🧠:<think>Alright, so I have this problem where an African American entrepreneur is looking at three successful Black-owned businesses in Atlanta, New York, and Los Angeles. They want to figure out when the combined revenue from all three cities will reach 50,000. Then, they also want to invest in a start-up whose revenue grows quadratically and match the combined growth rate at t=5 years. Hmm, okay, let me try to break this down step by step.First, for part 1, I need to find the time t when the combined revenue from Atlanta, New York, and Los Angeles equals 50,000. The revenue functions are given as:- Atlanta: ( R_A(t) = 5000e^{0.05t} )- New York: ( R_N(t) = 4500(1 + 0.04t) )- Los Angeles: ( R_L(t) = 4000left(frac{1.1^t - 1}{0.1}right) )So, the combined revenue is ( R_A(t) + R_N(t) + R_L(t) ). I need to set this equal to 50,000 and solve for t.Let me write that equation out:( 5000e^{0.05t} + 4500(1 + 0.04t) + 4000left(frac{1.1^t - 1}{0.1}right) = 50,000 )Hmm, okay. That looks a bit complicated because it has exponential terms with different bases and a linear term. I don't think I can solve this algebraically, so I might need to use numerical methods or graphing to approximate t.Let me first simplify each term to see if that helps.Starting with Atlanta's revenue: ( 5000e^{0.05t} ). That's straightforward, it's an exponential growth function.New York's revenue: ( 4500(1 + 0.04t) ). That simplifies to ( 4500 + 180t ). So that's a linear function.Los Angeles's revenue: ( 4000left(frac{1.1^t - 1}{0.1}right) ). Let me compute the denominator first: 1/0.1 is 10, so this becomes ( 4000 times 10 times (1.1^t - 1) ) which is ( 40,000(1.1^t - 1) ). So, simplifying, that's ( 40,000 times 1.1^t - 40,000 ).So putting it all together, the combined revenue is:( 5000e^{0.05t} + 4500 + 180t + 40,000 times 1.1^t - 40,000 )Simplify constants: 4500 - 40,000 is -35,500.So, the equation becomes:( 5000e^{0.05t} + 180t + 40,000 times 1.1^t - 35,500 = 50,000 )Bring the 50,000 to the left side:( 5000e^{0.05t} + 180t + 40,000 times 1.1^t - 35,500 - 50,000 = 0 )Simplify constants: -35,500 - 50,000 = -85,500So:( 5000e^{0.05t} + 180t + 40,000 times 1.1^t - 85,500 = 0 )Hmm, this is still a complex equation. Maybe I can compute the left-hand side for different values of t and see when it crosses zero.Alternatively, maybe I can approximate it numerically. Let me try plugging in some values of t to see where it crosses 50,000.Wait, actually, the original combined revenue is 5000e^{0.05t} + 4500(1 + 0.04t) + 4000*(1.1^t -1)/0.1. Let me compute each term at different t and sum them up.Let me make a table:t | R_A(t) | R_N(t) | R_L(t) | Total---|-------|-------|-------|-----0 | 5000 | 4500 | 0 | 95001 | 5000e^0.05 ≈5127.63 | 4500*(1.04)=4680 | 4000*(1.1 -1)/0.1=4000*0.1/0.1=4000 | 5127.63+4680+4000≈13807.632 | 5000e^0.1≈5127.63* e^0.05≈5127.63*1.05127≈5387.92 | 4500*(1.08)=4860 | 4000*(1.21 -1)/0.1=4000*0.21/0.1=8400 | 5387.92+4860+8400≈18647.923 | 5000e^0.15≈5000*1.1618≈5809 | 4500*(1.12)=5040 | 4000*(1.331 -1)/0.1=4000*0.331/0.1=13240 | 5809+5040+13240≈240894 | 5000e^0.2≈5000*1.2214≈6107 | 4500*(1.16)=5220 | 4000*(1.4641 -1)/0.1=4000*0.4641/0.1≈18564 | 6107+5220+18564≈298915 | 5000e^0.25≈5000*1.284≈6420 | 4500*(1.20)=5400 | 4000*(1.61051 -1)/0.1≈4000*0.61051/0.1≈24420.4 | 6420+5400+24420.4≈36240.46 | 5000e^0.3≈5000*1.34986≈6749.3 | 4500*(1.24)=5580 | 4000*(1.771561 -1)/0.1≈4000*0.771561/0.1≈30862.44 | 6749.3+5580+30862.44≈43191.747 | 5000e^0.35≈5000*1.41906≈7095.3 | 4500*(1.28)=5760 | 4000*(1.9487171 -1)/0.1≈4000*0.9487171/0.1≈37948.684 | 7095.3+5760+37948.684≈50803.984Wait, at t=7, the total revenue is approximately 50,804, which is just over 50,000. So, the time t is approximately 7 years.But let me check t=6.5 to see if it's closer.At t=6.5:Compute each term:R_A(6.5)=5000e^{0.05*6.5}=5000e^{0.325}≈5000*1.383≈6915R_N(6.5)=4500*(1 + 0.04*6.5)=4500*(1 + 0.26)=4500*1.26=5670R_L(6.5)=4000*(1.1^6.5 -1)/0.1First, compute 1.1^6.5. Let's see, 1.1^6 is approximately 1.771561, and 1.1^0.5 is sqrt(1.1)≈1.0488. So, 1.771561*1.0488≈1.858. So, 1.1^6.5≈1.858.Therefore, R_L(6.5)=4000*(1.858 -1)/0.1=4000*(0.858)/0.1=4000*8.58=34320So total revenue at t=6.5 is approximately 6915 + 5670 + 34320≈46905, which is still below 50,000.Wait, but at t=7, it was 50,804. So, the crossing point is between 6.5 and 7.Let me try t=6.8.Compute each term:R_A(6.8)=5000e^{0.05*6.8}=5000e^{0.34}≈5000*1.4049≈7024.5R_N(6.8)=4500*(1 + 0.04*6.8)=4500*(1 + 0.272)=4500*1.272≈5724R_L(6.8)=4000*(1.1^6.8 -1)/0.1Compute 1.1^6.8. Let's see, 1.1^6=1.771561, 1.1^0.8≈1.083 (since 1.1^0.8 is e^{0.8*ln1.1}≈e^{0.8*0.09531}≈e^{0.07625}≈1.079). So, 1.771561*1.079≈1.912.Therefore, R_L(6.8)=4000*(1.912 -1)/0.1=4000*(0.912)/0.1=4000*9.12=36480Total revenue≈7024.5 + 5724 + 36480≈49228.5, which is still below 50,000.Hmm, so at t=6.8, it's about 49,228.5. At t=7, it's 50,804. So, the crossing point is between 6.8 and 7.Let me try t=6.9.R_A(6.9)=5000e^{0.05*6.9}=5000e^{0.345}≈5000*1.4116≈7058R_N(6.9)=4500*(1 + 0.04*6.9)=4500*(1 + 0.276)=4500*1.276≈5742R_L(6.9)=4000*(1.1^6.9 -1)/0.1Compute 1.1^6.9. 1.1^6=1.771561, 1.1^0.9≈1.093 (since 1.1^0.9≈e^{0.9*0.09531}≈e^{0.0858}≈1.089). So, 1.771561*1.089≈1.933.Thus, R_L(6.9)=4000*(1.933 -1)/0.1=4000*(0.933)/0.1=4000*9.33≈37320Total revenue≈7058 + 5742 + 37320≈50120, which is just over 50,000.So, at t=6.9, total revenue≈50,120, which is just above 50,000. So, the time t is approximately 6.9 years.To get a more accurate estimate, let's use linear approximation between t=6.8 and t=6.9.At t=6.8: total≈49,228.5At t=6.9: total≈50,120The difference between t=6.8 and t=6.9 is 0.1 years, and the revenue increases by 50,120 - 49,228.5 = 891.5 over that interval.We need to find the t where total revenue=50,000. So, from t=6.8, we need an additional 50,000 - 49,228.5=771.5.So, the fraction is 771.5 / 891.5≈0.865.Therefore, t≈6.8 + 0.865*0.1≈6.8 + 0.0865≈6.8865 years.So, approximately 6.89 years.But since the question asks for the time t in years, and it's about 6.89 years, which is roughly 6 years and 10.68 months.But since it's a business problem, maybe they expect an approximate whole number or a decimal. So, 6.89 years is about 6.9 years.Alternatively, if we need to be precise, we can use more accurate calculations for 1.1^6.8 and 1.1^6.9.But for the sake of this problem, I think 6.9 years is a good approximation.So, the answer to part 1 is approximately 6.9 years.Now, moving on to part 2.The entrepreneur wants to invest in a start-up with revenue function ( R_S(t) = k cdot t^2 ). They want the rate of revenue growth at t=5 to match the combined rate of revenue growth of the established business at t=5.So, first, I need to find the combined rate of revenue growth at t=5 for the established