🧠:<think>Alright, so I've got this problem about a logarithmic spiral of a cephalopod shell. It's divided into two parts. Let me tackle them one by one.Problem 1: Finding constants ( a ) and ( b )The equation given is ( r = ae^{btheta} ). I know that at ( theta = 0 ), the radius ( r ) is 2 cm. Let me plug that in:( r(0) = ae^{b cdot 0} = ae^0 = a cdot 1 = a ).So, ( a = 2 ) cm. That was straightforward.Next, at ( theta = frac{pi}{2} ), the radius is 4 cm. Let's plug that into the equation:( rleft(frac{pi}{2}right) = 2e^{b cdot frac{pi}{2}} = 4 ).So, ( 2e^{frac{bpi}{2}} = 4 ).Divide both sides by 2:( e^{frac{bpi}{2}} = 2 ).To solve for ( b ), take the natural logarithm of both sides:( frac{bpi}{2} = ln 2 ).Multiply both sides by ( frac{2}{pi} ):( b = frac{2}{pi} ln 2 ).Okay, so ( a = 2 ) and ( b = frac{2}{pi} ln 2 ). That seems right.Problem 2: Calculating the arc length ( L ) from ( theta = 0 ) to ( theta = pi )The formula given is:( L = int_{0}^{pi} sqrt{r^2 + left( frac{dr}{dtheta} right)^2} , dtheta ).First, let's write down ( r ) and ( frac{dr}{dtheta} ).We have ( r = 2e^{btheta} ), and ( b = frac{2}{pi} ln 2 ). So,( r = 2e^{left( frac{2}{pi} ln 2 right) theta} ).Let me simplify ( frac{dr}{dtheta} ):( frac{dr}{dtheta} = 2 cdot frac{2}{pi} ln 2 cdot e^{left( frac{2}{pi} ln 2 right) theta} ).Simplify that:( frac{dr}{dtheta} = frac{4}{pi} ln 2 cdot e^{left( frac{2}{pi} ln 2 right) theta} ).So, both ( r ) and ( frac{dr}{dtheta} ) are expressed in terms of exponentials with the same exponent. Let me denote ( k = frac{2}{pi} ln 2 ) for simplicity.Then, ( r = 2e^{ktheta} ) and ( frac{dr}{dtheta} = 2k e^{ktheta} ).Now, let's compute the integrand:( sqrt{r^2 + left( frac{dr}{dtheta} right)^2} = sqrt{(2e^{ktheta})^2 + (2k e^{ktheta})^2} ).Factor out ( (2e^{ktheta})^2 ):( sqrt{(2e^{ktheta})^2 (1 + k^2)} = 2e^{ktheta} sqrt{1 + k^2} ).So, the integrand simplifies to ( 2sqrt{1 + k^2} cdot e^{ktheta} ).Therefore, the integral becomes:( L = int_{0}^{pi} 2sqrt{1 + k^2} cdot e^{ktheta} , dtheta ).Since ( 2sqrt{1 + k^2} ) is a constant with respect to ( theta ), we can factor it out:( L = 2sqrt{1 + k^2} int_{0}^{pi} e^{ktheta} , dtheta ).Compute the integral:( int e^{ktheta} dtheta = frac{1}{k} e^{ktheta} + C ).So, evaluating from 0 to ( pi ):( int_{0}^{pi} e^{ktheta} dtheta = frac{1}{k} (e^{kpi} - e^{0}) = frac{1}{k} (e^{kpi} - 1) ).Putting it all together:( L = 2sqrt{1 + k^2} cdot frac{1}{k} (e^{kpi} - 1) ).Simplify:( L = frac{2sqrt{1 + k^2}}{k} (e^{kpi} - 1) ).Now, let's substitute back ( k = frac{2}{pi} ln 2 ):First, compute ( k pi = frac{2}{pi} ln 2 cdot pi = 2 ln 2 ).So, ( e^{kpi} = e^{2 ln 2} = (e^{ln 2})^2 = 2^2 = 4 ).So, ( e^{kpi} - 1 = 4 - 1 = 3 ).Now, compute ( sqrt{1 + k^2} ):( k = frac{2}{pi} ln 2 ), so ( k^2 = left( frac{2}{pi} ln 2 right)^2 = frac{4}{pi^2} (ln 2)^2 ).Thus,( 1 + k^2 = 1 + frac{4}{pi^2} (ln 2)^2 ).So,( sqrt{1 + k^2} = sqrt{1 + frac{4}{pi^2} (ln 2)^2} ).Putting it all back into ( L ):( L = frac{2 cdot sqrt{1 + frac{4}{pi^2} (ln 2)^2}}{ frac{2}{pi} ln 2 } cdot 3 ).Simplify the denominator:( frac{2}{pi} ln 2 ) in the denominator can be written as multiplying by ( frac{pi}{2 ln 2} ).So,( L = 2 cdot sqrt{1 + frac{4}{pi^2} (ln 2)^2} cdot frac{pi}{2 ln 2} cdot 3 ).The 2 in the numerator and denominator cancel out:( L = sqrt{1 + frac{4}{pi^2} (ln 2)^2} cdot frac{pi}{ln 2} cdot 3 ).So,( L = 3 cdot frac{pi}{ln 2} cdot sqrt{1 + frac{4}{pi^2} (ln 2)^2} ).Let me see if this can be simplified further.Factor out ( frac{1}{pi^2} ) inside the square root:( sqrt{1 + frac{4 (ln 2)^2}{pi^2}} = sqrt{ frac{pi^2 + 4 (ln 2)^2}{pi^2} } = frac{ sqrt{pi^2 + 4 (ln 2)^2} }{ pi } ).So, substituting back:( L = 3 cdot frac{pi}{ln 2} cdot frac{ sqrt{pi^2 + 4 (ln 2)^2} }{ pi } ).The ( pi ) in the numerator and denominator cancel:( L = 3 cdot frac{ sqrt{pi^2 + 4 (ln 2)^2} }{ ln 2 } ).So, that's the simplified form.Let me compute the numerical value to check if it makes sense, but since the problem says to express it in terms of elementary functions, maybe we can leave it like that.But let me see if I can write it differently.Alternatively, note that ( sqrt{pi^2 + 4 (ln 2)^2} ) is just a constant, so the expression is:( L = frac{3 sqrt{pi^2 + 4 (ln 2)^2} }{ ln 2 } ).Alternatively, factor out 4 inside the square root:( sqrt{pi^2 + 4 (ln 2)^2} = sqrt{4 left( frac{pi^2}{4} + (ln 2)^2 right)} = 2 sqrt{ left( frac{pi}{2} right)^2 + (ln 2)^2 } ).So, substituting back:( L = frac{3 cdot 2 sqrt{ left( frac{pi}{2} right)^2 + (ln 2)^2 } }{ ln 2 } = frac{6 sqrt{ left( frac{pi}{2} right)^2 + (ln 2)^2 } }{ ln 2 } ).But I don't think that's necessarily simpler. So, perhaps the earlier expression is better.Wait, let me check my steps again to make sure I didn't make any mistakes.Starting from the integral:( L = int_{0}^{pi} sqrt{r^2 + (dr/dtheta)^2} dtheta ).We had ( r = 2 e^{ktheta} ), ( dr/dtheta = 2k e^{ktheta} ).So, ( r^2 + (dr/dtheta)^2 = 4 e^{2ktheta} + 4k^2 e^{2ktheta} = 4(1 + k^2) e^{2ktheta} ).Thus, the square root is ( 2 sqrt{1 + k^2} e^{ktheta} ).Therefore, the integral becomes ( 2 sqrt{1 + k^2} int_{0}^{pi} e^{ktheta} dtheta ).Which is correct.Then, integrating ( e^{ktheta} ) gives ( frac{1}{k} (e^{kpi} - 1) ).So, ( L = frac{2 sqrt{1 + k^2}}{k} (e^{kpi} - 1) ).Substituting ( k = frac{2}{pi} ln 2 ), so ( kpi = 2 ln 2 ), so ( e^{kpi} = 4 ).Thus, ( e^{kpi} - 1 = 3 ).So, ( L = frac{2 sqrt{1 + left( frac{2}{pi} ln 2 right)^2}}{ frac{2}{pi} ln 2 } cdot 3 ).Simplify denominator: ( frac{2}{pi} ln 2 ) in the denominator is multiplied by ( frac{pi}{2 ln 2} ).So, ( 2 cdot frac{pi}{2 ln 