🧠:<think>Okay, so I have this problem where an art history major wants to recreate a Renaissance-style room. They’re using two types of wallpaper: one with geometric designs and another with floral motifs. The room is 10 meters long, 8 meters wide, and 3 meters high. There's a door that's 2 meters high and 1 meter wide, and two windows each measuring 1.5 meters by 1.5 meters. The wallpaper comes in rolls that cover 5 square meters each.The first question is asking how many rolls of each type of wallpaper they need to buy if the geometric design is supposed to cover exactly 60% of the total wall area, excluding the door and windows. The second question is about a decorative border that repeats every 1.2 meters along the top edge of the walls. I need to figure out how many complete cycles can fit around the room, excluding the door and windows.Let me tackle the first question first.First, I need to calculate the total wall area, excluding the door and windows. The room is a rectangular prism, so the walls are made up of four rectangles: two longer walls and two shorter walls.The longer walls are each 10 meters long and 3 meters high. The shorter walls are 8 meters long and 3 meters high.So, the area of one longer wall is 10 * 3 = 30 square meters. Since there are two of these, that's 30 * 2 = 60 square meters.Similarly, the area of one shorter wall is 8 * 3 = 24 square meters. Two of these would be 24 * 2 = 48 square meters.So, the total wall area without subtracting anything is 60 + 48 = 108 square meters.Now, we need to subtract the areas of the door and windows.The door is 2 meters high and 1 meter wide, so its area is 2 * 1 = 2 square meters.Each window is 1.5 meters by 1.5 meters, so each has an area of 1.5 * 1.5 = 2.25 square meters. There are two windows, so that's 2.25 * 2 = 4.5 square meters.Adding the door and windows together, the total area to subtract is 2 + 4.5 = 6.5 square meters.Therefore, the total wall area to be covered with wallpaper is 108 - 6.5 = 101.5 square meters.Now, the geometric design is supposed to cover 60% of this area. So, let me calculate 60% of 101.5.60% is 0.6, so 0.6 * 101.5 = 60.9 square meters.The remaining 40% will be covered with floral motifs. Let me check: 0.4 * 101.5 = 40.6 square meters. Yep, that adds up to 101.5.So, the geometric wallpaper needs to cover 60.9 square meters, and the floral wallpaper needs to cover 40.6 square meters.Each roll of wallpaper covers 5 square meters. So, for the geometric wallpaper, we need 60.9 / 5 rolls. Let me compute that.60.9 divided by 5 is 12.18. Since you can't buy a fraction of a roll, we need to round up to the next whole number. So, 13 rolls of geometric wallpaper.Similarly, for the floral wallpaper, 40.6 / 5 = 8.12. Again, rounding up, that's 9 rolls.Wait, but let me double-check my calculations because 13 rolls of geometric would cover 13 * 5 = 65 square meters, which is more than 60.9. Similarly, 9 rolls of floral would cover 45 square meters, which is more than 40.6. So, in total, 13 + 9 = 22 rolls, covering 65 + 45 = 110 square meters. But the total wall area is 101.5, so there is some extra. But since you can't buy partial rolls, this is the minimum needed.Alternatively, maybe I should check if 12 rolls of geometric would be enough. 12 * 5 = 60, which is less than 60.9, so 12 rolls wouldn't be enough. So, 13 rolls are necessary.Similarly, 8 rolls of floral would give 40 square meters, which is less than 40.6, so 9 rolls are needed.So, the answer is 13 rolls of geometric and 9 rolls of floral.Wait, but let me make sure I didn't make a mistake in calculating the total wall area.The room is 10m x 8m x 3m. So, the walls: two walls are 10x3, two are 8x3.Total wall area: 2*(10*3) + 2*(8*3) = 60 + 48 = 108.Subtract door (2*1=2) and two windows (each 1.5*1.5=2.25, so 4.5 total). So, 108 - 6.5 = 101.5. That seems correct.60% of 101.5 is 60.9, 40% is 40.6. Divided by 5 per roll, so 12.18 and 8.12. Rounded up, 13 and 9. So, yes, 13 rolls of geometric and 9 rolls of floral.Now, the second