🧠:<think>First, I need to determine how many training sessions are required each day to ensure that all 120 youth members attend at least one session weekly. Since the sessions are spread across 6 days, I'll divide the total number of players by the number of days to find the number of sessions needed per day.Next, for the monthly tournaments, I'll calculate the total number of matches required so that each of the 120 players participates exactly once. Each match involves 7 players, so I'll divide the total number of players by the number of players per match to find the total number of matches needed in a month.Then, I'll distribute these matches evenly across 4 weekends. This will give me the number of matches per weekend. Additionally, if an extra match is added every third weekend to encourage competition, I'll adjust the distribution accordingly to ensure that all players still get to participate without any overlap.Finally, I'll summarize the findings to provide clear answers to both parts of the problem.</think>

🧠:<think>Okay, so I have this problem where a parent is organizing peace rallies to maximize their impact after losing their child in a war. They have a budget of 50,000, and the effectiveness of a rally is given by the function E(x, y) = 1000 ln(x) + 2000y. Here, x is the money spent on advertising, and y is the number of volunteers. The first part of the problem asks me to express the effectiveness function E(x, y) solely in terms of the number of volunteers y. They also mention that each volunteer costs 100 for training and materials, and the rest of the budget is spent on advertising. Alright, so let's break this down. The total budget is 50,000. Each volunteer costs 100, so if the parent hires y volunteers, the total cost for volunteers would be 100y dollars. That means the remaining money, which is 50,000 - 100y, is spent on advertising. Therefore, x, the amount spent on advertising, is equal to 50,000 - 100y. So, substituting x in the effectiveness function E(x, y), we get E(y) = 1000 ln(50,000 - 100y) + 2000y. That should be the effectiveness function in terms of y alone. Wait, let me make sure I did that correctly. The original function is E(x, y) = 1000 ln(x) + 2000y. Since x is the money spent on advertising, and the total budget is 50,000, then x = 50,000 - 100y. So, substituting x into E(x, y), we get E(y) = 1000 ln(50,000 - 100y) + 2000y. Yeah, that seems right.Moving on to the second part. I need to determine the number of volunteers y that maximizes the effectiveness E(y). Also, I have to find the corresponding amount of money allocated to advertising x and the maximum effectiveness E(y). So, to maximize E(y), I should take the derivative of E(y) with respect to y, set it equal to zero, and solve for y. That will give me the critical points, which could be maxima or minima. Then, I can check the second derivative or use some other method to confirm it's a maximum.Let me write down E(y) again: E(y) = 1000 ln(50,000 - 100y) + 2000y.First, let's compute the derivative E'(y). The derivative of ln(50,000 - 100y) with respect to y is (1/(50,000 - 100y)) * (-100), by the chain rule. So, the derivative of 1000 ln(50,000 - 100y) is 1000 * ( -100 / (50,000 - 100y) ). Then, the derivative of 2000y with respect to y is 2000. So, putting it all together, E'(y) = (1000 * -100) / (50,000 - 100y) + 2000. Simplify that: E'(y) = (-100,000) / (50,000 - 100y) + 2000.Now, set E'(y) equal to zero to find critical points:(-100,000) / (50,000 - 100y) + 2000 = 0.Let me solve for y.First, move the second term to the other side:(-100,000) / (50,000 - 100y) = -2000.Multiply both sides by (50,000 - 100y):-100,000 = -2000 * (50,000 - 100y).Simplify the right side:-100,000 = -2000*50,000 + (-2000)*(-100y).Calculate each term:-2000*50,000 = -100,000,000.-2000*(-100y) = +200,000y.So, putting it together:-100,000 = -100,000,000 + 200,000y.Now, let's add 100,000,000 to both sides:-100,000 + 100,000,000 = 200,000y.Calculate the left side:100,000,000 - 100,000 = 99,900,000.So, 99,900,000 = 200,000y.Now, solve for y:y = 99,900,000 / 200,000.Divide numerator and denominator by 1000:y = 99,900 / 200.Calculate that:99,900 divided by 200. Let's see, 200 goes into 99,900 how many times?200 * 499 = 99,800.So, 99,900 - 99,800 = 100.So, 100 / 200 = 0.5.Therefore, y = 499.5.Hmm, y is the number of volunteers, which should be an integer. So, we have y ≈ 499.5. Since we can't have half a volunteer, we might need to check y = 499 and y = 500 to see which gives a higher effectiveness.But before that, let's make sure that y = 499.5 is within the feasible region.Given that x = 50,000 - 100y, and x must be positive because you can't spend negative money on advertising.So, 50,000 - 100y > 0.Which implies y < 500.So, y must be less than 500. Since y = 499.5 is less than 500, it's feasible. But since y must be an integer, we can check y = 499 and y = 500.But wait, if y = 500, then x = 50,000 - 100*500 = 50,000 - 50,000 = 0. So, x would be zero, which might not be allowed because ln(0) is undefined. So, y cannot be 500 because that would make x zero, which is not feasible.Therefore, the maximum y can be is 499. So, y = 499 is the maximum number of volunteers possible without making x zero. Therefore, y = 499 is the critical point we need to consider.But wait, let me think again. The critical point was at y = 499.5, which is between 499 and 500. Since y must be an integer, we can check both y = 499 and y = 500, but y = 500 is not feasible because x becomes zero. So, y must be 499.Alternatively, maybe we can relax the integer constraint and take y = 499.5, but since volunteers are people, you can't have half a person. So, we have to stick with y = 499.But let me verify if y = 499 is indeed the maximum. Maybe I should check the second derivative to confirm if it's a maximum.So, let's compute the second derivative E''(y).We have E'(y) = (-100,000)/(50,000 - 100y) + 2000.Compute E''(y):The derivative of (-100,000)/(50,000 - 100y) is:Let me denote f(y) = (-100,000)/(50,000 - 100y).Then, f'(y) = (-100,000) * derivative of 1/(50,000 - 100y).The derivative of 1/(50,000 - 100y) is (0 - (-100))/(50,000 - 100y)^2) = 100/(50,000 - 100y)^2.Therefore, f'(y) = (-100,000) * (100)/(50,000 - 100y)^2) = (-10,000,000)/(50,000 - 100y)^2.So, E''(y) = (-10,000,000)/(50,000 - 100y)^2 + 0 (since the derivative of 2000 is zero).So, E''(y) = (-10,000,000)/(50,000 - 100y)^2.Since the denominator is squared, it's positive, and the numerator is negative, so E''(y) is negative. Therefore, the function is concave down at the critical point, which means it's a maximum.Therefore, y = 499.5 is a maximum, but since y must be an integer, we need to check y = 499 and y = 500.But as I thought earlier, y = 500 is not feasible because x would be zero. So, y must be 499.Wait, but let me compute E(y) at y = 499 and y = 499.5 to see how much difference it makes.But since y must be an integer, maybe y = 499 is the optimal.Alternatively, perhaps we can treat y as a continuous variable for the sake of optimization and then round to the nearest integer. Let's see.But let's proceed step by step.So, we have y ≈ 499.5, but since y must be less than 500, and an integer, y = 499 is the feasible solution.Therefore, the number of volunteers y that maximizes effectiveness is 499.Now, let's compute the corresponding x, which is the money spent on advertising.x = 50,000 - 100y = 50,000 - 100*499 = 50,000 - 49,900 = 100 dollars.So, x = 100 dollars.Wait, that seems very low. Only 100 on advertising? That's interesting.But let's check if that's correct.Given that the effectiveness function is E(y) = 1000 ln(x) + 2000y, and x = 50,000 - 100y.So, substituting y = 499, x = 100.Therefore, E(y) = 1000 ln(100) + 2000*499.Compute that:First, ln(100) is approximately 4.60517.So, 1000 * 4.60517 ≈ 4605.17.Then, 2000 * 499 = 998,000.So, total E(y) ≈ 4605.17 + 998,000 ≈ 1,002,605.17.Wait, that's a pretty high effectiveness. But let me check if y = 499 is indeed the maximum.Alternatively, if I take y = 499, x = 100.If I take y = 498, then x = 50,000 - 100*498 = 50,000 - 49,800 = 200.Compute E(y) for y = 498:E(y) = 1000 ln(200) + 2000*498.ln(200) ≈ 5.2983.So, 1000 * 5.2983 ≈ 5,298.3.2000 * 498 = 996,000.Total E(y) ≈ 5,298.3 + 996,000 ≈ 1,001,298.3.Compare that to y = 499, which was approximately 1,002,605.17.So, E(y) is higher at y = 499.Similarly, if I check y = 500, which is not feasible because x = 0, but let's see:E(y) = 1000 ln(0) + 2000*500. But ln(0) is undefined (approaches negative infinity), so E(y) would be negative infinity, which is worse.Therefore, y = 499 is indeed the maximum.Wait, but let me check y = 499.5, even though it's not an integer, just to see the value.x = 50,000 - 100*499.5 = 50,000 - 49,950 = 50.So, x = 50.E(y) = 1000 ln(50) + 2000*499.5.ln(50) ≈ 3.9120.So, 1000 * 3.9120 ≈ 3,912.2000 * 499.5 = 999,000.Total E(y) ≈ 3,912 + 999,000 ≈ 1,002,912.Which is slightly higher than y = 499, which was approximately 1,002,605.17.But since y must be an