2} = frac{pi}{ln 2} ).Thus, ( L = frac{pi}{ln 2} cdot sqrt{1 + left( frac{2}{pi} ln 2 right)^2} cdot 3 ).Which is the same as:( L = 3 cdot frac{pi}{ln 2} cdot sqrt{1 + frac{4 (ln 2)^2}{pi^2}} ).Which simplifies to:( L = 3 cdot frac{sqrt{pi^2 + 4 (ln 2)^2}}{ln 2} ).Yes, that seems correct.Alternatively, if I compute the numerical value, just to have an idea:Compute ( pi^2 approx 9.8696 ), ( ln 2 approx 0.6931 ), so ( 4 (ln 2)^2 approx 4 * 0.4804 = 1.9216 ).Thus, ( pi^2 + 4 (ln 2)^2 approx 9.8696 + 1.9216 = 11.7912 ).So, ( sqrt{11.7912} approx 3.434 ).Then, ( 3 * 3.434 / 0.6931 approx 3 * 4.95 approx 14.85 ) cm.But since the problem asks for an expression in terms of elementary functions, I should leave it in exact form.So, the final expression is:( L = frac{3 sqrt{pi^2 + 4 (ln 2)^2}}{ln 2} ).Alternatively, factor out the 4:( L = frac{3 sqrt{4 left( frac{pi^2}{4} + (ln 2)^2 right)}}{ln 2} = frac{6 sqrt{ left( frac{pi}{2} right)^2 + (ln 2)^2 }}{ln 2} ).But I think the first form is simpler.Final Answer1. The constants are ( a = boxed{2} ) cm and ( b = boxed{dfrac{2}{pi} ln 2} ).2. The arc length ( L ) is ( boxed{dfrac{3 sqrt{pi^2 + 4 (ln 2)^2}}{ln 2}} ) cm.</think>

🧠:<think>Okay, so I need to help Dr. Smith figure out the optimal study and training hours for student athletes. There are two sub-problems here, and I need to tackle them one by one. Let me start with Sub-problem 1.First, let me understand the problem. The student athlete has to balance studying and training. They have some constraints on minimum hours for each activity and a total maximum. The goal is to maximize the combined performance score P, which is the sum of academic performance A and athletic performance T.Given the equations:- A = 2S - 0.5T_s- T = 3T_s + 2SWhere S is study hours and T_s is training hours. The total hours S + T_s can't exceed 40. Also, S must be at least 15, and T_s must be at least 10.So, the combined performance score P is A + T, which would be:P = (2S - 0.5T_s) + (3T_s + 2S)Let me simplify that:P = 2S - 0.5T_s + 3T_s + 2SCombine like terms:2S + 2S = 4S-0.5T_s + 3T_s = 2.5T_sSo, P = 4S + 2.5T_sOkay, so we need to maximize P = 4S + 2.5T_s, subject to the constraints:1. S >= 152. T_s >= 103. S + T_s <= 40Additionally, since hours can't be negative, S >= 0 and T_s >= 0, but the first two constraints already cover that.So, this is a linear programming problem. To maximize P, we can use the graphical method since there are only two variables.Let me write down the constraints:1. S >= 152. T_s >= 103. S + T_s <= 40We can plot these on a graph with S on the x-axis and T_s on the y-axis.First, the feasible region is defined by these constraints. The maximum P will occur at one of the vertices of this feasible region.Let me find the vertices.The intersection points of the constraints:1. Intersection of S = 15 and T_s = 10: (15,10)2. Intersection of S = 15 and S + T_s = 40: (15, 25)3. Intersection of T_s = 10 and S + T_s = 40: (30,10)So, the feasible region is a polygon with vertices at (15,10), (15,25), and (30,10). Let me confirm that.Wait, actually, when S = 15, T_s can be from 10 up to 25 (since 15 + 25 = 40). Similarly, when T_s = 10, S can be from 15 up to 30. So, the feasible region is a triangle with those three vertices.So, to find the maximum P, we evaluate P at each of these three vertices.Let me compute P for each:1. At (15,10):P = 4*15 + 2.5*10 = 60 + 25 = 852. At (15,25):P = 4*15 + 2.5*25 = 60 + 62.5 = 122.53. At (30,10):P = 4*30 + 2.5*10 = 120 + 25 = 145So, comparing these, 145 is the highest. Therefore, the optimal allocation is S = 30 hours and T_s = 10 hours.Wait, but hold on. Let me make sure I didn't make a mistake. Because 30 hours studying and 10 hours training. But the total is 40, which is within the limit. And both meet the minimums: 30 >=15, 10 >=10.But let me think about the equations for A and T. A = 2S - 0.5T_s, so if S is 30, A = 60 - 5 = 55. T = 3T_s + 2S = 30 + 60 = 90. So P = 55 + 90 = 145. That's correct.Wait, but is this the maximum? Because when I increased S, P increased more because the coefficient of S in P is 4, which is higher than 2.5 for T_s. So, to maximize P, it's better to allocate as much as possible to S, which is why the maximum is at (30,10). That makes sense.So, Sub-problem 1 answer is S = 30, T_s = 10.Now, moving on to Sub-problem 2. The university introduces a new policy. It requires at least 20 hours per week dedicated to either studying or training, but allows a maximum of 35 hours per week in total.Wait, let me parse that. So, the new constraints are:- Either S or T_s must be at least 20? Or the sum of S and T_s must be at least 20? Wait, the wording is: "at least 20 hours per week to be dedicated to either studying or training". Hmm, that could be interpreted in two ways.But in the context, since they are talking about allocating hours, it's more likely that the total hours spent on either studying or training must be at least 20. But since they already have a total of 40, which is more than 20, but the new policy might be changing the total available hours.Wait, the original total was 40, but the new policy allows a maximum of 35 hours per week in total. So, perhaps the new constraints are:- S + T_s <= 35- And either S >=20 or T_s >=20? Or maybe the total time spent on studying or training is at least 20? Wait, the wording is: "at least 20 hours per week to be dedicated to either studying or training but allows a maximum of 35 hours per week in total".Hmm, perhaps it's that the student must spend at least 20 hours on