question is about the decorative border. The border is a repeating pattern that completes one full cycle every 1.2 meters. We need to find how many complete cycles can fit around the entire room, excluding the door and windows.So, first, I need to find the total length around the room where the border will be installed. Since it's along the top edge of the walls, it's the perimeter of the room, but excluding the areas where the door and windows are.Wait, but the door and windows are on the walls, so the border is along the top edge, which is a continuous line around the room, but the door and windows would have their own edges. However, the border is along the top edge of the walls, so it would go around the entire perimeter, but the door and windows are on the walls, so the border would still go along the top of the door and windows? Or does it stop at the door and windows?Wait, the problem says "excluding the door and windows." So, does that mean the border is installed around the entire room but not covering the door and window areas? So, the border would go along the top of the walls, but not over the door and windows.So, the length of the border would be the perimeter of the room minus the lengths of the door and windows.Wait, but the door and windows are on the walls, so their heights are subtracted from the walls, but the border is along the top edge, so it's a horizontal line. So, the length of the border is the perimeter of the room, but the door and windows are vertical, so their widths would be subtracted from the perimeter.Wait, perhaps I'm overcomplicating.The border is along the top edge of the walls, so it's a continuous strip around the room at the top. However, the door and windows are on the walls, so the border would have to go around them? Or does it just go along the top of the walls, not covering the door and window areas?Wait, the problem says "excluding the door and windows." So, the border is installed along the top edge of the walls, but not covering the door and window areas. So, the length of the border is the perimeter of the room minus the widths of the door and windows.Wait, but the door is 1 meter wide, and each window is 1.5 meters wide. So, the total width to subtract is 1 + 1.5 + 1.5 = 4 meters.But wait, the perimeter of the room is 2*(length + width) = 2*(10 + 8) = 36 meters.But if we exclude the door and windows, does that mean we subtract their widths from the perimeter? So, the door is 1m wide, each window is 1.5m wide, so total subtraction is 1 + 1.5 + 1.5 = 4 meters.Therefore, the length of the border is 36 - 4 = 32 meters.Now, the border pattern repeats every 1.2 meters. So, the number of complete cycles is 32 / 1.2.Let me compute that.32 divided by 1.2 is equal to 26.666...So, 26 complete cycles, with a remainder of 0.666... meters, which is 2/3 of a meter, so not enough for a complete cycle.Therefore, the number of complete cycles is 26.Wait, let me double-check.Perimeter is 2*(10+8)=36 meters.Door is 1m wide, two windows each 1.5m, so total subtraction is 1 + 1.5 + 1.5 = 4 meters.So, border length is 36 - 4 = 32 meters.32 / 1.2 = 26.666...So, 26 complete cycles.Alternatively, maybe the border is installed along the top edge, but the door and windows are on the walls, so the border would have to go around the door and windows, meaning that the border would have to go around the edges of the door and windows, effectively adding to the length.Wait, that might be another interpretation.If the border is along the top edge of the walls, but the door and windows are cut out, then the border would have to go around the top of the door and windows, effectively adding their widths to the perimeter.Wait, that might be a different way to think about it.Wait, no, the door and windows are on the walls, so the top edge of the door and windows are part of the wall's top edge. So, the border would go along the top of the door and windows as well, but since the door and windows are not covered by wallpaper, does that mean the border is