🧠:<think>Alright, so I have this problem about Captain Ivan planning his ship's route from Odessa to Shanghai. There are two parts: calculating total fuel consumption and figuring out the minimum number of weeks needed for the crew shifts. Let me tackle each part step by step.Starting with the first part: calculating total fuel consumption. The journey is divided into four segments with different distances and fuel rates. I need to find the fuel consumed in each segment and then sum them up. Also, I have to determine which segment uses the most fuel.First segment: Black Sea. Distance is 800 nautical miles, fuel rate is 0.05 gallons per nautical mile. So, fuel consumed here would be 800 multiplied by 0.05. Let me calculate that: 800 * 0.05 = 40 gallons.Second segment: Mediterranean Sea. Distance is 1500 nautical miles, fuel rate is 0.04 gallons per nautical mile. Fuel consumed is 1500 * 0.04. Let me do that multiplication: 1500 * 0.04 = 60 gallons.Third segment: Suez Canal. Distance is 120 nautical miles, fuel rate is 0.08 gallons per nautical mile. So, 120 * 0.08. That's 9.6 gallons.Fourth segment: Indian Ocean. Distance is 3000 nautical miles, fuel rate is 0.03 gallons per nautical mile. So, 3000 * 0.03 = 90 gallons.Now, adding up all these fuel consumptions: 40 + 60 + 9.6 + 90. Let me add them step by step. 40 + 60 is 100. 100 + 9.6 is 109.6. 109.6 + 90 is 199.6 gallons. So, total fuel consumption is 199.6 gallons.Now, to find the segment with the highest fuel consumption. Looking at each segment: Black Sea used 40, Mediterranean 60, Suez 9.6, and Indian Ocean 90. The highest is the Indian Ocean with 90 gallons.Moving on to the second part: determining the minimum number of weeks needed for the crew shifts. The ship requires at least 5 crew members on duty at any time. The total crew is 15, each working 8-hour shifts. No crew member can work more than 48 hours a week. The ship's average speed is 20 nautical miles per hour. I need to figure out how long the journey takes and then determine the shifts.First, let's find the total distance of the journey. The distances are 800 + 1500 + 120 + 3000. Let me add those up: 800 + 1500 is 2300. 2300 + 120 is 2420. 2420 + 3000 is 5420 nautical miles.Now, with an average speed of 20 nautical miles per hour, the total time required is total distance divided by speed. So, 5420 / 20. Let me calculate that: 5420 divided by 20 is 271 hours.Now, the ship needs 5 crew members on duty at all times. Each crew member works 8-hour shifts. So, each shift has 5 people working for 8 hours. But since the ship is moving continuously, we need to cover 271 hours with overlapping shifts.Wait, actually, since the crew needs to be present 24/7, the number of shifts required is total hours divided by shift length. So, 271 hours / 8 hours per shift. Let me compute that: 271 / 8 is approximately 33.875 shifts. Since you can't have a fraction of a shift, we'll need 34 shifts.Each shift requires 5 crew members, so total crew hours needed are 34 shifts * 5 crew members * 8 hours. Wait, no, that's not right. Each shift is 8 hours with 5 crew members, so each shift is 40 crew-hours. So, total crew-hours needed is 271 hours * 5 crew members = 1355 crew-hours.Each crew member can work up to 48 hours per week. There are 15 crew members. So, total crew-hours available per week is 15 * 48 = 720 crew-hours per week.To find the number of weeks needed, divide total crew-hours required by crew-hours available per week: 1355 / 720 ≈ 1.88 weeks. Since you can't have a fraction of a week, we'll need 2 weeks.But wait, let me double-check. If we have 2 weeks, total