studying or training, but the total time can't exceed 35. So, the constraints become:1. S >=152. T_s >=103. S + T_s >=204. S + T_s <=35Wait, but originally, S + T_s was <=40. Now it's <=35, and also S + T_s >=20. But since S >=15 and T_s >=10, the minimum S + T_s is 25, which is more than 20. So, the constraint S + T_s >=20 is redundant because S + T_s >=25.Therefore, the new constraints are:1. S >=152. T_s >=103. S + T_s <=35So, we need to maximize P =4S +2.5T_s subject to:S >=15T_s >=10S + T_s <=35Again, a linear programming problem.Let me find the feasible region.The constraints:1. S >=152. T_s >=103. S + T_s <=35So, the feasible region is a polygon defined by the intersection of these constraints.Let me find the vertices.First, the intersection of S =15 and T_s =10: (15,10). But S + T_s =25, which is less than 35, so this point is inside the feasible region.But the vertices will be where the constraints intersect.So, the vertices are:1. Intersection of S=15 and S + T_s=35: (15,20)2. Intersection of T_s=10 and S + T_s=35: (25,10)3. Intersection of S=15 and T_s=10: (15,10) but as above, this is inside.Wait, actually, in the feasible region, the vertices are (15,20) and (25,10), and also the point where S=15, T_s=10 is a vertex, but since S + T_s=25 is less than 35, it's not on the boundary.Wait, actually, the feasible region is bounded by S >=15, T_s >=10, and S + T_s <=35.So, the feasible region is a polygon with vertices at (15,20), (25,10), and (15,10). But (15,10) is not on the boundary of S + T_s <=35, because 15 +10=25 <35.Wait, actually, no. The feasible region is the area where all constraints are satisfied. So, it's the intersection of S >=15, T_s >=10, and S + T_s <=35.So, the vertices are:1. Where S=15 and T_s=10: (15,10)2. Where S=15 and S + T_s=35: (15,20)3. Where T_s=10 and S + T_s=35: (25,10)So, three vertices: (15,10), (15,20), (25,10)Wait, but (15,10) is a vertex, but it's inside the feasible region because S + T_s=25 <35. But in terms of the feasible region, the boundaries are S=15, T_s=10, and S + T_s=35.So, the feasible region is a triangle with vertices at (15,20), (25,10), and (15,10). But (15,10) is a vertex, but since S + T_s=25 is less than 35, it's not on the edge.Wait, maybe I'm overcomplicating. Let me just plot it mentally.The line S + T_s=35 intersects S=15 at T_s=20, and intersects T_s=10 at S=25.So, the feasible region is a polygon bounded by:- From (15,20) to (25,10), along S + T_s=35- From (25,10) to (15,10), along T_s=10- From (15,10) to (15,20), along S=15So, the three vertices are (15,20), (25,10), and (15,10). But (15,10) is a vertex, but it's not on the S + T_s=35 line.Wait, but in terms of maximizing P, we need to check all vertices.So, let's compute P at each vertex:1. At (15,20):P =4*15 +2.5*20 =60 +50=1102. At (25,10):P=4*25 +2.5*10=100 +25=1253. At (15,10):P=4*15 +2.5*10=60 +25=85So, the maximum P is 125 at (25,10).Wait, but let me check if this is correct.At (25,10), S=25, T_s=10. So, A=2*25 -0.5*10=50 -5=45T=3*10 +2*25=30 +50=80So, P=45+80=125. Correct.At (15,20):A=2*15 -0.5*20=30 -10=20T=3*20 +2*15=60 +30=90P=20+90=110. Correct.At (15,10):A=2*15 -0.5*10=30 -5=25T=3*10 +2*15=30 +30=60P=25+60=85. Correct.So, indeed, the maximum is at (25,10) with P=125.But wait, is there any other point on the edge that could give a higher P? Since P is linear, the maximum must be at a vertex, so no.Therefore, the optimal allocation under the new policy is S=25, T_s=10.But let me think again. The new policy says "at least 20 hours per week to be dedicated to either studying or training". Wait, does that mean that either S >=20 or T_s >=20? Or does it mean that the total time spent on studying or training is at least 20? Because in the original problem, the student was already spending 15+10=25 hours, which is more than 20. So, if the new policy is that the total time must be at least 20, it's redundant because they were already above that.But the problem says: "a new policy that requires at least 20 hours per week to be dedicated to either studying or training but allows a maximum of 35 hours per week in total".So, perhaps the new constraints are:Either S >=20 or T_s >=20, and S + T_s <=35.But that complicates things because it's an OR constraint, which is non-linear. But in the initial problem, the constraints were S >=15, T_s >=10, and S + T_s <=40.So, the new policy changes the total maximum to 35, and adds a requirement that either S or T_s must be at least 20.Wait, that's a different interpretation. So, the constraints would be:1. S >=152. T_s >=103. Either S >=20 or T_s >=204. S + T_s <=35This complicates the feasible region because it's not a convex polygon anymore due to the OR condition.But in linear programming, we can't handle OR constraints directly. So, perhaps we need to split it into two cases:Case 1: S >=20Case 2: T_s >=20And find the maximum P in each case, then take the higher one.Let me try that.Case 1: S >=20Constraints:S >=20T_s >=10S + T_s <=35So, the feasible region is S >=20, T_s >=10, S + T_s <=35.The vertices are:- Intersection of S=20 and T_s=10: (20,10)- Intersection of S=20 and S + T_s=35: (20,15)- Intersection of T_s=10 and S + T_s=35: (25,10)So, vertices are (20,10), (20,15), (25,10)Compute P at each:1. (20,10): P=4*20 +2.5*10=80 +25=1052. (20,15): P=4*20 +2.5*15=80 +37.5=117.53. (25,10): P=125 as beforeSo, maximum in Case 1 is 125 at (25,10)Case 2: T_s >=20Constraints:S >=15T_s >=20S + T_s <=35So, the feasible region is S >=15, T_s >=20, S + T_s <=35.The vertices are:- Intersection of S=15 