🧠:<think>First, I need to set up the system of equations based on the information provided. Let ( x ) represent the number of Limited Edition anthologies sold, and ( y ) represent the number of Standard Edition anthologies sold.From the total number of anthologies sold, I know that:[x + y = 5000]From the total revenue generated, I know that:[150x + 50y = 450000]Next, I'll solve this system of equations. I can express ( y ) in terms of ( x ) from the first equation:[y = 5000 - x]Substituting this into the revenue equation:[150x + 50(5000 - x) = 450000][150x + 250000 - 50x = 450000][100x + 250000 = 450000][100x = 200000][x = 2000]Now, substituting ( x = 2000 ) back into the equation for ( y ):[y = 5000 - 2000 = 3000]So, Amelia sold 2000 Limited Edition anthologies and 3000 Standard Edition anthologies last year.For the next year, the prices will increase by 10%, and the number of anthologies sold will change as follows:- Limited Edition: 2000 anthologies sold last year, decreasing by 5%: ( 2000 times 0.95 = 1900 )- Standard Edition: 3000 anthologies sold last year, increasing by 8%: ( 3000 times 1.08 = 3240 )The new prices will be:- Limited Edition: ( 150 times 1.10 = 165 ) dollars- Standard Edition: ( 50 times 1.10 = 55 ) dollarsFinally, calculating the expected revenue:[1900 times 165 + 3240 times 55 = 313500 + 178200 = 491700]</think>

🧠:<think>First, I need to determine the volume of Linda's room and the volume of each display case to see how many cases can fit inside.The room is a rectangular prism with a floor area of 180 square feet and a height of 10 feet. So, the volume of the room is 180 multiplied by 10, which equals 1800 cubic feet.Each display case has a base area of 15 square feet and a height of 6 feet. Therefore, the volume of one display case is 15 multiplied by 6, totaling 90 cubic feet.To find out how many display cases can fit into the room, I'll divide the room's volume by the volume of one display case: 1800 divided by 90 equals 20. This means Linda can fit 20 display cases in her room.Next, Linda has 120 movie posters that she wants to distribute evenly across all 20 display cases. I'll divide the total number of posters by the number of display cases: 120 divided by 20 equals 6. So, each display case will contain 6 posters.Since 120 divided by 20 is exactly 6 with no remainder, there will be no posters left to store separately.</think>

🧠:<think>Okay, so I have this problem about a marine geologist studying seismic activity and ocean currents. It's divided into two parts. Let me tackle them one by one.Starting with part 1: The seismic activity is modeled by the function f(x, y, z) = A * e^(-α(x² + y²)) * sin(βz). The goal is to find the integral of f over the domain D, which is defined by -180 ≤ x ≤ 180, -90 ≤ y ≤ 90, and 0 ≤ z ≤ 4000. This integral represents the total seismic energy.Hmm, so I need to compute a triple integral over x, y, and z. The function f is separable into functions of x, y, and z. That is, f(x, y, z) can be written as a product of a function of x, a function of y, and a function of z. Specifically, f(x, y, z) = [A * e^(-αx²)] * [e^(-αy²)] * [sin(βz)]. So, that should make the integral easier because I can separate the integrals for x, y, and z.So, the integral over D is the product of three integrals: the integral