business.The rate of revenue growth is the derivative of the revenue function with respect to t.So, for each city:- Atlanta: ( R_A(t) = 5000e^{0.05t} ). The derivative is ( R_A'(t) = 5000*0.05e^{0.05t} = 250e^{0.05t} )- New York: ( R_N(t) = 4500(1 + 0.04t) ). The derivative is ( R_N'(t) = 4500*0.04 = 180 )- Los Angeles: ( R_L(t) = 4000left(frac{1.1^t - 1}{0.1}right) ). Let's compute the derivative.First, rewrite R_L(t):( R_L(t) = 4000 times frac{1.1^t - 1}{0.1} = 40000 times (1.1^t - 1) )So, derivative is ( R_L'(t) = 40000 times ln(1.1) times 1.1^t )Compute ln(1.1): approximately 0.09531So, ( R_L'(t) = 40000 * 0.09531 * 1.1^t ≈ 3812.4 * 1.1^t )Now, the combined rate of revenue growth is:( R_A'(t) + R_N'(t) + R_L'(t) = 250e^{0.05t} + 180 + 3812.4 times 1.1^t )At t=5, we need to compute this.First, compute each term:- ( R_A'(5) = 250e^{0.25} ≈250*1.284≈321 )- ( R_N'(5) = 180 )- ( R_L'(5) = 3812.4 * 1.1^5 ≈3812.4 * 1.61051≈3812.4*1.61051≈6143.5 )So, total combined growth rate at t=5 is:321 + 180 + 6143.5≈6644.5 per year.Now, the start-up's revenue function is ( R_S(t) = k t^2 ). The rate of revenue growth is the derivative, which is ( R_S'(t) = 2k t ).At t=5, this is ( R_S'(5) = 2k*5 = 10k ).We need this to equal 6644.5, so:10k = 6644.5Therefore, k = 6644.5 / 10 = 664.45So, k≈664.45But let me double-check the calculations to make sure.First, R_A'(5):e^{0.25}= approximately 1.2840254, so 250*1.2840254≈321.00635R_N'(5)=180R_L'(5):1.1^5=1.61051, so 3812.4*1.61051≈3812.4*1.61051≈Let me compute 3812.4*1.6=6099.84 and 3812.4*0.01051≈3812.4*0.01=38.124, 3812.4*0.00051≈1.944, so total≈38.124+1.944≈40.068. So total R_L'(5)≈6099.84 + 40.068≈6139.908≈6140So, total combined growth rate≈321.00635 + 180 + 6140≈6641.00635≈6641So, 10k=6641 => k=664.1Wait, earlier I got 6644.5, but actually, it's more precise to say 6641.So, k≈664.1But to be precise, let's compute R_L'(5) more accurately.Compute 1.1^5:1.1^1=1.11.1^2=1.211.1^3=1.3311.1^4=1.46411.1^5=1.61051So, 1.1^5=1.61051 exactly.So, R_L'(5)=3812.4 *1.61051Compute 3812.4 *1.61051:First, 3812.4 *1=3812.43812.4 *0.6=2287.443812.4 *0.01=38.1243812.4 *0.00051≈1.944So, adding up:3812.4 + 2287.44=6100.846100.84 +38.124=6138.9646138.964 +1.944≈6140.908So, R_L'(5)=≈6140.908Therefore, total combined growth rate:321.00635 + 180 + 6140.908≈321.00635 + 180=501.00635 + 6140.908≈6641.91435≈6641.91So, 10k=6641.91 => k=664.191So, k≈664.19Therefore, k is approximately 664.19But since the problem says "the rate of revenue growth of the start-up at t=5 years matches the combined rate of revenue growth of the business...", so we need to set 10k=6641.91, so k≈664.19But let me check if I computed R_L'(t) correctly.R_L(t)=40000*(1.1^t -1)So, derivative is 40000*ln(1.1)*1.1^tln(1.1)=0.0953102So, 40000*0.0953102=3812.408So, R_L'(t)=3812.408*1.1^tAt t=5, 1.1^5=1.61051So, R_L'(5)=3812.408*1.61051≈3812.408*1.61051Compute 3812.408*1.6=6099.85283812.408*0.01051≈3812.408*0.01=38.12408 and 3812.408*0.00051≈1.944328So, total≈38.12408 +1.944328≈40.068408Thus, R_L'(5)=6099.8528 +40.068408≈6139.9212≈6139.92So, total combined growth rate:R_A'(5)=321.00635R_N'(5)=180R_L'(5)=6139.92Total≈321.00635 +180 +6139.92≈6640.92635≈6640.93So, 10k=6640.93 => k=664.093So, k≈664.09Therefore, k≈664.09But to be precise, let's carry out the exact calculation.Compute R_A'(5):250e^{0.25}=250* e^{0.25}e^{0.25}=1.2840254066So, 250*1.2840254066=321.00635165R_N'(5)=180R_L'(5)=3812.408*1.61051=3812.408*1.61051Compute 3812.408*1.61051:Let me compute 3812.408 *1.61051First, 3812.408 *1=3812.4083812.408 *0.6=2287.44483812.408 *0.01=38.124083812.408 *0.00051≈1.94432808So, total:3812.408 +2287.4448=6100.85286100.8528 +38.12408=6138.976886138.97688 +1.94432808≈6140.921208So, R_L'(5)=≈6140.921208Therefore, total combined growth rate:321.00635165 +180 +6140.921208≈321.00635165 +180=501.00635165 +6140.921208≈6641.92755965≈6641.93So, 10k=6641.93 => k=664.193So, k≈664.19Therefore, the value of k is approximately 664.19.But since the problem might expect an exact expression, let me see if I can write it in terms of exact values.Wait, R_A'(5)=250e^{0.25}R_N'(5)=180R_L'(5)=40000*ln(1.1)*1.1^5So, total growth rate:250e^{0.25} + 180 + 40000*ln(1.1)*1.1^5Thus, 10k=250e^{0.25} + 180 + 40000*ln(1.1)*1.1^5So, k=(250e^{0.25} + 180 + 40000*ln(1.1)*1.1^5)/10But unless they want an exact expression, which is complicated, I think the approximate value is fine.So, k≈664.19Therefore, the value of k is approximately 664.19.So, summarizing:1. The time t when combined revenue reaches 50,000 is approximately 6.89 years.2. The value of k is approximately 664.19.Final Answer1. The time when the combined revenue reaches 50,000 is boxed{6.9} years.2. The value of ( k ) is boxed{664.19}.</think>

🧠:<think>Alright, so I've got this problem about climate modeling, and I need to figure out the expressions for CO₂ concentration and global temperature anomaly over time. Let me try to break this down step by step.First, the problem gives me two differential equations:1. dC/dt = E(t) - αC(t)2. dT/dt = β(C(t) - C0) - γT(t)And there are some given parameters:- E(t) = E0 * e^(-δt) with E0 = 10 ppm/year and δ = 0.02 per year- α = 0.1 per year- β = 0.01 °C per ppm- C(0) = 400 ppm- C0 = 280 ppm- T(0) = 1 °C- γ = 0.05 per yearSo, I need to solve these differential equations to find C(t) and then T(t).Starting with the first equation: dC/dt = E(t) - αC(t). This is a linear first-order differential equation. I remember that for such equations, we can use an integrating factor to solve them.The standard form is dy/dt + P(t)y = Q(t). Let me rewrite the equation:dC/dt + αC(t) = E(t)So, P(t) = α and Q(t) = E(t) = 10e^(-0.02t).The integrating factor, μ(t), is e^(∫P(t)dt) = e^(αt). Multiplying both sides by μ(t):e^(αt) dC/dt + α e^(αt) C(t) = 10 e^(αt) e^(-0.02t)The left side is the derivative of [C(t) e^(αt)] with respect to t. So, integrating both sides:∫ d/dt [C(t) e^(αt)] dt = ∫ 10 e^(αt - 0.02t) dtSimplify the exponent on the right: αt - 0.02t = (0.1 - 0.02)t = 0.08tSo, the integral becomes:C(t) e^(αt) = 10 ∫ e^(0.08t) dt + KCompute the integral:∫ e^(0.08t) dt = (1/0.08) e^(0.08t) + CSo, plugging back in:C(t) e^(0.1t) = 10 * (1/0.08) e^(0.08t) + KSimplify 10 / 0.08: 10 / 0.08 = 125So,C(t) e^(0.1t) = 125 e^(0.08t) + KNow, solve for C(t):C(t) = 125 e^(0.08t) e^(-0.1t) + K e^(-0.1t)Simplify the exponents:0.08t - 0.1t = -0.02tSo,C(t) = 125 e^(-0.02t) + K e^(-0.1t)Now, apply the initial condition C(0) = 400 ppm.At t = 0:C(0) = 125 e^(0) + K e^(0) = 125 + K = 400So, K = 400 - 125 = 275Therefore, the expression for C(t) is:C(t) = 125 e^(-0.02t) + 275 e^(-0.1t)Okay, that should be the solution for the first part. Let me double-check my steps.1. Wrote the equation in standard linear form: correct.2. Calculated integrating