also not installed there? Or is the border installed along the entire top edge, including over the door and windows?The problem says "excluding the door and windows," so I think the border is installed along the top edge of the walls, but not covering the door and window areas. So, the length of the border is the perimeter of the room minus the widths of the door and windows.Therefore, 36 - 4 = 32 meters.32 / 1.2 = 26.666..., so 26 complete cycles.Alternatively, if the border goes around the door and windows, meaning that the border has to go along the top of the door and windows, effectively adding their widths to the perimeter.Wait, that would mean the border length is the perimeter plus the widths of the door and windows.But that seems counterintuitive because the door and windows are cut out, so the border would have to go around them, adding their widths.Wait, perhaps I need to visualize this.Imagine the room as a rectangle. The border is along the top edge of the walls. However, there's a door and two windows. The door is 1m wide, and each window is 1.5m wide.If the border is installed along the top edge, it would have to go around the door and windows, meaning that instead of a straight line along the wall, it would have to turn around the door and windows, effectively adding their widths to the perimeter.Wait, but that would complicate the calculation because the border would have to go around each door and window, which are on the walls.But the problem says "excluding the door and windows," so perhaps the border is installed along the top edge of the walls, but not covering the areas where the door and windows are. So, the border is installed on the walls, but not on the door and windows.Therefore, the length of the border is the perimeter of the room minus the widths of the door and windows.So, 36 - 4 = 32 meters.Therefore, 32 / 1.2 = 26.666..., so 26 complete cycles.Alternatively, if the border is installed along the top edge, including around the door and windows, then the length would be the perimeter plus the widths of the door and windows.But that seems less likely because the door and windows are openings, not part of the wall.Wait, perhaps the border is installed along the top edge of the walls, but the door and windows are on the walls, so the border would have to go around the edges of the door and windows, effectively adding their widths to the perimeter.Wait, let me think.If the border is along the top edge of the walls, it would go along the top of the door and windows as well, but since the door and windows are not covered by wallpaper, the border is installed on the walls, but not on the door and windows.Therefore, the border would have to go around the door and windows, meaning that the length of the border is the perimeter of the room minus the widths of the door and windows.Wait, no, because the door and windows are on the walls, so the border would have to go around them, effectively adding their widths to the perimeter.Wait, this is confusing.Let me try to visualize it.Imagine the room as a rectangle. The top edge of the walls is a continuous line around the room. However, there are a door and two windows. The door is 1m wide, and each window is 1.5m wide.If the border is installed along the top edge of the walls, it would have to go around the door and windows. So, instead of a straight line along the wall, it would have to go around the door and windows, effectively adding their widths to the perimeter.Wait, but that would mean the border length is the perimeter plus the widths of the door and windows.But that seems incorrect because the door and windows are cut out, so the border would have to go around them, but not cover them.Wait, perhaps the border is installed along the top edge of the walls, but not covering the door and window areas. So, the border is installed on the walls, but not on the door and windows.Therefore, the length of the border is the perimeter of the room minus the widths of the door and windows.So, 36 - 4 = 