🧠:<think>Okay, so I'm trying to solve these two problems related to opening a chain of high-end tea houses. Let's start with the first one about market dynamics.The differential equation given is:[ frac{dC}{dt} = k C (M - C) - frac{C}{T} ]Hmm, this looks like a modified logistic equation. The standard logistic equation is:[ frac{dC}{dt} = k C (M - C) ]But here, there's an additional term, which is subtracting ( frac{C}{T} ). So, it's like the logistic growth is being opposed by a linear term in C. That probably represents some sort of customer attrition or loss.I need to find the explicit solution for C(t) given the initial condition C(0) = C0. Let me write down the equation again:[ frac{dC}{dt} = k C (M - C) - frac{C}{T} ]Maybe I can rewrite this equation to make it easier to solve. Let's factor out C from the right-hand side:[ frac{dC}{dt} = C left[ k(M - C) - frac{1}{T} right] ]So, it's a separable differential equation. Let me try to separate the variables. I can write:[ frac{dC}{C left[ k(M - C) - frac{1}{T} right]} = dt ]Hmm, the denominator on the left side is a bit complicated. Let me simplify it:First, distribute the k:[ k(M - C) = kM - kC ]So, the denominator becomes:[ kM - kC - frac{1}{T} ]Let me rearrange terms:[ -kC + (kM - frac{1}{T}) ]So, the equation is:[ frac{dC}{-kC + (kM - frac{1}{T})} = dt ]Wait, that might not be the most helpful. Maybe I can write it as:[ frac{dC}{dt} = -k C^2 + (kM - frac{1}{T}) C ]Yes, that's a quadratic in C. So, it's a Bernoulli equation, but maybe it can be solved using partial fractions.Let me write it as:[ frac{dC}{dt} = -k C^2 + (kM - frac{1}{T}) C ]Which can be rewritten as:[ frac{dC}{dt} = -k C^2 + (kM - frac{1}{T}) C ]So, this is a Riccati equation, but perhaps it can be transformed into a linear differential equation by substitution.Alternatively, let's try to write it as:[ frac{dC}{dt} = -k C^2 + (kM - frac{1}{T}) C ]Let me factor out C:[ frac{dC}{dt} = C left( -k C + (kM - frac{1}{T}) right) ]So, it's separable. Let's separate variables:[ frac{dC}{C left( -k C + (kM - frac{1}{T}) right)} = dt ]Let me make a substitution to simplify this integral. Let me set:Let me denote:Let me write the denominator as:[ -k C + (kM - frac{1}{T}) = -k left( C - left( M - frac{1}{kT} right) right) ]So, the denominator becomes:[ -k left( C - left( M - frac{1}{kT} right) right) ]So, the equation becomes:[ frac{dC}{C cdot (-k) left( C - left( M - frac{1}{kT} right) right)} = dt ]Simplify the constants:[ frac{dC}{-k C left( C - left( M - frac{1}{kT} right) right)} = dt ]Multiply both sides by -1:[ frac{dC}{k C left( C - left( M - frac{1}{kT} right) right)} = -dt ]Now, let me set:Let me denote ( A = M - frac{1}{kT} ). So, the equation becomes:[ frac{dC}{k C (C - A)} = -dt ]Now, this is a rational function, so I can use partial fractions to integrate the left side.Let me write:[ frac{1}{k C (C - A)} = frac{1}{k} left( frac{1}{A C} + frac{1}{C - A} right) ]Wait, let me check that. Let me suppose:[ frac{1}{C (C - A)} = frac{B}{C} + frac{D}{C - A} ]Multiply both sides by C (C - A):[ 1 = B (C - A) + D C ]Let me solve for B and D.Set C = 0:1 = B (-A) => B = -1/ASet C = A:1 = D A => D = 1/ASo, we have:[ frac{1}{C (C - A)} = frac{-1/A}{C} + frac{1/A}{C - A} ]Therefore,[ frac{1}{k C (C - A)} = frac{1}{k} left( frac{-1/A}{C} + frac{1/A}{C - A} right) ]So, the integral becomes:[ int frac{1}{k} left( frac{-1/A}{C} + frac{1/A}{C - A} right) dC = int -dt ]Simplify:[ frac{1}{k A} int left( frac{-1}{C} + frac{1}{C - A} right) dC = - int dt ]Compute the integrals:Left side:[ frac{1}{k A} left( -ln |C| + ln |C - A| right) + C_1 ]Right side:[ -t + C_2 ]Combine constants:[ frac{1}{k A} left( ln |C - A| - ln |C| right) = -t + C ]Where C is the constant of integration.Simplify the left side:[ frac{1}{k A} ln left| frac{C - A}{C} right| = -t + C ]Multiply both sides by ( k A ):[ ln left| frac{C - A}{C} right| = -k A t + C' ]Exponentiate both sides:[ left| frac{C - A}{C} right| = e^{-k A t + C'} = e^{C'} e^{-k A t} ]Let me denote ( e^{C'} = K ), a positive constant.So,[ frac{C - A}{C} = pm K e^{-k A t} ]But since we're dealing with a physical system where C is positive, we can drop the absolute value and consider the sign based on initial conditions.Let me write:[ frac{C - A}{C} = K e^{-k A t} ]Where K is a constant to be determined by initial conditions.Simplify the left side:[ 1 - frac{A}{C} = K e^{-k A t} ]Rearrange:[ frac{A}{C} = 1 - K e^{-k A t} ]Therefore,[ C = frac{A}{1 - K e^{-k A t}} ]Now, apply the initial condition C(0) = C0.At t = 0,[ C0 = frac{A}{1 - K} ]Solve for K:[ 1 - K = frac{A}{C0} ][ K = 1 - frac{A}{C0} ]So, substitute back into the expression for C(t):[ C(t) = frac{A}{1 - left(1 - frac{A}{C0}right) e^{-k A t}} ]Simplify the denominator:[ 1 - left(1 - frac{A}{C0}right) e^{-k A t} = 1 - e^{-k A t} + frac{A}{C0} e^{-k A t} ]Factor out e^{-k A t}:[ 1 - e^{-k A t} left(1 - frac{A}{C0}right) ]But let me just leave it as is for now.So, the solution is:[ C(t) = frac{A}{1 - left(1 - frac{A}{C0}right) e^{-k A t}} ]But remember that A was defined as:[ A = M - frac{1}{kT} ]So, substitute back:[ C(t) = frac{M - frac{1}{kT}}{1 - left(1 - frac{M - frac{1}{kT}}{C0}right) e^{-k (M - frac{1}{kT}) t}} ]Simplify the exponent:[ k (M - frac{1}{kT}) = kM - frac{1}{T} ]So, the exponent becomes:[ - (kM - frac{1}{T}) t ]So, the solution is:[ C(t) = frac{M - frac{1}{kT}}{1 - left(1 - frac{M - frac{1}{kT}}{C0}right) e^{-(kM - frac{1}{T}) t}} ]This seems a bit messy, but it's the explicit solution.Let me check if this makes sense. As t approaches infinity, the exponential term goes to zero, so C(t) approaches:[ frac{M - frac{1}{kT}}{1 - 0} = M - frac{1}{kT} ]Which is A, as expected. So, the carrying capacity is M - 1/(kT), which makes sense because the logistic term is being opposed by the linear term.If I didn't make any mistakes in the algebra, this should be the solution.Now, moving on to the second problem: supply chain optimization.We have a cost function:[ f(x) = alpha x + beta x^2 ]And a profit function:[ P(x) = gamma ln(1 + x) - f(x) ]Given a budget B, we need to determine the optimal x to maximize P(x) while maintaining the budget constraint.Wait, the budget constraint is given, but how is it related? Is the budget B the total amount we can spend on distribution, so f(x) ≤ B? Or is B the total tea leaves we can distribute? The problem says "given a budget B for distribution", so I think f(x) ≤ B.But the profit function is P(x) = γ ln(1 + x) - f(x). So, we need to maximize P(x) subject to f(x) ≤ B.Alternatively, maybe the budget is the total amount we can spend, so f(x) = B, and we need to choose x to maximize P(x) given f(x) = B.Wait, the problem says "determine the optimal amount of tea leaves x to distribute to maximize the tea house's profit while maintaining the budget constraint."So, probably, the budget is the total cost, so f(x) = B, and we need to maximize P(x) given that f(x) = B.But let me read again: "Given a budget B for distribution, determine the optimal amount of tea leaves x to distribute to maximize the tea house's profit while maintaining the budget constraint."So, it's a constrained optimization problem where the cost f(x) must be less than or equal to B, and we need to maximize P(x).But since P(x) = γ ln(1 + x) - f(x), and f(x) is increasing in x, there might be a trade-off between increasing x (which increases ln(1+x) but also increases f(x)).But since the budget is B, perhaps we can set f(x) = B, and then maximize P(x) with f(x) = B.Alternatively, if we have f(x) ≤ B, then the maximum P(x) would occur at the maximum x such that f(x) = B, because P(x) is increasing until a certain point and then decreasing.Wait, let's analyze P(x):P(x) = γ ln(1 + x) - (α x + β x^2)To find the maximum, we can take the derivative of P(x) with respect to x and set it equal to zero.But since we have a budget constraint f(x) ≤ B, the maximum could be either at the critical point of P(x) or at the boundary where f(x) = B.So, first, let's find the critical point of P(x):dP/dx = γ / (1 + x) - α - 2 β xSet this equal to zero:γ / (1 + x) - α - 2 β x = 0So,γ / (1 + x) = α + 2 β xThis is a nonlinear equation in x. Let's denote y = x + 1, so x = y - 1.Substitute:γ / y = α + 2 β (y - 1)Multiply both sides by y:γ = (α + 2 β (y - 1)) yExpand:γ = α y + 2 β y (y - 1)= α y + 2 β y^2 - 2 β yCombine like terms:γ = 2 β y^2 + (α - 2 β) yBring all terms to one side:2 β y^2 + (α - 2 β) y - γ = 0This is a quadratic equation in y:2 β y^2 + (α - 2 β) y - γ = 0We can solve for y using the quadratic formula:y = [ - (α - 2 β) ± sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )Since y = x + 1 must be positive, we take the positive root:y = [ - (α - 2 β) + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )Simplify:Let me compute the discriminant:D = (α - 2 β)^2 + 8 β γSo,y = [ -α + 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )Then,x = y - 1 = [ -α + 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β ) - 1Simplify:x = [ -α + 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) - 4 β ] / (4 β )= [ -α - 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )So, that's the critical point x_c.Now, we need to check if this x_c satisfies f(x_c) ≤ B.If f(x_c) ≤ B, then the optimal x is x_c.If f(x_c) > B, then the optimal x is the x that satisfies f(x) = B.So, we need to solve for x in f(x) = B:α x + β x^2 = BWhich is a quadratic equation:β x^2 + α x - B = 0Solutions:x = [ -α ± sqrt(α^2 + 4 β B) ] / (2 β )Again, since x must be positive, we take the positive root:x = [ -α + sqrt(α^2 + 4 β B) ] / (2 β )So, let me denote this x_b.So, the optimal x is:If x_c ≤ x_b, then x = x_c.Else, x = x_b.But wait, actually, since f(x) is increasing in x, if x_c is such that f(x_c) ≤ B, then x_c is feasible, and since it's the critical point, it's the maximum.If x_c is such that f(x_c) > B, then the maximum occurs at x_b.But to determine which case we are in, we need to compare f(x_c) with B.But this might be complicated, so perhaps we can express the optimal x as the minimum of x_c and x_b.But let me think differently.Alternatively, since we have a budget constraint f(x) ≤ B, we can use Lagrange multipliers to maximize P(x) subject to f(x) ≤ B.But since P(x) is concave (because the second derivative is negative), the maximum will occur either at the critical point or at the boundary.So, compute x_c, check if f(x_c) ≤ B. If yes, x = x_c. If not, x = x_b.But without knowing the specific values of α, β, γ, B, we can't compute numerically, but we can express the optimal x as:x* = min{ x_c, x_b }But actually, it's more precise to say:If f(x_c) ≤ B, then x* = x_c.Else, x* = x_b.So, the optimal x is the smaller of x_c and x_b.But let me verify this.Wait, when f(x_c) ≤ B, then x_c is within the budget, so it's the optimal point.If f(x_c) > B, then we can't afford x_c, so we have to choose the maximum x such that f(x) = B, which is x_b.Therefore, the optimal x is:x* = x_c if f(x_c) ≤ B,otherwise x* = x_b.But since we can't compute f(x_c) without knowing the parameters, perhaps we can express the optimal x in terms of these expressions.Alternatively, maybe we can combine the two conditions.But perhaps the problem expects us to set up the Lagrangian and find the optimal x.Let me try that approach.We need to maximize P(x) = γ ln(1 + x) - α x - β x^2Subject to f(x) = α x + β x^2 ≤ BUsing Lagrange multipliers, we set up:L = γ ln(1 + x) - α x - β x^2 - λ (α x + β x^2 - B)Take derivative with respect to x:dL/dx = γ / (1 + x) - α - 2 β x - λ (α + 2 β x) = 0And the constraint:α x + β x^2 = BSo, we have two equations:1. γ / (1 + x) - α - 2 β x - λ (α + 2 β x) = 02. α x + β x^2 = BFrom equation 1, we can solve for λ:λ = [ γ / (1 + x) - α - 2 β x ] / (α + 2 β x )But this seems similar to the earlier approach.Alternatively, if we assume that the maximum occurs at the interior point (i.e., f(x) < B), then we can set the derivative of P(x) to zero, which gives us x_c.If the maximum occurs at the boundary (f(x) = B), then x = x_b.Therefore, the optimal x is the minimum of x_c and x_b.But since x_c is the critical point, and x_b is the maximum x allowed by the budget, the optimal x is x_c if x_c ≤ x_b, else x_b.But without knowing the relationship between x_c and x_b, we can't say for sure, but we can express it as:x* = min{ x_c, x_b }But perhaps the problem expects us to find x in terms of the given parameters, so we can write:If the critical point x_c satisfies f(x_c) ≤ B, then x* = x_c.Otherwise, x* = x_b.But to express this explicitly, we can write:x* = [ -α - 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )if [ α x* + β x*^2 ] ≤ B,else x* = [ -α + sqrt(α^2 + 4 β B) ] / (2 β )But since this is a bit involved, perhaps the answer is to set up the Lagrangian and find the critical point, then check against the budget.Alternatively, maybe the optimal x is found by setting the derivative of P(x) equal to zero, and if that x is within the budget, that's the answer, else the maximum x allowed by the budget.But in any case, the optimal x is either x_c or x_b, depending on whether x_c is within the budget.So, to summarize, the optimal x is:x* = min{ x_c, x_b }Where x_c is the critical point found by setting dP/dx = 0, and x_b is the solution to f(x) = B.But perhaps the problem expects us to express x* in terms of the parameters, so we can write:x* = [ -α - 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )provided that α x* + β x*^2 ≤ B.Otherwise, x* = [ -α + sqrt(α^2 + 4 β B) ] / (2 β )But since the problem doesn't specify whether x_c is within the budget, we can present both possibilities.Alternatively, perhaps the optimal x is found by solving the equation γ / (1 + x) = α + 2 β x, which is the critical point, and then ensuring that f(x) ≤ B.If f(x) at x_c is less than or equal to B, then x_c is optimal. Otherwise, the optimal x is the solution to f(x) = B.So, the optimal x is:x* = [ -α - 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )if [ α x* + β x*^2 ] ≤ B,otherwise,x* = [ -α + sqrt(α^2 + 4 β B) ] / (2 β )But perhaps the problem expects us to express the optimal x in terms of the parameters, so we can write:The optimal amount of tea leaves x is given by:x = [ -α - 2 β + sqrt( (α - 2 β)^2 + 8 β γ ) ] / (4 β )provided that α x + β x^2 ≤ B. If not, then x is the solution to α x + β x^2 = B, which is:x = [ -α + sqrt(α^2 + 4 β B) ] / (2 β )So, that's the optimal x.But to make it more precise, perhaps we can write:The optimal x is the smaller of the two solutions x_c and x_b, where x_c is the critical point and x_b is the boundary point.Alternatively, since x_c might not always be feasible, the optimal x is:x* = min{ x_c, x_b }But in terms of the given parameters, we can express it as above.I think that's as far as I can go without specific numerical values.</think>

🧠:<think>Okay, so I have these two problems to solve for John's business. Let me take them one at a time.Starting with the first problem about the conservative apparel line. The profit function is given as ( P(x) = 4x^3 - 15x^2 + 18x - 7 ). I need to find the number of units ( x ) that maximizes the profit. Hmm, since it's a profit function, I think I need to find its maximum point. I remember from calculus that to find maxima or minima, I should take the derivative of the function and set it equal to zero. So, let me compute the first derivative of ( P(x) ).The derivative of ( 4x^3 ) is ( 12x^2 ), the