and T_s=20: (15,20)- Intersection of S=15 and S + T_s=35: (15,20) same as above- Intersection of T_s=20 and S + T_s=35: (15,20) same as aboveWait, actually, let me think.Wait, if T_s >=20, and S >=15, and S + T_s <=35.So, the feasible region is bounded by:- S=15, T_s from 20 to 20 (since 15 +20=35)- T_s=20, S from15 to15 (since 15 +20=35)- S + T_s=35, with S >=15 and T_s >=20.Wait, actually, the intersection points:1. S=15 and T_s=20: (15,20)2. S=15 and S + T_s=35: (15,20)3. T_s=20 and S + T_s=35: (15,20)Wait, that can't be right. If T_s=20, then S=15 is the only point because 15 +20=35.Wait, no, if T_s=20, S can be from 15 up to 15, because 15 +20=35. So, the feasible region is just a single point (15,20).Wait, that doesn't make sense. Let me think again.If T_s >=20, and S >=15, and S + T_s <=35.So, the minimum S is15, minimum T_s is20, and their sum is35.So, the only feasible point is (15,20). Because if T_s is more than20, say21, then S must be <=14, which contradicts S >=15. Similarly, if S is more than15, T_s must be less than20, which contradicts T_s >=20.Therefore, the only feasible point in Case 2 is (15,20).So, compute P at (15,20): 4*15 +2.5*20=60 +50=110.So, in Case 2, the maximum P is110.Comparing Case1 and Case2, the maximum P is125 in Case1 at (25,10).Therefore, under the new policy, the optimal allocation is S=25, T_s=10.Wait, but let me check if the new policy is interpreted as either S >=20 or T_s >=20, or if it's the total time spent on studying or training is at least20. Because in the initial problem, the student was already spending 25 hours, so the new policy might not add any new constraints except the total time limit.But the problem says: "a new policy that requires at least 20 hours per week to be dedicated to either studying or training but allows a maximum of 35 hours per week in total".So, perhaps it's that the student must spend at least20 hours on studying or training, but the total can't exceed35. But since they were already spending25, which is more than20, the only new constraint is S + T_s <=35.But in that case, the feasible region is S >=15, T_s >=10, S + T_s <=35.Which is what I did earlier, leading to maximum at (25,10).But the wording is a bit ambiguous. If it's interpreted as either S >=20 or T_s >=20, then we have to consider both cases, but as we saw, the maximum is still at (25,10).Therefore, regardless of the interpretation, the optimal allocation is S=25, T_s=10.Wait, but let me think again. If the policy is that either S or T_s must be at least20, then the feasible region is the union of two regions: one where S >=20 and the other where T_s >=20. So, the maximum could be in either region.But as we saw, in the S >=20 region, the maximum is at (25,10), and in the T_s >=20 region, the maximum is at (15,20). So, the overall maximum is still at (25,10).Therefore, the optimal allocation is S=25, T_s=10.So, summarizing:Sub-problem1: S=30, T_s=10Sub-problem2: S=25, T_s=10Wait, but in Sub-problem1, the total hours were 40, but in Sub-problem2, the total is limited to35, so the student has to reduce 5 hours. Since the coefficient of S in P is higher, it's better to reduce training hours, which have a lower coefficient, so the optimal is to keep T_s at minimum10, and reduce S from30 to25, keeping T_s=10.Yes, that makes sense.So, final answers:Sub-problem1: S=30, T_s=10Sub-problem2: S=25, T_s=10I think that's it.</think>

🧠:<think>Alright, so I have this problem about gravitational microlensing events, which is something I remember hearing about in my astrophysics class. It's a method used to detect exoplanets, right? The first part asks me to calculate the mass of the lensing object given some parameters. Let me try to break this down step by step.First, the formula for the Einstein radius is given as:[ theta_E = sqrt{frac{4GM}{c^2} frac{D_{LS}}{D_L D_S}} ]I need to solve for the mass M. So, I should rearrange this formula to express M in terms of the other variables.Starting with the equation:[ theta_E = sqrt{frac{4GM}{c^2} frac{D_{LS}}{D_L D_S}} ]Let me square both sides to eliminate the square root:[ theta_E^2 = frac{4GM}{c^2} frac{D_{LS}}{D_L D_S} ]Now, I can solve for M by multiplying both sides by ( frac{c^2 D_L D_S}{4G D_{LS}} ):[ M = frac{theta_E^2 c^2 D_L D_S}{4G D_{LS}} ]Okay, so that's the expression for M. Now, let's plug in the given values.The angular Einstein radius ( theta_E ) is 1 milliarcsecond (mas). I need to convert this into radians because the gravitational constant G is in SI units, which use radians.I remember that 1 arcsecond is ( frac{pi}{180 times 3600} ) radians. So, 1 milliarcsecond is ( frac{pi}{180 times 3600 times 1000} ) radians.Calculating that:First, 180 degrees is ( pi ) radians, so 1 degree is ( frac{pi}{180} ) radians. Then, 1 arcsecond is ( frac{pi}{180 times 3600} ) radians, and 1 milliarcsecond is ( frac{pi}{180 times 3600 times 1000} ) radians.Let me compute that:[ 1 text{ mas} = frac{pi}{180 times 3600 times 1000} text{ radians} ]Calculating the denominator:180 * 3600 = 648,000648,000 * 1000 = 648,000,000So,[ 1 text{ mas} = frac{pi}{648,000,000} approx 4.848 times 10^{-9} text{ radians} ]So, ( theta_E = 1 text{ mas} = 4.848 times 10^{-9} text{ radians} ).Next, the distances are given as ( D_L = 4 ) kpc and ( D_S = 8 ) kpc. I need to find ( D_{LS} ), which is the distance between the lens and the source. Since ( D_S ) is the distance from the observer to the source, and ( D_L ) is the distance from the observer to the lens, ( D_{LS} = D_S - D_L ) if the lens is between the