crew-hours available is 2 * 720 = 1440 crew-hours. Since 1440 is more than 1355, 2 weeks should be sufficient.However, we also need to ensure that no crew member works more than 48 hours in a week. Since each shift is 8 hours, each crew member can work up to 6 shifts per week (6 * 8 = 48). With 15 crew members, the maximum number of shifts that can be covered per week is 15 * 6 = 90 shifts. Each shift is 8 hours, so 90 shifts * 8 hours = 720 crew-hours, which matches the earlier calculation.We need 34 shifts in total. Each week can cover 90 shifts, so 34 shifts would take less than a week. But wait, the shifts are spread over the 271 hours, which is about 11.29 days. So, it's more than a week. Therefore, we need to see if 2 weeks are enough.Wait, perhaps I confused shifts with weeks. Let me clarify.Total time is 271 hours. Each week has 168 hours (24*7). So, 271 / 168 ≈ 1.61 weeks. So, approximately 1.61 weeks. But since we can't have a fraction of a week, we need 2 weeks.But considering the crew hours, 2 weeks provide 1440 crew-hours, which is more than the required 1355. So, yes, 2 weeks are sufficient.But let me think again. If the journey takes about 11.29 days, which is roughly 1 week and 4 days, so 2 weeks would cover it. But the crew can work up to 48 hours in a week. So, in the first week, they can work 48 hours, and in the second week, they can work the remaining hours.Wait, but the total crew-hours needed is 1355. In two weeks, 15 crew members can work 15*48*2 = 1440 crew-hours, which is enough. So, 2 weeks are sufficient.But is 1 week enough? 1 week provides 720 crew-hours, which is less than 1355. So, 2 weeks are needed.Therefore, the minimum number of consecutive weeks required is 2.Final Answer1. The total fuel consumption is boxed{199.6} gallons, and the segment with the highest fuel consumption is the Indian Ocean.2. The minimum number of consecutive weeks required is boxed{2}.</think>

🧠:<think>Okay, so I've got these two calculus problems to solve, both related to the function ( f(x) = e^{-x^2} ). Let me take them one at a time.Starting with the first problem: calculating the area under the curve of ( f(x) ) from ( x = -infty ) to ( x = infty ). Hmm, I remember that the integral of ( e^{-x^2} ) over the entire real line is a classic result, but I don't remember the exact value off the top of my head. I think it involves pi somehow. Maybe it's related to the Gaussian integral?Let me recall. The Gaussian integral states that:[int_{-infty}^{infty} e^{-x^2} dx = sqrt{pi}]Yes, that's right! So, the area under the curve is ( sqrt{pi} ). But wait, let me make sure I'm not making a mistake here. I think the integral of ( e^{-ax^2} ) from ( -infty ) to ( infty ) is ( sqrt{frac{pi}{a}} ). In this case, ( a = 1 ), so it simplifies to ( sqrt{pi} ). Yeah, that seems correct.But just to be thorough, maybe I should derive it quickly. I remember the trick involves squaring the integral and converting to polar coordinates. Let me try that.Let ( I = int_{-infty}^{infty} e^{-x^2} dx ). Then,[I^2 = left( int_{-infty}^{infty} e^{-x^2} dx right) left( int_{-infty}^{infty} e^{-y^2} dy right) = int_{-infty}^{infty} int_{-infty}^{infty} e^{-(x^2 + y^2)} dx dy]Switching to polar coordinates where ( x^2 + y^2 = r^2 ) and ( dx dy = r dr dtheta ), we have:[I^2 = int_{0}^{2pi} int_{0}^{infty} e^{-r^2} r dr dtheta]Let me compute the radial integral first. Let ( u = r^2 ), so ( du = 2r dr ), which means ( r dr = frac{1}{2} du ). Then,[int_{0}^{infty} e^{-r^2} r dr = frac{1}{2} int_{0}^{infty} e^{-u} du = frac{1}{2} left[ -e^{-u} right]_0^{infty} = frac{1}{2} (0 - (-1)) = frac{1}{2}]So, plugging back into the expression for ( I^2 ):[I^2 = int_{0}^{2pi} frac{1}{2} dtheta = frac{1}{2} times 2pi = pi]Therefore, ( I = sqrt{pi} ). That confirms the result. So, the area under the curve is indeed ( sqrt{pi} ).Alright, moving on to the second problem. It involves the function ( g(x) = x^2 e^{-x^2} ). I need to find its maximum value and the x-value where this maximum occurs.To find the maximum, I should take the derivative of ( g(x) ) with respect to x, set it equal to zero, and solve for x. Then, I can check if it's a maximum using the second derivative test or some other method.Let's compute the first derivative ( g'(x) ). Since ( g(x) = x^2 e^{-x^2} ), I'll use the product rule. The product rule states that ( (uv)' = u'v + uv' ), where ( u = x^2 ) and ( v = e^{-x^2} ).First, find ( u' ) and ( v' ):- ( u = x^2 ) so ( u' = 2x )- ( v = e^{-x^2} ) so ( v' = e^{-x^2} times (-2x) = -2x e^{-x^2} )Now, applying the product rule:[g'(x) = u'v + uv' = 2x e^{-x^2} + x^2 (-2x e^{-x^2}) = 2x e^{-x^2} - 2x^3 e^{-x^2}]Factor out common terms:[g'(x) = 2x e^{-x^2} (1 - x^2)]To find critical points, set ( g'(x) = 0 ):[2x e^{-x^2} (1 - x^2) = 0]Since ( e^{-x^2} ) is never zero, we can ignore that term. So, the equation reduces to:[2x (1 - x^2) = 0]Which gives:1. ( 2x = 0 ) => ( x = 0 )2. ( 1 - x^2 = 0 ) => ( x^2 = 1 ) => ( x = pm 1 )So, critical points at ( x = 0 ), ( x = 1 ), and ( x = -1 ).Now, we need to determine which of these points is a maximum. Let's analyze the behavior of ( g'(x) ) around these points.Alternatively, we can use the second derivative test. Let's compute the second derivative ( g''(x) ).First, recall that ( g'(x) = 2x e^{-x^2} (1 - x^2) ). Let me rewrite this as:[g'(x) = 2x (1 - x^2) e^{-x^2}]To find ( g''(x) ), we'll need to differentiate this expression. Let me let ( h(x) = 2x (1 - x^2) ) and ( k(x) = e^{-x^2} ), so ( g'(x) = h(x) k(x) ). Then, ( g''(x) = h'(x) k(x) + h(x) k'(x) ).First, compute ( h(x) = 2x (1 - x^2) = 2x - 2x^3 ). Then, ( h'(x) = 2 - 6x^2 ).Next, ( k(x) = e^{-x^2} ), so ( k'(x) = -2x e^{-x^2} ).Now, compute ( g''(x) ):[g''(x) = (2 - 6x^2) e^{-x^2} + (2x - 2x^3)(-2x e^{-x^2})]Let me simplify each term:First term: ( (2 - 6x^2) e^{-x^2} )Second term: ( (2x - 2x^3)(-2x e^{-x^2}) = (-4x^2 + 4x^4) e^{-x^2} )Combine both terms:[g''(x) = (2 - 6x^2 - 4x^2 + 4x^4) e^{-x^2} = (2 - 10x^2 + 4x^4) e^{-x^2}]Simplify the polynomial:[4x^4 - 10x^2 + 2]So, ( g''(x) = (4x^4 - 10x^2 + 2) e^{-x^2} )Now, evaluate ( g''(x) ) at each critical point.1. At ( x = 0 ):[g''(0) = (0 - 0 + 2) e^{0} = 2 times 1 = 2 > 0]Since the second derivative is positive, ( x = 0 ) is a local minimum.2. At ( x = 1 ):Compute ( g''(1) ):[4(1)^4 - 10(1)^2 + 2 = 4 - 10 + 2 = -4]So, ( g''(1) = (-4) e^{-1} < 0 ). Therefore, ( x = 1 ) is a local maximum.3. At ( x = -1 ):Similarly, ( g''(-1) = 4(-1)^4 - 10(-1)^2 + 2 = 4 - 10 + 2 = -4 ). So, ( g''(-1) = (-4) e^{-1} < 0 ). Thus, ( x = -1 ) is also a local maximum.So, we have two local maxima at ( x = 1 ) and ( x = -1 ). Since the function ( g(x) ) is even (because ( g(-x) = (-x)^2 e^{-(-x)^2} = x^2 e^{-x^2} = g(x) )), it's symmetric about the y-axis. Therefore, the maximum values at ( x = 1 ) and ( x = -1 ) are equal.Now, let's compute the maximum value. Let's compute ( g(1) ):[g(1) = (1)^2 e^{-(1)^2} = 1 times e^{-1} = frac{1}{e}]Similarly, ( g(-1) = frac{1}{e} ). So, the maximum value of ( g(x) ) is ( frac{1}{e} ), occurring at ( x = pm 1 ).Wait, let me just double-check if these are indeed the global maxima. Since as ( x ) approaches infinity, ( x^2 e^{-x^2} ) tends to zero because the exponential decay