factor: e^(αt) is correct.3. Multiplied through: correct.4. Integrated both sides: correct, and the integral of e^(kt) is (1/k)e^(kt).5. Plugged in the integral and simplified: 10 / 0.08 is indeed 125.6. Applied initial condition: correct, 125 + K = 400, so K=275.Looks solid. So, moving on to the second part, finding T(t).The second differential equation is:dT/dt = β(C(t) - C0) - γT(t)We already have C(t), so we can plug that in.First, let's write the equation:dT/dt + γT(t) = β(C(t) - C0)This is also a linear first-order differential equation. So, again, we can use the integrating factor method.Let me write it in standard form:dT/dt + γT(t) = β(C(t) - C0)So, P(t) = γ, Q(t) = β(C(t) - C0)Compute the integrating factor: μ(t) = e^(∫γ dt) = e^(γt)Multiply both sides:e^(γt) dT/dt + γ e^(γt) T(t) = β e^(γt) (C(t) - C0)Left side is d/dt [T(t) e^(γt)]So, integrating both sides:∫ d/dt [T(t) e^(γt)] dt = ∫ β e^(γt) (C(t) - C0) dtThus,T(t) e^(γt) = β ∫ e^(γt) (C(t) - C0) dt + KWe need to compute this integral. Let's first write out C(t) - C0:C(t) - C0 = [125 e^(-0.02t) + 275 e^(-0.1t)] - 280So,C(t) - C0 = 125 e^(-0.02t) + 275 e^(-0.1t) - 280Therefore, the integral becomes:∫ e^(γt) [125 e^(-0.02t) + 275 e^(-0.1t) - 280] dtLet me split this into three separate integrals:125 ∫ e^(γt) e^(-0.02t) dt + 275 ∫ e^(γt) e^(-0.1t) dt - 280 ∫ e^(γt) dtSimplify each exponent:For the first integral: γt - 0.02t = (0.05 - 0.02)t = 0.03tSecond integral: γt - 0.1t = (0.05 - 0.1)t = -0.05tThird integral: just γt = 0.05tSo, the integrals become:125 ∫ e^(0.03t) dt + 275 ∫ e^(-0.05t) dt - 280 ∫ e^(0.05t) dtCompute each integral:First integral: ∫ e^(0.03t) dt = (1/0.03) e^(0.03t) + CSecond integral: ∫ e^(-0.05t) dt = (-1/0.05) e^(-0.05t) + CThird integral: ∫ e^(0.05t) dt = (1/0.05) e^(0.05t) + CPutting it all together:125*(1/0.03) e^(0.03t) + 275*(-1/0.05) e^(-0.05t) - 280*(1/0.05) e^(0.05t) + KCompute the constants:125 / 0.03 ≈ 4166.6667275 / 0.05 = 5500, but with a negative sign: -5500280 / 0.05 = 5600So, substituting:≈ 4166.6667 e^(0.03t) - 5500 e^(-0.05t) - 5600 e^(0.05t) + KTherefore, the integral is:≈ 4166.6667 e^(0.03t) - 5500 e^(-0.05t) - 5600 e^(0.05t) + KBut let's keep it exact instead of approximate:125 / 0.03 = 12500 / 3275 / 0.05 = 5500280 / 0.05 = 5600So,(12500/3) e^(0.03t) - 5500 e^(-0.05t) - 5600 e^(0.05t) + KTherefore, going back to the equation:T(t) e^(γt) = β [ (12500/3) e^(0.03t) - 5500 e^(-0.05t) - 5600 e^(0.05t) ] + KBut remember, γ is 0.05, so e^(γt) is e^(0.05t). So, we can write:T(t) e^(0.05t) = β [ (12500/3) e^(0.03t) - 5500 e^(-0.05t) - 5600 e^(0.05t) ] + KNow, solve for T(t):T(t) = β [ (12500/3) e^(0.03t) - 5500 e^(-0.05t) - 5600 e^(0.05t) ] e^(-0.05t) + K e^(-0.05t)Simplify each term:First term: (12500/3) e^(0.03t) e^(-0.05t) = (12500/3) e^(-0.02t)Second term: -5500 e^(-0.05t) e^(-0.05t) = -5500 e^(-0.1t)Third term: -5600 e^(0.05t) e^(-0.05t) = -5600 e^(0) = -5600So, putting it all together:T(t) = β [ (12500/3) e^(-0.02t) - 5500 e^(-0.1t) - 5600 ] + K e^(-0.05t)Now, plug in β = 0.01:T(t) = 0.01 [ (12500/3) e^(-0.02t) - 5500 e^(-0.1t) - 5600 ] + K e^(-0.05t)Compute each term:0.01 * (12500/3) = (12500/3) * 0.01 = 125/3 ≈ 41.66670.01 * (-5500) = -550.01 * (-5600) = -56So,T(t) = (125/3) e^(-0.02t) - 55 e^(-0.1t) - 56 + K e^(-0.05t)Now, apply the initial condition T(0) = 1 °C.At t = 0:T(0) = (125/3) e^(0) - 55 e^(0) - 56 + K e^(0) = (125/3) - 55 - 56 + K = 1Compute (125/3) ≈ 41.6667So,41.6667 - 55 - 56 + K = 1Calculate 41.6667 - 55 = -13.3333-13.3333 - 56 = -69.3333So,-69.3333 + K = 1Therefore, K = 1 + 69.3333 ≈ 70.3333But let's keep it exact:125/3 - 55 - 56 + K = 1Convert all to thirds:125/3 - 165/3 - 168/3 + K = 1Compute numerator: 125 - 165 - 168 = 125 - 333 = -208So,-208/3 + K = 1Thus, K = 1 + 208/3 = (3/3 + 208/3) = 211/3 ≈ 70.3333So, K = 211/3Therefore, the expression for T(t) is:T(t) = (125/3) e^(-0.02t) - 55 e^(-0.1t) - 56 + (211/3) e^(-0.05t)We can write this as:T(t) = (125/3) e^(-0.02t) - 55 e^(-0.1t) + (211/3) e^(-0.05t) - 56Alternatively, factor out 1/3 where possible:T(t) = (125 e^(-0.02t) + 211 e^(-0.05t))/3 - 55 e^(-0.1t) - 56But it's probably fine as is.Let me check my steps again.1. Wrote the equation in standard form: correct.2. Calculated integrating factor: e^(γt) is correct.3. Multiplied through and integrated: correct.4. Expanded C(t) - C0: correct.5. Split the integral into three parts: correct.6. Integrated each term: correct, with the exponents simplified.7. Plugged back into the equation: correct.8. Applied initial condition T(0)=1: correct, solved for K.9. Plugged K back into the expression: correct.Looks good. So, summarizing:1. C(t) = 125 e^(-0.02t) + 275 e^(-0.1t)2. T(t) = (125/3) e^(-0.02t) - 55 e^(-0.1t) + (211/3) e^(-0.05t) - 56I think that's it. Let me just write them neatly.For C(t):C(t) = 125 e^{-0.02t} + 275 e^{-0.1t}For T(t):T(t) = (125/3) e^{-0.02t} - 55 e^{-0.1t} + (211/3) e^{-0.05t} - 56Alternatively, if I want to write T(t) with all terms over 3:T(t) = (125 e^{-0.02t} + 211 e^{-0.05t} - 165 e^{-0.1t} - 168)/3Wait, let me see:(125/3) e^{-0.02t} is 125 e^{-0.02t}/3-55 e^{-0.1t} is -165 e^{-0.1t}/3(211/3) e^{-0.05t} is 211 e^{-0.05t}/3-56 is -168/3So, combining all over 3:[125 e^{-0.02t} + 211 e^{-0.05t} - 165 e^{-0.1t} - 168]/3But that might not necessarily be simpler. Maybe leave it as is.Alternatively, factor out 1/3:T(t) = (125 e^{-0.02t} + 211 e^{-0.05t} - 165 e^{-0.1t} - 168)/3But I think the initial form is clearer.So, I think I've got both expressions. Let me just recap:1. Solved the first ODE for C(t) using integrating factor, found C(t) = 125 e^{-0.02t} + 275 e^{-0.1t}2. Plugged C(t) into the second ODE, solved for T(t), got T(t) = (125/3) e^{-0.02t} - 55 e^{-0.1t} + (211/3) e^{-0.05t} - 56I think that's all. I don't see any mistakes in my calculations, so I feel confident about these results.Final Answer1. The expression for the CO₂ concentration is boxed{C(t) = 125 e^{-0.02t} + 275 e^{-0.1t}}.2. The expression for the global temperature anomaly is boxed{T(t) = frac{125}{3} e^{-0.02t} - 55 e^{-0.1t} + frac{211}{3} e^{-0.05t} - 56}.</think>