derivative of ( -15x^2 ) is ( -30x ), the derivative of ( 18x ) is 18, and the derivative of the constant term ( -7 ) is 0. So, putting it all together, the first derivative ( P'(x) ) is:( P'(x) = 12x^2 - 30x + 18 )Now, I need to set this equal to zero and solve for ( x ):( 12x^2 - 30x + 18 = 0 )This is a quadratic equation. Let me try to simplify it. All coefficients are divisible by 6, so dividing each term by 6:( 2x^2 - 5x + 3 = 0 )Now, let's solve this quadratic equation. I can use the quadratic formula:( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Here, ( a = 2 ), ( b = -5 ), and ( c = 3 ). Plugging these values in:Discriminant ( D = (-5)^2 - 4*2*3 = 25 - 24 = 1 )So, the solutions are:( x = frac{5 pm sqrt{1}}{4} )Which gives:( x = frac{5 + 1}{4} = frac{6}{4} = 1.5 )and( x = frac{5 - 1}{4} = frac{4}{4} = 1 )So, the critical points are at ( x = 1 ) and ( x = 1.5 ). Now, I need to determine which of these gives a maximum profit.To do that, I can use the second derivative test. Let me compute the second derivative of ( P(x) ). The first derivative is ( 12x^2 - 30x + 18 ), so the second derivative ( P''(x) ) is:( P''(x) = 24x - 30 )Now, evaluate the second derivative at each critical point.First, at ( x = 1 ):( P''(1) = 24*1 - 30 = 24 - 30 = -6 )Since this is negative, the function is concave down at ( x = 1 ), which means it's a local maximum.Next, at ( x = 1.5 ):( P''(1.5) = 24*1.5 - 30 = 36 - 30 = 6 )This is positive, so the function is concave up at ( x = 1.5 ), indicating a local minimum.Therefore, the profit is maximized at ( x = 1 ). But wait, ( x ) represents the number of units in hundreds. So, 1 unit in hundreds is 100 units. Hmm, but let me double-check my calculations because 1.5 is also a critical point, and sometimes cubic functions can have different behaviors.Wait, the original profit function is a cubic, which typically has one local maximum and one local minimum. Since the leading coefficient is positive (4), as ( x ) approaches infinity, ( P(x) ) goes to positive infinity, and as ( x ) approaches negative infinity, it goes to negative infinity. So, the function will have a local maximum at ( x = 1 ) and a local minimum at ( x = 1.5 ). Therefore, the maximum profit occurs at ( x = 1 ).But just to make sure, maybe I should plug these values back into the original profit function to see which one gives a higher profit.Calculating ( P(1) ):( P(1) = 4*(1)^3 - 15*(1)^2 + 18*(1) - 7 = 4 - 15 + 18 - 7 = (4 - 15) + (18 - 7) = (-11) + 11 = 0 )Hmm, profit is zero at ( x = 1 ). Let me check ( P(1.5) ):First, convert 1.5 to a fraction for easier calculation: 1.5 = 3/2.( P(3/2) = 4*(27/8) - 15*(9/4) + 18*(3/2) - 7 )Calculating each term:- ( 4*(27/8) = (4/1)*(27/8) = (108)/8 = 13.5 )- ( -15*(9/4) = (-135)/4 = -33.75 )- ( 18*(3/2) = 27 )- ( -7 ) remains as is.Adding them up:13.5 - 33.75 + 27 - 7 = (13.5 - 33.75) + (27 - 7) = (-20.25) + 20 = -0.25So, ( P(1.5) = -0.25 ). That's a loss. So, at ( x = 1 ), profit is zero, and at ( x = 1.5 ), it's a loss. Hmm, that's interesting.Wait, maybe I made a mistake in interpreting the critical points. Let me check the first derivative again.Wait, the first derivative was ( 12x^2 - 30x + 18 ), which we simplified to ( 2x^2 - 5x + 3 = 0 ). Solving that gave us x = 1 and x = 1.5. So, that seems correct.But the profit at x = 1 is zero, and at x = 1.5, it's negative. So, is there a higher profit somewhere else?Wait, maybe I should check the behavior of the function beyond these points. Since it's a cubic, it tends to infinity as x increases. So, maybe after a certain point, the profit becomes positive again.But since we are dealing with units produced and sold, x can't be negative. So, let's see what happens as x increases beyond 1.5.Let me compute P(2):( P(2) = 4*(8) - 15*(4) + 18*(2) - 7 = 32 - 60 + 36 - 7 = (32 - 60) + (36 - 7) = (-28) + 29 = 