over x from -180 to 180 of A * e^(-αx²) dx, the integral over y from -90 to 90 of e^(-αy²) dy, and the integral over z from 0 to 4000 of sin(βz) dz.Wait, actually, the A is a constant, so it can be factored out of the integral. So, the integral becomes A times the integral over x, y, z of e^(-αx²) * e^(-αy²) * sin(βz) dx dy dz.Since the variables are separable, I can write this as A times [integral from x=-180 to 180 of e^(-αx²) dx] times [integral from y=-90 to 90 of e^(-αy²) dy] times [integral from z=0 to 4000 of sin(βz) dz].Okay, so now I need to compute each of these integrals separately.First, the integral over x: ∫_{-180}^{180} e^(-αx²) dx. This is a Gaussian integral. The standard Gaussian integral is ∫_{-∞}^{∞} e^(-ax²) dx = sqrt(π/a). But here, the limits are finite, from -180 to 180. So, it's not the entire real line. Hmm, that complicates things. Similarly, the integral over y is from -90 to 90 of e^(-αy²) dy, which is also a finite Gaussian integral.I wonder if the problem expects me to use the error function (erf) for these integrals since they are finite. The error function is defined as erf(x) = (2/sqrt(π)) ∫_{0}^{x} e^(-t²) dt. So, for the integral of e^(-αx²) from -a to a, it can be written as sqrt(π/α) * erf(a sqrt(α)).Let me verify that. Let’s make a substitution: let u = sqrt(α) x, then du = sqrt(α) dx, so dx = du / sqrt(α). Then, ∫ e^(-αx²) dx = ∫ e^(-u²) * (du / sqrt(α)) = (1/sqrt(α)) ∫ e^(-u²) du. So, over the limits from -a sqrt(α) to a sqrt(α), this becomes (1/sqrt(α)) * sqrt(π) erf(a sqrt(α)).Wait, actually, the integral from -a to a is 2 times the integral from 0 to a. So, ∫_{-a}^{a} e^(-αx²) dx = 2 * ∫_{0}^{a} e^(-αx²) dx. Using substitution u = sqrt(α) x, then x = u / sqrt(α), dx = du / sqrt(α). So, the integral becomes 2 * ∫_{0}^{a sqrt(α)} e^(-u²) * (du / sqrt(α)) = (2 / sqrt(α)) ∫_{0}^{a sqrt(α)} e^(-u²) du = (2 / sqrt(α)) * (sqrt(π)/2) erf(a sqrt(α)) ) = sqrt(π / α) erf(a sqrt(α)).Yes, that seems right. So, the integral over x from -180 to 180 is sqrt(π / α) erf(180 sqrt(α)). Similarly, the integral over y from -90 to 90 is sqrt(π / α) erf(90 sqrt(α)).Now, the integral over z: ∫_{0}^{4000} sin(βz) dz. The integral of sin(βz) is (-1/β) cos(βz). Evaluating from 0 to 4000, it becomes (-1/β)[cos(4000β) - cos(0)] = (-1/β)[cos(4000β) - 1] = (1 - cos(4000β))/β.Putting it all together, the total integral is A times [sqrt(π / α) erf(180 sqrt(α))] times [sqrt(π / α) erf(90 sqrt(α))] times [(1 - cos(4000β))/β].Multiplying the constants, sqrt(π / α) * sqrt(π / α) = π / α. So, the expression becomes A * (π / α) * erf(180 sqrt(α)) * erf(90 sqrt(α)) * (1 - cos(4000β))/β.Therefore, the total seismic energy integral is:A * (π / α) * erf(180 sqrt(α)) * erf(90 sqrt(α)) * (1 - cos(4000β)) / β.I think that's the expression. Let me double-check the steps:1. Separated the triple integral into three single integrals because the function is separable.2. Recognized each integral as Gaussian for x and y, and a sine integral for z.3. Used the error function for the finite Gaussian integrals.4. Computed each integral separately and multiplied them together, including the constant A.5. Simplified the product of the Gaussian integrals to π / α.Seems correct. I don't see any mistakes here.Now, moving on to part 2: The ocean current velocity field is given by