🧠:<think>Alright, so I have this problem about a Beverly Hills resident investing in real estate. They want to maximize property prices based on distance from the city center (D) and historical appreciation rate (A). The function given is ( P(D, A) = 0.5D^2 + frac{100A}{D} ). The first part asks to find the critical points and determine which combination of D and A maximizes P, using Lagrange multipliers with the constraint that the total investment can't exceed 10 million, so ( P(D, A) leq 10 ).Hmm, okay. So I remember that critical points are where the gradient is zero or undefined, but since we have a constraint, Lagrange multipliers are the way to go. The constraint is ( P(D, A) = 10 ), right? Because we want to maximize P under the condition that it can't exceed 10. So we set up the Lagrangian function.Let me recall the method. The Lagrangian ( mathcal{L} ) is the function we want to maximize minus lambda times the constraint. But wait, actually, in this case, since we're maximizing P subject to P ≤ 10, the maximum will occur at P = 10, so the constraint is active. So we can set up the Lagrangian as ( mathcal{L}(D, A, lambda) = 0.5D^2 + frac{100A}{D} - lambda(0.5D^2 + frac{100A}{D} - 10) ).Wait, no, actually, the Lagrangian should be the function to maximize minus lambda times the constraint. But since we're maximizing P subject to P = 10, the Lagrangian is ( mathcal{L} = 0.5D^2 + frac{100A}{D} - lambda(0.5D^2 + frac{100A}{D} - 10) ). Hmm, but actually, that seems redundant because the constraint is the same as the function. Maybe I'm overcomplicating.Alternatively, perhaps I should set up the Lagrangian as ( mathcal{L} = 0.5D^2 + frac{100A}{D} - lambda(0.5D^2 + frac{100A}{D} - 10) ). Then take partial derivatives with respect to D, A, and lambda.Wait, but that would make the partial derivatives:( frac{partial mathcal{L}}{partial D} = D - lambda(D - frac{100A}{D^2}) = 0 )( frac{partial mathcal{L}}{partial A} = frac{100}{D} - lambda(frac{100}{D}) = 0 )( frac{partial mathcal{L}}{partial lambda} = -(0.5D^2 + frac{100A}{D} - 10) = 0 )So from the second equation, ( frac{100}{D} - lambda frac{100}{D} = 0 ). Factor out ( frac{100}{D} ):( frac{100}{D}(1 - lambda) = 0 )Since ( frac{100}{D} ) isn't zero (D can't be infinity), we have ( 1 - lambda = 0 ), so ( lambda = 1 ).Now plug lambda into the first equation:( D - 1*(D - frac{100A}{D^2}) = 0 )Simplify:( D - D + frac{100A}{D^2} = 0 )Which simplifies to ( frac{100A}{D^2} = 0 )But that implies A = 0, which doesn't make sense because appreciation rate can't be zero if we're trying to maximize P. Hmm, maybe I made a mistake in setting up the Lagrangian.Wait, perhaps I should think differently. Maybe the constraint is not P = 10, but P ≤ 10, so the maximum could be either at the critical point inside the feasible region or on the boundary. So first, find the critical points without the constraint, then check if they satisfy P ≤ 10. If not, then use Lagrange multipliers on the boundary.So let's try that approach.First, find the critical points of P(D, A) without any constraints. To do that, take partial derivatives with respect to D and A, set them equal to zero.Partial derivative with respect to D:( frac{partial P}{partial D} = D - frac{100A}{D^2} )Partial derivative with respect to A:( frac{partial P}{partial A} = frac{100}{D} )Set them equal to zero:1. ( D - frac{100A}{D^2} = 0 )2. ( frac{100}{D} = 0 )But equation 2 implies ( frac{100}{D} = 0 ), which is impossible because 100 divided by any finite D can't be zero. So there are no critical points inside the feasible region. Therefore, the maximum must occur on the boundary, which is P = 10.So now, we can use Lagrange multipliers on the boundary P = 10.So set up the Lagrangian as ( mathcal{L}(D, A, lambda) = 0.5D^2 + frac{100A}{D} - lambda(0.5D^2 + frac{100A}{D} - 10) )Wait, that's the same as before. So taking partial derivatives again:( frac{partial mathcal{L}}{partial D} = D - lambda(D - frac{100A}{D^2}) = 0 )( frac{partial mathcal{L}}{partial A} = frac{100}{D} - lambda(frac{100}{D}) = 0 )( frac{partial mathcal{L}}{partial lambda} = -(0.5D^2 + frac{100A}{D} - 10) = 0 )From the second equation, as before, ( frac{100}{D}(1 - lambda) = 0 ), so ( lambda = 1 ).Plugging back into the first equation:( D - (D - frac{100A}{D^2}) = 0 )Simplify:( D - D + frac{100A}{D^2} = 0 )So ( frac{100A}{D^2} = 0 ), which again implies A = 0. But A can't be zero because that would make P = 0.5D^2, which can be up to 10, but we need to maximize P. Wait, but if A = 0, then P = 0.5D^2, so to get P = 10, D would be sqrt(20) ≈ 4.472 miles. But is that the maximum?Wait, but if A = 0, then P is just 0.5D^2, which is a parabola opening upwards. So as D increases, P increases. But we have a constraint P ≤ 10, so the maximum P would be at D = sqrt(20), A = 0. But that seems counterintuitive because A is also a factor. Maybe I'm missing something.Alternatively, perhaps the maximum occurs when both D and A are positive. But from the Lagrangian, we ended up with A = 0, which suggests that the maximum on the boundary is at A = 0, D = sqrt(20). But let's check.If A = 0, then P = 0.5D^2, so to get P = 10, D = sqrt(20). If we try to increase A, we have to decrease D to keep P = 10. But does that make P larger? Wait, no, because P is fixed at 10. So maybe the maximum is indeed at A = 0, D = sqrt(20).But that seems odd because A is a positive term in P. Let me think again. The function P(D, A) = 0.5D^2 + 100A/D. So for a given D, increasing A increases P. So to maximize P, for a given D, we want A to be as large as possible. But subject to P = 10.Wait, but if we fix D, then A is determined by P = 10: 10 = 0.5D^2 + 100A/D => 100A/D = 10 - 0.5D^2 => A = (10 - 0.5D^2) * D / 100.So A = (10D - 0.5D^3)/100.So to maximize P, which is fixed at 10, but we need to find D and A such that P = 10 and the combination is optimal. But since P is fixed, maybe the question is to find the combination that allows the highest possible P, but since P is constrained to 10, maybe the critical point is where the gradient is parallel to the constraint, which is what Lagrange multipliers do.But earlier, we saw that the only solution is A = 0, which gives P = 10 when D = sqrt(20). But maybe that's the only critical point on the boundary. Alternatively, perhaps I made a mistake in setting up the Lagrangian.Wait, another approach: since we're maximizing P subject to P ≤ 10, the maximum is 10, achieved when P = 10. So we need to find the D and A that make P = 10 and also satisfy the condition for maximum. But since P is fixed, maybe we need to find where the function is increasing the most, but I'm not sure.Alternatively, perhaps the resident wants to maximize P, but since P is constrained to 10, the maximum is achieved at the boundary, and we need to find the D and A that satisfy P = 10 and also are critical points.But from the Lagrangian, we ended up with A = 0, which seems counterintuitive. Maybe I should try another method.Let me try to express A in terms of D from the constraint P = 10:10 = 0.5D^2 + 100A/D => 100A/D = 10 - 0.5D^2 => A = (10 - 0.5D^2) * D / 100 = (10D - 0.5D^3)/100.So A = (10D - 0.5D^3)/100.Now, to find the maximum of P, which is fixed at 10, but perhaps we need to find the D that maximizes some other function? Wait, no, P is fixed. Maybe the resident wants to maximize something else, but the problem says to maximize P under the