1 )So, P(2) = 1, which is positive. So, at x = 2, profit is 1. So, it seems that after x = 1.5, the profit starts increasing again.Wait, but the critical points are at x = 1 (local max) and x = 1.5 (local min). So, the function increases up to x = 1, then decreases until x = 1.5, then increases again beyond that.But since the profit at x = 1 is zero, and at x = 1.5 is negative, and at x = 2 is positive, it seems that the maximum profit occurs at x = 1, but it's zero. Hmm, that doesn't make much sense because if you can make a profit at x = 2, why would the maximum be at x = 1?Wait, maybe I made a mistake in calculating the profit at x = 1. Let me double-check:( P(1) = 4*(1)^3 - 15*(1)^2 + 18*(1) - 7 = 4 - 15 + 18 - 7 )Calculating step by step:4 - 15 = -11-11 + 18 = 77 - 7 = 0Yes, that's correct. So, P(1) is indeed zero.Wait, maybe the function is such that the maximum profit is zero, which is at x = 1, and beyond that, it goes negative, but then starts increasing again. But at x = 2, it's positive again. So, is there another critical point beyond x = 1.5?Wait, no, because the first derivative is a quadratic, which only has two roots. So, the function only has two critical points: a local max at x = 1 and a local min at x = 1.5. Beyond x = 1.5, the function increases towards infinity.So, in that case, the maximum profit occurs at x = 1, but it's zero. However, at x = 2, the profit is positive, which is higher than zero. So, that seems contradictory.Wait, maybe I need to consider the behavior of the function. Since it's a cubic with a positive leading coefficient, it goes to negative infinity as x approaches negative infinity and positive infinity as x approaches positive infinity. So, the function must have a local maximum at x = 1 and a local minimum at x = 1.5.But in the context of the problem, x represents the number of units produced and sold per month, so x must be a positive number. So, the function is increasing from x = 0 up to x = 1, then decreasing from x = 1 to x = 1.5, then increasing again beyond x = 1.5.So, if at x = 1, the profit is zero, and at x = 1.5, it's negative, and at x = 2, it's positive, that suggests that the function crosses the x-axis somewhere between x = 1.5 and x = 2.Wait, but the question is to find the number of units that maximizes the profit. So, even though at x = 2, the profit is positive, it's still increasing beyond that. So, does that mean that the maximum profit is actually at infinity? But that can't be practical.Wait, maybe I made a mistake in interpreting the problem. Let me check the original function again: ( P(x) = 4x^3 - 15x^2 + 18x - 7 ). So, it's a cubic function, which as x increases, the profit goes to infinity. So, technically, there is no maximum profit; it can be increased indefinitely by producing more units. But that doesn't make sense in a business context because there must be constraints like market demand, production capacity, etc.But the problem doesn't mention any constraints, so mathematically, the function has a local maximum at x = 1, but beyond that, it increases again. So, the local maximum is at x = 1, but the global maximum is at infinity. However, since the problem asks for the number of units that maximizes the profit, and given that it's a cubic, it's more likely that the question expects the local maximum, which is at x = 1.But in that case, the profit is zero, which seems odd. Maybe I need to check if I did the second derivative test correctly.Wait, the second derivative at x = 1 was -6, which is negative, so it's a local maximum. At x = 1.5, it's positive, so local minimum. So, mathematically, x = 1 is a local maximum, but the function continues to increase beyond x = 1.5. So, in the context of the problem, maybe John should produce more units beyond 1.5 to increase profit, but the local maximum is at x = 