V(x, y, z) = (v_x, v_y, v_z), where v_x = C sin(γx), v_y = C cos(γy), and v_z = C e^(-δz). We need to calculate the work done by the ocean currents along a path defined parametrically by r(t) = (t, t², 2000 + 50 sin(t)), where t varies from 0 to π.Work done by a vector field along a path is given by the line integral of the vector field dotted with the differential displacement vector dr. So, the work W is ∫_{C} V · dr, where C is the path from t=0 to t=π.First, let's express dr in terms of t. Since r(t) = (x(t), y(t), z(t)) = (t, t², 2000 + 50 sin(t)), then dr/dt = (dx/dt, dy/dt, dz/dt) = (1, 2t, 50 cos(t)). Therefore, dr = (1, 2t, 50 cos(t)) dt.So, the work integral becomes ∫_{0}^{π} V(r(t)) · (1, 2t, 50 cos(t)) dt.Now, we need to express V in terms of t. Since V(x, y, z) = (C sin(γx), C cos(γy), C e^(-δz)), and along the path, x = t, y = t², z = 2000 + 50 sin(t). So, substituting these into V:v_x = C sin(γ t),v_y = C cos(γ t²),v_z = C e^(-δ (2000 + 50 sin(t))).Therefore, V(r(t)) = (C sin(γ t), C cos(γ t²), C e^(-δ (2000 + 50 sin(t)))).Now, the dot product V · dr/dt is:v_x * dx/dt + v_y * dy/dt + v_z * dz/dt= C sin(γ t) * 1 + C cos(γ t²) * 2t + C e^(-δ (2000 + 50 sin(t))) * 50 cos(t).So, the integrand becomes:C sin(γ t) + 2 C t cos(γ t²) + 50 C cos(t) e^(-δ (2000 + 50 sin(t))).Therefore, the work W is the integral from t=0 to t=π of [C sin(γ t) + 2 C t cos(γ t²) + 50 C cos(t) e^(-δ (2000 + 50 sin(t)))] dt.Let me write that out:W = C ∫_{0}^{π} [sin(γ t) + 2 t cos(γ t²) + 50 cos(t) e^(-δ (2000 + 50 sin(t)))] dt.Now, let's see if we can evaluate each term separately.First term: ∫ sin(γ t) dt from 0 to π.The integral of sin(γ t) is (-1/γ) cos(γ t). Evaluated from 0 to π, it becomes (-1/γ)[cos(γ π) - cos(0)] = (-1/γ)(cos(γ π) - 1) = (1 - cos(γ π))/γ.Second term: ∫ 2 t cos(γ t²) dt from 0 to π.Let me make a substitution here. Let u = γ t², then du = 2 γ t dt. So, 2 t dt = du / γ. Therefore, the integral becomes ∫ cos(u) * (du / γ). The limits when t=0, u=0; t=π, u=γ π².So, the integral is (1/γ) ∫_{0}^{γ π²} cos(u) du = (1/γ)[sin(u)] from 0 to γ π² = (1/γ)(sin(γ π²) - sin(0)) = (sin(γ π²))/γ.Third term: ∫ 50 cos(t) e^(-δ (2000 + 50 sin(t))) dt from 0 to π.This looks more complicated. Let's see if substitution works here. Let me set u = -δ (2000 + 50 sin(t)). Then, du/dt = -δ * 50 cos(t). So, du = -50 δ cos(t) dt. Therefore, cos(t) dt = -du/(50 δ).So, the integral becomes 50 ∫ e^{u} * (-du/(50 δ)) = - (50 / (50 δ)) ∫ e^{u} du = - (1/δ) ∫ e^{u} du.But let's adjust the limits accordingly. When t=0, u = -δ (2000 + 50 sin(0)) = -2000 δ. When t=π, u = -δ (2000 + 50 sin(π)) = -2000 δ.Wait, both limits are the same? Because sin(0) = 0 and sin(π) = 0, so u starts and ends at -2000 δ. Therefore, the integral becomes - (1/δ) [e^{u}] from -2000 δ to -2000 δ = - (1/δ)(e^{-2000 δ} - e^{-2000 δ}) = 0.Wow, that's interesting. So, the third term integrates to zero because the limits are the same. That simplifies things.Therefore, putting it all together, the work W is:C [ (1 - cos(γ π))/γ + (sin(γ π²))/γ + 0 ].So, W = C [ (1 - cos(γ π) + sin(γ π²)) / γ ].Simplifying, W = (C / γ) [1 - cos(γ π) + sin(γ π²)].Let me double-check the substitution for the third term. I set u = -δ(2000 + 50 sin t), so du = -50 δ cos t dt. Then, the integral becomes 50 ∫ cos t e^{u} dt = 50 ∫ e^{u} (-du)/(50 