🧠:<think>Alright, so I have two problems to solve here. Let me take them one at a time.Starting with the first problem about Chef Alex's cake. He wants to create a cake with three layers: chocolate, vanilla, and strawberry. Each layer has specific ratios of flour, sugar, and eggs. The final cake needs to weigh exactly 600 grams, and the layers should be combined in the ratio 2:3:1 for chocolate, vanilla, and strawberry respectively. I need to find the exact amounts of flour, sugar, and eggs needed for each layer.Okay, let's break this down. First, the layers are in the ratio 2:3:1. That means for every 2 parts of chocolate, there are 3 parts of vanilla and 1 part of strawberry. So, the total number of parts is 2 + 3 + 1 = 6 parts. Since the total weight is 600 grams, each part must weigh 600 / 6 = 100 grams. Therefore, chocolate layer is 2 * 100 = 200 grams, vanilla is 3 * 100 = 300 grams, and strawberry is 1 * 100 = 100 grams.Now, each layer has its own ratio of flour, sugar, and eggs. Let's handle each layer one by one.Starting with the chocolate layer: 3 parts flour, 2 parts sugar, 1 part eggs. The total parts here are 3 + 2 + 1 = 6 parts. The total weight of the chocolate layer is 200 grams. So each part is 200 / 6 ≈ 33.333 grams. Therefore, flour is 3 * 33.333 ≈ 100 grams, sugar is 2 * 33.333 ≈ 66.666 grams, and eggs are 1 * 33.333 ≈ 33.333 grams.Next, the vanilla layer: 4 parts flour, 3 parts sugar, 2 parts eggs. Total parts: 4 + 3 + 2 = 9 parts. The vanilla layer is 300 grams. Each part is 300 / 9 ≈ 33.333 grams. So, flour is 4 * 33.333 ≈ 133.333 grams, sugar is 3 * 33.333 ≈ 100 grams, and eggs are 2 * 33.333 ≈ 66.666 grams.Lastly, the strawberry layer: 2 parts flour, 2 parts sugar, 1 part eggs. Total parts: 2 + 2 + 1 = 5 parts. The strawberry layer is 100 grams. Each part is 100 / 5 = 20 grams. So, flour is 2 * 20 = 40 grams, sugar is 2 * 20 = 40 grams, and eggs are 1 * 20 = 20 grams.Now, let me sum up the total amounts of each ingredient across all layers.Flour: Chocolate (100g) + Vanilla (133.333g) + Strawberry (40g) ≈ 100 + 133.333 + 40 = 273.333 grams.Sugar: Chocolate (66.666g) + Vanilla (100g) + Strawberry (40g) ≈ 66.666 + 100 + 40 = 206.666 grams.Eggs: Chocolate (33.333g) + Vanilla (66.666g) + Strawberry (20g) ≈ 33.333 + 66.666 + 20 = 120 grams.Wait, let me check if these totals add up to 600 grams. 273.333 + 206.666 + 120 ≈ 600 grams. Yes, that seems correct.But just to make sure, let me verify each layer's total weight.Chocolate layer: 100 + 66.666 + 33.333 ≈ 200 grams. Correct.Vanilla layer: 133.333 + 100 + 66.666 ≈ 300 grams. Correct.Strawberry layer: 40 + 40 + 20 = 100 grams. Correct.So, the calculations seem accurate.Moving on to the second problem. Chef Alex introduces a polynomial function P(x) = ax^4 + bx^3 + cx^2 + dx + e, representing the cake's taste profile. The constraints are:- P(0) = 5- P(10) = 0- P'(5) = 0- P''(5) < 0We need to find the coefficients a, b, c, d, and e.Alright, so let's note down the given conditions.1. P(0) = 5: When x = 0, the polynomial equals 5. Since all terms except the constant term e will be multiplied by 0, this directly gives e = 5.2. P(10) = 0: When x = 10, the polynomial equals 0. So, plugging x = 10 into P(x):a*(10)^4 + b*(10)^3 + c*(10)^2 + d*(10) + e = 0Which simplifies to:10000a + 1000b + 100c + 10d + e = 0But we already know e = 5, so:10000a + 1000b + 100c + 10d + 5 = 0Let me write that as equation (1):10000a + 1000b + 100c + 10d = -53. P'(5) = 0: The first derivative at x = 5 is zero. Let's compute P'(x):P'(x) = 4ax^3 + 3bx^2 + 2cx + dSo, P'(5) = 4a*(5)^3 + 3b*(5)^2 + 2c*(5) + d = 0Calculating each term:4a*125 = 500a3b*25 = 75b2c*5 = 10cSo, 500a + 75b + 10c + d = 0Let's call this equation (2):500a + 75b + 10c + d = 04. P''(5) < 0: The second derivative at x = 5 is negative. Let's compute P''(x):P''(x) = 12ax^2 + 6bx + 2cSo, P''(5) = 12a*(5)^2 + 6b*(5) + 2c = 12a*25 + 30b + 2c = 300a + 30b + 2cAnd this is less than 0:300a + 30b + 2c < 0Let me note this as inequality (3):300a + 30b + 2c < 0So, now we have:Equation (1): 10000a + 1000b + 100c + 10d = -5Equation (2): 500a + 75b + 10c + d = 0Inequality (3): 300a + 30b + 2c < 0We have five variables (a, b, c, d, e) but e is already known as 5. So, we need to solve for a, b, c, d.But we have only three equations and one inequality. Hmm, seems like we might need another condition or perhaps express variables in terms of others.Wait, maybe I missed something. Let's see. The problem says "find the coefficients a, b, c, d, and e of the polynomial function P(x)." So, perhaps there are more conditions or maybe we need to express in terms of some parameters? Or maybe the system is underdetermined, but perhaps we can find a specific solution.Wait, but the problem only gives four conditions: P(0)=5, P(10)=0, P'(5)=0, and P''(5)<0. So, four conditions for five variables. Hmm, but e is known, so effectively four equations for four variables a, b, c, d. Wait, no, equation (1) and (2) are two equations, and inequality (3) is another condition, but it's not an equation. So, perhaps we need to find a solution that satisfies the equations and the inequality.Alternatively, maybe I need to express the polynomial in terms of its roots or something else.Wait, let me think. The polynomial is of degree 4, so it can have up to 4 roots. We know that P(10) = 0, so x=10 is a root. Also, P'(5)=0, which suggests that x=5 is a critical point, and since P''(5) < 0, it's a local maximum. So, perhaps the polynomial can be expressed in a certain factored form?Alternatively, maybe we can set up a system of equations.We have:Equation (1): 10000a + 1000b + 100c + 10d = -5Equation (2): 500a + 75b + 10c + d = 0Inequality (3): 300a + 30b + 2c < 0We can try to solve equations (1) and (2) for d, and then express d in terms of a, b, c.From equation (2):d = -500a -75b -10cPlugging this into equation (1):10000a + 1000b + 100c + 10*(-500a -75b -10c) = -5Compute term by term:10000a - 5000a = 5000a1000b - 750b = 250b100c - 100c = 0So, 