently get 750 cm³ as the maximum volume.Therefore, I think the initial answer of 500 cm³ might be incorrect, and the correct maximum volume is 750 cm³.

💡First, I need to determine the initial number of oranges in the bin, which is 31.Next, the store throws away 9 old oranges. I'll subtract 9 from the initial amount to find out how many oranges remain:31 - 9 = 22 oranges.Then, the store adds 38 new oranges to the bin. I'll add these 38 oranges to the remaining 22 oranges to find the total number of oranges in the bin:22 + 38 = 60 oranges.Therefore, after throwing away the old oranges and adding the new ones, there will be 60 oranges in the bin.

💡Okay, so I have this problem where I need to figure out how many ways I can choose 60 unit squares from an 11x11 chessboard. The catch is that no two chosen squares can share a side. Hmm, that sounds a bit tricky, but let me try to break it down.First off, an 11x11 chessboard has 121 squares in total. That's straightforward—11 times 11. Now, I need to choose 60 of these squares, but with the condition that none of them are adjacent. So, no two squares can be next to each other either horizontally or vertically. Diagonally is okay, I think, because the problem only mentions sides, not corners.I remember that on a chessboard, the squares are alternately colored black and white. So, in an 11x11 board, there will be 61 squares of one color and 60 of the other. Let me confirm that: since 11 is odd, the number of squares of each color won't be equal. Specifically, there will be (11*11 + 1)/2 = 61 squares of one color and 60 of the other. Let's say there are 61 black squares and 60 white squares.Now, if I want to choose squares such that none are adjacent, I can only choose squares of the same color. Because on a chessboard, squares of the same color are never adjacent—they are always separated by squares of the opposite color. So, if I choose all the white squares, I can't choose any black squares, and vice versa.Wait, but I need to choose 60 squares. There are 60 white squares and 61 black squares. So, if I choose all the white squares, that's exactly 60 squares, and none of them are adjacent. That seems like one valid way.But what about choosing black squares? There are 61 black squares, and I need to choose 60. So, how many ways can I choose 60 black squares out of 61? That should be the combination of 61 choose 60, which is 61. Because for each black square, I can decide to leave it out or not, but since I need to leave out exactly one, there are 61 ways to do that.So, adding those together, there's 1 way to choose all the white squares and 61 ways to choose 60 black squares. That gives a total of 62 ways.Wait, let me make sure I didn't miss anything. Is there any other way to choose 60 squares without them being adjacent? For example, could I mix black and white squares somehow? But no, because if I choose both black and white squares, some of them would end up being adjacent since they're on the same row or column. So, mixing colors isn't an option here.Also, is there a way to choose more than 60 squares without them being adjacent? Well, the maximum number of non-adjacent squares you can choose on an 11x11 chessboard is 61, which is the number of squares of one color. But since we're only choosing 60, it's either all the white squares or 60 out of the black squares.So, I think my reasoning holds. There are 62 ways in total.