δ) = - (1/δ) ∫ e^{u} du. Since u starts and ends at -2000 δ, the integral is zero. That seems correct.So, the third term indeed cancels out. Therefore, the work done is just the sum of the first two terms.Therefore, the final expression for the work is:W = (C / γ) [1 - cos(γ π) + sin(γ π²)].I think that's the answer.Final Answer1. The total seismic energy is boxed{A cdot frac{pi}{alpha} cdot text{erf}(180sqrt{alpha}) cdot text{erf}(90sqrt{alpha}) cdot frac{1 - cos(4000beta)}{beta}}.2. The work done by the ocean currents is boxed{frac{C}{gamma} left(1 - cos(gamma pi) + sin(gamma pi^2)right)}.</think>

🧠:<think>Okay, so I have this problem where I need to help a startup optimize their marketing budget and also figure out their required growth rate. Let me try to break this down step by step.Starting with part 1: They have a total marketing budget of 50,000, which is 50 thousand dollars. They can split this into two channels, x1 and x2, such that x1 + x2 = 50. Each channel has its own customer acquisition function: A1(x1) = 100 ln(x1 + 1) and A2(x2) = 100 ln(x2 + 1). The goal is to maximize the total acquisition, which is A1 + A2.Hmm, so I need to maximize the sum of these two functions. Since both functions are in terms of natural logarithms, which are concave functions, I think this might be a problem where the maximum occurs at the point where the marginal gains from each channel are equal. That is, the derivative of A1 with respect to x1 should equal the derivative of A2 with respect to x2.Let me write that down. The total acquisition is:A_total = 100 ln(x1 + 1) + 100 ln(x2 + 1)Subject to the constraint x1 + x2 = 50.To maximize this, I can use the method of Lagrange multipliers or maybe substitution since it's a simple constraint.Let me try substitution. Since x2 = 50 - x1, I can substitute that into the total acquisition function:A_total = 100 ln(x1 + 1) + 100 ln(50 - x1 + 1)= 100 ln(x1 + 1) + 100 ln(51 - x1)Now, to find the maximum, I need to take the derivative of A_total with respect to x1 and set it equal to zero.Let's compute dA_total/dx1:dA_total/dx1 = 100 * [1/(x1 + 1)] + 100 * [-1/(51 - x1)]Set this equal to zero:100/(x1 + 1) - 100/(51 - x1) = 0Simplify:100/(x1 + 1) = 100/(51 - x1)Divide both sides by 100:1/(x1 + 1) = 1/(51 - x1)Cross-multiplied:51 - x1 = x1 + 1Combine like terms:51 - 1 = x1 + x150 = 2x1x1 = 25So x1 is 25, which means x2 is also 25. So the optimal allocation is to spend 25 thousand dollars on each channel.Wait, that seems straightforward. But let me verify if this is indeed a maximum. Since the second derivative would tell me about concavity.Compute the second derivative of A_total:d²A_total/dx1² = -100/(x1 + 1)^2 - 100/(51 - x1)^2Which is always negative because both terms are negative. So the function is concave, meaning this critical point is indeed a maximum.Okay, so part 1 seems solved with x1 = x2 = 25.Moving on to part 2: The startup's revenue function is given as R(t) = 5000 e^{0.1t} - 3000. They want to have a revenue of at least 10,000,000 in 5 years. But the function is supposed to change to R(t) = 5000 e^{rt} - 3000, and we need to find the required annual growth rate r.Wait, so originally, the exponent is 0.1t, which is a 10% growth rate. But they want to adjust it to r to achieve R(5) >= 10,000,000.So let's write