32 meters.Therefore, 32 / 1.2 = 26.666..., so 26 complete cycles.Alternatively, if the border is installed along the top edge, including around the door and windows, then the length would be the perimeter plus the widths of the door and windows.But that would be 36 + 4 = 40 meters, which would give 40 / 1.2 = 33.333..., so 33 cycles.But the problem says "excluding the door and windows," so I think the border is installed along the top edge of the walls, but not covering the door and window areas. So, the length is 36 - 4 = 32 meters.Therefore, 32 / 1.2 = 26.666..., so 26 complete cycles.I think that's the correct approach.</think>

🧠:<think>Okay, so I have this problem about a matrix A, and I need to find its eigenvalues and eigenvectors. Then, I also have to apply this matrix to a vector x and think about how eigenvalues and eigenvectors relate to neural networks. Hmm, let me start with the first part.First, eigenvalues. I remember that eigenvalues are scalars λ such that Ax = λx for some non-zero vector x. To find them, I need to solve the characteristic equation, which is det(A - λI) = 0. So, I need to compute the determinant of (A - λI). Let me write down matrix A:A = [4 1 2; 0 3 -1; 0 0 2]So, subtracting λ from the diagonal, A - λI becomes:[4-λ 1 2 ; 0 3-λ -1 ; 0 0 2-λ ]Now, the determinant of a triangular matrix (upper or lower) is just the product of the diagonal elements. Since this matrix is upper triangular, the determinant is (4 - λ)(3 - λ)(2 - λ). So, setting that equal to zero:(4 - λ)(3 - λ)(2 - λ) = 0So, the eigenvalues are λ = 4, 3, and 2. That was straightforward because the matrix is upper triangular, so eigenvalues are just the diagonal entries. Okay, that's the eigenvalues part done.Next, for each eigenvalue, find the corresponding eigenvectors. Starting with λ = 4.For λ = 4, we solve (A - 4I)v = 0. Let's compute A - 4I:[4-4 1 2 ; => [0 1 2; 0 3-4 -1 ; 0 -1 -1; 0 0 2-4 ] 0 0 -2]So, the matrix is:[0 1 2; 0 -1 -1; 0 0 -2]We can write the system of equations:0*v1 + 1*v2 + 2*v3 = 00*v1 -1*v2 -1*v3 = 00*v1 + 0*v2 -2*v3 = 0From the third equation: -2*v3 = 0 => v3 = 0.Plugging v3 = 0 into the second equation: -v2 = 0 => v2 = 0.Then, first equation: 0 + 0 + 0 = 0, which is always true. So, v1 is free. So, the eigenvectors are of the form [v1; 0; 0], where v1 ≠ 0. So, the eigenvector corresponding to λ=4 is any scalar multiple of [1; 0; 0]. So, we can write it as:v = [1; 0; 0]Okay, moving on to λ = 3.Compute A - 3I:[4-3 1 2 ; => [1 1 2; 0 3-3 -1 ; 0 0 -1; 0 0 2-3 ] 0 0 -1]So, the matrix is:[1 1 2; 0 0 -1; 0 0 -1]Now, the system is:1*v1 + 1*v2 + 2*v3 = 00*v1 + 0*v2 -1*v3 = 00*v1 + 0*v2 -1*v3 = 0From the second equation: -v3 = 0 => v3 = 0.From the third equation: same as second, so v3 = 0.Then, first equation: v1 + v2 + 0 = 0 => v1 = -v2.So, the eigenvectors are of the form [ -v2; v2; 0 ] = v2[-1; 1; 0]. So, we can take v2 = 1, so the eigenvector is:v = [-1; 1; 0]Alright, now for λ = 2.Compute A - 2I:[4-2 1 2 ; => [2 1 2; 0 3-2 -1 ; 0 1 -1; 0 0 2-2 ] 0 0 0]So, the matrix is:[2 1 2; 0 1 -1; 0 0 0]The system is:2*v1 + 1*v2 + 2*v3 = 00*v1 + 1*v2 -1*v3 = 00*v1 + 0*v2 + 0*v3 = 0Third equation is always true, so we can ignore it.From the second equation: v2 - v3 = 0 => v2 = v3.From the first equation: 2*v1 + v2 + 2*v3 = 0. But since v2 = v3, substitute:2*v1 + v3 + 2*v3 = 0 => 2*v1 + 3*v3 = 0 => v1 = (-3/2)*v3.So, let me express the eigenvector in terms of v3. Let v3 = t, then v2 = t, and v1 = (-3/2)t.So, the eigenvector is [ -3/2 t; t; t ] = t*[ -3/2; 1; 1 ]. To make it cleaner, we can multiply by 2 to eliminate the fraction:t*[ -3; 2; 2 ].So, taking t = 1, the eigenvector is:v = [-3; 2; 2]Wait, let me check that. If I plug back into A - 2I:First equation: 2*(-3) + 1*2 + 2*2 = -6 + 2 + 4 = 0. Good.Second equation: 0*(-3) + 1*2 -1*2 = 2 - 2 = 0. Good.Third equation: 0. So, yes, that works.So, the eigenvectors are:For λ=4: [1; 0; 0]For λ=3: [-1; 1; 0]For λ=2: [-3; 2; 2]Okay, that's part 1 done.Now, part 2: applying the matrix A to the vector x = [1; 0; 1].So, compute