5000a + 250b = -5Divide both sides by 250:20a + b = -0.02So, equation (4): b = -0.02 -20aNow, we can express b in terms of a.Now, let's express d from equation (2):d = -500a -75b -10cBut since b = -0.02 -20a, substitute:d = -500a -75*(-0.02 -20a) -10cCompute:-500a + 1.5 + 1500a -10cCombine like terms:(-500a + 1500a) + 1.5 -10c = 1000a + 1.5 -10cSo, d = 1000a -10c + 1.5Now, we have expressions for b and d in terms of a and c.Now, let's go back to inequality (3):300a + 30b + 2c < 0Substitute b = -0.02 -20a:300a + 30*(-0.02 -20a) + 2c < 0Compute:300a - 0.6 -600a + 2c < 0Combine like terms:(300a -600a) -0.6 + 2c < 0-300a -0.6 + 2c < 0Let me rearrange:2c < 300a + 0.6Divide both sides by 2:c < 150a + 0.3So, c must be less than 150a + 0.3Now, we have:b = -0.02 -20ad = 1000a -10c + 1.5c < 150a + 0.3But we still have two variables, a and c, with c dependent on a.Is there another condition? Hmm, perhaps we can choose a value for a and find c accordingly, but since it's a polynomial, we might need more conditions. Wait, but the problem only gives four conditions, so maybe we can express the polynomial in terms of a parameter.Alternatively, perhaps we can assume a value for a to simplify. Let me see.Wait, let me think about the system again.We have:Equation (4): b = -0.02 -20aEquation (5): d = 1000a -10c + 1.5Inequality (6): c < 150a + 0.3We need another equation or condition. Wait, perhaps we can use the fact that the polynomial is a quartic, so it tends to infinity or negative infinity as x approaches infinity, depending on the leading coefficient a.But without more conditions, perhaps we can express the polynomial in terms of a parameter.Alternatively, maybe we can set c as a function of a, such that c = 150a + 0.3 - k, where k > 0, to satisfy the inequality. But this might complicate things.Alternatively, perhaps we can set a specific value for a to find a particular solution. Let me try to choose a such that the coefficients are simple.Let me assume a = 0.0001 (a small positive number). Let's see what happens.If a = 0.0001,From equation (4): b = -0.02 -20*(0.0001) = -0.02 -0.002 = -0.022From inequality (6): c < 150*(0.0001) + 0.3 = 0.015 + 0.3 = 0.315Let's choose c = 0.3 (just below 0.315)Then, d = 1000*(0.0001) -10*(0.3) + 1.5 = 0.1 - 3 + 1.5 = -1.4So, e = 5Thus, the polynomial would be:P(x) = 0.0001x^4 -0.022x^3 + 0.3x^2 -1.4x + 5Let me check if this satisfies all conditions.1. P(0) = 5: Correct.2. P(10) = 0.0001*(10000) -0.022*(1000) + 0.3*(100) -1.4*(10) +5= 1 -22 + 30 -14 +5 = (1 -22) + (30 -14) +5 = (-21) +16 +5 = 0. Correct.3. P'(x) = 4*0.0001x^3 -3*0.022x^2 + 2*0.3x -1.4= 0.0004x^3 -0.066x^2 + 0.6x -1.4At x=5:P'(5) = 0.0004*(125) -0.066*(25) + 0.6*(5) -1.4= 0.05 -1.65 + 3 -1.4= (0.05 -1.65) + (3 -1.4) = (-1.6) +1.6 = 0. Correct.4. P''(x) = 12*0.0001x^2 -6*0.022x + 2*0.3= 0.0012x^2 -0.132x + 0.6At x=5:P''(5) = 0.0012*(25) -0.132*(5) + 0.6= 0.03 -0.66 + 0.6 = (0.03 -0.66) +0.6 = (-0.63) +0.6 = -0.03 < 0. Correct.So, this polynomial satisfies all the given conditions.But wait, is this the only solution? Probably not, since we had some freedom in choosing a and c. However, the problem asks to find the coefficients, so perhaps this is a valid solution.Alternatively, maybe we can express the coefficients in terms of a parameter. Let me see.From equation (4): b = -0.02 -20aFrom equation (5): d = 1000a -10c + 1.5From inequality (6): c < 150a + 0.3So, if we let c = 150a + 0.3 - k, where k > 0, then:d = 1000a -10*(150a + 0.3 -k) +1.5 = 1000a -1500a -3 +10k +1.5 = (-500a) -1.5 +10kBut this might complicate things further. Alternatively, perhaps we can express the polynomial in terms of a.But since the problem doesn't specify any additional conditions, I think the solution I found with a=0.0001 is acceptable. However, let me check if a can be zero.If a=0, then from equation (4): b = -0.02From equation (1): 10000*0 +1000*(-0.02) +100c +10d = -5Which is: -20 +100c +10d = -5 => 100c +10d =15 => 10c + d =1.5From equation (2): 500*0 +75*(-0.02) +10c +d =0 => -1.5 +10c +d =0 =>10c +d =1.5So, same as above. So, 10c + d =1.5From inequality (3): 300*0 +30*(-0.02) +2c <0 => -0.6 +2c <0 =>2c <0.6 =>c <0.3So, c must be less than 0.3.Let me choose c=0.2, then d=1.5 -10*0.2=1.5-2= -0.5So, the polynomial would be:P(x)=0x^4 + (-0.02)x^3 +0.2x^2 + (-0.5)x +5Simplify: P(x)= -0.02x^3 +0.2x^2 -0.5x +5Let me check the conditions.1. P(0)=5: Correct.2. P(10)= -0.02*(1000) +0.2*(100) -0.5*(10) +5= -20 +20 -5 +5=0. Correct.3. P'(x)= -0.06x^2 +0.4x -0.5At x=5: P'(5)= -0.06*(25) +0.4*(5) -0.5= -1.5 +2 -0.5=0. Correct.4. P''(x)= -0.12x +0.4At x=5: P''(5)= -0.6 +0.4= -0.2 <0. Correct.So, this is another valid solution with a=0.But wait, if a=0, the polynomial is a cubic, not quartic. The problem states it's a quartic, so a cannot be zero. Therefore, a must be non-zero. So, my first solution with a=0.0001 is valid as a quartic.But perhaps we can find a more general solution.Let me consider that the polynomial is quartic, so a ≠0.We have:From equation (4): b = -0.02 -20aFrom equation (5): d =1000a -10c +1.5From inequality (6): c <150a +0.3We can express c as c=150a +0.3 -k, where k>0.Then, d=1000a -10*(150a +0.3 -k) +1.5=1000a -1500a -3 +10k +1.5= (-500a) -1.5 +10kSo, d= -500a +10k -1.5But we still have two variables, a and k.Alternatively, perhaps we can express the polynomial in terms of a.But without more conditions, it's difficult to find a unique solution. However, the problem asks to find the coefficients, so perhaps we can express them in terms of a parameter.But maybe I can choose a specific value for a to simplify.Let me try a=0.0002Then, b= -0.02 -20*0.0002= -0.02 -0.004= -0.024From inequality (6): c <150*0.0002 +0.3=0.03 +0.3=0.33Let me choose c=0.3Then, d=1000*0.0002 -10*0.3 +1.5=0.2 -3 +1.5= -1.3So, the polynomial is:P(x)=0.0002x^4 -0.024x^3 +0.3x^2 -1.3x +5Check conditions:1. P(0)=5: Correct.2. P(10)=0.0002*10000 -0.024*1000 +0.3*100 -1.3*10 +5=0.2 -24 +30 -13 +5= (0.2 -24) + (30 -13) +5= (-23.8) +17 +5= (-23.8 +22)= -1.8 ≠0. Wait, that's not correct. Hmm, I must have made a mistake.Wait, let me recalculate P(10):0.0002*(10)^4=0.0002*10000=2-0.024*(10)^3= -0.024*1000= -240.3*(10)^2=0.3*100=30-1.3*(10)= -13+5So, total: 2 -24 +30 -13 +5= (2 -24)= -22 +30=8 -13= -5 +5=0. Correct.Wait, I think I miscalculated earlier. So, P(10)=0. Correct.3. P'(x)=4*0.0002x^3 -3*0.024x^2 +2*0.3x -1.3=0.0008x^3 -0.072x^2 +0.6x -1.3At x=5: 0.0008*125 -0.072*25 +0.6*5 -1.3=0.1 -1.8 +3 -1.3= (0.1 -1.8)= -1.7 +3=1.3 -1.3=0. Correct.4. P''(x)=12*0.0002x^2 -6*0.024x +2*0.3=0.0024x^2 -0.144x +0.6At x=5: 0.0024*25 -0.144*5 +0.6=0.06 -0.72 +0.6= (0.06 -0.72)= -0.66 +0.6= -0.06 <0. Correct.So, this works too.But again, this is just another solution. It seems like there are infinitely many solutions depending on the choice of a and c, as long as they satisfy the inequality.But the problem asks to "find the coefficients", so perhaps we need to express them in terms of a parameter or find a general solution.Alternatively, maybe the problem expects a specific solution, perhaps with integer coefficients or something. Let me see.Wait, in my first solution, a=0.0001, which is 1/10000, which is 0.0001. Similarly, b=-0.022, which is -22/1000, c=0.3, d=-1.4, e=5.Alternatively, perhaps we can express the polynomial with fractions.Let me try to express a as 1/10000, then:a=1/10000b= -0.02 -20*(1/10000)= -0.02 -0.002= -0.022= -22/1000= -11/500c=0.3=3/10d=1000*(1/10000) -10*(3/10) +1.5=0.1 -3 +1.5= -1.4= -14/10= -7/5e=5So, the polynomial is:P(x)= (1/10000)x^4 - (11/500)x^3 + (3/10)x^2 - (7/5)x +5This is a valid solution.Alternatively, if we want to eliminate denominators, multiply through by 10000:P(x)=x^4 -220x^3 +3000x^2 -14000x +50000But then, P(10)=10000 -22000 +30000 -140000 +50000= (10000 -22000)= -12000 +30000=18000 -140000= -122000 +50000= -72000 ≠0. Wait, that can't be right. I must have made a mistake.Wait, no, when I multiplied by 10000, I should have multiplied each term:P(x)= (1/10000)x^4 - (11/500)x^3 + (3/10)x^2 - (7/5)x +5Multiply by 10000:x^4 -220x^3 +3000x^2 -14000x +50000But then, P(10)=10000 -22000 +30000 -140000 +50000= (10000 -22000)= -12000 +30000=18000 -140000= -122000 +50000= -72000 ≠0. So, that's incorrect.Wait, that suggests that scaling the polynomial by 10000 changes the roots, which is not acceptable because P(10)=0 must hold. Therefore, scaling is not a valid approach here.Therefore, the polynomial must remain as is, with fractional coefficients.Alternatively, perhaps the problem expects a general solution, but I think it's more likely that a specific solution is expected, given the way the problem is phrased.Given that, I think the solution with a=0.0001, b=-0.022, c=0.3, d=-1.4, e=5 is acceptable.Alternatively, to make the coefficients more precise, perhaps we can express them as fractions.From equation (4): b= -0.02 -20aIf a=1/10000, then b= -1/50 -20*(1/10000)= -1/50 -1/500= (-10/500 -1/500)= -11/500Similarly, c=0.3=3/10d=1000a -10c +1.5=1000*(1/10000) -10*(3/10) +1.5=0.1 -3 +1.5= -1.4= -14/10= -7/5So, the polynomial is:P(x)= (1/10000)x^4 - (11/500)x^3 + (3/10)x^2 - (7/5)x +5Alternatively, to write all coefficients with denominator 10000:a=1/10000b= -220/10000c=3000/10000d= -14000/10000e=50000/10000But that's more complicated.Alternatively, perhaps we can write the polynomial as:P(x)= frac{1}{10000}x^4 - frac{11}{500}x^3 + frac{3}{10}x^2 - frac{7}{5}x +5This seems acceptable.Alternatively, to make it look cleaner, we can write:P(x)= frac{1}{10000}x^4 - frac{11}{500}x^3 + frac{3}{10}x^2 - frac{7}{5}x +5Yes, that's a valid polynomial that satisfies all the given conditions.So, summarizing:a=1/10000b=-11/500c=3/10d=-7/5e=5Alternatively, in decimal form:a=0.0001b=-0.022c=0.3d=-1.4e=5Either way is correct.So, to answer the second problem, the coefficients are:a=0.0001, b=-0.022, c=0.3, d=-1.4, e=5Or in fractions:a=1/10000, b=-11/500, c=3/10, d=-7/5, e=5I think either form is acceptable, but perhaps the fractional form is more precise.So, putting it all together.For the first problem, the amounts are:Chocolate layer:Flour: 100gSugar: 66.666g ≈66.67gEggs:33.333g≈33.33gVanilla layer:Flour:133.333g≈133.33gSugar:100gEggs:66.666g≈66.67gStrawberry layer:Flour:40gSugar:40gEggs:20gTotal:Flour:273.333g≈273.33gSugar:206.666g≈206.67gEggs:120gFor the second problem, the polynomial coefficients are:a=1/10000, b=-11/500, c=3/10, d=-7/5, e=5Or in decimal:a=0.0001, b=-0.022, c=0.3, d=-1.4, e=5I think that's it.</think>

🧠:<think>Alright, so I have two questions here about a bookshop owner curating a special section. Let me try to tackle them one by one. Starting with the first question: She wants to include 100 books, with exactly 40% being Cameroonian literature. That means 40 Cameroonian books and 60 from other countries. The database she has access to has 5000 titles, 200 of which are from Cameroon. She wants to know the probability that a randomly selected subset of 100 books will contain exactly 40 Cameroonian titles.Hmm, okay. So, probability of exactly 40 Cameroonian books out of 100 selected. This sounds like a hypergeometric distribution problem because she's sampling without replacement from a finite population. The hypergeometric distribution gives the probability of k successes (in this case, Cameroonian books) in n draws (100 books) without replacement from a finite population (5000 books) containing a specific number of successes (200 Cameroonian books).The formula for hypergeometric probability is:P(X = k) = [C(K, k) * C(N - K, n - k)] / C(N, n)Where:- N is the population size (5000)- K is the number of success states in the population (200)- n is the number of draws (100)- k is the number of observed successes (40)So plugging in the numbers:P(X = 40) = [C(200, 40) * C(5000 - 200, 100 - 40)] / C(5000, 100)Calculating this might be a bit tricky because the numbers are large, but I can write it out step by step.First, compute the combinations:C(200, 40) is the number of ways to choose 40 Cameroonian books from 200.C(4800, 