💡Okay, so I have this problem where I'm given a function f(x) = cos(x)(sin(x) + sqrt(3)cos(x)) - sqrt(3)/2, and I need to find the value of α such that when I shift the function f(x) by α, the resulting function g(x) = f(x + α) is an odd function. Also, it's given that α > 0. Hmm, okay, let me try to break this down step by step.First, I remember that an odd function satisfies the condition g(-x) = -g(x) for all x in its domain. So, if I can make sure that f(x + α) satisfies this condition, then I can find the appropriate α.Looking at the function f(x), it seems a bit complicated. Maybe I should try simplifying it first. Let me expand the terms:f(x) = cos(x)(sin(x) + sqrt(3)cos(x)) - sqrt(3)/2Let me distribute cos(x):f(x) = cos(x)sin(x) + sqrt(3)cos²(x) - sqrt(3)/2Hmm, okay. Now, I recall that cos(x)sin(x) can be written as (1/2)sin(2x) using the double-angle identity. So, let me rewrite that term:f(x) = (1/2)sin(2x) + sqrt(3)cos²(x) - sqrt(3)/2Now, for the cos²(x) term, I can use another identity: cos²(x) = (1 + cos(2x))/2. Let me substitute that in:f(x) = (1/2)sin(2x) + sqrt(3)*(1 + cos(2x))/2 - sqrt(3)/2Let me simplify this:First, distribute sqrt(3)/2:f(x) = (1/2)sin(2x) + sqrt(3)/2 + (sqrt(3)/2)cos(2x) - sqrt(3)/2Wait, the sqrt(3)/2 and -sqrt(3)/2 cancel each other out:f(x) = (1/2)sin(2x) + (sqrt(3)/2)cos(2x)Okay, so now f(x) simplifies to (1/2)sin(2x) + (sqrt(3)/2)cos(2x). Hmm, this looks like a linear combination of sine and cosine functions with the same argument, 2x. I remember that such expressions can be written as a single sine or cosine function with a phase shift.Specifically, A sin(θ) + B cos(θ) can be written as C sin(θ + φ), where C = sqrt(A² + B²) and tan(φ) = B/A. Let me apply that here.Here, A = 1/2 and B = sqrt(3)/2. So, C = sqrt((1/2)² + (sqrt(3)/2)²) = sqrt(1/4 + 3/4) = sqrt(1) = 1.Now, tan(φ) = B/A = (sqrt(3)/2)/(1/2) = sqrt(3). So, φ = arctan(sqrt(3)) = π/3.Therefore, f(x) can be written as:f(x) = sin(2x + π/3)Wait, let me check that. If I expand sin(2x + π/3), it should give me the same expression as before.Using the sine addition formula:sin(2x + π/3) = sin(2x)cos(π/3) + cos(2x)sin(π/3)We know that cos(π/3) = 1/2 and sin(π/3) = sqrt(3)/2. So,sin(2x + π/3) = sin(2x)*(1/2) + cos(2x)*(sqrt(3)/2) = (1/2)sin(2x) + (sqrt(3)/2)cos(2x)Which matches our simplified f(x). Great, so f(x) = sin(2x + π/3).Now, the function g(x) is defined as f(x + α), so:g(x) = f(x + α) = sin(2(x + α) + π/3) = sin(2x + 2α + π/3)We need g(x) to be an odd function, meaning that g(-x) = -g(x) for all x.Let's compute g(-x):g(-x) = sin(2*(-x) + 2α + π/3) = sin(-2x + 2α + π/3)But sin(-θ) = -sin(θ), so:g(-x) = sin(-2x + 2α + π/3) = -sin(2x - 2α - π/3)Wait, hold on. Let me make sure I apply the identity correctly. sin(-θ) = -sin(θ), so:g(-x) = sin(-2x + 2α + π/3) = sin(-(2x - 2α - π/3)) = -sin(2x - 2α - π/3)Hmm, okay. So, g(-x) = -sin(2x - 2α - π/3). On the other hand, -g(x) = -sin(2x + 2α + π/3).For g(x) to be odd, we must have:-sin(2x - 2α - π/3) = -sin(2x + 2α + π/3)Simplify both sides by multiplying by -1:sin(2x - 2α - π/3) = sin(2x + 2α + π/3)Now, when is sin(A) = sin(B)? Well, sin(A) = sin(B) implies that either:1. A = B + 2πk, for some integer k, or2. A = π - B + 2πk, for some integer k.So, applying this to our equation:Case 1: 2x - 2α - π/3 = 2x + 2α + π/3 + 2πkSimplify this:2x - 2α - π/3 = 2x + 2α + π/3 + 2πkSubtract 2x from both sides:-2α - π/3 = 2α + π/3 + 2πkBring like terms together:-2α - 2α = π/3 + π/3 + 2πk-4α = (2π/3) + 2πkMultiply both sides by -1:4α = -2π/3 - 2πkDivide both sides by 4:α = (-2π/3 - 2πk)/4 = (-π/6 - πk/2)But since α > 0, we need to find k such that α is positive. Let's see:α = (-π/6 - πk/2) > 0Multiply both sides by -1 (remember to reverse the inequality):π/6 + πk/2 < 0But π/6 is positive, and πk/2 is positive if k is positive, so the sum can't be negative. Therefore, this case doesn't give us a positive α. So, no solution from Case 1.Case 2: 2x - 2α - π/3 = π - (2x + 2α + π/3) + 2πkLet me simplify the right-hand side:π - (2x + 2α + π/3) = π - 2x - 2α - π/3 = (π - π/3) - 2x - 2α = (2π/3) - 2x - 2αSo, the equation becomes:2x - 2α - π/3 = 2π/3 - 2x - 2α + 2πkBring all terms to one side:2x - 2α - π/3 - 2π/3 + 2x + 2α - 2πk = 0Combine like terms:(2x + 2x) + (-2α + 2α) + (-π/3 - 2π/3) - 2πk = 04x + 0 + (-π) - 2πk = 0So, 4x - π - 2πk = 0But this must hold for all x, which is only possible if the coefficient of x is zero and the constant term is zero. However, the coefficient of x is 4, which isn't zero. Therefore, this equation can't hold for all x unless 4 = 0, which is impossible. Therefore, Case 2 also doesn't provide a solution.Wait, that seems problematic. Did I make a mistake in my reasoning?Let me double-check. So, we have:g(-x) = -sin(2x - 2α - π/3)And -g(x) = -sin(2x + 2α + π/3)We set them equal:-sin(2x - 2α - π/3) = -sin(2x + 2α + π/3)Which simplifies to:sin(2x - 2α - π/3) = sin(2x + 2α + π/3)Then, I considered the two cases where sin(A) = sin(B). But perhaps I should consider another approach.Alternatively, since g(x) must be odd, then g(-x) = -g(x). So, let's write that out:sin(2*(-x) + 2α + π/3) = -sin(2x + 2α + π/3)Simplify the left side:sin(-2x + 2α + π/3) = -sin(2x - 2α - π/3)So, we have:-sin(2x - 2α - π/3) = -sin(2x + 2α + π/3)Multiply both sides by -1:sin(2x - 2α - π/3) = sin(2x + 2α + π/3)Which is the same equation as before.So, perhaps another approach is needed. Maybe instead of trying to equate the arguments, I can use the property that for g(x) to be odd, all the even components must cancel out.Looking back at the expression for g(x):g(x) = sin(2x + 2α + π/3)We can write this as:g(x) = sin(2x + φ), where φ = 2α + π/3Now, for g(x) to be odd, sin(2x + φ) must be an odd function. But sin(2x + φ) is not necessarily odd unless φ is chosen such that the function becomes odd.Wait, but sin is an odd function, so sin(2x + φ) is odd if φ is zero or a multiple of π, but shifted by φ. Hmm, maybe not.Wait, let's recall that sin(θ) is odd, but sin(θ + φ) is not necessarily odd unless φ is chosen appropriately.Alternatively, maybe we can express sin(2x + φ) as a combination of sine and cosine, and then impose the condition for oddness.Let me expand sin(2x + φ):sin(2x + φ) = sin(2x)cos(φ) + cos(2x)sin(φ)For this function to be odd, the cosine term must be zero because cosine is an even function, and the sine term is odd. So, to eliminate the even component, we need the coefficient of cos(2x) to be zero.That is, sin(φ) = 0.So, sin(φ) = 0 implies φ = kπ, where k is an integer.But φ = 2α + π/3, so:2α + π/3 = kπSolving for α:2α = kπ - π/3α = (kπ - π/3)/2 = (kπ)/2 - π/6Since α > 0, we need:(kπ)/2 - π/6 > 0Multiply both sides by 6 to eliminate denominators:3kπ - π > 0π(3k - 1) > 0Since π > 0, this reduces to:3k - 1 > 0 => 3k > 1 => k > 1/3Since k is an integer, the smallest k satisfying this is k = 1.So, for k = 1:α = (π)/2 - π/6 = (3π/6 - π/6) = (2π/6) = π/3For k = 2:α = (2π)/2 - π/6 = π - π/6 = 5π/6And so on.Therefore, the general solution is α = (kπ)/2 - π/6 for k = 1, 2, 3, ...But since the problem asks for the value of α, and typically such problems expect the smallest positive solution, which would be when k = 1, giving α = π/3.Wait, let me verify this. If α = π/3, then:g(x) = f(x + π/3) = sin(2(x + π/3) + π/3) = sin(2x + 2π/3 + π/3) = sin(2x + π)But sin(2x + π) = -sin(2x), which is indeed an odd function because -sin(2x) is odd.Yes, that works. So, α = π/3 is a valid solution.Alternatively, for k = 2, α = 5π/6:g(x) = f(x + 5π/6) = sin(2(x + 5π/6) + π/3) = sin(2x + 5π/3 + π/3) = sin(2x + 6π/3) = sin(2x + 2π) = sin(2x), which is also odd.Wait, but sin(2x + 2π) is just sin(2x), which is odd. So, both α = π/3 and α = 5π/6 work, but since α > 0, both are valid, but the smallest positive α is π/3.But the problem doesn't specify to find the smallest α, just the value of α. However, in the context of such problems, usually, the smallest positive solution is expected unless otherwise specified.So, I think α = π/3 is the answer.Wait, but let me double-check with k = 0:If k = 0, then α = 0 - π/6 = -π/6, which is negative, so it's not acceptable since α > 0.So, the smallest positive α is indeed π/3.Therefore, the value of α is π/3.