observer and the source.So, ( D_{LS} = 8 text{ kpc} - 4 text{ kpc} = 4 text{ kpc} ).Now, I need to convert these distances into meters because the gravitational constant G is in meters cubed per kilogram per second squared.1 parsec (pc) is approximately ( 3.0857 times 10^{16} ) meters. So, 1 kiloparsec (kpc) is ( 3.0857 times 10^{19} ) meters.Therefore,( D_L = 4 text{ kpc} = 4 times 3.0857 times 10^{19} ) metersSimilarly,( D_S = 8 text{ kpc} = 8 times 3.0857 times 10^{19} ) metersAnd,( D_{LS} = 4 text{ kpc} = 4 times 3.0857 times 10^{19} ) metersLet me compute these:( D_L = 4 times 3.0857 times 10^{19} = 1.23428 times 10^{20} ) meters( D_S = 8 times 3.0857 times 10^{19} = 2.46856 times 10^{20} ) meters( D_{LS} = 4 times 3.0857 times 10^{19} = 1.23428 times 10^{20} ) metersNow, let's plug all these values into the formula for M.First, let's write down the formula again:[ M = frac{theta_E^2 c^2 D_L D_S}{4G D_{LS}} ]Plugging in the numbers:( theta_E = 4.848 times 10^{-9} ) radians( c = 3 times 10^8 ) m/s( G = 6.6743 times 10^{-11} ) m³ kg⁻¹ s⁻²So, let's compute each part step by step.First, compute ( theta_E^2 ):( (4.848 times 10^{-9})^2 = (4.848)^2 times 10^{-18} approx 23.5 times 10^{-18} = 2.35 times 10^{-17} )Next, compute ( c^2 ):( (3 times 10^8)^2 = 9 times 10^{16} )Now, multiply ( theta_E^2 ) and ( c^2 ):( 2.35 times 10^{-17} times 9 times 10^{16} = 2.35 times 9 times 10^{-1} = 21.15 times 10^{-1} = 2.115 )Wait, that seems too small. Let me check:Wait, 2.35e-17 * 9e16 = (2.35 * 9) * 10^(-17 +16) = 21.15 * 10^-1 = 2.115Yes, that's correct.Next, compute ( D_L times D_S ):( D_L = 1.23428 times 10^{20} ) m( D_S = 2.46856 times 10^{20} ) mMultiplying these:( 1.23428 times 2.46856 times 10^{40} )Calculating 1.23428 * 2.46856:Let me approximate this:1.23428 * 2.46856 ≈ 1.234 * 2.469 ≈ Let's compute 1.2 * 2.4 = 2.88, 1.2 * 0.069 ≈ 0.0828, 0.034 * 2.4 ≈ 0.0816, 0.034 * 0.069 ≈ 0.002346Adding up: 2.88 + 0.0828 + 0.0816 + 0.002346 ≈ 2.88 + 0.1667 ≈ 3.0467So approximately 3.0467e40 m²Now, multiply this by the previous result (2.115):2.115 * 3.0467e40 ≈ 2.115 * 3.0467 ≈ Let's compute 2 * 3.0467 = 6.0934, 0.115 * 3.0467 ≈ 0.3503, so total ≈ 6.0934 + 0.3503 ≈ 6.4437e40So, numerator is approximately 6.4437e40Now, the denominator is 4G D_{LS}Compute 4G:4 * 6.6743e-11 ≈ 26.6972e-11 = 2.66972e-10Compute D_{LS} = 1.23428e20 mSo, denominator = 2.66972e-10 * 1.23428e20Multiply these:2.66972e-10 * 1.23428e20 = (2.66972 * 1.23428) * 10^( -10 +20 ) = (approx 3.303) * 10^10Wait, let's compute 2.66972 * 1.23428:2 * 1.23428 = 2.468560.66972 * 1.23428 ≈ Let's compute 0.6 * 1.23428 = 0.740568, 0.06972 * 1.23428 ≈ 0.0858So total ≈ 0.740568 + 0.0858 ≈ 0.826368So total ≈ 2.46856 + 0.826368 ≈ 3.294928So, denominator ≈ 3.294928e10Now, the entire expression is numerator / denominator:6.4437e40 / 3.294928e10 ≈ (6.4437 / 3.294928) * 10^(40 -10) ≈ (1.955) * 10^30So, M ≈ 1.955e30 kgNow, I need to convert this mass into solar masses. 1 solar mass ( M_odot ) is approximately 1.9885e30 kg.So,M = 1.955e30 kg / 1.9885e30 kg/M_odot ≈ (1.955 / 1.9885) M_odot ≈ 0.983 M_odotSo, approximately 0.983 solar masses.Wait, that seems a bit low. Let me double-check my calculations.First, let's go back to the numerator:We had 2.115 * 3.0467e40 ≈ 6.4437e40Denominator: 4G D_{LS} = 4 * 6.6743e-11 * 1.23428e20Compute 4 * 6.6743e-11 = 2.66972e-102.66972e-10 * 1.23428e20 = 2.66972 * 1.23428 * 10^( -10 +20 ) = 3.2949e10So, 6.4437e40 / 3.2949e10 = (6.4437 / 3.2949) * 10^30 ≈ 1.955 * 10^30 kgWhich is about 0.983 M_odotHmm, that seems correct. So the mass is approximately 0.983 solar masses.Wait, but I thought microlensing events typically involve lensing masses like stars, which are around 1 solar mass. So, 0.983 is close to 1, which makes sense.So, maybe that's correct.Now, moving on to part 2.Given the magnification A = 1.34, and the formula:[ A = frac{u^2 + 2}{u sqrt{u^2 + 4}} ]We need to solve for u.This seems a bit tricky. Let me denote u as x for simplicity.So,[ A = frac{x^2 + 2}{x sqrt{x^2 + 4}} ]Given A = 1.34, so:[ 1.34 = frac{x^2 + 2}{x sqrt{x^2 + 4}} ]Let me denote ( y = x^2 ), so x = sqrt(y). Maybe that substitution will help.But perhaps it's better to square both sides to eliminate the square root.So, square both sides:[ (1.34)^2 = left( frac{x^2 + 2}{x sqrt{x^2 + 4}} right)^2 ]Compute left side:1.34^2 ≈ 1.7956Right side:[ frac{(x^2 + 2)^2}{x^2 (x^2 + 4)} ]So,1.7956 = (x^4 + 4x^2 + 4) / (x^4 + 4x^2)Let me write that as:1.7956 = (x^4 + 4x^2 + 4) / (x^4 + 4x^2)Let me denote z = x^4 + 4x^2Then, numerator is z + 4, denominator is z.So,1.7956 = (z + 4)/z = 1 + 4/zSubtract 1:0.7956 = 4/zSo,z = 4 / 0.7956 ≈ 5.027But z = x^4 + 4x^2 = 5.027So,x^4 + 4x^2 - 5.027 = 0This is a quadratic in terms of y = x^2:y^2 + 4y - 5.027 = 0Solve for y:Using quadratic formula:y = [-4 ± sqrt(16 + 4 * 5.027)] / 2Compute discriminant:16 + 4 * 5.027 = 16 + 20.108 = 36.108sqrt(36.108) ≈ 6.009So,y = [-4 ± 6.009]/2We discard the negative root because y = x^2 cannot be negative.So,y = (-4 + 6.009)/2 ≈ (2.009)/2 ≈ 1.0045So, y ≈ 1.0045Thus, x^2 ≈ 1.0045, so x ≈ sqrt(1.0045) ≈ 1.0022So, u ≈ 1.0022Wait, that seems very close to 1. Let me check if this makes sense.If u = 1, then plugging into the magnification formula:A = (1 + 2)/(1 * sqrt(1 + 4)) = 3 / sqrt(5) ≈ 3 / 2.236 ≈ 1.3416Which is very close to 1.34. So, u ≈ 1.002 is a good approximation.Therefore, the impact parameter u is approximately 1.002.But let me verify the calculations step by step to ensure accuracy.Starting from:1.7956 = (x^4 + 4x^2 + 4)/(x^4 + 4x^2)Let me denote x^4 + 4x^2 = zSo,1.7956 = (z + 4)/zMultiply both sides by z:1.7956 z = z + 4Subtract z:0.7956 z = 4So,z = 4 / 0.7956 ≈ 5.027So, x^4 + 4x^2 = 5.027Let me write this as:x^4 + 4x^2 - 5.027 = 0Let y = x^2:y^2 + 4y - 5.027 = 0Solutions:y = [-4 ± sqrt(16 + 20.108)] / 2 = [-4 ± sqrt(36.108)] / 2sqrt(36.108) ≈ 6.009So,y = (-4 + 6.009)/2 ≈ 2.009/2 ≈ 1.0045Thus, x = sqrt(1.0045) ≈ 1.0022So, yes, u ≈ 1.002Therefore, the impact parameter u is approximately 1.002.But let me check if plugging u = 1.002 into the magnification formula gives A ≈1.34.Compute numerator: u² + 2 = (1.002)^2 + 2 ≈ 1.004 + 2 = 3.004Denominator: u * sqrt(u² + 4) = 1.002 * sqrt(1.004 + 4) = 1.002 * sqrt(5.004) ≈ 1.002 * 2.237 ≈ 2.242So, A ≈ 3.004 / 2.242 ≈ 1.34Yes, that's correct.So, the impact parameter u is approximately 1.002.Therefore, the answers are:1. The mass M is approximately 0.983 solar masses.2. The impact parameter u is approximately 1.002.But let me express them with more precise decimals if possible.For part 1, M ≈ 0.983 M_odot, which is roughly 0.98 M_odot.For part 2, u ≈ 1.002, which is very close to 1.I think that's it.</think>

🧠:<think>Alright, so I've got this problem where I need to allocate a 10 million police department budget across four areas: personnel, equipment, training, and community outreach. The goal is to maximize both efficiency ROI and public satisfaction while sticking to certain constraints. Let me try to break this down step by step.First, let me list out the constraints to make sure I don't miss anything:1. Personnel costs must be at least 55% of the total budget. Since the total budget is 10 million, that means personnel needs to get at least 5.5 million.2. Equipment and training combined can't exceed 25% of the budget. 25% of 10 million is 2.5 million, so together, equipment and training must be less than or equal to 2.5 million.3. Community outreach needs to be at least 10% of the budget. 10% of 10 million is 1 million, so community outreach should get at least 1 million.Now, the total budget is 10 million, so all these allocations should add up to exactly 10 million. Let me write that as an equation:Personnel (P) + Equipment (E) + Training (T) + Community Outreach (C) = 10,000,000From the constraints, we have:P ≥ 5,500,000E + T ≤ 2,500,000C ≥ 1,000,000So, if I add up the minimums for personnel and community outreach, that's 5.5 million + 1 million = 6.5 million. That leaves 3.5 million for equipment and training, but they can't exceed 2.5 million. Hmm, that seems contradictory. Wait, no, actually, the 2.5 million is the maximum for equipment and training combined. So, if I have 3.5 million left after personnel and community outreach, but equipment and training can only take up to 2.5 million, that means I have an extra 1 million that needs to be allocated somewhere. But where?Wait, maybe I'm miscalculating. Let me think again. The total budget is 10 million. Personnel must be at least 55%, which is 5.5 million. Community outreach must be at least 10%, which is 1 million. So, the minimum for personnel and community outreach together is 6.5 million. That leaves 3.5 million for equipment and training. However, the constraint says that equipment and training combined must not exceed 25%, which is 2.5 million. So, 3.5 million is more than 2.5 million. That doesn't add up. There must be something wrong here.Wait, perhaps I misread the constraints. Let me check again. The problem says:1. Personnel must account for at least 55%.2. Equipment and training combined must not exceed 25%.3. Community outreach must be at least 10%.So, adding up the minimums: 55% + 10% = 65%, which is 6.5 million. The remaining 35% is 3.5 million, but equipment and training can only take up to 25%, which is 2.5 million. So, that leaves an extra 1 million that needs to be allocated, but where?Wait, perhaps the 25% is the maximum for equipment and training, so they can be less than or equal to 25%. So, if I have 3.5 million left after personnel and community outreach, but equipment and training can only take up to 2.5 million, then I have an extra 1 million that needs to be allocated to either personnel, equipment, training, or community outreach. But personnel already has a minimum, and community outreach also has a minimum. So, perhaps I can increase personnel or community outreach beyond their minimums, but that might not be optimal.Alternatively, maybe I need to adjust the allocations to meet all constraints. Let me try to set up equations.Let me denote:P = PersonnelE = EquipmentT = TrainingC = Community OutreachWe have:P + E + T + C = 10,000,000Constraints:P ≥ 5,500,000E + T ≤ 2,500,000C ≥ 1,000,000We also want to maximize the efficiency ROI and public satisfaction. The ROI from training is 10%, and the public satisfaction index increases by 2 points per percentage point allocated to community outreach.So, the objective is to maximize both ROI and public satisfaction. However, since these are two different objectives, we might need to combine them into a single objective function or find a way to prioritize them.But the problem says "maximize the efficiency ROI and public satisfaction index," so perhaps we need to maximize both. However, since they are separate, maybe we can maximize one while ensuring the other is as high as possible within the constraints.Alternatively, perhaps we can create a combined objective function. For example, since ROI is 10% per dollar spent on training, and public satisfaction is 2 points per percentage point allocated to community outreach, we can