1.Hmm, this is confusing. Maybe I need to consider the economic interpretation. If at x = 1, the profit is zero, and at x = 1.5, it's negative, and at x = 2, it's positive, then the maximum profit in the feasible region (where profit is positive) would be at x approaching infinity, but that's not practical.Alternatively, maybe the function is supposed to have a global maximum at x = 1, but given the cubic nature, that's not the case. Perhaps there's a mistake in the problem setup or my calculations.Wait, let me double-check the derivative:Original function: ( P(x) = 4x^3 - 15x^2 + 18x - 7 )First derivative: ( P'(x) = 12x^2 - 30x + 18 ). That seems correct.Setting it equal to zero: ( 12x^2 - 30x + 18 = 0 ). Dividing by 6: ( 2x^2 - 5x + 3 = 0 ). Solutions: x = 1 and x = 1.5. Correct.Second derivative: ( P''(x) = 24x - 30 ). At x = 1: 24 - 30 = -6. At x = 1.5: 36 - 30 = 6. Correct.So, mathematically, x = 1 is a local maximum, but the function increases beyond x = 1.5. So, in the context of the problem, maybe John should produce at x = 1 to avoid incurring losses, but if he can handle the initial loss, he can increase production to x = 2 and beyond to make a profit.But the question specifically asks for the number of units that maximizes the profit. Since the function doesn't have a global maximum, but only a local maximum at x = 1, I think the answer is x = 1, even though the profit is zero there.Alternatively, maybe the problem expects us to consider the feasible region where profit is positive. So, beyond x = 1.5, the profit becomes positive again, and since the function increases without bound, the maximum profit is unbounded. But that doesn't make sense in a real-world scenario.Wait, perhaps I need to consider the profit function more carefully. Let me plot it mentally. At x = 0, P(0) = -7. At x = 1, P(1) = 0. At x = 1.5, P(1.5) = -0.25. At x = 2, P(2) = 1. So, it goes from -7 at x=0, up to 0 at x=1, down to -0.25 at x=1.5, then up to 1 at x=2, and continues increasing.So, the function has a local maximum at x=1 (profit=0), then a local minimum at x=1.5 (profit=-0.25), then increases again. So, the maximum profit in the positive region is achieved as x approaches infinity, but that's not practical.Therefore, the only local maximum is at x=1, which gives zero profit. So, perhaps John should produce 100 units to break even, but beyond that, he incurs a loss until x=1.5, after which he starts making a profit again.But the question is to find the number of units that maximizes the profit. So, if we consider the entire domain, the function doesn't have a global maximum, but if we consider only up to the local maximum, then x=1 is the answer.Alternatively, maybe the problem expects us to consider the maximum profit in the feasible region where profit is positive. But since the function increases beyond x=1.5, the maximum profit is unbounded. So, perhaps the answer is that there is no maximum profit, but the function increases indefinitely.But the problem says "find the number of units x that maximizes the profit," so I think it expects the local maximum, which is at x=1.So, tentatively, I think the answer is x=1, which is 100 units, and it's a maximum because the second derivative is negative there.Now, moving on to the second problem about the patriotic home décor line. The price per unit is given by ( p = 50 - 2d ), where d is the demand. The production cost per unit is ( C(d) = 10 + d^2 ). We need to find the price per unit that maximizes John's profit.First, let's recall that profit is equal to total revenue minus total cost. So, we need to express profit in terms of d, then find its maximum.Total revenue (R) is price per unit (p) multiplied by the number of units sold (d). So, R = p * d.Given p = 50 - 2d, so R = (50 - 2d) * d = 50d - 2d^2.Total cost (C_total) is the production cost per unit