constraint P ≤ 10, which is a bit confusing because P is the function to maximize, but it's constrained by itself.Wait, perhaps the resident wants to maximize P, but subject to some other constraint, but the problem says the total investment cannot exceed 10 million, i.e., P(D, A) ≤ 10. So the maximum P is 10, achieved when P = 10. So the question is, what combination of D and A gives P = 10, and perhaps also satisfies some other condition, like being a critical point.But earlier, when trying to find critical points without constraints, we saw that there are none because the partial derivative with respect to A is always positive, meaning P increases with A for any fixed D. So to maximize P, for any D, we want A to be as large as possible, but constrained by P = 10.So perhaps the maximum occurs when A is as large as possible, which would require D to be as small as possible. But D can't be zero because then P would be undefined (division by zero). So as D approaches zero, A approaches infinity, but P approaches infinity as well, which contradicts the constraint P ≤ 10.Wait, that can't be. Let me think again. If D approaches zero, then 0.5D^2 approaches zero, and 100A/D approaches infinity if A is positive. So to have P = 10, as D approaches zero, A must approach zero as well, but in such a way that 100A/D approaches 10. So A = (10D)/100 = D/10. So as D approaches zero, A approaches zero as well, but very slowly.But that seems like P can be 10 even with very small D and A, but that doesn't necessarily mean it's a maximum. Wait, but since P is fixed at 10, maybe the question is to find the D and A that make P = 10 and also satisfy the condition for a critical point, which would be where the gradient is zero.But earlier, we saw that the gradient can't be zero because the partial derivative with respect to A is always positive. So perhaps there are no critical points on the boundary, meaning the maximum is achieved at the boundary, but the critical point approach doesn't yield a solution. Maybe the maximum is achieved when A is as large as possible, which would require D to be as small as possible, but D can't be zero.Alternatively, perhaps the maximum occurs when the partial derivatives are proportional, which is what Lagrange multipliers enforce. But we ended up with A = 0, which seems contradictory.Wait, maybe I should try to parameterize A in terms of D and then find the maximum of P, but since P is fixed, perhaps we need to find the D that maximizes some other function related to P. But I'm getting confused.Let me try a different approach. Since P = 10, and we have A = (10D - 0.5D^3)/100, perhaps we can express P in terms of D and then find the D that maximizes P, but P is fixed. Hmm, not helpful.Alternatively, maybe the resident wants to maximize the rate of change of P, but that's not what the problem says. The problem says to maximize P under the constraint P ≤ 10, so the maximum is 10, achieved when P = 10. So the critical point would be where the gradient of P is parallel to the gradient of the constraint, which is the same as the Lagrangian method.But from that, we got A = 0, which seems odd. Maybe I made a mistake in the partial derivatives.Let me recompute the partial derivatives for the Lagrangian.Given ( mathcal{L} = 0.5D^2 + frac{100A}{D} - lambda(0.5D^2 + frac{100A}{D} - 10) )Partial derivative with respect to D:( frac{partial mathcal{L}}{partial D} = D - lambda(D - frac{100A}{D^2}) )Partial derivative with respect to A:( frac{partial mathcal{L}}{partial A} = frac{100}{D} - lambda(frac{100}{D}) )Partial derivative with respect to lambda:( frac{partial mathcal{L}}{partial lambda} = -(0.5D^2 + frac{100A}{D} - 10) )So from the A derivative:( frac{100}{D}(1 - lambda) = 0 )Which gives ( lambda = 1 )From the D derivative:( D - (D - frac{100A}{D^2}) = 0 )Simplify:( D - D + frac{100A}{D^2} = 0 )So ( frac{100A}{D^2} = 0 ) => A = 0So indeed, A must be zero. Then, from the constraint P = 10:10 = 0.5D^2 + 0 => D^2 = 20 => D = sqrt(20) ≈ 4.472 miles.So the critical point is at D = sqrt(20), A = 0.But wait, if A = 0, then the appreciation rate is zero, which might not be desirable because the resident is looking to invest for appreciation. So maybe the maximum under the constraint is at A = 0, but that might not be the best investment. However, mathematically, that's the critical point.Alternatively, perhaps the resident wants to maximize P, but also consider the trade-off between D and A. Maybe the maximum occurs when the marginal increase in P from increasing D equals the marginal increase from increasing A, but under the constraint.Wait, but since P is fixed at 10, maybe the resident wants to choose D and A such that the rate of change of P with respect to D and A is balanced. But I'm not sure.Alternatively, perhaps the problem is to maximize P without the constraint, but that's not what it says. It says to maximize P under the constraint P ≤ 10.Wait, maybe I'm overcomplicating. The critical point on the boundary is at A = 0, D = sqrt(20). So that's the answer for part 1.Now, moving on to part 2. The resident decides to purchase two properties instead of one, with distances D1 and D2, and appreciation rates A1 and A2. The total price is P_total = P(D1, A1) + P(D2, A2). We need to maximize P_total under the constraint that the total investment doesn't exceed 10 million, so P(D1, A1) + P(D2, A2) ≤ 10.So now, we have four variables: D1, A1, D2, A2, and the constraint is P1 + P2 ≤ 10.To maximize P_total, we can use the method of Lagrange multipliers again, but now with four variables and one constraint.Alternatively, since the problem is symmetric, maybe the optimal solution is to have both properties at the same D and A. Let's test that.Assume D1 = D2 = D and A1 = A2 = A. Then P_total = 2*(0.5D^2 + 100A/D) = D^2 + 200A/D.Subject to D^2 + 200A/D ≤ 10.But we can also consider that the maximum might occur when both properties are at the same critical point as in part 1, which was D = sqrt(20), A = 0. But that would give P_total = 2*10 = 20, which exceeds the constraint of 10. So that's not allowed.Wait, but the total investment can't exceed 10, so if each property is 10, the total would be 20, which is over. So we need to find D1, A1, D2, A2 such that P1 + P2 = 10, and P_total is maximized. But since P_total is the sum, and we're subject to P1 + P2 ≤ 10, the maximum P_total is 10, achieved when P1 + P2 = 10.But how to distribute the investment between the two properties to maximize the total P, which is fixed at 10. Wait, but if we have two properties, maybe we can have a higher total P by optimizing D1, A1, D2, A2 such that their sum is 10, but individually, each is contributing to the total.But since P is a function of D and A, maybe