express the total benefit as:Total Benefit = 0.10*T + 2*(C/10,000,000)*100 (since C is in dollars, converting to percentage by dividing by 10 million and then multiplying by 100 to get percentage points)Wait, actually, the public satisfaction index improves by 2 points per percentage of the budget allocated. So, if C is the amount allocated to community outreach, the percentage is (C / 10,000,000)*100, and the public satisfaction index is 2*(C / 10,000,000)*100 = 2*(C / 100,000) = (C / 50,000). So, the public satisfaction index is C / 50,000.Similarly, the ROI from training is 10% per dollar spent on training. So, the ROI is 0.10*T.Therefore, the total benefit is 0.10*T + (C / 50,000).We need to maximize this total benefit subject to the constraints.So, the problem becomes:Maximize 0.10*T + (C / 50,000)Subject to:P + E + T + C = 10,000,000P ≥ 5,500,000E + T ≤ 2,500,000C ≥ 1,000,000And all variables are non-negative.Now, let's see how to approach this. Since we have a linear objective function and linear constraints, this is a linear programming problem. We can solve it using the simplex method or by analyzing the constraints.First, let's express all variables in terms of P, E, T, C.But since we have equality P + E + T + C = 10,000,000, we can express one variable in terms of the others. Let's express C as:C = 10,000,000 - P - E - TBut since C must be at least 1,000,000, we have:10,000,000 - P - E - T ≥ 1,000,000Which simplifies to:P + E + T ≤ 9,000,000But we also have P ≥ 5,500,000, so:5,500,000 ≤ P ≤ 9,000,000 - E - TBut E + T ≤ 2,500,000, so the maximum P can be is 10,000,000 - 1,000,000 - 2,500,000 = 6,500,000.Wait, that doesn't make sense because P must be at least 5,500,000. So, P can range from 5,500,000 to 6,500,000.Wait, let me think again. If C must be at least 1,000,000, and E + T must be at most 2,500,000, then the minimum amount allocated to P is 5,500,000, and the maximum is 10,000,000 - 1,000,000 - 2,500,000 = 6,500,000. So, P can be between 5,500,000 and 6,500,000.Therefore, P is between 5.5 million and 6.5 million.Now, our objective is to maximize 0.10*T + (C / 50,000). Let's express C in terms of P, E, T:C = 10,000,000 - P - E - TSo, substituting into the objective function:Total Benefit = 0.10*T + (10,000,000 - P - E - T) / 50,000Simplify this:= 0.10*T + (10,000,000 / 50,000) - (P / 50,000) - (E / 50,000) - (T / 50,000)= 0.10*T + 200 - (P / 50,000) - (E / 50,000) - (T / 50,000)Now, let's combine the T terms:0.10*T - (T / 50,000) = T*(0.10 - 1/50,000)Calculate 1/50,000 = 0.00002So, 0.10 - 0.00002 = 0.09998Therefore, Total Benefit = 200 + 0.09998*T - (P + E)/50,000So, the objective function is:Maximize 200 + 0.09998*T - (P + E)/50,000Since 200 is a constant, we can focus on maximizing 0.09998*T - (P + E)/50,000Now, since P and E are in the negative term, to maximize the expression, we need to minimize (P + E) and maximize T.But we have constraints:P ≥ 5,500,000E + T ≤ 2,500,000C = 10,000,000 - P - E - T ≥ 1,000,000So, let's see. To maximize T, we need to set E as low as possible, but E can't be negative. So, the minimum E is 0, which would allow T to be as high as 2,500,000. However, we also have to ensure that C is at least 1,000,000.If E is 0, then T can be up to 2,500,000, but then C = 10,000,000 - P - 0 - T.We need C ≥ 1,000,000, so:10,000,000 - P - T ≥ 1,000,000Which simplifies to:P + T ≤ 9,000,000But P is at least 5,500,000, so T can be at most 9,000,000 - P.Since P is at least 5,500,000, the maximum T can be is 9,000,000 - 5,500,000 = 3,500,000. But wait, earlier we had E + T ≤ 2,500,000, so T can't exceed 2,500,000 if E is 0.Wait, that seems conflicting. Let me clarify.If E is 0, then T can be up to 2,500,000. But we also have the constraint that P + T ≤ 9,000,000 because C must be at least 1,000,000.So, if P is 5,500,000, then T can be up to 9,000,000 - 5,500,000 = 3,500,000. But since E + T ≤ 2,500,000 and E is 0, T can only be up to 2,500,000.Therefore, the maximum T can be is 2,500,000 when E is 0.So, to maximize T, set E=0, T=2,500,000.Then, P + C = 10,000,000 - E - T = 10,000,000 - 0 - 2,500,000 = 7,500,000But P must be at least 5,500,000, so C can be at most 7,500,000 - 5,500,000 = 2,000,000.But C must be at least 1,000,000, so in this case, C would be 2,000,000.Wait, but if we set E=0, T=2,500,000, then P + C = 7,500,000. To satisfy P ≥ 5,500,000, P can be 5,500,000, which would make C = 2,000,000.So, in this case, the allocation would be:P = 5,500,000E = 0T = 2,500,000C = 2,000,000But let's check if this satisfies all constraints:P = 5,500,000 ≥ 5,500,000 ✔️E + T = 0 + 2,500,000 = 2,500,000 ≤ 2,500,000 ✔️C = 2,000,000 ≥ 1,000,000 ✔️Total = 5,500,000 + 0 + 2,500,000 + 2,000,000 = 10,000,000 ✔️So, this allocation satisfies all constraints.Now, let's calculate the total benefit:ROI from training = 0.10*T = 0.10*2,500,000 = 250,000Public satisfaction index = 2*(C / 10,000,000)*100 = 2*(2,000,000 / 10,000,000)*100 = 2*(0.2)*100 = 40 pointsWait, actually, earlier I thought the public satisfaction index was C / 50,000, but let me double-check.The problem says: "community outreach improves public satisfaction, measured by an index, by 2 points per percentage of the budget allocated."So, if C is the amount allocated, the percentage is (C / 10,000,000)*100. So, the public satisfaction index increases by 2*(C / 10,000,000)*100 = 2*(C / 100,000) = C / 50,000.So, in this case, C = 2,000,000, so public satisfaction index = 2,000,000 / 50,000 = 40 points.And ROI from training is 250,000.So, total benefit is 250,000 + 40 = 250,040.But