🧠:<think>Alright, so I have this differential equation to solve for the agricultural subsidies, S(t). The equation is given as:[ frac{dS}{dt} = rS + alpha L(t) ]And L(t) is defined as:[ L(t) = L_0 e^{-beta t} ]They also gave me the initial condition S(0) = S_0. Okay, so this is a linear first-order differential equation. I remember that for such equations, we can use an integrating factor to solve them. Let me recall the standard form:[ frac{dy}{dt} + P(t)y = Q(t) ]In this case, comparing to the standard form, I can rewrite the given equation as:[ frac{dS}{dt} - rS = alpha L(t) ]So here, P(t) is -r and Q(t) is α L(t). The integrating factor, μ(t), is given by:[ mu(t) = e^{int P(t) dt} = e^{int -r dt} = e^{-rt} ]Multiplying both sides of the differential equation by the integrating factor:[ e^{-rt} frac{dS}{dt} - r e^{-rt} S = alpha e^{-rt} L(t) ]The left side should now be the derivative of (S(t) * integrating factor). Let me check:[ frac{d}{dt} [S(t) e^{-rt}] = alpha e^{-rt} L(t) ]Yes, that looks right. So, integrating both sides with respect to t:[ int frac{d}{dt} [S(t) e^{-rt}] dt = int alpha e^{-rt} L(t) dt ]Which simplifies to:[ S(t) e^{-rt} = alpha int e^{-rt} L(t) dt + C ]Now, substituting L(t) = L_0 e^{-β t} into the integral:[ S(t) e^{-rt} = alpha L_0 int e^{-rt} e^{-beta t} dt + C ]Combine the exponents:[ S(t) e^{-rt} = alpha L_0 int e^{-(r + beta) t} dt + C ]The integral of e^{kt} is (1/k) e^{kt}, so:[ S(t) e^{-rt} = alpha L_0 left( frac{e^{-(r + beta) t}}{-(r + beta)} right) + C ]Simplify:[ S(t) e^{-rt} = -frac{alpha L_0}{r + beta} e^{-(r + beta) t} + C ]Now, solve for S(t):[ S(t) = -frac{alpha L_0}{r + beta} e^{-beta t} + C e^{rt} ]Wait, let me make sure I did that correctly. When I multiply both sides by e^{rt}, the first term becomes:- (α L_0)/(r + β) e^{-β t}, because e^{-rt} * e^{-(r + β)t} = e^{-rt - rt - β t}? Wait, no. Wait, hold on.Wait, no. Let me double-check. The integral was:∫ e^{-(r + β) t} dt = (-1)/(r + β) e^{-(r + β) t} + C.So when I multiply by e^{-rt}, it's:S(t) e^{-rt} = - (α L_0)/(r + β) e^{-(r + β) t} + CThen, multiplying both sides by e^{rt}:S(t) = - (α L_0)/(r + β) e^{-β t} + C e^{rt}Yes, that seems correct.Now, apply the initial condition S(0) = S_0.At t = 0:S(0) = - (α L_0)/(r + β) e^{0} + C e^{0} = - (α L_0)/(r + β) + C = S_0So, solving for C:C = S_0 + (α L_0)/(r + β)Therefore, the solution is:S(t) = - (α L_0)/(r + β) e^{-β t} + [S_0 + (α L_0)/(r + β)] e^{rt}We can write this as:S(t) = S_0 e^{rt} + (α L_0)/(r + β) [e^{rt} - e^{-β t}]Hmm, that seems a bit messy, but I think it's correct.Let me check the steps again:1. Wrote the DE in standard linear form.2. Found integrating factor e^{-rt}.3. Multiplied through, recognized the left side as derivative of S e^{-rt}.4. Integrated both sides.5. Substituted L(t) and integrated, getting the exponential term.6. Solved for S(t), applied initial condition.Yes, seems consistent.So, part 1 is done. Now, part 2 asks for the total amount of subsidies over 5 years, which is the integral from 0 to 5 of S(t) dt.So, need to compute:[ int_0^5 S(t) dt ]Given that S(t) is:S(t) = S_0 e^{rt} + (α L_0)/(r + β) [e^{rt} - e^{-β t}]So, let's write the integral as:[ int_0^5 left[ S_0 e^{rt} + frac{alpha L_0}{r + beta} e^{rt} - frac{alpha L_0}{r + beta} e^{-beta t} right] dt ]We can split this into three separate integrals:1. S_0 ∫ e^{rt} dt from 0 to 52. (α L_0)/(r + β) ∫ e^{rt} dt from 0 to 53. - (α L_0)/(r + β) ∫ e^{-β t} dt from 0 to 5Compute each integral separately.First integral:∫ e^{rt} dt = (1/r) e^{rt} + CSo, evaluated from 0 to 5:(1/r)(e^{5r} - 1)Second integral is the same as the first, multiplied by (α L_0)/(r + β):(α L_0)/(r + β) * (1/r)(e^{5r} - 1)Third integral:∫ e^{-β t} dt = (-1/β) e^{-β t} + CEvaluated from 0 to 5:(-1/β)(e^{-5β} - 1) = (1/β)(1 - e^{-5β})But since it's multiplied by - (α L_0)/(r + β):- (α L_0)/(r + β) * (1/β)(1 - e^{-5β}) = - (α L_0)/(β(r + β))(1 - e^{-5β})Putting it all together:Total subsidies = S_0*(1/r)(e^{5r} - 1) + (α L_0)/(r + β)*(1/r)(e^{5r} - 1) - (α L_0)/(β(r + β))(1 - e^{-5β})We can factor out common terms:Let me factor out (1/r)(e^{5r} - 1) from the first two terms:[ S_0 + (α L_0)/(r + β) ] * (1/r)(e^{5r} - 1) - (α L_0)/(β(r + β))(1 - e^{-5β})Alternatively, we can write it as:Total subsidies = (S_0 + (α L_0)/(r + β))*(e^{5r} - 1)/r - (α L_0)/(β(r + β))*(1 - e^{-5β})That seems to be the expression for the total subsidies over 5 years.Let me see if I can combine these terms or simplify further.Alternatively, we can write it as:Total subsidies = (S_0 + (α L_0)/(r + β))*(e^{5r} - 1)/r - (α L_0)/(β(r + β))*(1 - e^{-5β})I think that's as simplified as it gets unless we have specific values for the parameters.So, summarizing:1. The solution to the differential equation is:S(t) = S_0 e^{rt} + (α L_0)/(r + β) (e^{rt} - e^{-β t})2. The total subsidies over 5 years is:Total = (S_0 + (α L_0)/(r + β))*(e^{5r} - 1)/r - (α L_0)/(β(r + β))*(1 - e^{-5β})I think that's the answer.Final Answer1. The solution for ( S(t) ) is (boxed{S(t) = S_0 e^{rt} + frac{alpha L_0}{r + beta} left( e^{rt} - e^{-beta t} right)}).2. The total amount of subsidies over 5 years is (boxed{left( S_0 + frac{alpha L_0}{r + beta} right) frac{e^{5r} - 1}{r} - frac{alpha L_0}{beta(r + beta)} left( 1 - e^{-5beta} right)}).</think>

🧠:<think>Okay, so I have this problem about ancient chariots and warriors. Let me try to figure it out step by step. First, part 1 says there are two types of chariots: Type A and Type B. The total number of chariots is 480. Each Type A carries 3 warriors, and each Type B carries 2 warriors. The total number of warriors is 1140. I need to set up a system of equations to find out how many of each type there are.Alright, let's define variables. Let me call the number of Type A chariots as 'a' and Type B as 'b'. So, the total number of chariots is a + b = 480. That's my first equation.Now, for the warriors. Each Type A has 3 warriors, so that's 3a, and each Type B has 2 warriors, which is 2b. The total warriors are 1140, so 3a + 2b = 1140. That's my second equation.So, my system of equations is:1. a + b = 4802. 3a + 2b = 1140I need to solve this system. Maybe I can use substitution or elimination. Let me try elimination. If I multiply the first equation by 2, I'll have:2a + 2b = 960Then subtract this from the second equation:(3a + 2b) - (2a + 2b) = 1140 - 960That simplifies to:a = 180So, a is 180. Then, plugging back into the first equation, 180 + b = 480, so b = 300.Wait, let me check that. If a is 180, then 3a is 540. And b is 300, so 2b is 600. 540 + 600 is 1140. Yeah, that matches. So, Type A is 180, Type B is 300.Okay, moving on to part 2. Each Type A requires 4 horses, and Type B requires 3 horses. So, total horses needed would be 4a + 3b.From part 1, a is 180 and b is 300. So, let's compute that.4 * 180 = 7203 * 300 = 900Adding those together: 720 + 900 = 1620Hmm, the civilization can only muster 1600 horses. So, 1620 is more than 1600. That means they don't have enough horses. So, their fleet deployment isn't sustainable because they need 1620 horses but only have 1600.Wait, let me double-check my calculations. Type A: 180 chariots * 4 horses = 720Type B: 300 chariots * 3 horses = 900Total: 720 + 900 = 1620Yes, that's correct. So, 1620 - 1600 = 20 horses short. So, they can't sustain the deployment.I think that's it. So, the answers are 180 Type A, 300 Type B, and 1620 horses needed, which exceeds their limit.Final Answer1. The number of Type A chariots is boxed{180} and the number of Type B chariots is boxed{300}.2. The total number of horses needed is boxed{1620}, which exceeds the maximum of 1600 horses, so the fleet deployment was not sustainable.</think>