y = A*x.Let me compute that:First component: 4*1 + 1*0 + 2*1 = 4 + 0 + 2 = 6Second component: 0*1 + 3*0 + (-1)*1 = 0 + 0 -1 = -1Third component: 0*1 + 0*0 + 2*1 = 0 + 0 + 2 = 2So, y = [6; -1; 2]So, that's the output vector.Now, thinking about the implications in machine learning, especially neural networks. The matrix A is a weight matrix in a neural network layer. The eigenvalues and eigenvectors can tell us about the behavior of the transformation.Eigenvalues greater than 1 can cause the activations to grow when the network is applied multiple times, which might lead to exploding gradients or unstable training. Eigenvalues less than 1 can cause the activations to shrink, potentially leading to vanishing gradients.In this case, the eigenvalues are 4, 3, and 2, all greater than 1. So, if this transformation is applied multiple times, the outputs could grow exponentially, which might cause instability in training. However, in a single layer, it's just a linear transformation, so maybe not too bad, but in deep networks, repeated applications could be problematic.Eigenvectors tell us the directions in which the transformation acts by scaling. So, any input vector that's an eigenvector will just be scaled by the corresponding eigenvalue. For example, if the input is [1; 0; 0], it gets scaled by 4. Similarly, [-1; 1; 0] gets scaled by 3, and [-3; 2; 2] gets scaled by 2.In the context of neural networks, if the input lies along an eigenvector direction, the transformation is just a scaling, which might make learning easier or harder depending on the eigenvalue. If eigenvalues are too large or too small, it can affect the learning dynamics.Also, the condition number of the matrix, which is the ratio of the largest to smallest eigenvalue, is 4/2 = 2. A lower condition number implies better numerical stability, so this matrix isn't too bad, but if eigenvalues were spread out more, it could cause issues.In summary, the eigenvalues being greater than 1 might contribute to instability if the transformation is applied multiple times, but in a single layer, it's manageable. The eigenvectors give specific directions where the transformation is just a scaling, which could be useful for understanding how different inputs are transformed.Final Answer1. The eigenvalues of matrix (A) are (boxed{4}), (boxed{3}), and (boxed{2}). The corresponding eigenvectors are (boxed{begin{pmatrix} 1 0 0 end{pmatrix}}), (boxed{begin{pmatrix} -1 1 0 end{pmatrix}}), and (boxed{begin{pmatrix} -3 2 2 end{pmatrix}}).2. The output vector ( mathbf{y} ) after applying the linear transformation (A) to ( mathbf{x} ) is (boxed{begin{pmatrix} 6 -1 2 end{pmatrix}}).</think>

🧠:<think>Okay, so I have this problem where an investor is looking to allocate 10 million across several tech products. The goal is to maximize expected returns while considering the risks involved. Let me try to break this down step by step.First, the problem is divided into two parts. Part 1 is about formulating an optimization problem to maximize the total expected utility, considering the risk factors. Part 2 adds another constraint related to the probability of achieving a certain total revenue. I'll tackle them one by one.Starting with Part 1:The investor has n products, each with its own revenue potential modeled as a random variable R_i. Each R_i has an expected value μ_i and variance σ_i². The investor also assigns a risk factor β_i to each product, which is between 0 and 1. The utility function for each dollar invested in product i is given by U_i(x) = x(μ_i - β_i σ_i²). So, the utility is a linear function of the investment x, with the coefficient being the expected return minus a risk penalty term.The investor wants to maximize the total expected utility, which would be the sum of U_i(x_i) for all products. So, the objective function is:Maximize Σ [x_i (μ_i - β_i σ_i²)] for i = 1 to n.Subject to the constraint that the total investment is 10 million:Σ x_i = 10,000,000.Additionally, each x_i should be non-negative because you