distributing the investment in a way that each property is at its optimal point.Wait, but in part 1, the optimal point was A = 0, D = sqrt(20). But if we have two properties, each would have P = 5, so D^2 = 10, D = sqrt(10), and A = 0. So each property would be at D = sqrt(10), A = 0, giving P = 5 each, total 10.But is that the maximum? Or can we have a higher total P by having different D and A for each property?Wait, but since P_total is fixed at 10, the maximum is 10, so the question is to find the distribution of D1, A1, D2, A2 such that P1 + P2 = 10 and the combination is optimal.But perhaps the optimal is to have both properties at the same D and A as in part 1, but scaled down. So each property would have D = sqrt(10), A = 0, giving P = 5 each, total 10.But maybe there's a better distribution. Let's consider the Lagrangian for two variables.Let me set up the Lagrangian for two properties:( mathcal{L} = 0.5D1^2 + frac{100A1}{D1} + 0.5D2^2 + frac{100A2}{D2} - lambda(0.5D1^2 + frac{100A1}{D1} + 0.5D2^2 + frac{100A2}{D2} - 10) )Taking partial derivatives:For D1:( D1 - lambda(D1 - frac{100A1}{D1^2}) = 0 )For A1:( frac{100}{D1} - lambda(frac{100}{D1}) = 0 )Similarly for D2 and A2:( D2 - lambda(D2 - frac{100A2}{D2^2}) = 0 )( frac{100}{D2} - lambda(frac{100}{D2}) = 0 )From the A1 and A2 partial derivatives:( frac{100}{D1}(1 - lambda) = 0 ) => ( lambda = 1 )Similarly for A2: ( lambda = 1 )So lambda is 1 for both.Now, plug lambda = 1 into the D1 equation:( D1 - (D1 - frac{100A1}{D1^2}) = 0 )Simplify:( D1 - D1 + frac{100A1}{D1^2} = 0 ) => ( frac{100A1}{D1^2} = 0 ) => A1 = 0Similarly for D2: A2 = 0So again, we get A1 = A2 = 0, and then from the constraint:P1 + P2 = 0.5D1^2 + 0.5D2^2 = 10So 0.5(D1^2 + D2^2) = 10 => D1^2 + D2^2 = 20To maximize P_total, which is fixed at 10, but we need to find D1 and D2 such that D1^2 + D2^2 = 20.But since P_total is fixed, the maximum is achieved when P_total = 10, and the distribution of D1 and D2 can be anything that satisfies D1^2 + D2^2 = 20. However, to maximize the individual P's, we might want to have each property at its optimal point, but since A1 and A2 are zero, each P is 0.5D^2.So to maximize the sum, which is fixed, but perhaps the optimal is to have both D1 and D2 as equal as possible, but since A1 and A2 are zero, it doesn't matter. So D1 = D2 = sqrt(10), because then D1^2 + D2^2 = 10 + 10 = 20.So the optimal values are D1 = D2 = sqrt(10), A1 = A2 = 0.But wait, that seems similar to part 1, but scaled down. So each property is at D = sqrt(10), A = 0, giving P = 5 each, total 10.Alternatively, maybe we can have one property with higher D and lower A, and the other with lower D and higher A, but given that A must be zero, it's not possible.Wait, but from the Lagrangian, we found that A1 and A2 must be zero, so the optimal is indeed D1 = D2 = sqrt(10), A1 = A2 = 0.But that seems a bit strange because in part 1, the optimal was D = sqrt(20), A = 0, but when splitting into two properties, each has D = sqrt(10), A = 0.So, in conclusion, for part 1, the critical point is D = sqrt(20), A = 0, and for part 2, the optimal is D1 = D2 = sqrt(10), A1 = A2 = 0.But let me double-check. If I set A1 and A2 to zero, then P1 = 0.5D1^2 and P2 = 0.5D2^2, so P_total = 0.5(D1^2 + D2^2) = 10 => D1^2 + D2^2 = 20.To maximize P_total, which is fixed, but if we consider the individual P's, they are each maximized when D is as large as possible, but constrained by the total D1^2 + D2^2 = 20.Wait, but if we set D1 = sqrt(20) and D2 = 0, then P1 = 10 and P2 = 0, but D2 can't be zero because A2 would be undefined. So the maximum occurs when D1 and D2 are such that their squares sum to 20, and each is as large as possible. But since A1 and A2 are zero, it's just about distributing D1 and D2.But since the function P is convex in D when A is zero, the maximum sum occurs when one D is as large as possible and the other as small as possible, but subject to D2 > 0.Wait, but if we set D1 = sqrt(20 - ε^2) and D2 = ε, as ε approaches zero, then P1 approaches 10 and P2 approaches 0. So the maximum P_total is achieved when one property is at D = sqrt(20), A = 0, and the other is at D approaching zero, A approaching zero. But since we can't have D = 0, maybe the optimal is to have one property at D = sqrt(20), A = 0, and the other at D approaching zero, but that's not practical.Alternatively, perhaps the optimal is to have both properties at D = sqrt(10), A = 0, because that's symmetric and satisfies the constraint.But mathematically, the maximum P_total is achieved when one property is at D = sqrt(20), A = 0, and the other is at D approaching zero, A approaching zero, giving P_total approaching 10. But since we can't have D = 0, maybe the optimal is to have both properties at D = sqrt(10), A = 0, giving P_total = 10.But I'm not sure. Let me think again.If we have two properties, each with D = sqrt(10), A = 0, then P_total = 10, which is the maximum allowed. Alternatively, if we have one property with D = sqrt(20), A = 0, and the other with D approaching zero, A approaching zero, then P_total approaches 10 as well. So both distributions give the same total P, but the first is more balanced.However, since the problem asks for optimal values, which likely means the symmetric solution where both properties are identical, so D1 = D2 = sqrt(10), A1 = A2 = 0.So, in summary:1. For one property, the critical point is D = sqrt(20), A = 0.2. For two properties, the optimal is D1 = D2 = sqrt(10), A1 = A2 = 0.But let me verify if there's a better distribution where A1 and A2 are positive. Suppose we set A1 and A2 to some positive values, then D1 and D2 would be smaller, but would that allow for a higher total P? Wait, no, because P_total is constrained to 10. So even if we have positive A's, the total P can't exceed 10. So the maximum is still 10, achieved when P1 + P2 = 10.But perhaps by having positive A's, we can have a higher P_total, but that's not possible because P_total is fixed at 10. So the maximum is 10, and the distribution of D and A is such that P1 + P2 = 10.But from the Lagrangian, we saw that A1 and A2 must be zero, so the optimal is indeed D1 = D2 = sqrt(10), A1 = A2 = 0.Wait, but if A1 and A2 are zero, then P1 = 0.5D1^2 and P2 = 0.5D2^2, so P_total = 0.5(D1^2 + D2^2) = 10 => D1^2 + D2^2 = 20.So the optimal is when D1 and D2 are such that their squares sum to 20, and since we want to maximize P_total, which is fixed, but perhaps the individual P's are maximized when D1 and D2 are as large as possible. But since D1 and D2 are constrained by their squares summing to 20, the maximum occurs when one is sqrt(20) and the other is zero, but D can't be zero. So the optimal is to have one property at D = sqrt(20), A = 0, and the other at D approaching zero, A approaching zero.But since D can't be zero, maybe the optimal is to have both properties at D = sqrt(10), A = 0, which is a balanced approach.I think the answer expects the symmetric solution, so D1 = D2 = sqrt(10), A1 = A2 = 0.So, to conclude:1. The critical point for one property is D = sqrt(20), A = 0.2. For two properties, the optimal is D1 = D2 = sqrt(10), A1 = A2 = 0.</think>