🧠:<think>Okay, so Alex is working on this big circular mural, right? It's a 12-sector circle, each sector a different color. The radius is 6 meters. First, I need to find the area of one sector. Hmm, sectors in a circle are like slices of a pie, so each sector has an equal angle since there are 12 of them. A full circle is 360 degrees, so each sector must be 360 divided by 12, which is 30 degrees. So each sector is a 30-degree slice. Now, the area of a sector is given by the formula (θ/360) * π * r², where θ is the central angle in degrees. Plugging in the numbers, θ is 30 degrees, and the radius r is 6 meters. Let me compute that. So, (30/360) is 1/12. Then, π * 6² is π * 36, which is 36π. Multiply that by 1/12, so 36π divided by 12 is 3π. So, the area of one sector is 3π square meters. That seems right.Now, moving on to the triangular pattern inside each sector. The triangle has one vertex at the center, and the base is along the arc. The base forms a 30-degree angle with the center. Wait, so the triangle is an isosceles triangle with two sides equal to the radius, which is 6 meters, and the angle between them is 30 degrees. The area of a triangle with two sides and the included angle can be found using the formula: (1/2)*a*b*sinθ, where a and b are the sides, and θ is the included angle. Here, a and b are both 6 meters, and θ is 30 degrees. So, plugging in, it's (1/2)*6*6*sin(30). Sin(30) is 0.5, so that becomes (1/2)*36*0.5. Let's compute that: (1/2)*36 is 18, and 18*0.5 is 9. So, the area of the triangle is 9 square meters. Wait, hold on. The sector area was 3π, which is approximately 9.4248, and the triangle is 9. So, the triangle is slightly smaller than the sector, which makes sense because the triangle is inside the sector. That seems reasonable.Moving on to the second part. Alex wants to inscribe a regular hexagon within the same circle. The radius is 6 meters. A regular hexagon inscribed in a circle has all its vertices on the circumference. In a regular hexagon, all sides are equal, and each internal angle is 120 degrees. But to find the side length, I remember that in a regular hexagon inscribed in a circle, the side length is equal to the radius of the circle. So, if the radius is 6 meters, each side of the hexagon is also 6 meters. That seems straightforward.Now, Alex is going to paint each alternating triangle of the hexagon with different shading. Wait, a regular hexagon can be divided into six equilateral triangles, each with side length equal to the radius, which is 6 meters. So, each triangle is equilateral with sides of 6 meters. If Alex is painting each alternating triangle, that means every other triangle is shaded. Since there are six triangles, alternating would mean three are shaded and three are not. But the problem says each shaded triangle occupies exactly one-third of its area for a special textural effect. Hmm, so each shaded triangle isn't entirely shaded, but only one-third of its area is shaded.Wait, so each triangle has an area, and only one-third of that area is shaded. So, first, let's find the area of one equilateral triangle with side length 6 meters. The formula for the area of an equilateral triangle is (√3/4)*a², where a is the side length. Plugging in, a is 6, so (√3/4)*36 is 9√3. So, each triangle has an area of 9√3 square meters. If each shaded triangle has one-third of its area shaded, then the shaded area per triangle is (1/3)*9√3, which is 3√3. Since there are three shaded triangles (alternating), the total shaded area is 3*3√3, which is 9√3 square meters. Let me double-check that. Each triangle is 9√3, one-third is 3√3, times three shaded triangles gives 9√3. Yep, that seems correct.So, summarizing:1. The area of one sector is 3π square meters, and the area of the triangle is 9 square meters.2. The side length of the hexagon is 6 meters, and the total shaded area is 9√3 square meters.Final Answer1. The area of one sector is boxed{3pi} square meters, and the area of the triangle is boxed{9} square meters.2. The side length of the hexagon is boxed{6} meters, and the total shaded area is boxed{9sqrt{3}} square meters.</think>

🧠:<think>Alright, so I have this problem where I need to help a marketing director optimize their budget allocation between social media advertising and email marketing to maximize ROI. Let me try to break this down step by step.First, I need to understand the problem clearly. The company has two marketing channels: social media and email. Each has its own ROI function based on the budget allocated. The total budget is 100,000, which is 100,000, so in thousands of dollars, that's 100. Also, there's a constraint that the budget for social media must be at least twice that of email marketing. So, I need to find how much to allocate to each channel to maximize the total ROI.Let me write down the given functions:- ROI for social media: ( R_s(x) = -0.02x^2 + 2.4x )- ROI for email: ( R_e(y) = -0.03y^2 + 3.5y )Where ( x ) is the budget in thousands for social media, and ( y ) is the budget in thousands for email.The total budget constraint is ( x + y leq 100 ). But actually, since they want to maximize ROI, I think the optimal solution will use the entire budget, so ( x + y = 100 ).Additionally, there's a constraint that ( x geq 2y ). So, the budget for social media must be at least twice that of email.So, summarizing the constraints:1. ( x + y = 100 )2. ( x geq 2y )3. ( x geq 0 )4. ( y geq 0 )Our goal is to maximize the total ROI, which is ( R_s(x) + R_e(y) ).So, first, maybe I can express everything in terms of one variable. Since ( x + y = 100 ), I can express ( y = 100 - x ). Then, substitute this into the constraint ( x geq 2y ):( x geq 2(100 - x) )Let me solve this inequality:( x geq 200 - 2x )Adding 2x to both sides:( 3x geq 200 )Divide both sides by 3:( x geq frac{200}{3} approx 66.6667 )So, ( x ) must be at least approximately 66.6667 thousand dollars, which is about 66,666.67.Also, since ( x + y = 100 ), if ( x ) is 66.6667, then ( y ) is 33.3333.But let's keep it exact. ( x geq frac{200}{3} ) and ( y = 100 - x leq frac{100}{3} approx 33.3333 ).So, our feasible region is ( x ) from ( frac{200}{3} ) to 100, and ( y ) from 0 to ( frac{100}{3} ).Now, to maximize the total ROI, let's express the total ROI as a function of ( x ):Total ROI ( R = R_s(x) + R_e(y) = (-0.02x^2 + 2.4x) + (-0.03y^2 + 3.5y) )But since ( y = 100 - x ), substitute that in:( R = -0.02x^2 + 2.4x - 0.03(100 - x)^2 + 3.5(100 - x) )Now, let's expand this expression step by step.First, expand ( (100 - x)^2 ):( (100 - x)^2 = 10000 - 200x + x^2 )So, substitute back into R:( R = -0.02x^2 + 2.4x - 0.03(10000 - 200x + x^2) + 3.5(100 - x) )Now, distribute the -0.03 and 3.5:First term: -0.03 * 10000 = -300Second term: -0.03 * (-200x) = +6xThird term: -0.03 * x^2 = -0.03x^2Fourth term: 3.5 * 100 = 350Fifth term: 3.5 * (-x) = -3.5xSo, putting it all together:( R = -0.02x^2 + 2.4x - 300 + 6x - 0.03x^2 + 350 - 3.5x )Now, let's combine like terms.First, the ( x^2 ) terms:-0.02x^2 - 0.03x^2 = -0.05x^2Next, the x terms:2.4x + 6x - 3.5x = (2.4 + 6 - 3.5)x = (8.4 - 3.5)x = 4.9xConstant terms:-300 + 350 = 50So, the total ROI function simplifies to:( R = -0.05x^2 + 4.9x + 50 )Now, this is a quadratic function in terms of x, and since the coefficient of ( x^2 ) is negative (-0.05), the parabola opens downward, meaning the vertex is the maximum point.To find the maximum, we can find the vertex of this parabola. The x-coordinate of the vertex is given by ( x = -b/(2a) ), where a = -0.05 and b = 4.9.So,( x = -4.9 / (2 * -0.05) = -4.9 / (-0.1) = 49 )Wait, that's interesting. So, the maximum occurs at x = 49.But hold on, earlier we found that x must be at least 200/3 ≈ 66.6667. So, 49 is less than 66.6667, which is outside our feasible region.That means the maximum of this quadratic function is at x = 49, but since our feasible region starts at x = 66.6667, the maximum within the feasible region would be at the boundary, which is x = 66.6667.Wait, but let me double-check my calculations because sometimes when substituting, I might have made a mistake.Let me go back to the total ROI function:( R = -0.05x^2 + 4.9x + 50 )Yes, that seems correct.So, the vertex is at x = 49, which is less than 66.6667, so indeed, the maximum within our feasible region is at x = 66.6667.But wait, is that necessarily the case? Because sometimes, depending on the function, the maximum could be at a different point. Let me verify.Alternatively, maybe I made a mistake in the substitution or expansion. Let me go back and check.Original functions:( R_s(x) = -0.02x^2 + 2.4x )( R_e(y) = -0.03y^2 + 3.5y )With y = 100 - x.So, substituting y:( R_e(y) = -0.03(100 - x)^2 + 3.5(100 - x) )Expanding ( (100 - x)^2 = 10000 - 200x + x^2 )So,( R_e(y) = -0.03*(10000 - 200x + x^2) + 3.5*(100 - x) )Calculating each term:-0.03*10000 = -300-0.03*(-200x) = +6x-0.03*x^2 = -0.03x^23.5*100 = 3503.5*(-x) = -3.5xSo, combining:-300 + 6x -0.03x^2 + 350 -3.5xCombine constants: -300 + 350 = 50Combine x terms: 6x -3.5x = 2.5xSo, R_e(y) = -0.03x^2 + 2.5x + 50Wait, but earlier I had:Total ROI R = R_s(x) + R_e(y) = (-0.02x^2 + 2.4x) + (-0.03x^2 + 2.5x + 50)So, combining these:-0.02x^2 -0.03x^2 = -0.05x^22.4x + 2.5x = 4.9xAnd the constant is +50So, yes, R = -0.05x^2 + 4.9x + 50So, that seems correct.Therefore, the vertex is at x = 49, but our feasible x starts at 66.6667. So, the maximum within the feasible region is at x = 66.6667.But wait, let me think again. Is that necessarily the case? Because sometimes, when the maximum is outside the feasible region, the function could be increasing or decreasing beyond that point.Since the parabola opens downward, the function increases up to x = 49 and then decreases after that. So, if our feasible region starts at x = 66.6667, which is to the right of the vertex, then the function is decreasing in that interval. Therefore, the maximum in the feasible region would be at the leftmost point, which is x = 66.6667.Therefore, the optimal allocation is x = 200/3 ≈ 66.6667, and y = 100 - x ≈ 33.3333.But let me confirm this by evaluating the total ROI at x = 66.6667 and also checking the endpoints.Wait, the feasible region is x from 66.6667 to 100. So, we should check the ROI at x = 66.6667 and at x = 100, and see which one is higher.Wait, but if the function is decreasing beyond x = 49, then as x increases beyond 49, ROI decreases. So, at x = 66.6667, ROI is higher than at x = 100.But let me compute the ROI at x = 66.6667 and x = 100 to be sure.First, at x = 66.6667 (which is 200/3):Compute