60) is the number of ways to choose the remaining 60 books from the non-Cameroonian titles.C(5000, 100) is the total number of ways to choose any 100 books from the database.So, the probability is the product of the first two combinations divided by the third.I think that's the correct approach. Maybe I should double-check if it's hypergeometric or if it's something else. Since she's selecting without replacement, hypergeometric is appropriate. If it were with replacement, it would be binomial, but since the database is finite and she's selecting a significant portion (100 out of 5000), hypergeometric is better.Moving on to the second question: She wants to arrange the 40 Cameroonian books in a pyramid shape where each row has one more book than the row above, starting with one book at the top. So, the first row has 1 book, the second has 2, the third has 3, and so on.She wants to know how many complete rows she can build with 40 books and how many will remain unplaced.This is a problem of finding the largest integer n such that the sum from 1 to n is less than or equal to 40. The sum of the first n natural numbers is given by the formula:Sum = n(n + 1)/2So, we need to solve for n in the inequality:n(n + 1)/2 ≤ 40Multiplying both sides by 2:n(n + 1) ≤ 80So, we can solve the quadratic equation n² + n - 80 = 0.Using the quadratic formula:n = [-1 ± sqrt(1 + 320)] / 2Because sqrt(321) is approximately 17.916, so:n = (-1 + 17.916)/2 ≈ 16.916 / 2 ≈ 8.458Since n must be an integer, the largest complete row is 8. Let's check the sum:8*9/2 = 36. So, 36 books are used in 8 rows. Then, 40 - 36 = 4 books remain unplaced.Wait, but if she tries to make a 9th row, that would require 9 books, but she only has 4 left. So, yes, 8 complete rows and 4 books remaining.Let me verify:Row 1: 1Row 2: 2 (total 3)Row 3: 3 (total 6)Row 4: 4 (total 10)Row 5: 5 (total 15)Row 6: 6 (total 21)Row 7: 7 (total 28)Row 8: 8 (total 36)Remaining: 4Yes, that's correct.So, summarizing:1. The probability is calculated using the hypergeometric distribution formula.2. She can build 8 complete rows with 4 books remaining.Final Answer1. The probability is boxed{dfrac{dbinom{200}{40} dbinom{4800}{60}}{dbinom{5000}{100}}}.2. She can build boxed{8} complete rows with boxed{4} books remaining.</think>

🧠:<think>Alright, so I have two problems here to solve. Let me take them one at a time.Starting with the first problem: A coal miner's production over time is given by the quadratic function ( P(t) = -2t^2 + 12t + 40 ). I need to find the maximum amount of coal produced in a month and the specific month when this maximum occurs.Hmm, okay. Quadratic functions have the form ( at^2 + bt + c ), and since the coefficient of ( t^2 ) here is negative (-2), the parabola opens downward. That means the vertex of the parabola is the maximum point. So, the vertex will give me both the time ( t ) when the maximum production occurs and the maximum production value.I remember that the time ( t ) at the vertex of a parabola is given by ( t = -frac{b}{2a} ). In this equation, ( a = -2 ) and ( b = 12 ). Plugging those values in:( t = -frac{12}{2*(-2)} = -frac{12}{-4} = 3 ).So, the maximum production occurs at ( t = 3 ) months. Now, to find the maximum production, I need to plug this value back into the original function ( P(t) ).Calculating ( P(3) ):( P(3) = -2*(3)^2 + 12*(3) + 40 ).Breaking it down:First, ( (3)^2 = 9 ), so ( -2*9 = -18 ).Then, ( 12*3 = 36 ).Adding it all up: ( -18 + 36 + 40 ).Let me compute that step by step:-18 + 36 is 18, and 18 + 40 is 58.So, the maximum production is 58 tons in the third month.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Calculating ( P(3) ) again:( -2*(9) = -18 ), correct.( 12*3 = 36 ), correct.Adding: -18 + 36 = 18, then 18 + 40 = 58. Yep, that seems right.Alternatively, I can also use the vertex formula for the maximum value. The vertex form of a quadratic is ( a(t - h)^2 + k ), where ( (h, k) ) is the vertex. But since I already found ( t = 3 ) and calculated ( P(3) = 58 ), I think that's solid.So, for the first problem, the maximum coal production is 58 tons, occurring in the 3rd month.Moving on to the second problem: The coal miner wants to become a footballer and his skill level improves with practice. The skill function is ( S(x) = 5sqrt{x} ), where ( x ) is the total hours of practice. He can practice at most 3 hours a day. Currently, his skill level is 20, and he wants to reach 30. I need to find the minimum number of additional hours he needs to practice.Alright, let's parse this. His current skill is 20, which is ( S(x) = 20 ). He wants to reach 30, so we need to find the value of ( x ) such that ( S(x) = 30 ). But since he's already practiced some hours to get to 20, we need to find the additional hours beyond that.Wait, actually, let me think. The function ( S(x) = 5sqrt{x} ) gives his skill level based on total hours practiced. If his current skill is 20, that means ( 5sqrt{x} = 20 ). So, we can solve for ( x ) to find how many hours he has already practiced.Let me write that down:( 5sqrt{x} = 20 )Divide both sides by 5:( sqrt{x} = 4 )Square both sides:( x = 16 )So, he has already practiced 16 hours to reach a skill level of 20.Now, he wants to reach 30. So, set ( S(x) = 30 ):( 5sqrt{x} = 30 )Divide both sides by 5:( sqrt{x} = 6 )Square both sides:( x = 36 )So, to reach a skill level of 30, he needs a total of 36 hours of practice.Since he has already practiced 16 hours, the additional hours needed are ( 36 - 16 = 20 ) hours.But wait, the problem says he can practice at most 3 hours a day. So, we need to find the minimum number of days required to accumulate 20 hours, given he can do up to 3 hours each day.So, 20 hours divided by 3 hours per day is approximately 6.666... days. But since he can't practice a fraction of a day, he needs to round up to the next whole day, which is 7 days.But hold on, the question asks for the minimum number of additional hours, not days. So, maybe I misinterpreted.Wait, let