integer, y = 499 is the closest feasible point, giving a slightly lower effectiveness than y = 499.5, but it's the best we can do.Alternatively, maybe the parent can consider y = 499.5, but since volunteers are people, they can't have half a volunteer. So, y must be 499.Therefore, the optimal number of volunteers is 499, with 100 spent on advertising, giving an effectiveness of approximately 1,002,605.Wait, but let me compute the exact value of E(y) at y = 499.E(y) = 1000 ln(100) + 2000*499.ln(100) is exactly ln(10^2) = 2 ln(10) ≈ 2*2.302585093 ≈ 4.605170186.So, 1000 * 4.605170186 ≈ 4,605.170186.2000 * 499 = 998,000.So, total E(y) ≈ 4,605.170186 + 998,000 ≈ 1,002,605.170186.Similarly, at y = 499.5, x = 50.E(y) = 1000 ln(50) + 2000*499.5.ln(50) = ln(5*10) = ln(5) + ln(10) ≈ 1.609437912 + 2.302585093 ≈ 3.912023005.So, 1000 * 3.912023005 ≈ 3,912.023005.2000 * 499.5 = 999,000.Total E(y) ≈ 3,912.023005 + 999,000 ≈ 1,002,912.023.So, indeed, y = 499.5 gives a slightly higher effectiveness, but since y must be an integer, y = 499 is the optimal.Alternatively, perhaps the parent can consider y = 499 and y = 500, but y = 500 is not feasible because x = 0, which would make the effectiveness undefined (since ln(0) is undefined). So, y must be 499.Therefore, the optimal number of volunteers is 499, with 100 spent on advertising, giving a maximum effectiveness of approximately 1,002,605.Wait, but let me double-check the calculations to make sure I didn't make any errors.First, the derivative:E'(y) = (-100,000)/(50,000 - 100y) + 2000.Set to zero:(-100,000)/(50,000 - 100y) + 2000 = 0.Move 2000 to the other side:(-100,000)/(50,000 - 100y) = -2000.Multiply both sides by (50,000 - 100y):-100,000 = -2000*(50,000 - 100y).Divide both sides by -1000:100 = 2*(50,000 - 100y).Wait, wait, that's a different approach. Let me try that.Alternatively, let's solve the equation again:(-100,000)/(50,000 - 100y) = -2000.Multiply both sides by (50,000 - 100y):-100,000 = -2000*(50,000 - 100y).Divide both sides by -1000:100 = 2*(50,000 - 100y).Wait, that's a different approach. Let me see:-100,000 = -2000*(50,000 - 100y).Divide both sides by -1000:100 = 2*(50,000 - 100y).So, 100 = 100,000 - 200y.Then, subtract 100,000 from both sides:100 - 100,000 = -200y.-99,900 = -200y.Divide both sides by -200:y = 99,900 / 200 = 499.5.Yes, same result. So, y = 499.5.Therefore, the calculations are correct.So, in conclusion, the optimal number of volunteers is 499, with 100 spent on advertising, giving a maximum effectiveness of approximately 1,002,605.But let me just think about this result again. It seems that the parent is spending almost the entire budget on volunteers, with only 100 left for advertising. That seems counterintuitive because advertising is also a significant factor in the effectiveness function.Wait, the effectiveness function is E(x, y) = 1000 ln(x) + 2000y. So, the marginal effectiveness of advertising is 1000/x, and the marginal effectiveness of volunteers is 2000.So, to maximize effectiveness, we set the marginal effectiveness per dollar equal for both.Wait, that's another way to approach this problem. Let me try that.The idea is that the parent should allocate their budget such that the marginal effectiveness per dollar spent on advertising is equal to the marginal effectiveness per dollar spent on volunteers.So, the marginal effectiveness of advertising is dE/dx = 1000 / x.The marginal effectiveness of volunteers is dE/dy = 2000.But since each volunteer costs 100, the marginal effectiveness per dollar spent on volunteers is (dE/dy) / (cost per volunteer) = 2000 / 100 = 20.Similarly, the marginal effectiveness per dollar spent on advertising is (dE/dx) / 1 = 1000 / x.To maximize effectiveness, set these equal:1000 / x = 20.Solve for x:x = 1000 / 20 = 50.So, x = 50.Therefore, the optimal x is 50, and the corresponding y is (50,000 - x)/100 = (50,000 - 50)/100 = 49,950 / 100 = 499.5.Which is the same result as before.Therefore, the optimal allocation is x = 50 and y = 499.5.But since y must be an integer, y = 499 or 500.But y = 500 would require x = 0, which is not feasible because ln(0) is undefined.Therefore, y = 499 and x = 100.Wait, but according to the marginal effectiveness approach, the optimal x is 50, but since y must be an integer, we have to adjust.Wait, perhaps I made a mistake in interpreting the marginal effectiveness per dollar.Wait, let me think again.The marginal effectiveness of advertising is dE/dx = 1000 / x.The marginal effectiveness of volunteers is dE/dy = 2000.But since each volunteer costs 100, the parent has to spend 100 to get one more volunteer. Therefore, the marginal effectiveness per dollar spent on volunteers is 2000 / 100 = 20.Similarly, the marginal effectiveness per dollar spent on advertising is 1000 / x.To maximize effectiveness, set these equal:1000 / x = 20.So, x = 50.Therefore, the optimal x is 50, and y = (50,000 - 50)/100 = 499.5.But since y must be an integer, we have to choose y = 499 or 500.But y = 500 would require x = 0, which is not feasible because ln(0) is undefined.Therefore, the optimal y is 499, with x = 100.So, that's consistent with our previous result.Therefore, the optimal number of volunteers is 499, with 100 spent on advertising, giving a maximum effectiveness of approximately 1,002,605.Wait, but let me compute the exact value of E(y) at y = 499.E(y) = 1000 ln(100) + 2000*499.ln(100) = 4.605170186.So, 1000 * 4.605170186 = 4,605.170186.2000 * 499 = 998,000.Total E(y) = 4,605.170186 + 998,000 = 1,002,605.170186.Similarly, at y = 499.5, x = 50.E(y) = 1000 ln(50) + 2000*499.5.ln(50) ≈ 3.912023005.1000 * 3.912023005 = 3,912.023005.2000 * 499.5 = 999,000.Total E(y) = 3,912.023005 + 999,000 = 1,002,912.023005.So, indeed, y = 499.5 gives a slightly higher effectiveness, but since y must be an integer, y = 499 is the optimal.Therefore, the parent should hire 499 volunteers and spend 100 on advertising, resulting in a maximum effectiveness of approximately 1,002,605.Wait, but let me just think about this again. If the parent spends 100 on advertising, that seems very little, especially considering that the marginal effectiveness of advertising is 1000 / x, which would be 1000 / 100 = 10 per dollar. Whereas the marginal effectiveness of volunteers is 20 per dollar. So, actually, the marginal effectiveness per dollar is higher for volunteers, which is why the optimal solution is to spend almost all the budget on volunteers.But in this case, the parent is trying to maximize effectiveness, so it makes sense to allocate more to the factor with higher marginal effectiveness per dollar.Therefore, the result seems consistent.So, to summarize:1. The effectiveness function in terms of y alone is E(y) = 1000 ln(50,000 - 100y) + 2000y.2. The number of volunteers y that maximizes effectiveness is 499, with x = 100, giving a maximum effectiveness of approximately 1,002,605.Wait, but let me just check if y = 499 is indeed the maximum by checking the second derivative.We had E''(y) = (-10,000,000)/(50,000 - 100y)^2.At y = 499, the denominator is (50,000 - 100*499)^2 = (50,000 - 49,900)^2 = (100)^2 = 10,000.So, E''(499) = (-10,000,000)/10,000 = -1,000.Since E''(y) is negative, the function is concave down, confirming that y = 499 is a local maximum.Therefore, the calculations are correct.So, the final answer is:1. E(y) = 1000 ln(50,000 - 100y) + 2000y.2. The optimal number of volunteers is 499, with 100 spent on advertising, resulting in a maximum effectiveness of approximately 1,002,605.But let me write the exact value of E(y) at y = 499.E(y) = 1000 ln(100) + 2000*499.ln(100) = 4.605170186.So, 1000 * 4.605170186 = 4,605.170186.2000 * 499 = 998,000.Total E(y) = 4,605.170186 + 998,000 = 1,002,605.170186.So, approximately 1,002,605.17.Therefore, the maximum effectiveness is approximately 1,002,605.But perhaps we can write it more precisely.Alternatively, we can express it in terms of exact logarithms.E(y) = 1000 ln(100) + 2000*499.ln(100) = 2 ln(10).So, E(y) = 1000 * 2 ln(10) + 998,000.Which is 2000 ln(10) + 998,000.But ln(10) is approximately 2.302585093.So, 2000 * 2.302585093 ≈ 4,605.170186.Therefore, E(y) ≈ 4,605.17 + 998,000 ≈ 1,002,605.17.So, that's the exact value.Therefore, the maximum effectiveness is approximately 1,002,605.17.But since the problem doesn't specify rounding, we can present it as is.So, in conclusion:1. The effectiveness function in terms of y is E(y) = 1000 ln(50,000 - 100y) + 2000y.2. The optimal number of volunteers is 499, with 100 spent on advertising, resulting in a maximum effectiveness of approximately 1,002,605.17.But let me just check if I can express E(y) in a more simplified form.E(y) = 1000 ln(50,000 - 100y) + 2000y.We can factor out 100 from the argument of the logarithm:E(y) = 1000 ln(100*(500 - y)) + 2000y.Which is 1000 [ln(100) + ln(500 - y)] + 2000y.Which simplifies to 1000 ln(100) + 1000 ln(500 - y) + 2000y.But that might not be necessary unless the problem asks for it.Therefore, the final answers are:1. E(y) = 1000 ln(50,000 - 100y) + 2000y.2. y = 499 volunteers, x = 100, and maximum effectiveness E(y) ≈ 1,002,605.17.I think that's it.</think>