dominates the polynomial growth. Similarly, as ( x ) approaches negative infinity, the same happens. Therefore, the function tends to zero at both ends, so the local maxima at ( x = pm 1 ) must be the global maxima.Hence, the maximum value is ( frac{1}{e} ) at ( x = pm 1 ).Just to recap, for problem 1, the area under the curve is ( sqrt{pi} ), and for problem 2, the maximum of ( g(x) ) is ( frac{1}{e} ) at ( x = pm 1 ).I think that's all. I don't see any mistakes in my reasoning, but let me just quickly verify the derivative calculations.For ( g'(x) ), I had ( 2x e^{-x^2} (1 - x^2) ). Plugging in ( x = 1 ), we get ( 2(1) e^{-1} (1 - 1) = 0 ), which is correct. Similarly, at ( x = 0 ), it's zero. So, the critical points are correctly identified.And the second derivative at ( x = 1 ) was negative, confirming a maximum. At ( x = 0 ), positive, confirming a minimum. So, all steps check out.Final Answer1. The area under the curve is boxed{sqrt{pi}}.2. The maximum value of ( g(x) ) is boxed{dfrac{1}{e}} and it occurs at ( x = boxed{pm 1} ).</think>

🧠:<think>Okay, so I have this problem about budget allocation and then projecting net revenue. Let me try to break it down step by step.First, for part 1, I need to allocate 10 million between three departments: Research, Operations, and Development. There are several constraints given, so I need to make sure I satisfy all of them.Let me define variables for each department:Let R = budget allocated to ResearchO = budget allocated to OperationsD = budget allocated to DevelopmentTotal budget is 10 million, so R + O + D = 10,000,000.Now, the constraints:1. Research must receive at least 40% of the total budget. So, R ≥ 0.4 * 10,000,000 = 4,000,000.2. Operations must receive between 25% and 35% of the total budget. So, 2,500,000 ≤ O ≤ 3,500,000.3. Development receives the remaining budget, which must be at least 1.5 million. So, D ≥ 1,500,000. Since D = 10,000,000 - R - O, this translates to 10,000,000 - R - O ≥ 1,500,000, which simplifies to R + O ≤ 8,500,000.4. Additionally, the sum of Research and Operations should not exceed 75% of the total budget. 75% of 10 million is 7,500,000. So, R + O ≤ 7,500,000.Wait, hold on. There's a conflict here. From constraint 3, R + O ≤ 8,500,000, but from constraint 4, R + O ≤ 7,500,000. So, the stricter constraint is R + O ≤ 7,500,000. So, we can ignore the 8,500,000 constraint because 7,500,000 is more restrictive.So, summarizing the constraints:1. R ≥ 4,000,0002. 2,500,000 ≤ O ≤ 3,500,0003. R + O ≤ 7,500,0004. D = 10,000,000 - R - O ≥ 1,500,000, which is automatically satisfied if R + O ≤ 8,500,000, but since R + O is already ≤7,500,000, D will be at least 2,500,000, which is more than the required 1,500,000. So, that's fine.So, our constraints are:- R ≥ 4,000,000- 2,500,000 ≤ O ≤ 3,500,000- R + O ≤ 7,500,000And R + O + D = 10,000,000But since we have R, O, and D, we can express D as 10,000,000 - R - O.Now, the problem is to find R, O, D that satisfy these constraints. But wait, is this a linear programming problem where we need to maximize or minimize something? The problem says "formulate and solve a linear programming problem to determine the optimal budget allocation for each department."But it doesn't specify an objective function. Hmm. Maybe the goal is to just find a feasible allocation? Or perhaps to minimize or maximize something. Since it's about budget allocation, maybe the objective is to maximize the budget for a particular department? Or perhaps it's just to find any feasible solution.Wait, the problem says "the optimal budget allocation." So, maybe we need to define an