can't invest a negative amount. So, x_i ≥ 0 for all i.So, putting it all together, the optimization problem is:Maximize Σ [x_i (μ_i - β_i σ_i²)]Subject to:Σ x_i = 10,000,000x_i ≥ 0 for all i.That seems straightforward. It's a linear optimization problem because both the objective function and the constraints are linear in terms of x_i. So, we can use linear programming techniques to solve this.Moving on to Part 2:The investor now wants to ensure that the probability of achieving at least 15 million in total revenue is at least 95%. The total revenue is modeled as a normal distribution with mean equal to the sum of μ_i x_i and variance equal to the sum of σ_i² x_i². So, the total revenue R_total is N(Σ μ_i x_i, Σ σ_i² x_i²). The investor wants P(R_total ≥ 15,000,000) ≥ 0.95.In terms of the standard normal distribution, we can express this probability using the cumulative distribution function (CDF). Let me recall that for a normal variable X ~ N(μ, σ²), the probability P(X ≥ a) can be written as 1 - Φ((a - μ)/σ), where Φ is the CDF of the standard normal distribution.So, in our case, we have:P(R_total ≥ 15,000,000) = 1 - Φ((15,000,000 - Σ μ_i x_i) / sqrt(Σ σ_i² x_i²)) ≥ 0.95.To satisfy this inequality, we can rearrange it:1 - Φ((15,000,000 - Σ μ_i x_i) / sqrt(Σ σ_i² x_i²)) ≥ 0.95=> Φ((15,000,000 - Σ μ_i x_i) / sqrt(Σ σ_i² x_i²)) ≤ 0.05.Looking up the standard normal distribution table, the z-score corresponding to Φ(z) = 0.05 is approximately -1.645. So, we have:(15,000,000 - Σ μ_i x_i) / sqrt(Σ σ_i² x_i²) ≤ -1.645.Multiplying both sides by the denominator (which is positive, so inequality direction remains the same):15,000,000 - Σ μ_i x_i ≤ -1.645 * sqrt(Σ σ_i² x_i²).Let me rearrange this:Σ μ_i x_i - 1.645 * sqrt(Σ σ_i² x_i²) ≥ 15,000,000.So, this becomes another constraint in our optimization problem. Therefore, the updated optimization problem now includes this constraint in addition to the budget constraint and non-negativity constraints.Wait, let me double-check the direction of the inequality. If Φ(z) ≤ 0.05, then z ≤ -1.645. So, (15,000,000 - Σ μ_i x_i) / sqrt(Σ σ_i² x_i²) ≤ -1.645.Multiplying both sides by sqrt(Σ σ_i² x_i²), which is positive, so the inequality remains:15,000,000 - Σ μ_i x_i ≤ -1.645 * sqrt(Σ σ_i² x_i²).Then, bringing the sqrt term to the left and the rest to the right:15,000,000 ≤ Σ μ_i x_i - 1.645 * sqrt(Σ σ_i² x_i²).Yes, that seems correct.So, the constraint is:Σ μ_i x_i - 1.645 * sqrt(Σ σ_i² x_i²) ≥ 15,000,000.But wait, in optimization, it's often preferred to have constraints in a certain form. This constraint is a bit tricky because it involves both linear and square root terms. It might complicate the optimization problem because it's not linear or convex in a straightforward way.However, since the original problem in part 1 was linear, adding this constraint may make it a nonlinear optimization problem. But given that the total revenue is normally distributed, this is the correct way to model the probability constraint.So, summarizing, the optimization problem now has:Objective: Maximize Σ [x_i (μ_i - β_i σ_i²)]Subject to:1. Σ x_i = 10,000,0002. Σ μ_i x_i - 1.645 * sqrt(Σ σ_i² x_i²) ≥ 15,000,0003. x_i ≥ 0 for all i.This is a more complex optimization problem because of the nonlinear constraint. It might require nonlinear programming techniques to solve, which could be more computationally intensive.But for the purpose of formulating the problem, we just need to include this constraint as part of the optimization model.Let me recap:1. The first part is a linear program with the objective to maximize the sum of utilities, subject to the budget and non-negativity constraints.2. The second part adds a probabilistic constraint, which translates into a nonlinear constraint involving the mean and standard deviation of the total revenue, ensuring that the probability of meeting the revenue target is at least 95%.I think that covers both parts. I should make sure I didn't miss any details.Wait, in part 1, the utility function is given per dollar