🧠:<think>Alright, so I'm trying to figure out how to organize this exhibition with artifacts from three different ancient civilizations. The goal is to maximize visitor engagement, which means I need to balance both the spatial constraints and the historical significance of the artifacts. Let me break this down step by step.First, the exhibition hall is a rectangle measuring 50 meters by 30 meters. That gives a total area of 50 * 30 = 1500 square meters. The artifacts can't take up more than 70% of this space. So, 70% of 1500 is 0.7 * 1500 = 1050 square meters. That means the total area required by all artifacts must be less than or equal to 1050 square meters.Now, each civilization's artifacts have their own area functions. For civilization 1, it's ( A_1(x) = 3x^2 + 2x + 1 ), for civilization 2, ( A_2(y) = 4y^2 + 5y + 3 ), and for civilization 3, ( A_3(z) = 2z^2 + 3z + 4 ). Here, x, y, z are the number of artifacts from each civilization respectively.So, the first task is to find the maximum number of artifacts (x, y, z) such that the sum of their areas doesn't exceed 1050. That gives me the inequality:( 3x^2 + 2x + 1 + 4y^2 + 5y + 3 + 2z^2 + 3z + 4 leq 1050 )Simplifying the constants: 1 + 3 + 4 = 8. So,( 3x^2 + 2x + 4y^2 + 5y + 2z^2 + 3z + 8 leq 1050 )Subtracting 8 from both sides:( 3x^2 + 2x + 4y^2 + 5y + 2z^2 + 3z leq 1042 )Okay, so I need to maximize x, y, z such that this inequality holds. But since x, y, z are integers (you can't have a fraction of an artifact), I need to find the maximum integer values for each variable without exceeding the total area.But wait, this seems a bit tricky because it's a three-variable equation. Maybe I can approach this by trying to find the maximum possible for each variable individually, assuming the others are zero, and then see how they can be combined.Let me start with x. If I set y = 0 and z = 0, then:( 3x^2 + 2x leq 1042 )Let me solve for x:( 3x^2 + 2x - 1042 = 0 )Using the quadratic formula:x = [-2 ± sqrt(4 + 4*3*1042)] / (2*3)Calculating discriminant:sqrt(4 + 12504) = sqrt(12508) ≈ 111.8So,x ≈ (-2 + 111.8)/6 ≈ 109.8/6 ≈ 18.3Since x must be an integer, x_max ≈ 18. Let me check:A1(18) = 3*(18)^2 + 2*18 + 1 = 3*324 + 36 + 1 = 972 + 36 + 1 = 1009That's way over 1042. Wait, that can't be right. Wait, no, when I set y and z to zero, the total area is just A1(x). So, 3x² + 2x +1 ≤ 1050. Wait, no, earlier I subtracted 8, but actually, the total area is A1 + A2 + A3, which is 3x² + 2x +1 + 4y² +5y +3 + 2z² +3z +4. So, the total is 3x² +2x +4y² +5y +2z² +3z +8. So, when I set y and z to zero, the total area is 3x² +2x +8. So, 3x² +2x +8 ≤ 1050, which is 3x² +2x ≤ 1042.Wait, but when I plug x=18, A1(18) is 3*(18)^2 +2*18 +1 = 972 +36 +1=1009. So, 1009 +8=1017, which is less than 1050. So, x=18 gives total area 1017, which is under 1050. What about x=19?A1(19)=3*(361)+38+1=1083+38+1=1122. 1122 +8=1130 >1050. So, x can be at most 18.Similarly, let's find y_max when x=0 and z=0.A2(y)=4y² +5y +3. So, 4y² +5y +3 ≤1050.4y² +5y ≤1047.Solving 4y² +5y -1047=0.Discriminant: 25 + 4*4*1047=25 +16752=16777.sqrt(16777)≈129.53.So, y=(-5 +129.53)/8≈124.53/8≈15.56. So, y_max=15.Check A2(15)=4*(225)+75+3=900+75+3=978. 978 +8=986 ≤1050. y=16: A2(16)=4*256 +80 +3=1024+80+3=1107. 1107+8=1115>1050. So, y_max=15.Similarly for z.A3(z)=2z² +3z +4. So, 2z² +3z +4 ≤1050.2z² +3z ≤1046.Solving 2z² +3z -1046=0.Discriminant:9 + 8*1046=9+8368=8377.sqrt(8377)≈91.53.z=(-3 +91.53)/4≈88.53/4≈22.13. So, z_max=22.Check A3(22)=2*(484)+66 +4=968+66+4=1038. 1038 +8=1046 ≤1050. z=23: A3(23)=2*529 +69 +4=1058+69+4=1131. 1131 +8=1139>1050. So, z_max=22.So, individually, x can be up to 18, y up to15, z up to22.But since we have to combine them, we need to find x, y, z such that 3x² +2x +4y² +5y +2z² +3z ≤1042.Now, the second part is to maximize the total historical significance, which is S1(x) + S2(y) + S3(z) =5x² +4x +6y² +7y +4z² +6z.So, we need to maximize 5x² +4x +6y² +7y +4z² +6z subject to 3x² +2x +4y² +5y +2z² +3z ≤1042, with x,y,z non-negative integers.This seems like a quadratic optimization problem with integer variables. It might be complex, but perhaps we can approach it by trying to maximize each variable in turn, considering the trade-offs.Alternatively, since the significance functions are quadratic and increasing, we might want to maximize the variables as much as possible, but constrained by the area.But since the coefficients of x² in area and significance are different, we need to see which variables give more significance per unit area.Let me calculate the ratio of significance to area for each variable.For x: Significance per unit area is (5x² +4x)/(3x² +2x). As x increases, this ratio approaches 5/3 ≈1.666.For y: (6y² +7y)/(4y² +5y). As y increases, ratio approaches 6/4=1.5.For z: (4z² +6z)/(2z² +3z). As z increases, ratio approaches 4/2=2.So, z gives the highest significance per area, followed by x, then y.Therefore, to maximize significance, we should prioritize increasing z first, then x, then y.So, perhaps we can start by setting z as high as possible, then x, then y, within the area constraint.Let me try this approach.First, set z=22 (max possible). Then, calculate the remaining area.A3(22)=2*(22)^2 +3*22 +4=2*484 +66 +4=968 +66 +4=1038.So, total area used by z=22 is 1038. But wait, the total area is A1 + A2 + A3, which is 3x² +2x +1 +4y² +5y +3 +2z² +3z +4. So, when z=22, the area is 2*(22)^2 +3*22 +4=1038. So, the remaining area for x and y is 1050 -1038=12.So, 3x² +2x +4y² +5y +8 ≤12.Wait, because A1 + A2 + A3 +8= total area. So, if z=22, then 3x² +2x +4y² +5y +8 ≤1050 -1038=12.So, 3x² +2x +4y² +5y ≤4.Since x and y are non-negative integers, let's see possible combinations.Possible x=0:Then 4y² +5y ≤4.Try y=0: 0 ≤4. Okay.y=1:4 +5=9>4. Not allowed.So, y=0.Thus, x=0, y=0, z=22. Total significance S=0 +0 +4*(22)^2 +6*22=4*484 +132=1936 +132=2068.Alternatively, x=1:3 +2 +4y² +5y ≤4.So, 5 +4y² +5y ≤4. Not possible.Similarly, x=0, y=0 is the only possibility.So, total significance is 2068.Now, let's try z=21.A3(21)=2*(441)+63 +4=882 +63 +4=949.So, remaining area:1050 -949=101.Thus, 3x² +2x +4y² +5y +8 ≤101.So, 3x² +2x +4y² +5y ≤93.Now, we need to maximize S=5x² +4x +6y² +7y.We can try different combinations.Let me try to maximize x first, since x has a higher significance per area ratio than y.Find x_max such that 3x² +2x ≤93.Solve 3x² +2x -93=0.Discriminant=4 + 1116=1120.sqrt(1120)=33.466.x=(-2 +33.466)/6≈31.466/6≈5.24. So, x=5.Check A1(5)=3*25 +10 +1=75+10+1=86. So, 86 +8=94. But 94 >101? Wait, no, the remaining area is 101, but we have to subtract 8, so 3x² +2x +4y² +5y ≤93.Wait, when z=21, the remaining area is 101, so 3x² +2x +4y² +5y +8 ≤101, so 3x² +2x +4y² +5y ≤93.So, x=5: 3*25 +10=75 +10=85. So, 85 +4y² +5y ≤93.Thus, 4y² +5y ≤8.Possible y:y=0:0 ≤8.y=1:4 +5=9>8.So, y=0.Thus, x=5, y=0. Total significance S=5*(25) +4*5 +6*0 +7*0=125 +20=145. Plus S3(z=21)=4*(441)+6*21=1764 +126=1890. Total S=145 +1890=2035.Alternatively, x=4:3*16 +8=48 +8=56. So, 4y² +5y ≤93 -56=37.Find y_max such that 4y² +5y ≤37.Solve 4y² +5y -37=0.Discriminant=25 + 592=617.sqrt(617)=24.84.y=(-5 +24.84)/8≈19.84/8≈2.48. So, y=2.Check y=2:4*4 +10=16 +10=26 ≤37.y=3:4*9 +15=36 +15=51>37.So, y=2.Thus, x=4, y=2.Significance:5*(16)+4*4 +6*(4)+7*2=80 +16 +24 +14=134.Total S=134 +1890=2024.Compare with x=5, y=0:145 +1890=2035. So, x=5, y=0 is better.Alternatively, x=3:3*9 +6=27 +6=33. So, 4y² +5y ≤93 -33=60.Solve 4y² +5y -60=0.Discriminant=25 +960=985.sqrt(985)=31.4.y=(-5 +31.4)/8≈26.4/8≈3.3. So, y=3.Check y=3:4*9 +15=36 +15=51 ≤60.y=4:4*16 +20=64 +20=84>60.So, y=3.Significance:5*9 +4*3 +6*9 +7*3=45 +12 +54 +21=132.Total S=132 +1890=2022.Less than 2035.Similarly, x=2:3*4 +4=12 +4=16. 4y² +5y ≤93 -16=77.Solve 4y² +5y -77=0.Discriminant=25 +1232=1257.sqrt(1257)=35.45.y=(-5 +35.45)/8≈30.45/8≈3.8. So, y=3.Check y=3:51 ≤77. y=4:84>77.So, y=3.Significance:5*4 +4*2 +6*9 +7*3=20 +8 +54 +21=103.Total S=103 +1890=1993.Less than 2035.x=1:3 +2=5. 4y² +5y ≤93 -5=88.Solve 4y² +5y -88=0.Discriminant=25 +1408=1433.sqrt(1433)=37.85.y=(-5 +37.85)/8≈32.85/8≈4.1. So, y=4.Check y=4:4*16 +20=64 +20=84 ≤88.y=5:4*25 +25=100 +25=125>88.So, y=4.Significance:5*1 +4*1 +6*16 +7*4=5 +4 +96 +28=133.Total S=133 +1890=2023.Still less than 2035.x=0:4y² +5y ≤93.Solve 4y² +5y -93=0.Discriminant=25 +1488=1513.sqrt(1513)=38.9.y=(-5 +38.9)/8≈33.9/8≈4.23. So, y=4.Check y=4:84 ≤93.y=5:125>93.So, y=4.Significance:6*16 +7*4=96 +28=124.Total S=124 +1890=2014.Less than 2035.So, the best when z=21 is x=5, y=0, z=21 with S=2035.Compare with z=22, which gave S=2068. So, z=22 is better.Now, let's try z=20.A3(20)=2*400 +60 +4=800 +60 +4=864.Remaining area:1050 -864=186.So, 3x² +2x +4y² +5y +8 ≤186.Thus, 3x² +2x +4y² +5y ≤178.Now, we need to maximize S=5x² +4x +6y² +7y.Again, prioritize x first.Find x_max:3x² +2x ≤178.Solve 3x² +2x -178=0.Discriminant=4 +2136=2140.sqrt(2140)=46.26.x=(-2 +46.26)/6≈44.26/6≈7.37. So, x=7.Check A1(7)=3*49 +14 +1=147 +14 +1=162. 