objective. Since it's not given, perhaps it's just to find any feasible allocation, but given that it's called a linear programming problem, it's more likely that we need to define an objective function.But the problem doesn't specify whether to maximize or minimize something. Maybe the default is to just find a feasible solution? Or perhaps it's implied that we need to maximize something.Wait, maybe the problem is just to allocate the budget according to the constraints, so it's more of a system of inequalities rather than an optimization problem. But the question says "formulate and solve a linear programming problem," so I think we need to define an objective function.Perhaps the objective is to maximize the budget for Development, which is the remaining. Or maybe to minimize the budget for Operations. Since it's not specified, maybe we can assume that the goal is to maximize the budget for Research, since it's the most constrained with a minimum of 40%.Alternatively, maybe the goal is to maximize the budget for Development, as that's the one with a minimum of 1.5 million.Wait, let me read the problem again."Formulate and solve a linear programming problem to determine the optimal budget allocation for each department."It doesn't specify what to optimize, just to allocate optimally. Maybe in such cases, the objective is to maximize the budget for a particular department, but since it's not specified, perhaps the problem is just to find a feasible allocation.But in linear programming, you need an objective function. So, maybe the problem is to maximize the budget for Development, which is the only one without a specified maximum, except for the sum constraint.Alternatively, perhaps the problem is to minimize the budget for Operations, given that it's between 25% and 35%.Wait, maybe the problem is just to find the minimal possible budget for Operations, which would be 25%, but let's see.Wait, if we set R to its minimum, which is 4,000,000, and O to its minimum, which is 2,500,000, then R + O = 6,500,000, which is less than 7,500,000. So, D would be 3,500,000, which is above the 1.5 million requirement. So, that's a feasible solution.Alternatively, if we set R to 4,000,000 and O to 3,500,000, then R + O = 7,500,000, which is exactly the maximum allowed, so D would be 2,500,000, which is still above the 1.5 million.But since the problem says "optimal," perhaps we need to define an objective. Maybe the goal is to maximize the budget for Development, which would mean minimizing R + O.So, to maximize D, we need to minimize R + O.Given that R has a minimum of 4,000,000, and O has a minimum of 2,500,000, the minimal R + O is 6,500,000, which would make D = 3,500,000.Alternatively, if we want to minimize D, we need to maximize R + O, which is 7,500,000, so D would be 2,500,000.But since the problem says "the optimal budget allocation," without specifying, perhaps we need to assume that the objective is to maximize the budget for Development, which is the only one without a specified maximum beyond the sum constraint.Alternatively, maybe the problem is just to find a feasible allocation, but in that case, it's not an optimization problem.Wait, perhaps the problem is to maximize the budget for Research, which is already at its minimum. So, maybe the problem is just to find any feasible allocation, but given that it's called a linear programming problem, I think we need to define an objective.Alternatively, perhaps the problem is to minimize the budget for Operations, which would be 2,500,000, given that R is at its minimum.But without a specified objective, it's a bit