R = -0.05*(200/3)^2 + 4.9*(200/3) + 50First, calculate (200/3)^2 = (40000/9) ≈ 4444.4444So,-0.05*(4444.4444) ≈ -222.22224.9*(200/3) ≈ 4.9*(66.6667) ≈ 326.6667Adding the constant 50.So, total R ≈ -222.2222 + 326.6667 + 50 ≈ (-222.2222 + 326.6667) = 104.4445 + 50 = 154.4445So, approximately 154.4445 thousand dollars ROI.Now, at x = 100:R = -0.05*(100)^2 + 4.9*(100) + 50= -0.05*10000 + 490 + 50= -500 + 490 + 50= (-500 + 490) = -10 + 50 = 40So, ROI is 40 thousand dollars, which is much lower.Therefore, the maximum ROI within the feasible region is at x = 66.6667, y = 33.3333.But wait, let me also check if there's a possibility that the maximum could be somewhere else. Since the function is quadratic, and we've checked the endpoints, I think that's sufficient.Alternatively, maybe I should consider the possibility that the maximum could be at the vertex if the vertex is within the feasible region, but in this case, it's not.So, conclusion: the optimal allocation is x = 200/3 ≈ 66.6667 thousand dollars to social media, and y = 100 - 200/3 = 100/3 ≈ 33.3333 thousand dollars to email.But let me also check if the constraint x ≥ 2y is satisfied.x = 200/3 ≈ 66.6667y = 100/3 ≈ 33.3333Is 66.6667 ≥ 2*33.3333?2*33.3333 = 66.6666, so yes, 66.6667 ≥ 66.6666, which is just barely satisfied.So, that's correct.Therefore, the optimal budget allocation is approximately 66,666.67 to social media and 33,333.33 to email marketing.Now, for part b), calculate the maximum total ROI.From earlier, at x = 200/3, the ROI was approximately 154.4445 thousand dollars.But let me compute it more precisely.Compute R = -0.05*(200/3)^2 + 4.9*(200/3) + 50First, (200/3)^2 = 40000/9So,-0.05*(40000/9) = -2000/9 ≈ -222.22224.9*(200/3) = (4.9*200)/3 = 980/3 ≈ 326.6667Adding 50:Total R = (-2000/9) + (980/3) + 50Convert all to ninths:-2000/9 + (980/3)*(3/3) = 2940/9 + 50*(9/9) = 450/9So,-2000/9 + 2940/9 + 450/9 = (-2000 + 2940 + 450)/9 = (1390)/9 ≈ 154.4444So, exactly, it's 1390/9 ≈ 154.4444 thousand dollars.Therefore, the maximum total ROI is approximately 154,444.44.Wait, but let me check the calculation again.Wait, 4.9*(200/3) = 4.9*66.6667 ≈ 326.6667-0.05*(200/3)^2 = -0.05*(4444.4444) ≈ -222.2222Adding 50:-222.2222 + 326.6667 + 50 = 154.4445Yes, that's correct.So, the maximum ROI is approximately 154,444.44.But let me express it as a fraction.1390/9 is equal to 154 and 4/9, which is approximately 154.4444.So, that's the maximum ROI.Therefore, the answers are:a) Optimal budget allocation: x = 200/3 ≈ 66.6667 thousand dollars to social media, and y = 100/3 ≈ 33.3333 thousand dollars to email.b) Maximum total ROI: 1390/9 ≈ 154.4444 thousand dollars.But let me express these in exact terms rather than decimals.So, x = 200/3, y = 100/3, and ROI = 1390/9.Alternatively, we can write these as fractions:x = 66 2/3 thousand dollars, y = 33 1/3 thousand dollars, ROI = 154 4/9 thousand dollars.But since the question says to express the answer in terms of x and y, which are in thousands, I think it's acceptable to present them as fractions or decimals.But perhaps the exact fractional form is better.So, x = 200/3, y = 100/3.And ROI = 1390/9.Alternatively, simplifying 1390/9:Divide 1390 by 9:9*154 = 1386, so 1390 - 1386 = 4, so 154 and 4/9.So, 154 4/9 thousand dollars.Therefore, the maximum ROI is 154 4/9 thousand dollars, which is approximately 154,444.44.So, summarizing:a) Allocate 200/3 thousand dollars (approximately 66,666.67) to social media and 100/3 thousand dollars (approximately 33,333.33) to email marketing.b) The maximum total ROI is 1390/9 thousand dollars, which is approximately 154,444.44.I think that's the solution.</think>