me read again: "calculate the minimum number of additional hours he needs to practice to reach a skill level of 30."So, it's just 20 additional hours, regardless of the daily limit. But wait, the problem says he can spend at most 3 hours a day practicing. So, does that affect the total hours? Or is it just a constraint on how he practices, but the total hours needed is still 20?Wait, perhaps I need to consider that he can only practice 3 hours a day, so the minimum number of days is 7, but the question is about hours, not days. So, the additional hours needed is 20, regardless of how he schedules them. So, maybe the answer is 20 hours.But let me think again. The problem says he can spend at most 3 hours a day. So, is there a constraint on the number of hours per day? But the question is asking for the minimum number of additional hours, not days. So, perhaps the answer is 20 hours, regardless of how he distributes them over days.Alternatively, maybe the question is implying that he can only practice 3 hours each day, so the total hours needed is 20, but he can't do more than 3 per day, so the minimum number of hours is still 20, but spread over multiple days.Wait, the question is a bit ambiguous. It says, "he can spend at most 3 hours a day practicing," but it's asking for the minimum number of additional hours, not days. So, perhaps the answer is 20 hours, regardless of the daily limit.But let me check the wording again: "calculate the minimum number of additional hours he needs to practice to reach a skill level of 30." So, it's about the total hours, not the number of days. So, the answer is 20 hours.But wait, let me double-check my calculations.Current skill: 20 = 5√x => √x = 4 => x = 16.Desired skill: 30 = 5√x => √x = 6 => x = 36.Additional hours: 36 - 16 = 20.Yes, that seems correct. So, he needs 20 additional hours of practice.But the problem mentions he can practice at most 3 hours a day. Does that affect the answer? If the question was asking for the minimum number of days, then yes, it would be 7 days. But since it's asking for hours, I think 20 is the answer.Wait, but maybe the question is trying to trick me. Maybe it's not about total hours, but considering the daily limit, how many hours he can accumulate. But no, the function S(x) depends on total hours, not on the rate. So, regardless of how he distributes the practice, the total hours needed is 20. So, the minimum number of additional hours is 20.Alternatively, maybe the question is implying that he can only practice 3 hours a day, so he can't practice more than that, so the total hours he can add is limited by that. But no, the function doesn't have a time constraint, just a total hours constraint.Wait, perhaps the question is just about solving for x, regardless of the daily limit. So, the answer is 20 hours.But let me think again. The problem says, "he can spend at most 3 hours a day practicing," which might be a constraint on the rate, but the question is about the total hours needed. So, unless it's asking for the minimum number of days, which it's not, the answer is 20 hours.Alternatively, maybe the question is trying to say that he can only practice 3 hours a day, so the total hours he can accumulate is limited, but that doesn't make sense because he can practice over multiple days.Wait, perhaps the problem is just giving extra information that he can practice up to 3 hours a day, but the question is about the total hours needed, so 20 hours is the answer.I think I'm overcomplicating this. The key is that the function S(x) depends on total hours, so regardless of how he practices, he needs 20 more hours. So, the answer is 20 hours.But just to be thorough, let me consider both interpretations.If the question is asking for the number of hours, it's 20.If it's asking for the number of days, it's 7.But the question specifically says, "the minimum number of additional hours," so it's 20 hours.Therefore, the answer is 20 hours.Wait, but let me check the math again.Current skill: 20 = 5√x => x = 16.Desired skill: 30 = 5√x => x = 36.Difference: 36 - 16 = 20.Yes, that's correct.So, the minimum number of additional hours is 20.Alright, I think I've got both problems figured out.Final Answer1. The maximum coal production is boxed{58} tons in the boxed{3}rd month.2. The minimum number of additional hours needed is boxed{20}.</think>

🧠:<think>Okay, so I have this problem where I need to solve a system of differential equations modeling the GDP growth of two countries, A and B. The equations are given as:[ frac{dG_A}{dt} = aG_A + bG_B + ccos(omega t) ][ frac{dG_B}{dt} = dG_B + eG_A + hsin(omega t) ]And I need to solve this system with initial conditions ( G_A(0) = G_{A0} ) and ( G_B(0) = G_{B0} ). Then, I also have to analyze the stability of the solutions by looking at the eigenvalues of the Jacobian matrix at an equilibrium point.Alright, let's start with part 1: solving the system of differential equations.First, I notice that these are linear differential equations with constant coefficients and some time-dependent forcing terms (the cosine and sine terms). So, this is a nonhomogeneous system. I think the standard approach is to solve the homogeneous system first and then find a particular solution for the nonhomogeneous part.The system can be written in matrix form as:[ frac{d}{dt} begin{pmatrix} G_A G_B end{pmatrix} = begin{pmatrix} a & b e & d end{pmatrix} begin{pmatrix} G_A G_B end{pmatrix} + begin{pmatrix} ccos(omega t) hsin(omega t) end{pmatrix} ]So, if I denote the vector ( mathbf{G} = begin{pmatrix} G_A G_B end{pmatrix} ), then the system becomes:[ frac{dmathbf{G}}{dt} = M mathbf{G} + mathbf{F}(t) ]where ( M ) is the matrix ( begin{pmatrix} a & b e & d end{pmatrix} ) and ( mathbf{F}(t) = begin{pmatrix} ccos(omega t) hsin(omega t) end{pmatrix} ).To solve this, I can use the method of integrating factors or variation of parameters. Since it's a linear system, I think variation of parameters might be the way to go.First, I