unclear. Maybe the problem is to find the allocation that satisfies all constraints, and since it's a linear programming problem, perhaps the objective is to maximize the budget for Development.So, let's proceed with that assumption: maximize D, which is equivalent to minimizing R + O.So, our objective function is to maximize D = 10,000,000 - R - O, which is equivalent to minimizing R + O.Subject to:R ≥ 4,000,000O ≥ 2,500,000R + O ≤ 7,500,000And R, O ≥ 0So, let's set up the problem.Minimize R + OSubject to:R ≥ 4,000,000O ≥ 2,500,000R + O ≤ 7,500,000And R, O ≥ 0But since R and O are already constrained to be above certain values, the feasible region is defined by R ≥4, O ≥2.5, and R + O ≤7.5.So, to minimize R + O, we set R and O to their minimums, but we have to check if R + O ≤7.5.So, R =4, O=2.5, R + O=6.5 ≤7.5, so that's feasible.Therefore, the minimal R + O is 6.5, so D=3.5.So, the optimal allocation would be R=4, O=2.5, D=3.5.But let's check if that's the case.Alternatively, if we set R=4, O=3.5, then R + O=7.5, D=2.5.But that would minimize D, not maximize.So, to maximize D, we need to minimize R + O, which is achieved by setting R and O to their minimums.So, R=4, O=2.5, D=3.5.But let's verify if this satisfies all constraints.R=4 ≥4: yes.O=2.5 ≥2.5 and ≤3.5: yes.R + O=6.5 ≤7.5: yes.D=3.5 ≥1.5: yes.So, that's a feasible solution.Alternatively, is there a way to have R + O less than 6.5? No, because R can't be less than 4, and O can't be less than 2.5.So, 6.5 is the minimal R + O, hence D=3.5 is the maximal D.Therefore, the optimal allocation is R=4, O=2.5, D=3.5.Wait, but let me think again. If the objective is to maximize D, then yes, that's correct. But if the objective is different, say, to minimize D, then we would set R + O to its maximum, which is 7.5, making D=2.5.But since the problem says "optimal," and without a specified objective, perhaps the problem is just to find any feasible allocation. But given that it's a linear programming problem, I think we need to define an objective.Alternatively, maybe the problem is to maximize the budget for Research, which is already at its minimum. So, perhaps the problem is to find the allocation that satisfies all constraints, and since it's a linear programming problem, perhaps the objective is to maximize the budget for Development.Therefore, I think the optimal allocation is R=4, O=2.5, D=3.5.Now, moving to part 2.Given the optimal budget allocation from problem 1, which is R=4, O=2.5, D=3.5, so total budget is 10 million.Now, the institution's revenue grows at a continuous annual growth rate of 5%, while expenses grow at a continuous annual growth rate of 3%. We need to calculate the projected net revenue after 5 years, starting with an initial revenue of 15 million and initial expenses equal to the total annual budget from problem 1, which is 10 million.So, net revenue is revenue minus expenses.Given that both revenue and expenses grow continuously, we can model them using exponential growth.The formula for continuous growth is:Final amount = Initial amount * e^(rt)Where r is the growth rate, t is time in years.So, revenue after 5 years: R(t) = 15,000,000 * e^(0.05*5)Expenses after 5 years: E(t) = 10,000,000 * e^(0.03*5)Net revenue = R(t) - E(t)So, let's compute these.First, compute R(t):R(t) = 15,000,000 * e^(0.25)Similarly, E(t) = 10,000,000 * e^(0.15)Compute e^0.25 and e^0.15.We know that e^0.25 ≈ 1.2840254e^0.15 ≈ 1.1618342So,R(t) ≈ 15,000,000 * 1.2840254 ≈ 19,260,381E(t) ≈ 10,000,000 * 1.1618342 ≈ 11,618,342Net revenue ≈ 19,260,381 - 11,618,342 ≈ 7,642,039So, approximately 7,642,039.But let me compute