wait, is this the maximum? Let's see if we can get a higher total benefit by adjusting the allocations.Suppose we increase C beyond 2,000,000. Since C is in the numerator for public satisfaction, increasing C would increase the public satisfaction index. However, increasing C would require decreasing either P, E, or T. But P is already at its minimum, so we can't decrease P. E is already at 0, so we can't decrease E further. Therefore, to increase C, we would have to decrease T.But T is contributing to the ROI, which is 0.10*T. So, if we decrease T, we lose some ROI but gain some public satisfaction.We need to find the optimal balance where the marginal gain in public satisfaction from increasing C equals the marginal loss in ROI from decreasing T.Let me set up the trade-off.Suppose we decrease T by x dollars and increase C by x dollars. Then, the change in total benefit would be:ΔBenefit = (ΔROI) + (ΔPublic Satisfaction)ΔROI = -0.10*xΔPublic Satisfaction = 2*(x / 10,000,000)*100 = 2*(x / 100,000) = x / 50,000So, ΔBenefit = -0.10*x + x / 50,000We want to find if this ΔBenefit is positive or negative.Let's calculate:-0.10*x + x / 50,000 = x*(-0.10 + 1/50,000) = x*(-0.10 + 0.00002) = x*(-0.09998)Since x is positive, ΔBenefit is negative. Therefore, any decrease in T to increase C would result in a net loss in total benefit. Therefore, it's better not to decrease T.Therefore, the optimal allocation is to set T as high as possible, which is 2,500,000, with E=0, P=5,500,000, and C=2,000,000.But wait, let me check if there's another way to allocate. Suppose we don't set E=0, but instead allocate some amount to E and less to T. Would that help?Suppose we set E=1,000,000, then T can be up to 1,500,000 (since E + T ≤ 2,500,000). Then, P + C = 10,000,000 - 1,000,000 - 1,500,000 = 7,500,000. Since P must be at least 5,500,000, C can be up to 2,000,000.So, in this case, P=5,500,000, E=1,000,000, T=1,500,000, C=2,000,000.Calculating the total benefit:ROI from training = 0.10*1,500,000 = 150,000Public satisfaction index = 2,000,000 / 50,000 = 40Total benefit = 150,000 + 40 = 150,040Which is less than the previous total benefit of 250,040. So, this allocation is worse.Similarly, if we set E=2,500,000, then T=0, and P + C = 7,500,000. P=5,500,000, C=2,000,000.ROI from training = 0Public satisfaction index = 40Total benefit = 0 + 40 = 40Which is much worse.Therefore, the optimal allocation is to set E=0, T=2,500,000, P=5,500,000, and C=2,000,000.But wait, let me check if we can increase C beyond 2,000,000 by decreasing P. But P must be at least 5,500,000, so we can't decrease P further. Therefore, C can't be increased beyond 2,000,000 without violating the P constraint.Alternatively, if we set P higher than 5,500,000, say P=6,000,000, then E + T + C = 4,000,000. But E + T must be ≤ 2,500,000, so C would be at least 1,500,000.Wait, let's see:If P=6,000,000, then E + T + C = 4,000,000.But E + T ≤ 2,500,000, so C ≥ 1,500,000.So, in this case, we can set E=0, T=2,500,000, and C=1,500,000.Calculating the total benefit:ROI from training = 0.10*2,500,000 = 250,000Public satisfaction index = 1,500,000 / 50,000 = 30Total benefit = 250,000 + 30 = 250,030Which is slightly less than the previous total benefit of 250,040 when C=2,000,000.Therefore, it's better to have P=5,500,000, T=2,500,000, C=2,000,000.Alternatively, if we set P=5,500,000, E=0, T=2,500,000, C=2,000,000, we get the highest total benefit.Therefore, the optimal allocation is:Personnel: 5,500,000Equipment: 0Training: 2,500,000Community Outreach: 2,000,000But wait, is it realistic to allocate 0 to equipment? That seems extreme. Maybe the department needs some equipment. However, according to the constraints and the objective function, since equipment doesn't contribute to either ROI or public satisfaction, it's better to allocate as much as possible to training and community outreach, which do contribute.But let me check if the problem allows E=0. The constraints only say that E + T ≤ 2,500,000, so E can be 0 as long as T is ≤2,500,000. So, yes, E=0 is allowed.Therefore, the optimal allocation is as above.But let me double-check the calculations.Total budget: 5,500,000 + 0 + 2,500,000 + 2,000,000 = 10,000,000 ✔️Personnel ≥5,500,000 ✔️Equipment + Training = 0 + 2,500,000 = 2,500,000 ≤2,500,000 ✔️Community Outreach =2,000,000 ≥1,000,000 ✔️ROI from training: 2,500,000 *0.10=250,000Public satisfaction index: 2,000,000 /50,000=40Total benefit:250,000 +40=250,040If we try to allocate any amount to E, we have to decrease T or C, which would decrease the total benefit. For example, if E=1,000,000, then T=1,500,000, C=2,000,000, as before, giving a total benefit of 150,000 +40=190,040, which is less.Alternatively, if E=1,000,000, T=1,500,000, and C=2,000,000, total benefit is 150,000 +40=190,040.If E=2,500,000, T=0, C=2,000,000, total benefit=0 +40=40.Therefore, the maximum total benefit is achieved when E=0, T=2,500,000, C=2,000,000, and P=5,500,000.So, the optimal allocation is:Personnel: 5,500,000Equipment: 0Training: 2,500,000Community Outreach: 2,000,000But let me consider if there's a way to have both higher ROI and higher public satisfaction by adjusting the allocations differently. For example, if we increase C beyond 2,000,000 by decreasing P, but P can't go below 5,500,000. So, we can't do that.Alternatively, if we decrease T to increase C, but as we saw earlier, the marginal loss in ROI is greater than the gain in public satisfaction, so it's not beneficial.Therefore, the optimal allocation is as calculated.</think>