162 +8=170 ≤186. So, x=7.Remaining area:186 -162=24.So, 4y² +5y ≤24 -8=16? Wait, no, the total area is 3x² +2x +4y² +5y +8 ≤186. So, if x=7, 3*49 +14=147 +14=161. So, 4y² +5y ≤186 -161=25.So, 4y² +5y ≤25.Find y_max:4y² +5y -25=0.Discriminant=25 +400=425.sqrt(425)=20.615.y=(-5 +20.615)/8≈15.615/8≈1.95. So, y=1.Check y=1:4 +5=9 ≤25.y=2:16 +10=26>25.So, y=1.Thus, x=7, y=1.Significance:5*49 +4*7 +6*1 +7*1=245 +28 +6 +7=286.Plus S3(z=20)=4*400 +6*20=1600 +120=1720.Total S=286 +1720=2006.Alternatively, x=6:3*36 +12=108 +12=120. So, 4y² +5y ≤178 -120=58.Solve 4y² +5y -58=0.Discriminant=25 +928=953.sqrt(953)=30.87.y=(-5 +30.87)/8≈25.87/8≈3.23. So, y=3.Check y=3:4*9 +15=36 +15=51 ≤58.y=4:64 +20=84>58.So, y=3.Significance:5*36 +4*6 +6*9 +7*3=180 +24 +54 +21=279.Total S=279 +1720=1999.Less than 2006.x=5:3*25 +10=75 +10=85. So, 4y² +5y ≤178 -85=93.Wait, earlier when z=21, we had similar calculations. Let me see.Wait, no, z=20, so remaining area after x=5 is 178 -85=93.So, 4y² +5y ≤93.Solve 4y² +5y -93=0.Discriminant=25 +1488=1513.sqrt(1513)=38.9.y=(-5 +38.9)/8≈33.9/8≈4.23. So, y=4.Check y=4:4*16 +20=64 +20=84 ≤93.y=5:100 +25=125>93.So, y=4.Significance:5*25 +4*5 +6*16 +7*4=125 +20 +96 +28=269.Total S=269 +1720=1989.Less than 2006.x=8:Wait, x=7 was the max, x=8 would give 3*64 +16=192 +16=208>178. So, x=7 is max.Alternatively, x=7, y=1: S=286.Alternatively, x=7, y=0: 4y² +5y=0. So, y=0.Significance:5*49 +4*7=245 +28=273.Total S=273 +1720=1993.Less than 2006.So, x=7, y=1 is better.Now, let's see if we can get a higher S by reducing z further.z=19:A3(19)=2*361 +57 +4=722 +57 +4=783.Remaining area:1050 -783=267.So, 3x² +2x +4y² +5y +8 ≤267.Thus, 3x² +2x +4y² +5y ≤259.Maximize S=5x² +4x +6y² +7y.Again, prioritize x.Find x_max:3x² +2x ≤259.Solve 3x² +2x -259=0.Discriminant=4 +3108=3112.sqrt(3112)=55.78.x=(-2 +55.78)/6≈53.78/6≈8.96. So, x=8.Check A1(8)=3*64 +16 +1=192 +16 +1=209. 209 +8=217 ≤267. So, x=8.Remaining area:267 -209=58.So, 4y² +5y ≤58.Solve 4y² +5y -58=0.Discriminant=25 +928=953.sqrt(953)=30.87.y=(-5 +30.87)/8≈25.87/8≈3.23. So, y=3.Check y=3:4*9 +15=36 +15=51 ≤58.y=4:64 +20=84>58.So, y=3.Significance:5*64 +4*8 +6*9 +7*3=320 +32 +54 +21=427.Plus S3(z=19)=4*361 +6*19=1444 +114=1558.Total S=427 +1558=1985.Alternatively, x=7:3*49 +14=147 +14=161. So, 4y² +5y ≤259 -161=98.Solve 4y² +5y -98=0.Discriminant=25 +1568=1593.sqrt(1593)=39.91.y=(-5 +39.91)/8≈34.91/8≈4.36. So, y=4.Check y=4:64 +20=84 ≤98.y=5:100 +25=125>98.So, y=4.Significance:5*49 +4*7 +6*16 +7*4=245 +28 +96 +28=397.Total S=397 +1558=1955.Less than 1985.x=9:3*81 +18=243 +18=261>259. So, x=8 is max.Alternatively, x=8, y=3: S=427.Alternatively, x=8, y=2:4*4 +10=16 +10=26 ≤58.Significance:5*64 +4*8 +6*4 +7*2=320 +32 +24 +14=390.Total S=390 +1558=1948.Less than 427.So, x=8, y=3 is better.Compare with z=20, which gave S=2006. So, z=20 is better.Similarly, z=18:A3(18)=2*324 +54 +4=648 +54 +4=706.Remaining area:1050 -706=344.So, 3x² +2x +4y² +5y +8 ≤344.Thus, 3x² +2x +4y² +5y ≤336.Maximize S=5x² +4x +6y² +7y.Find x_max:3x² +2x ≤336.Solve 3x² +2x -336=0.Discriminant=4 +4032=4036.sqrt(4036)=63.53.x=(-2 +63.53)/6≈61.53/6≈10.25. So, x=10.Check A1(10)=3*100 +20 +1=300 +20 +1=321. 321 +8=329 ≤344. So, x=10.Remaining area:344 -321=23.So, 4y² +5y ≤23 -8=15? Wait, no, the total area is 3x² +2x +4y² +5y +8 ≤344. So, if x=10, 3*100 +20=300 +20=320. So, 4y² +5y ≤344 -320=24.Thus, 4y² +5y ≤24.Solve 4y² +5y -24=0.Discriminant=25 +384=409.sqrt(409)=20.22.y=(-5 +20.22)/8≈15.22/8≈1.90. So, y=1.Check y=1:4 +5=9 ≤24.y=2:16 +10=26>24.So, y=1.Significance:5*100 +4*10 +6*1 +7*1=500 +40 +6 +7=553.Plus S3(z=18)=4*324 +6*18=1296 +108=1404.Total S=553 +1404=1957.Alternatively, x=9:3*81 +18=243 +18=261. So, 4y² +5y ≤336 -261=75.Solve 4y² +5y -75=0.Discriminant=25 +1200=1225.sqrt(1225)=35.y=(-5 +35)/8=30/8=3.75. So, y=3.Check y=3:36 +15=51 ≤75.y=4:64 +20=84>75.So, y=3.Significance:5*81 +4*9 +6*9 +7*3=405 +36 +54 +21=516.Total S=516 +1404=1920.Less than 1957.x=11:3*121 +22=363 +22=385>336. So, x=10 is max.Alternatively, x=10, y=1: S=553.Alternatively, x=10, y=0: Significance=500 +40=540.Total S=540 +1404=1944.Less than 553.So, x=10, y=1 is better.Compare with z=20, which gave S=2006. So, z=20 is better.Continuing this way would take a lot of time, but it seems that as we decrease z, the total significance decreases. So, the maximum S we found so far is when z=22, x=0, y=0, S=2068.But wait, when z=22, x=0, y=0, the area used is A3(22)=1038, plus 8=1046, which is under 1050. So, we have 1050 -1046=4 extra area. Can we add any artifacts?Looking back, when z=22, x=0, y=0, the area used is 1038 +8=1046. So, 4 units left. Can we add x=1?A1(1)=3 +2 +1=6. But 1046 +6=1052>1050. Not allowed.Similarly, y=1: A2(1)=4 +5 +3=12. 1046 +12=1058>1050.So, no, can't add any more artifacts.Alternatively, maybe reduce z by 1 to 21, and see if we can add more x and y to get higher S.When z=21, we had x=5, y=0, S=2035.But 2035 <2068.Alternatively, maybe z=22, x=0, y=0 is the best.Wait, but let's check if we can have z=22, x=0, y=0, and then see if we can add any other artifacts without exceeding the area.Wait, the total area is 1038 +8=1046. So, 4 units left. Maybe we can add a fraction of an artifact, but since x, y, z must be integers, we can't. So, no.Alternatively, maybe reduce z by 1 to 21, and use the extra area to add more x and y.When z=21, A3=949, so total area used is 949 +8=957. Remaining area:1050 -957=93.So, 3x² +2x +4y² +5y ≤93.Earlier, we found that x=5, y=0 gives S=145, total S=145 +1890=2035.But when z=22, S=2068.So, 2068 >2035.Thus, z=22 is better.Alternatively, maybe z=22, x=0, y=0 is the optimal.But let's check another approach: maybe not setting z to max, but a combination where we can have higher x and y, leading to higher S.For example, suppose z=20, x=8, y=3, S=1985.Less than 2068.Alternatively, z=19, x=8, y=3, S=1985.No, same.Alternatively, z=15, x=10, y=5.Wait, let's try z=15.A3(15)=2*225 +45 +4=450 +45 +4=499.Remaining area:1050 -499=551.So, 3x² +2x +4y² +5y +8 ≤551.Thus, 3x² +2x +4y² +5y ≤543.Maximize S=5x² +4x +6y² +7y.Find x_max:3x² +2x ≤543.Solve 3x² +2x -543=0.Discriminant=4 +6516=6520.sqrt(6520)=80.75.x=(-2 +80.75)/6≈78.75/6≈13.125. So, x=13.Check A1(13)=3*169 +26 +1=507 +26 +1=534. 534 +8=542 ≤551. So, x=13.Remaining area:551 -534=17.So, 4y² +5y ≤17 -8=9? Wait, no, the total area is 3x² +2x +4y² +5y +8 ≤551. So, if x=13, 3*169 +26=507 +26=533. So, 4y² +5y ≤551 -533=18.Thus, 4y² +5y ≤18.Solve 4y² +5y -18=0.Discriminant=25 +288=313.sqrt(313)=17.69.y=(-5 +17.69)/8≈12.69/8≈1.58. So, y=1.Check y=1:4 +5=9 ≤18.y=2:16 +10=26>18.So, y=1.Significance:5*169 +4*13 +6*1 +7*1=845 +52 +6 +7=910.Plus S3(z=15)=4*225 +6*15=900 +90=990.Total S=910 +990=1900.Alternatively, x=12:3*144 +24=432 +24=456. So, 4y² +5y ≤543 -456=87.Solve 4y² +5y -87=0.Discriminant=25 +1392=1417.sqrt(1417)=37.66.y=(-5 +37.66)/8≈32.66/8≈4.08. So, y=4.Check y=4:64 +20=84 ≤87.y=5:100 +25=125>87.So, y=4.Significance:5*144 +4*12 +6*16 +7*4=720 +48 +96 +28=892.Total S=892 +990=1882.Less than 1900.x=14:3*196 +28=588 +28=616>543. So, x=13 is max.Alternatively, x=13, y=1: S=910.Alternatively, x=13, y=0: Significance=845 +52=897.Total S=897 +990=1887.Less than 910.So, x=13, y=1 is better.But 1900 <2068.Thus, z=22, x=0, y=0 still gives higher S.Another approach: Maybe instead of setting z to max, we can have a combination where z is slightly less, but x and y are higher, leading to higher total S.For example, let's try z=20, x=8, y=3: S=1985.Less than 2068.Alternatively, z=18, x=10, y=3: S=1957.Still less.Alternatively, z=17:A3(17)=2*289 +51 +4=578 +51 +4=633.Remaining area:1050 -633=417.So, 3x² +2x +4y² +5y +8 ≤417.Thus, 3x² +2x +4y² +5y ≤409.Maximize S=5x² +4x +6y² +7y.Find x_max:3x² +2x ≤409.Solve 3x² +2x -409=0.Discriminant=4 +4908=4912.sqrt(4912)=70.08.x=(-2 +70.08)/6≈68.08/6≈11.34. So, x=11.Check A1(11)=3*121 +22 +1=363 +22 +1=386. 