it more accurately.Compute e^0.25:We can use a calculator:e^0.25 ≈ 1.2840254066e^0.15 ≈ 1.1618342428So,R(t) = 15,000,000 * 1.2840254066 = 15,000,000 * 1.2840254066Let's compute 15,000,000 * 1.2840254066:15,000,000 * 1 = 15,000,00015,000,000 * 0.2840254066 = 15,000,000 * 0.2840254066Compute 15,000,000 * 0.2 = 3,000,00015,000,000 * 0.0840254066 ≈ 15,000,000 * 0.08 = 1,200,000; 15,000,000 * 0.0040254066 ≈ 60,381.099So, total ≈ 3,000,000 + 1,200,000 + 60,381.099 ≈ 4,260,381.099So, R(t) ≈ 15,000,000 + 4,260,381.099 ≈ 19,260,381.099 ≈ 19,260,381.10Similarly, E(t) = 10,000,000 * 1.1618342428 ≈ 11,618,342.43So, Net revenue ≈ 19,260,381.10 - 11,618,342.43 ≈ 7,642,038.67So, approximately 7,642,038.67, which we can round to 7,642,039.Therefore, the projected net revenue after 5 years is approximately 7,642,039.But let me double-check the calculations.Alternatively, using more precise calculations:Compute R(t):15,000,000 * e^(0.05*5) = 15,000,000 * e^0.25e^0.25 ≈ 1.284025406615,000,000 * 1.2840254066 = 15,000,000 * 1 + 15,000,000 * 0.2840254066= 15,000,000 + 4,260,381.099 ≈ 19,260,381.099Similarly, E(t):10,000,000 * e^(0.03*5) = 10,000,000 * e^0.15 ≈ 10,000,000 * 1.1618342428 ≈ 11,618,342.43Net revenue: 19,260,381.099 - 11,618,342.43 ≈ 7,642,038.67Yes, that's correct.So, the projected net revenue after 5 years is approximately 7,642,039.</think>

🧠:<think>Okay, so I have two optimization problems to solve for Alex, the farmer. Let me tackle them one by one.Starting with the first problem: Field A can be planted with either Crop X or Crop Y. The yield functions are given as Y_X(x) = 100x - 0.5x² and Y_Y(y) = 80y - 0.3y². The total resources allocated to Field A can't exceed 120 units. I need to find the optimal allocation between x and y to maximize the total yield.Hmm, so since both crops are in the same field, the resources allocated to each must add up to 120. That is, x + y = 120. So, I can express y as 120 - x. Then, the total yield would be Y_X(x) + Y_Y(y) = 100x - 0.5x² + 80y - 0.3y². Substituting y = 120 - x, the total yield becomes a function of x alone.Let me write that out:Total Yield, Y_total(x) = 100x - 0.5x² + 80(120 - x) - 0.3(120 - x)².I need to expand this expression and then find the value of x that maximizes Y_total.First, let's compute each term:100x is straightforward.-0.5x² is also straightforward.80*(120 - x) = 80*120 - 80x = 9600 - 80x.-0.3*(120 - x)²: Let's expand (120 - x)² first. That's 120² - 2*120*x + x² = 14400 - 240x + x². Multiply by -0.3: -0.3*14400 + 0.3*240x - 0.3x² = -4320 + 72x - 0.3x².Now, let's combine all these terms:Y_total(x) = 100x - 0.5x² + 9600 - 80x - 4320 + 72x - 0.3x².Combine like terms:First, the constants: 9600 - 4320 = 5280.Next, the x terms: 100x - 80x + 72x = (100 - 80 + 72)x = 92x.Then, the x² terms: -0.5x² - 0.3x² = -0.8x².So, Y_total(x) = -0.8x² + 92x + 5280.This is a quadratic function in terms of x, and since the coefficient of x² is negative (-0.8), the parabola opens downward, meaning the maximum is at the vertex.The vertex of a parabola given by ax² + bx + c is at x = -b/(2a). So here, a = -0.8, b = 92.Calculating x:x = -92 / (2*(-0.8)) = -92 / (-1.6) = 57.5.So, x = 57.5 units. Then y = 120 - x = 120 - 57.5 = 62.5 units.Wait, but resources are in arbitrary units, so fractional units are acceptable? I think so, unless specified otherwise.Let me double-check my calculations to make sure I didn't make a mistake.First, expanding Y_Y(y):80y - 0.3y², with y = 120 - x.So, substituting:80*(120 - x) = 9600 - 80x.