need to find the general solution to the homogeneous equation:[ frac{dmathbf{G}}{dt} = M mathbf{G} ]The solution to this is:[ mathbf{G}_h(t) = Phi(t) mathbf{C} ]where ( Phi(t) ) is the fundamental matrix solution, which can be found by exponentiating the matrix ( M ). However, exponentiating a matrix can be complicated unless we can diagonalize it or find its eigenvalues and eigenvectors.So, maybe I should first find the eigenvalues and eigenvectors of matrix ( M ). The eigenvalues ( lambda ) satisfy the characteristic equation:[ det(M - lambda I) = 0 ][ detleft( begin{pmatrix} a - lambda & b e & d - lambda end{pmatrix} right) = 0 ][ (a - lambda)(d - lambda) - be = 0 ][ lambda^2 - (a + d)lambda + (ad - be) = 0 ]So, the eigenvalues are:[ lambda = frac{(a + d) pm sqrt{(a + d)^2 - 4(ad - be)}}{2} ][ lambda = frac{(a + d) pm sqrt{(a - d)^2 + 4be}}{2} ]Hmm, that's the standard quadratic formula. So, depending on the discriminant, the eigenvalues can be real and distinct, repeated, or complex.Once I have the eigenvalues, I can find the eigenvectors and construct the fundamental matrix ( Phi(t) ). Then, using variation of parameters, the particular solution ( mathbf{G}_p(t) ) can be found by integrating:[ mathbf{G}_p(t) = Phi(t) int Phi^{-1}(s) mathbf{F}(s) ds ]But this seems a bit involved. Maybe there's a more straightforward way, especially since the forcing functions are sinusoidal.Alternatively, I can look for a particular solution in the form of a combination of sine and cosine functions. Since the nonhomogeneous terms are ( ccos(omega t) ) and ( hsin(omega t) ), I can assume a particular solution of the form:[ mathbf{G}_p(t) = begin{pmatrix} G_{A_p} G_{B_p} end{pmatrix} cos(omega t) + begin{pmatrix} tilde{G}_{A_p} tilde{G}_{B_p} end{pmatrix} sin(omega t) ]So, let me denote:[ mathbf{G}_p(t) = mathbf{P} cos(omega t) + mathbf{Q} sin(omega t) ]where ( mathbf{P} ) and ( mathbf{Q} ) are constant vectors to be determined.Then, the derivative of ( mathbf{G}_p(t) ) is:[ frac{dmathbf{G}_p}{dt} = -omega mathbf{P} sin(omega t) + omega mathbf{Q} cos(omega t) ]Substituting ( mathbf{G}_p(t) ) and its derivative into the original differential equation:[ -omega mathbf{P} sin(omega t) + omega mathbf{Q} cos(omega t) = M (mathbf{P} cos(omega t) + mathbf{Q} sin(omega t)) + mathbf{F}(t) ]Expanding the right-hand side:[ Mmathbf{P} cos(omega t) + Mmathbf{Q} sin(omega t) + begin{pmatrix} ccos(omega t) hsin(omega t) end{pmatrix} ]Now, equate the coefficients of ( cos(omega t) ) and ( sin(omega t) ) on both sides.For the ( cos(omega t) ) terms:[ omega mathbf{Q} = M mathbf{P} + begin{pmatrix} c 0 end{pmatrix} ]Wait, hold on. The forcing function ( mathbf{F}(t) ) is ( begin{pmatrix} ccos(omega t) hsin(omega t) end{pmatrix} ), so when we expand, it's ( Mmathbf{P} cos(omega t) + Mmathbf{Q} sin(omega t) + ccos(omega t) + hsin(omega t) ). So, grouping terms:For ( cos(omega t) ):[ omega mathbf{Q} = M mathbf{P} + begin{pmatrix} c 0 end{pmatrix} ]Wait, no. Let me write it properly.Left-hand side (LHS):- Coefficient of ( cos(omega t) ): ( omega mathbf{Q} )- Coefficient of ( sin(omega t) ): ( -omega mathbf{P} )Right-hand side (RHS):- Coefficient of ( cos(omega t) ): ( M mathbf{P} + begin{pmatrix} c 0 end{pmatrix} )- Coefficient of ( sin(omega t) ): ( M mathbf{Q} + begin{pmatrix} 0 h end{pmatrix} )Therefore, we have two vector equations:1. ( omega mathbf{Q} = M mathbf{P} + begin{pmatrix} c 0 end{pmatrix} )2. ( -omega mathbf{P} = M mathbf{Q} + begin{pmatrix} 0 h end{pmatrix} )So, now we have a system of two equations with two unknown vectors ( mathbf{P} ) and ( mathbf{Q} ). Let me write these equations component-wise.Let ( mathbf{P} = begin{pmatrix} P_A P_B end{pmatrix} ) and ( mathbf{Q} = begin{pmatrix} Q_A Q_B end{pmatrix} ).Then, equation 1 becomes:1. ( omega Q_A = a P_A + b P_B + c )2. ( omega Q_B = e P_A + d P_B )Equation 2 becomes:3. ( -omega P_A = a Q_A + b Q_B )4. ( -omega P_B = e Q_A + d Q_B + h )So, now we have four equations:1. ( omega Q_A = a P_A + b P_B + c ) --- (1)2. ( omega Q_B = e P_A + d P_B ) --- (2)3. ( -omega P_A = a Q_A + b Q_B ) --- (3)4. ( -omega P_B = e Q_A + d Q_B + h ) --- (4)Our unknowns are ( P_A, P_B, Q_A, Q_B ). So, we need to solve this system.Let me write equations (3) and (4) in terms of ( Q_A ) and ( Q_B ):From (3): ( P_A = -frac{1}{omega}(a Q_A + b Q_B) ) --- (3a)From (4): ( P_B = -frac{1}{omega}(e Q_A + d Q_B + h) ) --- (4a)Now, substitute (3a) and (4a) into equations (1) and (2):Equation (1):( omega Q_A = a P_A + b P_B + c )Substitute P_A and P_B:( omega Q_A = a left( -frac{1}{omega}(a Q_A + b Q_B) right) + b left( -frac{1}{omega}(e Q_A + d Q_B + h) right) + c )Simplify:( omega Q_A = -frac{a}{omega}(a Q_A + b Q_B) - frac{b}{omega}(e Q_A + d Q_B + h) + c )Multiply through by ( omega ) to eliminate denominators:( omega^2 Q_A = -a(a Q_A + b Q_B) - b(e Q_A + d Q_B + h) + c omega )Expand the right-hand side:( omega^2 Q_A = -a^2 Q_A - a b Q_B - b e Q_A - b d Q_B - b h + c omega )Combine like terms:( omega^2 Q_A + a^2 Q_A + b e Q_A + a b Q_B + b d Q_B = -b h + c omega )Factor terms:( Q_A ( omega^2 + a^2 + b e ) + Q_B (a b + b d ) = c omega - b h ) --- (1a)Similarly, let's process equation (2):Equation (2):( omega Q_B = e P_A + d P_B )Substitute P_A and P_B from (3a) and (4a):( omega Q_B = e left( -frac{1}{omega}(a Q_A + b Q_B) right) + d left( -frac{1}{omega}(e Q_A + d Q_B + h) right) )Simplify:( omega Q_B = -frac{e}{omega}(a Q_A + b Q_B) - frac{d}{omega}(e Q_A + d Q_B + h) )Multiply through by ( omega ):( omega^2 Q_B = -e(a Q_A + b Q_B) - d(e Q_A + d Q_B + h) )Expand:( omega^2 Q_B = -a e Q_A - b e Q_B - d e Q_A - d^2 Q_B - d h )Combine like terms:( omega^2 Q_B + a e Q_A + b e Q_B + d e Q_A + d^2 Q_B = -d h )Factor terms:( Q_A (a e + d e ) + Q_B ( omega^2 + b e + d^2 ) = -d h ) --- (2a)So, now we have two equations (1a) and (2a):(1a): ( Q_A ( omega^2 + a^2 + b e ) + Q_B (a b + b d ) = c omega - b h )(2a): ( Q_A (e(a + d)) + Q_B ( omega^2 + b e + d^2 ) = -d h )Let me write these in matrix form:[ begin{pmatrix} omega^2 + a^2 + b e & b(a + d) e(a + d) & omega^2 + b e + d^2 end{pmatrix} begin{pmatrix} Q_A Q_B end{pmatrix} = begin{pmatrix} c omega - b h -d h end{pmatrix} ]So, this is a linear system for ( Q_A ) and ( Q_B ). Let me denote the coefficient matrix as ( N ):[ N = begin{pmatrix} omega^2 + a^2 + b e & b(a + d) e(a + d) & omega^2 + b e + d^2 end{pmatrix} ]And the right-hand side vector as ( mathbf{K} = begin{pmatrix} c omega - b h -d h end{pmatrix} )To solve for ( Q_A ) and ( Q_B ), we can compute ( N^{-1} mathbf{K} ), provided that ( N ) is invertible (i.e., determinant is non-zero).First, let's compute the determinant of ( N ):[ det(N) = (omega^2 + a^2 + b e)(omega^2 + b e + d^2) - [b(a + d)] [e(a + d)] ]Let me expand this:First term: ( (omega^2 + a^2 + b e)(omega^2 + b e + d^2) )Let me denote ( X = omega^2 + b e ), so the first term becomes ( (X + a^2)(X + d^2) = X^2 + X(a^2 + d^2) + a^2 d^2 )Second term: ( b(a + d) e(a + d) = b e (a + d)^2 )So, determinant:[ det(N) = X^2 + X(a^2 + d^2) + a^2 d^2 - b e (a + d)^2 ]Substituting back ( X = omega^2 + b e ):[ det(N) = (omega^2 + b e)^2 + (omega^2 + b e)(a^2 + d^2) + a^2 d^2 - b e (a + d)^2 ]This seems complicated, but perhaps we can factor or simplify it.Alternatively, maybe it's better to just keep it as is for now.Assuming that ( det(N) neq 0 ), we can write:[ Q_A = frac{ det(N_{Q_A}) }{ det(N) } ][ Q_B = frac{ det(N_{Q_B}) }{ det(N) } ]Where ( N_{Q_A} ) is the matrix formed by replacing the first column of ( N ) with ( mathbf{K} ), and ( N_{Q_B} ) is the matrix formed by replacing the second column with ( mathbf{K} ).So, compute ( N_{Q_A} ):[ N_{Q_A} = begin{pmatrix} c omega - b h & b(a + d) -d h & omega^2 + b e + d^2 end{pmatrix} ]Compute ( det(N_{Q_A}) ):[ (c omega - b h)(omega^2 + b e + d^2) - (-d h) b(a + d) ][ = (c omega - b h)(omega^2 + b e + d^2) + b d h (a + d) ]Similarly, compute ( N_{Q_B} ):[ N_{Q_B} = begin{pmatrix} omega^2 + a^2 + b e & c omega - b h e(a + d) & -d h end{pmatrix} ]Compute ( det(N_{Q_B}) ):[ (omega^2 + a^2 + b e)(-d h) - (c omega - b h) e(a + d) ][ = -d h (omega^2 + a^2 + b e) - e(a + d)(c omega - b h) ]So, now, ( Q_A ) and ( Q_B ) can be expressed as:[ Q_A = frac{ (c omega - b h)(omega^2 + b e + d^2) + b d h (a + d) }{ det(N) } ][ Q_B = frac{ -d h (omega^2 + a^2 + b e) - e(a + d)(c omega - b h) }{ det(N) } ]Once we have ( Q_A ) and ( Q_B ), we can find ( P_A ) and ( P_B ) using equations (3a) and (4a):[ P_A = -frac{1}{omega}(a Q_A + b Q_B) ][ P_B = -frac{1}{omega}(e Q_A + d Q_B + h) ]This will give us the particular solution ( mathbf{G}_p(t) = mathbf{P} cos(omega t) + mathbf{Q} sin(omega t) ).Now, the general solution to the system is the sum of the homogeneous solution and the particular solution:[ mathbf{G}(t) = mathbf{G}_h(t) + mathbf{G}_p(t) ][ mathbf{G}(t) = Phi(t) mathbf{C} + mathbf{G}_p(t) ]Where ( mathbf{C} ) is a constant vector determined by the initial conditions.But to write ( Phi(t) ), we need to find the eigenvalues and eigenvectors of matrix ( M ). As I started earlier, the eigenvalues are:[ lambda = frac{(a + d) pm sqrt{(a - d)^2 + 4be}}{2} ]Depending on whether the discriminant ( D = (a - d)^2 + 4be ) is positive, zero, or negative, we have real distinct, repeated, or complex eigenvalues.Case 1: ( D > 0 ) (Real and distinct eigenvalues)In this case, the matrix ( M ) can be diagonalized, and the fundamental matrix ( Phi(t) ) is:[ Phi(t) = begin{pmatrix} frac{e^{lambda_1 t} - e^{lambda_2 t}}{lambda_1 - lambda_2} & frac{b (e^{lambda_1 t} - e^{lambda_2 t})}{lambda_1 - lambda_2} frac{e (e^{lambda_1 t} - e^{lambda_2 t})}{lambda_1 - lambda_2} & frac{e^{lambda_1 t} - e^{lambda_2 t}}{lambda_1 - lambda_2} end{pmatrix} ]Wait, actually, I think I need to find the eigenvectors corresponding to each eigenvalue to construct ( Phi(t) ). Let me recall that for each eigenvalue ( lambda ), the eigenvector ( mathbf{v} ) satisfies ( (M - lambda I)mathbf{v} = 0 ).For eigenvalue ( lambda_1 ), the eigenvector ( mathbf{v}_1 ) can be found by solving:[ (a - lambda_1) v_{1A} + b v_{1B} = 0 ][ e v_{1A} + (d - lambda_1) v_{1B} = 0 ]Similarly for ( lambda_2 ), eigenvector ( mathbf{v}_2 ):[ (a - lambda_2) v_{2A} + b v_{2B} = 0 ][ e v_{2A} + (d - lambda_2) v_{2B} = 0 ]Assuming ( lambda_1 neq lambda_2 ), the fundamental matrix is:[ Phi(t) = begin{pmatrix} v_{1A} e^{lambda_1 t} & v_{2A} e^{lambda_2 t} v_{1B} e^{lambda_1 t} & v_{2B} e^{lambda_2 t} end{pmatrix} ]But since the eigenvectors can be scaled, we can choose them such that ( v_{1A} = 1 ) or something similar for simplicity.Alternatively, since the system is linear, the homogeneous solution can be written as:[ mathbf{G}_h(t) = C_1 mathbf{v}_1 e^{lambda_1 t} + C_2 mathbf{v}_2 e^{lambda_2 t} ]Where ( C_1 ) and ( C_2 ) are constants determined by initial conditions.Case 2: ( D = 0 ) (Repeated eigenvalues)In this case, the eigenvalue is ( lambda = frac{a + d}{2} ), and we need to find the eigenvector and a generalized eigenvector.Case 3: ( D < 0 ) (Complex eigenvalues)Here, the eigenvalues are complex: ( lambda = alpha pm i beta ), where ( alpha = frac{a + d}{2} ) and ( beta = frac{sqrt{4be - (a - d)^2}}{2} ). The solution will involve exponential functions multiplied by sine and cosine terms.Given the complexity, maybe it's better to express the homogeneous solution in terms of the eigenvalues and eigenvectors without explicitly computing them, unless specific values are given.But since the problem doesn't specify particular values for ( a, b, d, e ), I think the solution will have to remain in terms of these parameters.Therefore, the general solution is:[ mathbf{G}(t) = Phi(t) mathbf{C} + mathbf{G}_p(t) ]Where ( Phi(t) ) is the fundamental matrix solution, ( mathbf{C} ) is determined by initial conditions, and ( mathbf{G}_p(t) ) is the particular solution we found earlier.To find ( mathbf{C} ), we apply the initial conditions at ( t = 0 ):[ mathbf{G}(0) = Phi(0) mathbf{C} + mathbf{G}_p(0) = begin{pmatrix} G_{A0} G_{B0} end{pmatrix} ]Since ( Phi(0) ) is the identity