the equation:5000 e^{r*5} - 3000 >= 10,000,000First, add 3000 to both sides:5000 e^{5r} >= 10,003,000Divide both sides by 5000:e^{5r} >= 10,003,000 / 5000Calculate the right side:10,003,000 / 5000 = 2000.6So e^{5r} >= 2000.6Take the natural logarithm of both sides:5r >= ln(2000.6)Compute ln(2000.6):I know that ln(2000) is approximately ln(2000). Let me recall that ln(1000) is about 6.9078, so ln(2000) is ln(2) + ln(1000) ≈ 0.6931 + 6.9078 ≈ 7.6009. Since 2000.6 is slightly more than 2000, ln(2000.6) is approximately 7.601.So 5r >= 7.601Therefore, r >= 7.601 / 5 ≈ 1.5202So r is approximately 1.5202, which is 152.02%.Wait, that seems extremely high. Let me check my calculations.Wait, the revenue function is R(t) = 5000 e^{rt} - 3000. They want R(5) >= 10,000,000.So, plugging t=5:5000 e^{5r} - 3000 >= 10,000,000Adding 3000:5000 e^{5r} >= 10,003,000Divide by 5000:e^{5r} >= 2000.6Yes, that's correct.Then ln(2000.6) ≈ 7.601So 5r = 7.601 => r ≈ 1.5202, which is 152.02%.Hmm, that's a very high growth rate. Let me see if I interpreted the problem correctly.Wait, is the revenue function R(t) = 5000 e^{rt} - 3000, or is it R(t) = 5000 e^{0.1t} - 3000, and they want to change the exponent to r? The problem says: "If the startup aims to have a revenue of at least 10,000,000 in 5 years, determine the required annual growth rate r (as a percentage) that they must achieve. Assume the growth rate impacts the exponent in the revenue function, changing it to R(t) = 5000 e^{rt} - 3000."So yes, they are changing the exponent from 0.1t to rt, so r is the new growth rate. So the calculation seems correct, but 152% seems too high. Maybe I made a mistake in the units? Let me check.Wait, is the revenue in dollars? The original function is 5000 e^{0.1t} - 3000. So 5000 is in thousands? Wait, no, the original problem says the marketing budget is in thousands, but the revenue function is given as 5000 e^{0.1t} - 3000. So 5000 is in dollars? Or is it in thousands?Wait, the problem says: "revenue R(t) over time t (in years) follows the function: R(t) = 5000 e^{0.1t} - 3000". So 5000 is in dollars, as is 3000. So the function is in dollars.They want R(5) >= 10,000,000 dollars.So plugging t=5:5000 e^{5r} - 3000 >= 10,000,000Yes, that's correct.So e^{5r} >= (10,000,000 + 3000)/5000 = 10,003,000 / 5000 = 2000.6So ln(2000.6) ≈ 7.601Thus, r ≈ 7.601 / 5 ≈ 1.5202, which is 152.02%.That's a growth rate of over 150% per year, which is extremely high. Maybe I misread the problem.Wait, let me check the problem statement again: "If the startup aims to have a revenue of at least 10,000,000 in 5 years, determine the required annual growth rate r (as a percentage) that they must achieve. Assume the growth rate impacts the exponent in the revenue function, changing it to R(t) = 5000 e^{rt} - 3000."So yes, it's correct. The exponential term is rt, so r is the annual growth rate in the exponent. So to get R(5) = 10,000,000, we need r ≈ 152%.Alternatively, maybe the original function is in thousands? Let me check.Wait, the problem says: "revenue R(t) over time t (in years) follows the function: R(t) = 5000 e^{0.1t} - 3000". It doesn't specify units, but given that in part 1, x is in thousands, but here it's just R(t). So perhaps R(t) is in thousands of dollars? Let me check.If R(t) is in thousands, then 5000 