386 +8=394 ≤417. So, x=11.Remaining area:417 -386=31.So, 4y² +5y ≤31 -8=23? Wait, no, the total area is 3x² +2x +4y² +5y +8 ≤417. So, if x=11, 3*121 +22=363 +22=385. So, 4y² +5y ≤417 -385=32.Thus, 4y² +5y ≤32.Solve 4y² +5y -32=0.Discriminant=25 +512=537.sqrt(537)=23.17.y=(-5 +23.17)/8≈18.17/8≈2.27. So, y=2.Check y=2:16 +10=26 ≤32.y=3:36 +15=51>32.So, y=2.Significance:5*121 +4*11 +6*4 +7*2=605 +44 +24 +14=687.Plus S3(z=17)=4*289 +6*17=1156 +102=1258.Total S=687 +1258=1945.Alternatively, x=10:3*100 +20=300 +20=320. So, 4y² +5y ≤409 -320=89.Solve 4y² +5y -89=0.Discriminant=25 +1424=1449.sqrt(1449)=38.07.y=(-5 +38.07)/8≈33.07/8≈4.13. So, y=4.Check y=4:64 +20=84 ≤89.y=5:100 +25=125>89.So, y=4.Significance:5*100 +4*10 +6*16 +7*4=500 +40 +96 +28=664.Total S=664 +1258=1922.Less than 1945.x=12:3*144 +24=432 +24=456>409. So, x=11 is max.Alternatively, x=11, y=2: S=687.Alternatively, x=11, y=1:4 +5=9 ≤32.Significance:5*121 +4*11 +6*1 +7*1=605 +44 +6 +7=662.Total S=662 +1258=1920.Less than 687.So, x=11, y=2 is better.But still, 1945 <2068.Thus, it seems that the maximum S is achieved when z=22, x=0, y=0, giving S=2068.But let's check another possibility: maybe z=22, x=0, y=0, but with some other combination.Wait, when z=22, x=0, y=0, the area used is 1038 +8=1046. So, 4 units left. Can we add any artifact?If we add x=1, area becomes 1046 +6=1052>1050. Not allowed.If we add y=1, area becomes 1046 +12=1058>1050. Not allowed.So, no, can't add any.Alternatively, maybe reduce z by 1 to 21, and use the extra area to add x and y.When z=21, A3=949, so total area used is 949 +8=957. Remaining area:1050 -957=93.So, 3x² +2x +4y² +5y ≤93.Earlier, we found that x=5, y=0 gives S=145, total S=145 +1890=2035.But 2035 <2068.Alternatively, maybe x=4, y=2: S=134, total S=134 +1890=2024.Still less.Alternatively, x=3, y=3: S=132, total S=132 +1890=2022.No.Thus, z=22, x=0, y=0 is the best.Wait, but let's check if we can have z=22, x=0, y=0, and then see if we can add any other artifacts by reducing z and adding x and y.But as we saw, reducing z by 1 allows us to add x=5, y=0, but the total S decreases.Alternatively, maybe reduce z by 2 to 20, and add more x and y.When z=20, A3=864, so total area used is 864 +8=872. Remaining area:1050 -872=178.So, 3x² +2x +4y² +5y ≤178.Earlier, we found that x=7, y=1 gives S=286, total S=286 +1720=2006.Still less than 2068.Thus, the maximum S is achieved when z=22, x=0, y=0.But wait, let's check if we can have z=22, x=0, y=0, and then see if we can add any other artifacts by reducing z and adding x and y in a way that the total S increases.But as we saw, any reduction in z leads to a decrease in S, because the significance per area of z is higher than x and y.Thus, the optimal solution is z=22, x=0, y=0.But wait, let me double-check the area calculation.When z=22, A3=2*(22)^2 +3*22 +4=2*484 +66 +4=968 +66 +4=1038.Total area used:1038 +8=1046 ≤1050.Yes, correct.So, the maximum number of artifacts is x=0, y=0, z=22.But wait, the first part of the problem asks for the maximum number of artifacts from each civilization without exceeding the spatial constraints.So, the answer for part 1 is x=18, y=15, z=22.But in part 2, to maximize S, we need to find the combination of x, y, z that maximizes S under the area constraint.So, the answer for part 2 is x=0, y=0, z=22.But let me confirm.Wait, when z=22, x=0, y=0, the area used is 1038 +8=1046, which is under 1050.But can we add any other artifacts without exceeding the area?As we saw, adding x=1 would require 6 units, making total area 1052>1050.Similarly, y=1 requires 12 units.Thus, no.Alternatively, maybe reduce z by 1 to 21, and add x=5, y=0, but S decreases.Thus, the optimal is z=22, x=0, y=0.But wait, let me check if there's a combination where z=21, x=5, y=0, which gives S=2035, which is less than 2068.Alternatively, maybe z=22, x=0, y=0 is indeed the optimal.Thus, the answers are:1. Maximum number of artifacts: x=18, y=15, z=22.2. Optimal combination for maximum S: x=0, y=0, z=22.But wait, in part 1, it's the maximum number of artifacts from each civilization without exceeding the spatial constraints. So, it's the maximum possible x, y, z individually, not necessarily all at the same time.Wait, no, the problem says "the maximum number of artifacts that can be displayed from each civilization without exceeding the spatial constraints."So, it's the maximum x, y, z such that A1(x) + A2(y) + A3(z) ≤1050.But in part 1, it's not necessarily that all x, y, z are used at the same time. It's just the maximum possible x, y, z individually.Wait, no, the problem says "the maximum number of artifacts that can be displayed from each civilization without exceeding the spatial constraints."So, it's the maximum x, y, z such that A1(x) + A2(y) + A3(z) ≤1050.But that's a bit ambiguous. It could mean the maximum x, y, z individually, or the maximum combination.But in the first part, it's likely asking for the maximum possible x, y, z individually, assuming the others are zero.So, x_max=18, y_max=15, z_max=22.Then, in part 2, find the combination of x, y, z that maximizes S under the same area constraint.Thus, the answers are:1. x=18, y=15, z=22.2. x=0, y=0, z=22.But let me confirm.In part 1, it's about the maximum number of artifacts from each civilization, so individually, not combined.Thus, x=18, y=15, z=22.In part 2, it's the combination that maximizes S, which is x=0, y=0, z=22.Yes, that makes sense.</think>

🧠:<think>Alright, so I have this problem about designing artificial dunes to protect a beachfront property. It's split into two parts. Let me tackle them one by one.Starting with part 1: The dunes are modeled by the equation ( y = -ax^2 + bx + c ). I need to find the values of ( a ), ( b ), and ( c ) given that the highest point is 10 meters above sea level and the base extends 50 meters horizontally.First, let's visualize this. The dune is a parabola opening downward because the coefficient of ( x^2 ) is negative. The vertex of this parabola will be at the highest point, which is 10 meters. The base of the dune is 50 meters wide, so the roots of the parabola (where ( y = 0 )) are at ( x = 0 ) and ( x = 50 ).Wait, actually, the problem says the base extends 50 meters horizontally. So, does that mean the entire width from one end to the other is 50 meters? If so, the roots are at ( x = 0 ) and ( x = 50 ). So, the parabola crosses the x-axis at these points.Given that, I can write the equation in its factored form. Since it's a downward opening parabola with roots at 0 and 50, it can be written as:( y = -kx(x - 50) )Where ( k ) is a positive constant. Expanding this, we get:( y = -kx^2 + 50kx )Comparing this to the given equation ( y = -ax^2 + bx + c ), we can see that:- ( a = k )- ( b = 50k )- ( c = 0 )But wait, the equation given in the problem has a constant term ( c ). In my factored form, ( c ) is zero because the parabola passes through the origin. However, the problem says the highest point is 10 meters above sea level. So, maybe the vertex is at some point, not necessarily at the origin.Wait, perhaps I need to adjust my approach. Let me think again.The standard form of a parabola is ( y = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. Since the parabola opens downward, ( a ) is negative.Given that the highest point is 10 meters, the vertex is at ( (h, 10) ). The base is 50 meters wide, so the roots are at ( x = h - 25 ) and ( x = h + 25 ). That way, the distance between the roots is 50 meters.So, the equation can be written as:( y = -a(x - (h - 25))(x - (h + 25)) )Simplifying this, it becomes:( y = -a(x - h + 25)(x - h - 25) )( y = -a[(x - h)^2 - (25)^2] )( y = -a(x - h)^2 + 625a )But we also know that the vertex is at ( (h, 10) ), so substituting ( x = h ) into the equation gives:( y = -a(0) + 625a = 625a = 10 )Therefore, ( 625a = 10 ) => ( a = 10 / 625 = 2/125 )So, ( a = 2/125 ). But since the parabola opens downward, the coefficient in front is negative, so actually, ( a = -2/125 ).Wait, hold on. Let me clarify. In the standard form, ( y = a(x - h)^2 + k ), if it opens downward, ( a ) is negative. In my previous step, I had ( y = -a(x - h)^2 + 625a ), so comparing to standard form, the coefficient is ( -a ), which should be negative. Therefore, ( -a ) is negative, so ( a ) is positive. Then, ( 625a = 10 ) gives ( a = 10/625 = 2/125 ). So, the equation becomes:( y = - (2/125)(x - h)^2 + 10 )But we need to express this in the form ( y = -ax^2 + bx + c ). So, let's expand it.First, let's note that the roots are at ( x = h - 25 ) and ( x = h + 25 ). So, the parabola is symmetric around ( x = h ). Without loss of generality, we can set ( h = 25 ) meters, so that the roots are at ( x = 0 ) and ( x = 50 ). This simplifies the equation because the vertex is at ( (25, 10) ).So, substituting ( h = 25 ):( y = - (2/125)(x - 25)^2 + 10 )Expanding ( (x - 25)^2 ):( (x - 25)^2 = x^2 - 50x + 625 )So,( y = - (2/125)(x^2 - 50x + 625) + 10 )( y = - (2/125)x^2 + (100/125)x - (1250/125) + 10 )Simplify each term:- ( - (2/125)x^2 ) remains as is.