matrix (assuming it's constructed from eigenvectors normalized at t=0), we have:[ mathbf{C} = mathbf{G}(0) - mathbf{G}_p(0) ]But ( mathbf{G}_p(0) = mathbf{P} cos(0) + mathbf{Q} sin(0) = mathbf{P} ). So,[ mathbf{C} = begin{pmatrix} G_{A0} - P_A G_{B0} - P_B end{pmatrix} ]Therefore, the complete solution is:[ mathbf{G}(t) = Phi(t) left( begin{pmatrix} G_{A0} - P_A G_{B0} - P_B end{pmatrix} right) + mathbf{P} cos(omega t) + mathbf{Q} sin(omega t) ]This is the general solution for ( G_A(t) ) and ( G_B(t) ).Moving on to part 2: Analyzing the stability of the solutions by determining the eigenvalues of the Jacobian matrix at an equilibrium point.First, an equilibrium point ( (G_A^*, G_B^*) ) is a constant solution where ( frac{dG_A}{dt} = 0 ) and ( frac{dG_B}{dt} = 0 ).So, setting the derivatives to zero:[ 0 = a G_A^* + b G_B^* + c cos(omega t) ][ 0 = d G_B^* + e G_A^* + h sin(omega t) ]Wait, but ( cos(omega t) ) and ( sin(omega t) ) are time-dependent. For an equilibrium point, these terms must also be constant, which is only possible if ( c = 0 ) and ( h = 0 ). Otherwise, the equilibrium point would have to vary with time, which contradicts the definition of an equilibrium.Therefore, unless ( c = 0 ) and ( h = 0 ), there are no fixed equilibrium points because the forcing functions are time-dependent. So, perhaps the question assumes that we're looking for equilibrium points in the absence of the forcing terms, i.e., considering the homogeneous system.Alternatively, maybe the analyst is considering the average behavior over time, but that's not a standard equilibrium point.Wait, perhaps the question is referring to equilibrium points of the autonomous system, which would require setting the forcing terms to zero. So, let's assume that ( c = 0 ) and ( h = 0 ). Then, the system becomes:[ frac{dG_A}{dt} = a G_A + b G_B ][ frac{dG_B}{dt} = e G_A + d G_B ]Then, the equilibrium points are solutions to:[ a G_A^* + b G_B^* = 0 ][ e G_A^* + d G_B^* = 0 ]Which can be written as:[ begin{pmatrix} a & b e & d end{pmatrix} begin{pmatrix} G_A^* G_B^* end{pmatrix} = begin{pmatrix} 0 0 end{pmatrix} ]So, unless ( G_A^* = G_B^* = 0 ), the only equilibrium point is the trivial solution. But in economic terms, a GDP of zero isn't really an equilibrium unless both countries collapse, which is probably not the case.Alternatively, maybe the analyst is considering a different kind of equilibrium where the time-dependent terms balance out. But in standard dynamical systems, equilibrium points are constant solutions, so they must satisfy the equations without the forcing terms.Therefore, perhaps the question assumes that ( c = 0 ) and ( h = 0 ), and we're analyzing the stability of the trivial equilibrium ( (0, 0) ).Alternatively, if ( c ) and ( h ) are non-zero, then the system doesn't have fixed equilibrium points, but rather exhibits forced oscillations. So, the stability analysis would be different, perhaps looking at the stability of the particular solution.But given that the question mentions "equilibrium point ( (G_A^*, G_B^*) )", I think it's safe to assume that ( c = 0 ) and ( h = 0 ), so the system is autonomous, and we can analyze the stability of the equilibrium at the origin.So, the Jacobian matrix at the equilibrium point is just the matrix ( M ):[ J = begin{pmatrix} a & b e & d end{pmatrix} ]The eigenvalues of this matrix are the same as the eigenvalues ( lambda ) we found earlier:[ lambda = frac{(a + d) pm sqrt{(a - d)^2 + 4be}}{2} ]The stability of the equilibrium point depends on the real parts of these eigenvalues.- If both eigenvalues have negative real parts, the equilibrium is asymptotically stable.- If at least one eigenvalue has a positive real part, the equilibrium is unstable.- If eigenvalues have zero real parts, the stability is inconclusive (could be stable, unstable, or neutral).So, the conditions for stability are that both eigenvalues have negative real parts.Case 1: Real eigenvalues.If ( D = (a - d)^2 + 4be > 0 ), then eigenvalues are real and distinct.For both eigenvalues to be negative, we need:1. The sum of the eigenvalues ( lambda_1 + lambda_2 = a + d < 0 )2. The product of the eigenvalues ( lambda_1 lambda_2 = ad - be > 0 )These are the Routh-Hurwitz conditions for a 2x2 system.Case 2: Complex eigenvalues.If ( D < 0 ), then eigenvalues are complex with real part ( alpha = frac{a + d}{2} ).For stability, the real part must be negative:[ alpha = frac{a + d}{2} < 0 implies a + d < 0 ]Additionally, for complex eigenvalues, the product ( lambda_1 lambda_2 = ad - be ) must be positive (as it's equal to the determinant, which is the product of eigenvalues). So, same as above, ( ad - be > 0 ).Therefore, regardless of whether the eigenvalues are real or complex, the equilibrium is asymptotically stable if:1. ( a + d < 0 )2. ( ad - be > 0 )If these conditions are met, the equilibrium point is stable.So, summarizing:The equilibrium point ( (G_A^*, G_B^*) ) is asymptotically stable if the trace of the Jacobian matrix ( a + d < 0 ) and the determinant ( ad - be > 0 ). If either the trace is non-negative or the determinant is non-positive, the equilibrium is unstable.Therefore, the conditions on the constants are:- ( a + d < 0 )- ( ad - be > 0 )These ensure that the equilibrium point is stable.Final Answer1. The solutions for ( G_A(t) ) and ( G_B(t) ) are given by the general solution of the system, which combines the homogeneous and particular solutions: [ boxed{G_A(t) = Phi_{11}(t) C_1 + Phi_{12}(t) C_2 + P_A cos(omega t) + Q_A sin(omega t)} ] [ boxed{G_B(t) = Phi_{21}(t) C_1 + Phi_{22}(t) C_2 + P_B cos(omega t) + Q_B sin(omega t)} ] where ( Phi(t) ) is the fundamental matrix solution, ( C_1 ) and ( C_2 ) are constants determined by initial conditions, and ( P_A, P_B, Q_A, Q_B ) are the particular solution components derived above.2. The equilibrium point ( (G_A^*, G_B^*) ) is asymptotically stable if the following conditions are satisfied: [ boxed{a + d < 0 quad text{and} quad ad - be > 0} ]</think>