invested. So, for each product, the utility is x_i*(μ_i - β_i σ_i²). So, the total utility is the sum over all products. That makes sense because each dollar contributes its own utility based on the product's expected return and risk.In part 2, the total revenue is the sum of each product's revenue, which is a random variable. The total revenue is normally distributed because it's the sum of normally distributed variables (assuming each R_i is normal, but actually, the problem says the total revenue is normally distributed, so maybe they are assuming that even if individual R_i aren't normal, the sum is, perhaps by the Central Limit Theorem? Or maybe each R_i is normal? The problem doesn't specify, but it says the total revenue is normally distributed, so we can proceed with that.Therefore, the constraint is correctly formulated using the normal CDF.I think I have covered all the steps. So, the optimization problem is correctly formulated with the objective and constraints as above.Final Answer1. The optimization problem is formulated as:Maximize ( sum_{i=1}^{n} x_i (mu_i - beta_i sigma_i^2) )subject to:[sum_{i=1}^{n} x_i = 10,000,000][x_i geq 0 quad text{for all } i]2. The additional constraint is:[sum_{i=1}^{n} mu_i x_i - 1.645 sqrt{sum_{i=1}^{n} sigma_i^2 x_i^2} geq 15,000,000]Thus, the final answers are:1. The optimization problem is:boxed{max sum_{i=1}^{n} x_i (mu_i - beta_i sigma_i^2) text{ subject to } sum_{i=1}^{n} x_i = 10,000,000 text{ and } x_i geq 0 text{ for all } i}2. The constraint is:boxed{sum_{i=1}^{n} mu_i x_i - 1.645 sqrt{sum_{i=1}^{n} sigma_i^2 x_i^2} geq 15,000,000}</think>

🧠:<think>Okay, so I've got this problem about Alex and their social media likes. It's a logistic growth model, which I remember is used to describe how populations grow when there's a carrying capacity. The equation given is a differential equation: dL/dt = rL(1 - L/K). Alright, the first part asks me to solve this differential equation given r = 0.04 per hour and K = 200, with the initial condition L(0) = 10. Hmm, solving a logistic equation. I think the general solution involves integrating both sides, maybe using separation of variables. Let me recall the steps.So, the logistic equation is dL/dt = rL(1 - L/K). To solve this, I can rewrite it as dL / [L(1 - L/K)] = r dt. Then, I can integrate both sides. The left side integral is a bit tricky, but I remember it can be done using partial fractions. Let me set it up:∫ [1 / (L(1 - L/K))] dL = ∫ r dtLet me make a substitution to simplify the left integral. Let me set u = L/K, so L = Ku, and dL = K du. Then, substituting into the integral:∫ [1 / (Ku(1 - u))] * K du = ∫ r dtThe K cancels out, so it becomes:∫ [1 / (u(1 - u))] du = ∫ r dtNow, I can split the fraction 1/(u(1 - u)) into partial fractions. Let me write it as A/u + B/(1 - u). So,1/(u(1 - u)) = A/u + B/(1 - u)Multiplying both sides by u(1 - u):1 = A(1 - u) + B uExpanding:1 = A - Au + BuGrouping terms:1 = A + (B - A)uSince this must hold for all u, the coefficients of like terms must be equal. So,A = 1B - A = 0 => B = A = 1So, the integral becomes:∫ [1/u + 1/(1 - u)] du = ∫ r dtIntegrating both sides:ln|u| - ln|1 - u| = rt + CCombining the logs:ln|u / (1 - u)| = rt + CExponentiating both sides:u / (1 - u) = e^{rt + C} = e^C e^{rt}Let me denote e^C as another constant, say, C'. So,u / (1 - u) = C' e^{rt}But u = L/K, so substituting back:(L/K) / (1 - L/K) = C' e^{rt}Simplify the left side:(L/K) / [(K - L)/K] = L / (K - L) = C' e^{rt}So,L / (K - L) = C' e^{rt}Now, solve for L:L = (K - L) C' e^{rt}L = K C' e^{rt} - L C' e^{rt}Bring the L terms to one side:L + L C' e^{rt} = K C' e^{rt}Factor L:L (1 + C' e^{rt}) = K C' e^{rt}So,L = (K C' e^{rt}) / (1 + C' e^{rt})Now, apply the initial condition L(0) = 10. At t = 0,10 = (K C' e^{0}) / (1 + C' e^{0}) = (K C') / (1 + C')So,10 = (200 C') / (1 + C')Multiply both sides by (1 + C'):10 (1 + C') = 200 C'10 + 10 C' = 200 C'10 = 190 C'C' = 10 / 190 = 1/19So, C' = 1/19. Therefore, the solution is:L(t) = (200 * (1/19) e^{0.04 t}) / (1 + (1/19) e^{0.04 t})Simplify numerator and denominator:Numerator: (200/19) e^{0.04 t}Denominator: 1 + (1/19) e^{0.04 t} = (19 + e^{0.04 t}) / 19So, L(t) = (200/19 e^{0.04 t}) / [(19 + e^{0.04 t}) / 19] = (200 e^{0.04 t}) / (19 + e^{0.04 t})Alternatively, factor out e^{0.04 t} in the denominator:L(t) = 200 / (19 e^{-0.04 t} + 1)But both forms are correct. Maybe the first form is better. Let me write it as:L(t) = (200 e^{0.04 t}) / (19 + e^{0.04 t})That's the solution to the differential equation. So, part 1 is done.Now, part 2: Alex wants to maintain the number of likes at 80% of K. K is 200, so 80% of 200 is 160. So, find t when L(t) = 160.So, set L(t) = 160:160 = (200 e^{0.04 t}) / (19 + e^{0.04 t})Multiply both sides by denominator:160 (19 + e^{0.04 t}) = 200 e^{0.04 t}Compute 160*19: 160*20=3200, so 160*19=3040So,3040 + 160 e^{0.04 t} = 200 e^{0.04 t}Bring terms with e^{0.04 t} to one side:3040 = 200 e^{0.04 t} - 160 e^{0.04 t} = 40 e^{0.04 t}So,40 e^{0.04 t} = 3040Divide both sides by 40:e^{0.04 t} = 3040 / 40 = 76Take natural log of both sides:0.04 t = ln(76)Compute ln(76). Let me calculate that. ln(76) is approximately... since ln(75) is about 4.317, and ln(76) is a bit more. Let me check:e^4 = 54.598, e^4.3 = e^4 * e^0.3 ≈ 54.598 * 1.3499 ≈ 73.7, which is close to 76. So, 4.3 gives about 73.7, so 76 is a bit higher. Let me compute ln(76):Using calculator approximation: ln(76) ≈ 4.3307So,0.04 t ≈ 4.3307Thus,t ≈ 4.3307 / 0.04 ≈ 108.2675 hours.So, approximately 108.27 hours. Let me check the calculation again to make sure.From L(t) = 160:160 = (200 e^{0.04 t}) / (19 + e^{0.04 t})Multiply both sides by (19 + e^{0.04 t}):160*19 + 160 e^{0.04 t} = 200 e^{0.04 t}3040 + 160 e^{0.04 t} = 200 e^{0.04 t}3040 = 40 e^{0.04 t}e^{0.04 t} = 76Yes, that's correct. So, t = ln(76)/0.04 ≈ 4.3307 / 0.04 ≈ 108.27 hours.So, about 108.27 hours. To be precise, maybe I should carry more decimal places for ln(76). Let me compute ln(76) more accurately.We know that ln(75) ≈ 4.3175, ln(76) is ln(75) + ln(76/75) ≈ 4.3175 + ln(1.013333). ln(1.013333) ≈ 0.0132 (since ln(1+x) ≈ x - x^2/2 + x^3/3... for small x). So, 0.0132. So, ln(76) ≈ 4.3175 + 0.0132 ≈ 4.3307, which matches the earlier approximation.So, t ≈ 4.3307 / 0.04 ≈ 108.2675 hours, which is approximately 108.27 hours.To express this in days, since 24 hours is a day, 108 / 24 = 4.5 days, so 108.27 hours is about 4.51 days. But the question just asks for the time t, so 108.27 hours is fine.Wait, but let me check if I did everything correctly. Let me plug t = 108.27 back into L(t) to see if it's approximately 160.Compute e^{0.04 * 108.27} ≈ e^{4.3308} ≈ 76, as before.So, L(t) = 200 * 76 / (19 + 76) = 200 * 76 / 95 = (200 / 95) * 76.200 / 95 is approximately 2.10526, and 2.10526 * 76 ≈ 160. So, yes, that checks out.Therefore, the time t is approximately 108.27 hours.But let me see if I can express it exactly. Since e^{0.04 t} = 76, then t = (ln 76)/0.04. So, exact form is t = (ln 76)/0.04. But maybe we can write it as (ln 76)/(1/25) since 0.04 is 1/25. So, t = 25 ln 76. Let me compute 25 ln 76:25 * 4.3307 ≈ 108.2675, same as before.Alternatively, if we leave it in terms of ln, it's exact, but the question probably expects a numerical value. So, 108.27 hours is fine.So, summarizing:1. The solution to the differential equation is L(t) = (200 e^{0.04 t}) / (19 + e^{0.04 t}).2. The time when L(t) reaches 80% of K (which is 160) is approximately 108.27 hours.I think that's it. Let me just double-check the algebra steps to make sure I didn't make any mistakes.Starting from the logistic equation, separation of variables, partial fractions, integrating, exponentiating, solving for L(t), applying initial condition. All steps seem correct. Then, setting L(t) = 160, solving for t. The algebra looks good. So, I think the answers are correct.Final Answer1. The function ( L(t) ) is ( boxed{dfrac{200 e^{0.04 t}}{19 + e^{0.04 t}}} ).2. The time ( t ) when the likes reach 80% of the carrying capacity is approximately ( boxed{108.27} ) hours.</think>