- ( (100/125)x = (4/5)x )- ( - (1250/125) = -10 )- So, ( -10 + 10 = 0 )Therefore, the equation simplifies to:( y = - (2/125)x^2 + (4/5)x )So, comparing this to ( y = -ax^2 + bx + c ):- ( a = 2/125 )- ( b = 4/5 )- ( c = 0 )Wait, but earlier I thought ( c ) might not be zero, but in this case, it is zero because when ( x = 0 ), ( y = 0 ). That makes sense because the dune starts at the base, which is at sea level.But let me double-check. The vertex is at (25, 10). Plugging x=25 into the equation:( y = - (2/125)(625) + (4/5)(25) )( y = -10 + 20 = 10 ). Correct.And at x=0:( y = 0 + 0 + 0 = 0 ). Correct.At x=50:( y = - (2/125)(2500) + (4/5)(50) )( y = -40 + 40 = 0 ). Correct.So, part 1 is solved with ( a = 2/125 ), ( b = 4/5 ), and ( c = 0 ).Moving on to part 2: The property owner wants the total volume of sand used per dune to not exceed 2000 cubic meters. I need to set up and solve the integral to find the total volume and determine the necessary conditions on ( a ), ( b ), and ( c ).First, volume is calculated by integrating the cross-sectional area along the length of the dune. Since the dune is a 2D parabola, if we assume it's extruded along the length (say, the y-axis), the volume would be the integral of the area under the curve multiplied by the depth (which is not given, but perhaps we can assume unit depth or that the volume is the integral of y dx over the base).Wait, actually, in typical volume calculations for such structures, if the dune is a parabolic shape in cross-section, and assuming it's extended along the beach, the volume would be the integral of the height function over the horizontal distance, multiplied by the depth (which is perpendicular to the cross-section). However, since the problem mentions "cross-sectional area function," perhaps it's considering the area under the curve as the cross-section, and then integrating over the length? Hmm, I might need to clarify.Wait, no. The cross-sectional area is given by the integral of y dx from x=0 to x=50. Then, if we assume that the dune is extended along the beach with some length L, the volume would be the cross-sectional area multiplied by L. However, the problem says "using the cross-sectional area function derived from the parabolic equation, set up and solve the integral to find the total volume of one dune."Wait, maybe it's considering the cross-sectional area as the area under the parabola, which is a 2D area, and then the volume is that area multiplied by the depth (which is not given). Hmm, but the problem doesn't specify the depth, so perhaps it's assuming unit depth, or that the volume is just the integral of y dx over the base.Alternatively, perhaps the dune is considered as a 3D object where the cross-section is the parabola, and the length is 1 meter, so the volume is just the area under the curve.Wait, the problem says "using the cross-sectional area function derived from the parabolic equation, set up and solve the integral to find the total volume of one dune." So, cross-sectional area function is the area under the parabola, which is the integral of y dx from 0 to 50. Then, the volume would be that area multiplied by the length of the dune. But since the problem doesn't specify the length, perhaps it's considering the dune as a 2D shape, and the volume is just the area? That doesn't make much sense because volume is 3D.Alternatively, maybe the cross-sectional area is given, and the volume is the integral over some other dimension. Hmm, this is a bit confusing.Wait, perhaps the cross-sectional area is the area under the parabola, which is 2D, and then the volume is that area multiplied by the depth, which is not given. But since the problem mentions "total volume of sand used," which is 3D, I think we need to consider the cross-sectional area and then multiply by the length of the dune. But since the problem doesn't specify the length, maybe it's assuming the dune is a single cross-section, so the volume is just the area under the curve.Wait, let me read the problem again: "using the cross-sectional area function derived from the parabolic equation, set up and solve the integral to find the total volume of one dune." So, cross-sectional area function is the area under the parabola, which is a function of x. Wait, no, the cross-sectional area is the integral of y dx from 0 to 50, which is a single value, not a function. So, maybe the problem is referring to the cross-sectional area as the integral, which is a number, and then the volume is that number multiplied by the length of the dune. But since the length isn't given, perhaps it's considering the dune as a 2D shape, and the "volume" is just the area? That seems inconsistent.Alternatively, perhaps the dune is a 3D object where the cross-section is the parabola, and the length is the same as the horizontal distance, so 50 meters. Then, the volume would be the area under the parabola multiplied by 50 meters. But that would be a 3D volume.Wait, let's think differently. Maybe the dune is modeled as a paraboloid, where the height is given by ( y = -ax^2 + bx + c ), and the volume is the integral over x and z (assuming z is the depth). But without more information, it's hard to say.Alternatively, perhaps the problem is considering the dune as a 2D shape, and the "volume" is just the area, which is 2D. But the problem mentions cubic meters, which is 3D. So, I think the correct approach is to consider the cross-sectional area (which is 2D) and then multiply by the length (which is given as 50 meters? Or is it 50 meters wide? Wait, the base extends 50 meters horizontally, so perhaps the length is 50 meters.Wait, actually, the base is 50 meters wide, so the cross-sectional area is the area under the parabola from x=0 to x=50, which is a 2D area. Then, if the dune is extended along the beach with some length, say L meters, then the volume would be that area multiplied by L. But the problem doesn't specify L, so perhaps it's assuming that the dune is 1 meter deep, making the volume equal to the cross-sectional area.Alternatively, perhaps the dune is a 3D paraboloid, and the volume is calculated by integrating over x and y. But without more information, it's unclear.Wait, maybe I'm overcomplicating. Let's read the problem again: "using the cross-sectional area function derived from the parabolic equation, set up and solve the integral to find the total volume of one dune." So, cross-sectional area function is the integral of y dx from 0 to 50, which gives the area under the curve. Then, the volume is that area multiplied by the depth, but since the problem doesn't specify depth, perhaps it's considering the dune as a 2D shape, and the volume is just the area. But that contradicts the units (cubic meters). So, maybe the cross-sectional area is the area under the curve, and the volume is that area multiplied by the length, which is 50 meters? But that would make the volume 50 times the cross-sectional area.Wait, let's think about it. If the cross-sectional area is A, and the length is L, then volume V = A * L. If the cross-sectional area is the area under the parabola from x=0 to x=50, and the length is 50 meters, then V = A * 50. But the problem says "total volume of sand used does not exceed 2000 cubic meters per dune." So, perhaps V = A * L <= 2000. But without knowing L, we can't solve for A. Alternatively, maybe the dune is considered as a 2D shape, so the "volume" is just the area, which would be in square meters, but the problem says cubic meters. Hmm.Wait, perhaps the cross-sectional area is the integral of y dx, which is in square meters, and then the volume is that area multiplied by the depth, which is 1 meter, making it cubic meters. So, if we assume the depth is 1 meter, then the volume is equal to the cross-sectional area.But the problem doesn't specify the depth, so maybe it's considering the dune as a 2D shape, and the volume is just the area, but that would be in square meters, not cubic. This is confusing.Alternatively, perhaps the dune is a 3D shape where the cross-section is the parabola, and the length is the same as the horizontal distance, so 50 meters. Then, the volume would be the integral of y dx from 0 to 50 multiplied by 50 meters, giving cubic meters.Wait, let's try that approach. So, cross-sectional area A = ∫₀^50 y dx. Then, volume V = A * 50. But the problem says "using the cross-sectional area function derived from the parabolic equation, set up and solve the integral to find the total volume of one dune." So, maybe the cross-sectional area is A = ∫₀^50 y dx, and then the volume is A * L, where L is the length. But since the problem doesn't specify L, perhaps it's considering L=1, making V = A.But the problem says "total volume of sand used does not exceed 2000 cubic meters per dune." So, perhaps the volume is just the cross-sectional area multiplied by 1 meter, making it cubic meters. So, V = A * 1 = A.Alternatively, maybe the dune is a 3D paraboloid, and the volume is calculated by integrating y over x and z, but that would require knowing the depth in the z-direction, which isn't given.Wait, perhaps the problem is simpler. Since the dune is a parabola in cross-section, and assuming it's extended along the beach with unit depth, the volume would be the area under the parabola. So, V = ∫₀^50 y dx.Given that, let's compute the integral.From part 1, we have y = - (2/125)x^2 + (4/5)x.So, V = ∫₀^50 [ - (2/125)x^2 + (4/5)x ] dxLet's compute this integral.First, integrate term by term:∫ [ - (2/125)x^2 ] dx = - (2/125) * (x^3)/3 = - (2/375)x^3∫ [ (4/5)x ] dx = (4/5) * (x^2)/2 = (2/5)x^2So, the integral becomes:V = [ - (2/375)x^3 + (2/5)x^2 ] from 0 to 50Evaluate at x=50:- (2/375)*(50)^3 + (2/5)*(50)^2Compute each term:First term: (50)^3 = 125,000So, - (2/375)*125,000 = - (2 * 125,000)/375 = - (250,000)/375 = -666.666...Second term: (2/5)*(2500) = (2/5)*2500 = 1000So, total at x=50: -666.666... + 1000 = 333.333...At x=0, both terms are zero, so the integral from 0 to 50 is 333.333... cubic meters.But the problem says the volume must not exceed 2000 cubic meters. So, 333.333... is much less than 2000. Therefore, the current parameters satisfy the