🧠:<think>Alright, so I have this problem where a former NFL player is running a training academy and wants to optimize the number of physical conditioning sessions (x) and tactical drills (y) to maximize overall performance. The performance is given by the function P(x, y) = 3x² + 2xy + 5y². The constraints are that the total number of sessions can't exceed 20 in a week, and there's also a budget constraint of 2500, with physical sessions costing 100 each and tactical drills costing 150 each.Starting with part 1: I need to maximize P(x, y) subject to x + y ≤ 20. Since this is a quadratic function, I think I should use methods from multivariable calculus, maybe finding critical points and checking the boundaries.First, let's find the critical points by taking partial derivatives. The partial derivative of P with respect to x is 6x + 2y, and with respect to y is 2x + 10y. Setting these equal to zero:6x + 2y = 0 2x + 10y = 0Hmm, solving these equations. Let me write them:1) 6x + 2y = 0 2) 2x + 10y = 0From equation 1: 6x + 2y = 0 => 3x + y = 0 => y = -3xPlugging into equation 2: 2x + 10*(-3x) = 0 => 2x - 30x = 0 => -28x = 0 => x = 0Then y = -3*0 = 0. So the only critical point is at (0,0), which is a minimum, not a maximum, since the function is quadratic and the coefficients are positive.Therefore, the maximum must occur on the boundary of the feasible region, which is x + y = 20. So we can express y as 20 - x and substitute into P(x, y).So P(x) = 3x² + 2x(20 - x) + 5(20 - x)² Let me expand this:First, 3x² Then, 2x(20 - x) = 40x - 2x² Then, 5(20 - x)² = 5*(400 - 40x + x²) = 2000 - 200x + 5x²Adding all together: 3x² + 40x - 2x² + 2000 - 200x + 5x² Combine like terms:3x² - 2x² + 5x² = 6x² 40x - 200x = -160x Constant term: 2000So P(x) = 6x² - 160x + 2000Now, to find the maximum of this quadratic function. Since the coefficient of x² is positive, it opens upwards, so the vertex is a minimum. Wait, that can't be right because we're looking for a maximum on the boundary. Hmm, maybe I made a mistake.Wait, no, actually, the function P(x, y) is quadratic, but when we substitute y = 20 - x, we get a quadratic in x which is also opening upwards, meaning it has a minimum, not a maximum. So does that mean that the maximum occurs at the endpoints of the interval?Yes, that must be it. So the maximum of P(x) on the interval x ∈ [0,20] will occur at one of the endpoints, either x=0 or x=20.Calculating P(0): y=20 P(0,20) = 3*(0)^2 + 2*(0)*(20) + 5*(20)^2 = 0 + 0 + 5*400 = 2000Calculating P(20): y=0 P(20,0) = 3*(20)^2 + 2*(20)*(0) + 5*(0)^2 = 3*400 + 0 + 0 = 1200Wait, so P(0,20) is 2000 and P(20,0) is 1200. So 2000 is larger, so the maximum is at (0,20). But that seems counterintuitive because both x and y contribute positively to P. Maybe I need to check my calculations.Wait, let me recalculate P(0,20):3*(0)^2 = 0 2*(0)*(20) = 0 5*(20)^2 = 5*400 = 2000 Total: 2000P(20,0):3*(20)^2 = 3*400 = 1200 2*(20)*(0) = 0 5*(0)^2 = 0 Total: 1200So yes, 2000 is larger. So according to this, the maximum occurs at (0,20). But that seems odd because if you don't do any physical conditioning, just tactical drills, you get higher performance. Maybe because the coefficient for y² is higher (5 vs 3), so each tactical drill contributes more per session.But let me think again. Maybe I should have considered the Lagrangian method for constrained optimization. Let me try that.The Lagrangian is L = 3x² + 2xy + 5y² - λ(x + y - 20)Taking partial derivatives:dL/dx = 6x + 2y - λ = 0 dL/dy = 2x + 10y - λ = 0 dL/dλ = x + y - 20 = 0So we have the system:1) 6x + 2y = λ 2) 2x + 10y = λ 3) x + y = 20From equations 1 and 2: 6x + 2y = 2x + 10y Subtract 2x + 2y from both sides: 4x = 8y => x = 2yFrom equation 3: x + y = 20 => 2y + y = 20 => 3y = 20 => y = 20/3 ≈ 6.666...Then x = 2*(20/3) = 40/3 ≈ 13.333...But x and y have to be integers since you can't have a fraction of a session. So we need to check the integer points around (13.333, 6.666). So possible candidates are (13,7), (14,6), (12,8), etc.Let me compute P for these points.First, (13,7):P = 3*(13)^2 + 2*(13)*(7) + 5*(7)^2 = 3*169 + 2*91 + 5*49 = 507 + 182 + 245 = 507 + 182 = 689; 689 + 245 = 934Wait, that can't be right because earlier at (0,20) we had 2000. Wait, no, hold on, I think I messed up the substitution earlier.Wait, no, when I substituted y = 20 - x, I got P(x) = 6x² - 160x + 2000, which at x=0 is 2000, but when I plug in (13,7), it's 3*(13)^2 + 2*(13)*(7) + 5*(7)^2.Wait, 3*169 is 507, 2*91 is 182, 5*49 is 245. So 507 + 182 is 689, plus 245 is 934. That's way lower than 2000. So that can't be right.Wait, maybe I made a mistake in the substitution earlier. Let me check that again.Original P(x, y) = 3x² + 2xy + 5y²When y = 20 - x,P(x) = 3x² + 2x(20 - x) + 5(20 - x)² = 3x² + 40x - 2x² + 5*(400 - 40x + x²) = 3x² + 40x - 2x² + 2000 - 200x + 5x² Combine like terms:3x² - 2x² + 5x² = 6x² 40x - 200x = -160x Constant term: 2000So P(x) = 6x² - 160x + 2000Wait, but when x=0, P=2000, which is correct. When x=20, P=6*(400) -160*20 +2000=2400 -3200 +2000=1200, which matches.But when I plug in x=13, y=7, P=934, which is way less than 2000. So that suggests that the maximum is indeed at (0,20). But that seems strange because the Lagrangian method gave me a critical point at (40/3, 20/3), which is approximately (13.333,6.666), but that point is a minimum because the function is convex.Wait, no, the function is convex, so the critical point is a minimum. Therefore, the maximum must be at the endpoints. So the maximum is at (0,20) with P=2000.But that seems counterintuitive because both x and y contribute positively. Maybe the function is such that y has a higher weight, so more y gives higher P.But let me confirm with another point. Let's try x=10, y=10.P=3*100 + 2*100 +5*100=300+200+500=1000, which is less than 2000.x=5, y=15:P=3*25 +2*75 +5*225=75 +150 +1125=1350 <2000x=15, y=5:P=3*225 +2*75 +5*25=675 +150 +125=950 <2000So yes, it seems that (0,20) gives the highest P. So the optimal is x=0, y=20.But that seems odd because you wouldn't have any physical conditioning. Maybe the model is such that y is more impactful.Okay, moving on to part 2: budget constraint. The cost is 100x +150y ≤2500. Also, x + y ≤20.So the system of inequalities is:x + y ≤20 100x +150y ≤2500 x ≥0, y ≥0We need to check if the optimal solution from part 1, which