-0.3*(120 - x)²: as above, 14400 - 240x + x², multiplied by -0.3 gives -4320 + 72x - 0.3x².Adding all terms:100x - 0.5x² + 9600 - 80x - 4320 + 72x - 0.3x².Combine constants: 9600 - 4320 = 5280.Combine x terms: 100x - 80x + 72x = 92x.Combine x² terms: -0.5x² - 0.3x² = -0.8x².So, yes, that seems correct. Then, vertex at x = -92/(2*(-0.8)) = 57.5.Therefore, the optimal allocation is 57.5 units to Crop X and 62.5 units to Crop Y in Field A.Moving on to the second problem: Field B is planted entirely with Crop Z, whose yield is given by Y_Z(u, v) = 50u + 60v - 0.4u² - 0.6v² + 0.2uv. The total resources allocated to Field B are 200 units, so u + v = 200. I need to find the allocation (u, v) that maximizes Y_Z.This is a constrained optimization problem. The function Y_Z is quadratic in u and v, and the constraint is u + v = 200.I can use substitution to reduce it to a single variable. Let me express v as 200 - u, then substitute into Y_Z.So, Y_Z(u, 200 - u) = 50u + 60*(200 - u) - 0.4u² - 0.6*(200 - u)² + 0.2u*(200 - u).Let me compute each term step by step.First, 50u is straightforward.60*(200 - u) = 12000 - 60u.-0.4u² is straightforward.-0.6*(200 - u)²: Let's expand (200 - u)² first. That's 40000 - 400u + u². Multiply by -0.6: -24000 + 240u - 0.6u².0.2u*(200 - u) = 40u - 0.2u².Now, let's combine all these terms:Y_Z(u) = 50u + 12000 - 60u - 0.4u² - 24000 + 240u - 0.6u² + 40u - 0.2u².Let me combine like terms:Constants: 12000 - 24000 = -12000.u terms: 50u - 60u + 240u + 40u = (50 - 60 + 240 + 40)u = 270u.u² terms: -0.4u² - 0.6u² - 0.2u² = (-0.4 - 0.6 - 0.2)u² = -1.2u².So, Y_Z(u) = -1.2u² + 270u - 12000.Again, this is a quadratic function in u, opening downward (since coefficient of u² is negative), so the maximum is at the vertex.Vertex at u = -b/(2a). Here, a = -1.2, b = 270.Calculating u:u = -270 / (2*(-1.2)) = -270 / (-2.4) = 112.5.Therefore, u = 112.5 units, and v = 200 - u = 87.5 units.Let me verify my calculations.Starting with substitution:Y_Z(u, v) = 50u + 60v - 0.4u² - 0.6v² + 0.2uv.Substituting v = 200 - u:50u + 60*(200 - u) - 0.4u² - 0.6*(200 - u)² + 0.2u*(200 - u).Compute each term:50u.60*(200 - u) = 12000 - 60u.-0.4u².-0.6*(200 - u)²: 200² = 40000, so (200 - u)² = 40000 - 400u + u². Multiply by -0.6: -24000 + 240u - 0.6u².0.2u*(200 - u) = 40u - 0.2u².Combine all terms:50u + 12000 - 60u - 0.4u² - 24000 + 240u - 0.6u² + 40u - 0.2u².Constants: 12000 - 24000 = -12000.u terms: 50u - 60u + 240u + 40u = 270u.u² terms: -0.4u² - 0.6u² - 0.2u² = -1.2u².So, Y_Z(u) = -1.2u² + 270u - 12000.Vertex at u = -270/(2*(-1.2)) = 112.5.Yes, that's correct. So, u = 112.5, v = 87.5.Wait, let me just make sure that when I substituted, I didn't make any sign errors.Looking back:-0.6*(200 - u)²: that's correct, expanding to -24000 + 240u - 0.6u².0.2u*(200 - u): 40u - 0.2u², correct.Adding all together:50u -60u +240u +40u = 270u.-0.4u² -0.6u² -0.2u² = -1.2u².Constants: 12000 -24000 = -12000.Yes, so the quadratic is correct. Therefore, u = 112.5, v = 87.5.So, summarizing:For Field A, allocate 57.5 units to Crop X and 62.5 units to Crop Y.For Field B, allocate 112.5 units to u and 87.5 units to v.I think that's it. Let me just make sure I didn't mix up any coefficients.In the first problem, the yields were Y_X(x) = 100x -0.5x² and Y_Y(y) =80y -0.3y². Then, with x + y =120, we substituted and found x=57.5, y=62.5.In the second problem, Y_Z(u, v) =50u +60v -0.4u² -0.6v² +0.2uv, with u + v=200. Substituted v=200 -u, expanded, and found u=112.5, v=87.5.Yes, that seems consistent.Final AnswerFor Field A, the optimal allocation is (boxed{57.5}) units to Crop X and (boxed{62.5}) units to Crop Y. For Field B, the optimal allocation is (boxed{112.5}) units to (u) and (boxed{87.5}) units to (v).</think>