volume constraint.Wait, but this seems too straightforward. Maybe I misunderstood the problem. Let me double-check.Wait, in part 1, we found a specific parabola with a=2/125, b=4/5, c=0, which gives a volume of approximately 333.33 cubic meters. But the problem says the volume must not exceed 2000 cubic meters. So, in this case, the volume is already under the limit, so no further conditions are needed. But that seems odd because the problem says "set up and solve the integral to find the total volume of one dune. Determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint."Wait, perhaps I misinterpreted the problem. Maybe the volume is not just the area under the parabola, but the volume of the dune as a 3D object, which would require integrating over both x and another dimension, say z, which is the depth. If the dune is extruded along the z-axis, then the volume would be the cross-sectional area multiplied by the length in the z-direction. But since the problem doesn't specify the length, perhaps it's assuming the dune is 1 meter deep, making the volume equal to the cross-sectional area.But in that case, the volume is 333.33 cubic meters, which is way below 2000. So, perhaps the problem is considering the dune as a 3D paraboloid, where the height is given by the parabola, and the volume is the integral over x and y (assuming y is the height and x is the horizontal). Wait, that might not make sense.Alternatively, perhaps the dune is a parabolic cylinder, extending infinitely in the z-direction, but that doesn't make sense for a dune.Wait, maybe the problem is considering the dune as a 3D shape where the cross-section is the parabola, and the length is 50 meters, so the volume is the cross-sectional area multiplied by 50 meters. So, V = A * 50.In that case, A = 333.33 m², so V = 333.33 * 50 = 16,666.5 cubic meters, which is way above 2000. So, that can't be.Wait, perhaps the dune is a 3D paraboloid, and the volume is calculated by integrating over x and y. But without more information, it's hard to set up the integral.Alternatively, maybe the problem is simpler, and the volume is just the area under the parabola, which is 333.33 cubic meters, and since that's less than 2000, no conditions are needed. But the problem says "set up and solve the integral to find the total volume of one dune. Determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint."Wait, perhaps the problem is considering the dune as a 3D object where the cross-sectional area is the integral of y dx, and then the volume is that area multiplied by the depth, which is another parameter. But since the problem doesn't specify the depth, maybe it's considering the depth as 1 meter, making the volume equal to the cross-sectional area.But in that case, the volume is 333.33 cubic meters, which is way below 2000. So, perhaps the problem is considering the dune as a 3D object with a certain depth, and we need to find the conditions on a, b, c such that the volume is <= 2000.Wait, but without knowing the depth, we can't set up the integral properly. Maybe the problem is considering the dune as a 2D shape, and the volume is just the area, but that contradicts the units.Alternatively, perhaps the problem is considering the dune as a 3D shape where the height is given by the parabola, and the length is 50 meters, so the volume is the integral of y dx from 0 to 50 multiplied by 50 meters. So, V = (∫₀^50 y dx) * 50.In that case, V = 333.33 * 50 = 16,666.5 cubic meters, which is way above 2000. So, that can't be.Wait, maybe the problem is considering the dune as a 3D shape where the cross-sectional area is the integral of y dx, and the length is 1 meter, making the volume equal to the cross-sectional area. So, V = 333.33 cubic meters, which is below 2000. So, no conditions are needed.But the problem says "determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint." So, perhaps the volume is not fixed by the given parameters, and we need to find a relationship between a, b, c such that the volume is <= 2000.Wait, but in part 1, we found specific values for a, b, c based on the highest point and the base width. So, if we change a, b, c, the volume will change. So, perhaps the problem is asking, given the general form y = -ax^2 + bx + c, with the highest point at 10 meters and base at 50 meters, what conditions on a, b, c ensure that the volume is <= 2000.Wait, but in part 1, we already fixed a, b, c based on the highest point and base width. So, maybe the volume is fixed, and it's already under 2000, so no conditions are needed. But that seems unlikely because the problem specifically asks to set up the integral and find the conditions.Alternatively, perhaps the problem is considering that the dune can have different shapes while still having the highest point at 10 meters and base at 50 meters, and we need to find the conditions on a, b, c such that the volume is <= 2000.Wait, but in part 1, we derived a specific parabola. So, maybe the problem is considering that the dune can have different shapes (different a, b, c) while still having the same highest point and base width, and we need to find the conditions on a, b, c such that the volume is <= 2000.Wait, but in part 1, we found that the parabola is uniquely determined by the highest point and the base width. So, a, b, c are fixed. Therefore, the volume is fixed, and it's 333.33 cubic meters, which is way below 2000. So, no conditions are needed.But the problem says "set up and solve the integral to find the total volume of one dune. Determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint."Hmm, maybe I'm missing something. Let me think again.Wait, perhaps the problem is considering that the dune can have different shapes, not necessarily symmetric. So, in part 1, we assumed the vertex is at the midpoint, but maybe it's not. So, the dune could be asymmetric, with the highest point not at 25 meters, but somewhere else, as long as the base is 50 meters wide and the highest point is 10 meters.In that case, the equation would still be y = -ax^2 + bx + c, but the vertex could be at any x between 0 and 50, as long as y=10 at that point, and y=0 at x=0 and x=50.So, in that case, the integral for the volume would depend on a, b, c, and we need to find the conditions such that the integral is <= 2000.Wait, but in part 1, we found a specific case where the vertex is at x=25, giving a specific volume. If the vertex is elsewhere, the volume might be different.Wait, let me consider that. Let's assume that the dune is a parabola with roots at x=0 and x=50, and vertex at (h, 10), where h is between 0 and 50. Then, the equation can be written as y = -a(x - h)^2 + 10, and since it passes through (0,0) and (50,0), we can find a in terms of h.So, plugging x=0, y=0:0 = -a(h)^2 + 10 => a = 10 / h^2Similarly, plugging x=50, y=0:0 = -a(50 - h)^2 + 10 => a = 10 / (50 - h)^2Therefore, 10 / h^2 = 10 / (50 - h)^2 => h^2 = (50 - h)^2Which implies h = 50 - h => 2h = 50 => h=25So, the vertex must be at x=25, meaning the parabola is symmetric. Therefore, the volume is uniquely determined, and it's 333.33 cubic meters, which is way below 2000. So, the volume constraint is automatically satisfied.But the problem says "determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint." So, perhaps the problem is considering that the dune can have different shapes, not necessarily with the vertex at 10 meters, but with the maximum height being 10 meters. Wait, no, the highest point must be 10 meters.Wait, maybe the problem is considering that the dune can have different shapes as long as the maximum height is 10 meters and the base is 50 meters, but the volume can vary depending on the shape. So, we need to find the conditions on a, b, c such that the volume is <= 2000.But from part 1, we saw that the volume is fixed once the highest point and base are fixed. So, perhaps the problem is considering that the dune can have different shapes, but with the same maximum height and base width, but varying the position of the vertex, which would change the volume.Wait, but earlier, we saw that the vertex must be at x=25, so the volume is fixed. Therefore, the volume is uniquely determined, and it's 333.33 cubic meters, which is way below 2000. So, no conditions are needed.But the problem says "set up and solve the integral to find the total volume of one dune. Determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint."Wait, maybe the problem is considering that the dune can have different heights and base widths, but in this case, the height is fixed at 10 meters and base at 50 meters, so the volume is fixed. Therefore, the conditions are automatically satisfied.Alternatively, perhaps the problem is considering that the dune can have different heights and base widths, and we need to find the conditions on a, b, c such that the volume is <= 2000, given that the maximum height is 10 meters and base width is 50 meters.But in that case, the volume is fixed, so the condition is automatically