is x=0, y=20, satisfies the budget constraint.Calculating cost: 100*0 +150*20=0 +3000=3000>2500. So it exceeds the budget.Therefore, we need to find the adjusted number of sessions that maximize P(x,y) within the budget.So now, we have two constraints: x + y ≤20 and 100x +150y ≤2500.We need to maximize P(x,y)=3x² +2xy +5y² subject to these constraints.First, let's find the feasible region defined by the constraints.The budget constraint: 100x +150y ≤2500 can be simplified by dividing by 50: 2x +3y ≤50.So the constraints are:x + y ≤20 2x +3y ≤50 x ≥0, y ≥0We can graph these inequalities to find the feasible region.First, find the intersection points.1. Intersection of x + y =20 and 2x +3y=50.Solve the system:x + y =20 2x +3y=50Multiply the first equation by 2: 2x +2y=40Subtract from the second equation: (2x +3y) - (2x +2y)=50 -40 => y=10Then x=20 - y=10So intersection at (10,10)2. Intersection of 2x +3y=50 with axes:x=0: 3y=50 => y≈16.666 y=0: 2x=50 =>x=25But since x + y ≤20, the feasible region is bounded by x=0, y=0, x + y=20, and 2x +3y=50.So the vertices of the feasible region are:(0,0), (0,16.666), (10,10), (20,0)But wait, when x=0, from 2x +3y=50, y=50/3≈16.666, which is less than 20, so that point is (0,16.666). Similarly, when y=0, x=25, but since x + y ≤20, the point (25,0) is outside the feasible region, so the feasible region's vertices are (0,0), (0,16.666), (10,10), and (20,0).Wait, but (20,0) is on x + y=20, and 2x +3y=40 +0=40 ≤50, so it's within the budget.So the feasible region is a polygon with vertices at (0,0), (0,16.666), (10,10), and (20,0).Now, to maximize P(x,y), we need to evaluate P at each of these vertices and also check if the maximum occurs on the edges.But since P is a quadratic function, it's convex, so the maximum will occur at one of the vertices.Let's compute P at each vertex:1. (0,0): P=0 2. (0,16.666): y=50/3≈16.666 P=3*(0)^2 +2*(0)*(50/3) +5*(50/3)^2=0 +0 +5*(2500/9)=12500/9≈1388.893. (10,10): P=3*100 +2*100 +5*100=300 +200 +500=10004. (20,0): P=3*400 +0 +0=1200So the maximum among these is at (0,16.666) with P≈1388.89.But since x and y must be integers, we need to check the integer points around (0,16.666). Since y must be an integer, y=16 or y=17.But y=17 would require x=0, but 100*0 +150*17=2550>2500, which is over budget. So y=16: cost=150*16=2400, which leaves 100 left, but x must be 0 since x + y=16, but wait, x can be up to 20 - y=4. But wait, if y=16, x can be up to 4, but the budget allows 100x +150*16=100x +2400 ≤2500 =>100x ≤100 =>x ≤1.So x can be 0 or 1.So let's check (0,16) and (1,16).Compute P(0,16)=3*0 +0 +5*(16)^2=5*256=1280P(1,16)=3*1 +2*1*16 +5*(16)^2=3 +32 +1280=1315Similarly, check (0,17): but 150*17=2550>2500, so not allowed.Alternatively, maybe other points near (10,10) or (20,0) could give higher P.Wait, but (0,16) gives 1280, (1,16) gives 1315, which is higher.Is there a point with y=15, x=5: 100*5 +150*15=500 +2250=2750>2500, too much.Wait, let's see: 2x +3y ≤50.If y=16, 2x +48 ≤50 =>2x ≤2 =>x=1.So (1,16) is the only integer point near (0,16.666) that's within budget.Similarly, let's check other points.What about (5,10): 100*5 +150*10=500 +1500=2000 ≤2500.P(5,10)=3*25 +2*50 +5*100=75 +100 +500=675Less than 1315.What about (2,16): but 2x +3y=4 +48=52>50, so not allowed.(1,16): allowed.What about (0,16): allowed, gives 1280.What about (10,10): P=1000, less than 1315.What about (15,5): 100*15 +150*5=1500 +750=2250 ≤2500.P=3*225 +2*75 +5*25=675 +150 +125=950 <1315.What about (2,14): 2x +3y=4 +42=46 ≤50.P=3*4 +2*28 +5*196=12 +56 +980=1048 <1315.What about (3,13): 2*3 +3*13=6 +39=45 ≤50.P=3*9 +2*39 +5*169=27 +78 +845=950 <1315.What about (4,12): 2*4 +3*12=8 +36=44 ≤50.P=3*16 +2*48 +5*144=48 +96 +720=864 <1315.What about (5,10): already checked, 675.What about (6,8): 2*6 +3*8=12 +24=36 ≤50.P=3*36 +2*48 +5*64=108 +96 +320=524 <1315.What about (7,7): 2*7 +3*7=14 +21=35 ≤50.P=3*49 +2*49 +5*49=147 +98 +245=490 <1315.What about (8,6): 2*8 +3*6=16 +18=34 ≤50.P=3*64 +2*48 +5*36=192 +96 +180=468 <1315.What about (9,5): 2*9 +3*5=18 +15=33 ≤50.P=3*81 +2*45 +5*25=243 +90 +125=458 <1315.What about (10,4): 2*10 +3*4=20 +12=32 ≤50.P=3*100 +2*40 +5*16=300 +80 +80=460 <1315.What about (11,3): 2*11 +3*3=22 +9=31 ≤50.P=3*121 +2*33 +5*9=363 +66 +45=474 <1315.What about (12,2): 2*12 +3*2=24 +6=30 ≤50.P=3*144 +2*24 +5*4=432 +48 +20=500 <1315.What about (13,1): 2*13 +3*1=26 +3=29 ≤50.P=3*169 +2*13 +5*1=507 +26 +5=538 <1315.What about (14,0): 2*14 +0=28 ≤50.P=3*196 +0 +0=588 <1315.So among all these integer points, the maximum P is at (1,16) with P=1315.But wait, let's check if there are other points near (10,10) that might give higher P.Wait, (10,10) gives P=1000, which is less than 1315.Alternatively, maybe points where y is higher than 16 but x is more than 1, but that would exceed the budget.Wait, y=17 would require x=0, but 150*17=2550>2500, so not allowed.y=16, x=1: total cost=100 +2400=2500, exactly the budget.So (1,16) is the optimal integer point within the budget.But wait, let me check if there are other points with y=16 and x=1, but maybe x=1, y=16 is the only one.Alternatively, maybe x=2, y=16: 2x +3y=4 +48=52>50, so not allowed.So yes, (1,16) is the optimal.But wait, let me check if there are other points with y=15 and x=5: 100*5 +150*15=500 +2250=2750>2500, so not allowed.y=14, x= (50 -3*14)/2=(50 -42)/2=4. So x=4, y=14.Check P(4,14)=3*16 +2*56 +5*196=48 +112 +980=1140 <1315.Similarly, y=13, x=(50 -39)/2=5.5, but x must be integer, so x=5, y=13: P=3*25 +2*65 +5*169=75 +130 +845=1050 <1315.So yes, (1,16) is the best.But wait, let me check if there's a point with y=16 and x=1, which is allowed, and gives P=1315.Is there a way to get a higher P by having x=1, y=16?Yes, because P(1,16)=1315, which is higher than P(0,16)=1280.Alternatively, maybe x=1, y=16 is the optimal.But let me also check if there's a point with y=16 and x=1, which is allowed, and gives P=1315.Alternatively, maybe x=2, y=16 is not allowed, as it exceeds the budget.So, the optimal solution within the budget is x=1, y=16, giving P=1315.But wait, let me confirm the calculation for P(1,16):3*(1)^2 +2*(1)*(16) +5*(16)^2=3 +32 +1280=1315. Yes, correct.So, the optimal solution within the budget is x=1, y=16.But wait, let me check if there's a point with y=16 and x=1, which is allowed, and gives P=1315.Alternatively, maybe x=1, y=16 is the optimal.But let me also check if there's a point with y=16 and x=1, which is allowed, and gives P=1315.Alternatively, maybe x=1, y=16 is the optimal.Wait, but let me check if there's a point with y=16 and x=1, which is allowed, and gives P=1315.Alternatively, maybe x=1, y=16 is the optimal.Wait, I think I've confirmed enough. So the optimal solution within the budget is x=1, y=16.</think>