would be 5,000,000 dollars, which seems high. Wait, no, if R(t) is in thousands, then 5000 would be 5,000,000 dollars, which is 5 million. But the target is 10,000,000 dollars, which is 10 million. So if R(t) is in thousands, then 10,000,000 dollars is 10,000 thousand, which is 10,000.Wait, that doesn't make sense. Let me clarify.Wait, the problem says: "revenue R(t) over time t (in years) follows the function: R(t) = 5000 e^{0.1t} - 3000". It doesn't specify units, but in part 1, x is in thousands. So perhaps R(t) is in dollars, meaning 5000 is 5000 dollars, and 3000 is 3000 dollars. So R(t) is in dollars.But then, 5000 e^{0.1*5} - 3000 = 5000 e^{0.5} - 3000 ≈ 5000*1.6487 - 3000 ≈ 8243.5 - 3000 ≈ 5243.5 dollars. But they want R(5) >= 10,000,000 dollars, which is way higher. So that would require a much higher growth rate.Alternatively, maybe R(t) is in thousands of dollars. So 5000 is 5,000,000 dollars, and 3000 is 3,000,000 dollars. Then R(t) is in thousands.Wait, but 5000 e^{0.1t} - 3000 would be in thousands, so R(t) in thousands of dollars. So R(5) = 5000 e^{0.5} - 3000 ≈ 5000*1.6487 - 3000 ≈ 8243.5 - 3000 ≈ 5243.5 thousand dollars, which is 5,243,500 dollars. But they want 10,000,000 dollars, which is 10,000 thousand dollars. So R(5) needs to be 10,000.So if R(t) is in thousands, then:5000 e^{5r} - 3000 >= 10,000So 5000 e^{5r} >= 13,000Divide by 5000:e^{5r} >= 2.6Then ln(2.6) ≈ 0.9555So 5r >= 0.9555 => r >= 0.1911, which is 19.11%.That seems more reasonable. So perhaps I misinterpreted the units earlier.Wait, the problem didn't specify units for R(t), but in part 1, x is in thousands. So maybe R(t) is in thousands as well? Let me see.If R(t) is in thousands, then 5000 is 5,000,000 dollars, which is 5 million. But the target is 10,000,000 dollars, which is 10 million, so in thousands, that's 10,000.So let me recast the problem with R(t) in thousands:R(t) = 5000 e^{rt} - 3000They want R(5) >= 10,000 (thousand dollars)So:5000 e^{5r} - 3000 >= 10,000Add 3000:5000 e^{5r} >= 13,000Divide by 5000:e^{5r} >= 2.6Take ln:5r >= ln(2.6) ≈ 0.9555So r >= 0.9555 / 5 ≈ 0.1911, which is 19.11%.So approximately 19.11% annual growth rate.That makes more sense. So I think I initially misinterpreted the units. The problem didn't specify, but given that in part 1, x is in thousands, perhaps R(t) is also in thousands. So the correct answer is approximately 19.11%.Wait, but the problem says "revenue of at least 10,000,000". If R(t) is in thousands, then 10,000,000 dollars is 10,000 thousand, so R(t) needs to be 10,000. So yes, that's correct.Alternatively, if R(t) is in dollars, then 10,000,000 is 10 million, and the calculation would be as I did first, leading to 152%, which is unrealistic. So I think the correct interpretation is that R(t) is in thousands, so the answer is approximately 19.11%.Let me confirm:If R(t) is in thousands, then R(5) = 5000 e^{5r} - 3000 >= 10,000So 5000 e^{5r} >= 13,000e^{5r} >= 2.65r >= ln(2.6) ≈ 0.9555r >= 0.1911 or 19.11%Yes, that seems correct.So to summarize:1. Optimal allocation is x1 = x2 = 25 thousand dollars.2. Required annual growth rate is approximately 19.11%.I think that's it.Final Answer1. The optimal allocation is ( x_1 = boxed{25} ) and ( x_2 = boxed{25} ) thousand dollars.2. The required annual growth rate is ( boxed{19.11%} ).</think>