satisfied.Wait, maybe I'm overcomplicating. Let's proceed with the integral.From part 1, we have y = - (2/125)x^2 + (4/5)x.So, the volume V = ∫₀^50 y dx = 333.33 cubic meters.Since 333.33 < 2000, the volume constraint is satisfied. Therefore, no additional conditions are needed on a, b, c beyond what was found in part 1.But the problem says "determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint." So, perhaps the problem is considering that the dune can have different shapes, and we need to express the volume in terms of a, b, c and then set up the inequality V <= 2000.So, let's do that.Given y = -ax^2 + bx + c.We know that the dune has a maximum height of 10 meters, so the vertex is at (h, 10). Also, the roots are at x=0 and x=50, so y=0 at x=0 and x=50.From part 1, we found that the equation can be written as y = - (2/125)x^2 + (4/5)x, which is a specific case where the vertex is at x=25.But if we consider a general case where the vertex is at (h, 10), then the equation is y = -a(x - h)^2 + 10.Since it passes through (0,0) and (50,0), we can find a in terms of h.From x=0:0 = -a(h)^2 + 10 => a = 10 / h^2From x=50:0 = -a(50 - h)^2 + 10 => a = 10 / (50 - h)^2Therefore, 10 / h^2 = 10 / (50 - h)^2 => h^2 = (50 - h)^2 => h = 25.So, the vertex must be at x=25, making the parabola symmetric. Therefore, the equation is uniquely determined, and the volume is fixed.Thus, the volume is 333.33 cubic meters, which is less than 2000, so no conditions are needed beyond what was found in part 1.But the problem says "determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint." So, perhaps the problem is considering that the dune can have different shapes, but with the same maximum height and base width, but varying the position of the vertex, which would change the volume. But as we saw, the vertex must be at x=25, so the volume is fixed.Therefore, the necessary condition is that the volume is 333.33 cubic meters, which is less than 2000, so no additional conditions are needed.But perhaps the problem is considering that the dune can have different shapes, not necessarily symmetric, but still with maximum height 10 meters and base width 50 meters. Wait, but as we saw, the vertex must be at x=25, so the parabola is symmetric. Therefore, the volume is uniquely determined.So, in conclusion, the volume is 333.33 cubic meters, which is less than 2000, so the necessary condition is automatically satisfied.But the problem says "set up and solve the integral to find the total volume of one dune. Determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint."So, perhaps the problem is expecting us to express the volume in terms of a, b, c and then set up the inequality V <= 2000.Let's try that.Given y = -ax^2 + bx + c.We know that the roots are at x=0 and x=50, so y=0 at x=0 and x=50.At x=0: y=0 => c=0.At x=50: y=0 => -a*(50)^2 + b*(50) + c = 0 => -2500a + 50b + c = 0. But since c=0, we have -2500a + 50b = 0 => 50b = 2500a => b = 50a.Also, the maximum height is 10 meters. The vertex occurs at x = -b/(2a). Since the parabola opens downward, the vertex is at x = -b/(2a). But since the roots are at x=0 and x=50, the vertex is at x=25, so:25 = -b/(2a) => b = -50a.Wait, but earlier we found that b = 50a. So, combining these two:From roots: b = 50aFrom vertex: b = -50aThis implies 50a = -50a => 100a = 0 => a=0, which is a contradiction because a cannot be zero (otherwise, it's not a parabola).Wait, this suggests that my earlier assumption is wrong. Let me re-examine.Wait, the vertex x-coordinate is given by x = -b/(2a). Since the parabola opens downward, a is negative. So, x = -b/(2a). We know the vertex is at x=25, so:25 = -b/(2a) => b = -50a.But from the roots, we have b = 50a.So, combining these:b = 50a and b = -50a => 50a = -50a => 100a = 0 => a=0, which is not possible.Wait, this is a contradiction. So, where did I go wrong?Wait, in part 1, I assumed that the vertex is at x=25, which is correct because the parabola is symmetric with roots at 0 and 50. So, the vertex must be at x=25.But when I derived the equation, I got y = - (2/125)x^2 + (4/5)x, which has a vertex at x=25.Let me check the vertex formula:For y = ax^2 + bx + c, the vertex is at x = -b/(2a).In our case, y = - (2/125)x^2 + (4/5)x.So, a = -2/125, b = 4/5.Thus, x = -b/(2a) = -(4/5)/(2*(-2/125)) = -(4/5)/(-4/125) = (4/5)*(125/4) = 25. Correct.So, in this case, b = 4/5, a = -2/125.So, from the roots, we have b = 50a.But in this case, b = 4/5, a = -2/125.So, 50a = 50*(-2/125) = -100/125 = -4/5.But b = 4/5, which is not equal to -4/5.Wait, this suggests that my earlier assumption that b = 50a is incorrect.Wait, let's re-examine.From the roots, at x=50, y=0:y = -a*(50)^2 + b*(50) + c = 0.But c=0, so:-2500a + 50b = 0 => 50b = 2500a => b = 50a.But in our specific case, b = 4/5, a = -2/125.So, 50a = 50*(-2/125) = -100/125 = -4/5.But b = 4/5, which is not equal to -4/5.So, this is a contradiction.Wait, so where is the mistake?Wait, in the equation y = -ax^2 + bx + c, a is positive because the parabola opens downward, so the coefficient of x^2 is negative. Therefore, in the standard form, a is negative.Wait, no, in the equation y = -ax^2 + bx + c, the coefficient of x^2 is -a, which is negative, so a must be positive.So, in our case, a = 2/125, which is positive, and the equation is y = - (2/125)x^2 + (4/5)x.So, in this case, b = 4/5, a = 2/125.So, from the roots, we have:At x=50, y=0:- (2/125)*(50)^2 + (4/5)*(50) + 0 = 0Compute:- (2/125)*2500 + (4/5)*50 = -40 + 40 = 0. Correct.So, from the roots, we have:-2500a + 50b = 0 => 50b = 2500a => b = 50a.But in our case, a = 2/125, so b = 50*(2/125) = 100/125 = 4/5. Correct.So, b = 50a.From the vertex, x = -b/(2a) = 25.So, x = -b/(2a) = 25.But b = 50a, so:x = -50a/(2a) = -25.Wait, that can't be right because the vertex is at x=25, not x=-25.Wait, this suggests that x = -b/(2a) = -50a/(2a) = -25.But the vertex is at x=25, so:-25 = 25 => Contradiction.Wait, this is confusing. Let me re-examine.Wait, in the equation y = -ax^2 + bx + c, the coefficient of x^2 is -a, which is negative, so a is positive.The vertex x-coordinate is given by x = -b/(2*(-a)) = b/(2a).So, x = b/(2a).Given that the vertex is at x=25, we have:b/(2a) = 25 => b = 50a.Which matches the earlier result from the roots.So, in our case, b = 50a, and a = 2/125, so b = 50*(2/125) = 4/5. Correct.So, the vertex is at x=25, which is correct.Therefore, the integral for the volume is:V = ∫₀^50 y dx = ∫₀^50 (-ax^2 + bx) dx, since c=0.Compute the integral:∫ (-ax^2 + bx) dx = [ -a*(x^3)/3 + b*(x^2)/2 ] from 0 to 50.At x=50:- a*(125000)/3 + b*(2500)/2At x=0: 0So, V = - (125000/3)a + (2500/2)b.But since b = 50a, substitute:V = - (125000/3)a + (2500/2)*(50a)Simplify:V = - (125000/3)a + (125000/2)aConvert to common denominator:V = - (250000/6)a + (375000/6)a = (125000/6)a = (62500/3)a ≈ 20833.33aWait, but in our specific case, a = 2/125, so:V = (62500/3)*(2/125) = (62500*2)/(3*125) = (125000)/(375) = 333.33 cubic meters.So, V = (62500/3)a.We need V <= 2000.So,(62500/3)a <= 2000Multiply both sides by 3:62500a <= 6000Divide both sides by 62500:a <= 6000 / 62500 = 60/625 = 12/125 = 0.096But in our specific case, a = 2/125 ≈ 0.016, which is less than 0.096, so the volume is 333.33 < 2000.Therefore, the necessary condition is that a <= 12/125.But wait, in our specific case, a = 2/125, which is much less than 12/125. So, if we increase a, the volume increases.Wait, but a is positive, and increasing a would make the parabola steeper, which would increase the volume.Wait, let me think. The integral V = (62500/3)a. So, as a increases, V increases.Therefore, to have V <= 2000, we need a <= 2000 * 3 / 62500 = 6000 / 62500 = 0.096, as above.But in our specific case, a = 2/125 ≈ 0.016, which is much less than 0.096.Therefore, the necessary condition is that a <= 12/125.But wait, in our specific case, a = 2/125, which is less than 12/125, so the volume is 333.33 < 2000.Therefore, the condition is satisfied.But the problem says "determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint."So, the condition is that a <= 12/125.But since b = 50a, then b <= 50*(12/125) = 600/125 = 4.8.And since c=0, there's no condition on c.Therefore, the necessary conditions are:a <= 12/125,b <= 4.8,c = 0.But in our specific case, a = 2/125, b = 4/5 = 0.8, which is much less than 4.8.So, the volume can be increased by increasing a and b, but not beyond a=12/125 and b=4.8.But the problem says "determine the necessary conditions on the parameters a, b, and c to satisfy this volume constraint."So, the conditions are:a <= 12/125,b = 50a,c = 0.Therefore, as long as a <= 12/125, b = 50a, and c=0, the volume will be <= 2000.But in our specific case, a=2/125, which is less than 12/125, so the condition is satisfied.Therefore, the necessary conditions are:a <= 12/125,b = 50a,c = 0.So, summarizing:Part 1:a = 2/125,b = 4/5,c = 0.Part 2:The volume is V = (62500/3)a.To satisfy V <= 2000,a <= 12/125,b = 50a,c = 0.Therefore, the necessary conditions are a <= 12/125, b = 50a, and c=0.</think>