multiplied by the number of units, so C_total = C(d) * d = (10 + d^2) * d = 10d + d^3.Therefore, profit (π) is R - C_total:π = (50d - 2d^2) - (10d + d^3) = 50d - 2d^2 - 10d - d^3 = (50d - 10d) + (-2d^2) + (-d^3) = 40d - 2d^2 - d^3.So, the profit function is π(d) = -d^3 - 2d^2 + 40d.To find the maximum profit, we need to take the derivative of π with respect to d, set it equal to zero, and solve for d.First derivative π'(d):π'(d) = derivative of (-d^3) is -3d^2, derivative of (-2d^2) is -4d, derivative of 40d is 40. So,π'(d) = -3d^2 - 4d + 40.Set this equal to zero:-3d^2 - 4d + 40 = 0.Multiply both sides by -1 to make it easier:3d^2 + 4d - 40 = 0.Now, solve this quadratic equation. Using the quadratic formula:d = [-b ± sqrt(b^2 - 4ac)] / (2a)Here, a = 3, b = 4, c = -40.Discriminant D = 4^2 - 4*3*(-40) = 16 + 480 = 496.So,d = [-4 ± sqrt(496)] / (2*3) = [-4 ± sqrt(16*31)] / 6 = [-4 ± 4*sqrt(31)] / 6.Simplify:Factor out 4 in numerator:d = [4*(-1 ± sqrt(31))]/6 = [2*(-1 ± sqrt(31))]/3.So, two solutions:d = [2*(-1 + sqrt(31))]/3 and d = [2*(-1 - sqrt(31))]/3.Since demand d cannot be negative, we discard the negative solution. So,d = [2*(-1 + sqrt(31))]/3.Compute sqrt(31): approximately 5.56776.So,d ≈ [2*(-1 + 5.56776)] / 3 = [2*(4.56776)] / 3 ≈ (9.13552)/3 ≈ 3.045.So, d ≈ 3.045 units.But since d represents demand, which is in units, and the price is given per unit, we need to find the price per unit p that corresponds to this d.Given p = 50 - 2d, so plugging d ≈ 3.045:p ≈ 50 - 2*(3.045) ≈ 50 - 6.09 ≈ 43.91.So, approximately 43.91 per unit.But let's do this more accurately without approximating sqrt(31).sqrt(31) is irrational, so we can leave it in exact form or compute a more precise decimal.But for the purposes of this problem, maybe we can express the exact value.So, d = [2*(-1 + sqrt(31))]/3.Therefore, p = 50 - 2d = 50 - 2*[2*(-1 + sqrt(31))]/3 = 50 - [4*(-1 + sqrt(31))]/3.Simplify:p = 50 + [4*(1 - sqrt(31))]/3.But that might not be the most simplified form. Alternatively, let's compute it step by step.Alternatively, let's express p in terms of d:p = 50 - 2d.We have d = [2*(-1 + sqrt(31))]/3.So,p = 50 - 2*[2*(-1 + sqrt(31))]/3 = 50 - [4*(-1 + sqrt(31))]/3.Distribute the 4:p = 50 - [ -4 + 4sqrt(31) ] / 3 = 50 + 4/3 - (4sqrt(31))/3.Combine terms:50 + 4/3 = (150/3 + 4/3) = 154/3.So,p = 154/3 - (4sqrt(31))/3 = (154 - 4sqrt(31))/3.We can factor out 2:p = 2*(77 - 2sqrt(31))/3.But maybe it's better to leave it as (154 - 4sqrt(31))/3.Alternatively, we can rationalize or approximate.But perhaps the problem expects an exact value, so we can leave it in terms of sqrt(31).Alternatively, if we need a decimal approximation, let's compute it more accurately.sqrt(31) ≈ 5.5677643628.So,4sqrt(31) ≈ 4*5.5677643628 ≈ 22.271057451.So,154 - 22.271057451 ≈ 131.72894255.Divide by 3:≈ 131.72894255 / 3 ≈ 43.909647516.So, approximately 43.91 per unit.But let's check if this is indeed a maximum. We can use the second derivative test.First, compute the second derivative of π(d):π'(d) = -3d^2 - 4d + 40.So, π''(d) = -6d - 4.Evaluate π''(d) at d ≈ 3.045:π''(3.045) ≈ -6*(3.045) - 4 ≈ -18.27 - 4 ≈ -22.27.Since this is negative, the function is concave down at this point, indicating a local maximum. So, this is indeed the price that maximizes profit.Therefore, the optimal price per unit is approximately 43.91, or exactly (154 - 4sqrt(31))/3 dollars.But let me check if I did everything correctly.Starting from the profit function:π = R - C = (50d - 2d^2) - (10d + d^3) = 40d - 2d^2 - d^3.First derivative: π' = 40 - 4d - 3d^2.Wait, hold on, I think I made a mistake in computing the first derivative earlier.Wait, π = -d^3 - 2d^2 + 40d.So, derivative is:π' = -3d^2 - 4d + 40.Yes, that's correct.Setting to zero: -3d^2 - 4d + 40 = 0.Multiply by -1: 3d^2 + 4d - 40 = 0.Solutions: d = [-4 ± sqrt(16 + 480)] / 6 = [-4 ± sqrt(496)] / 6.sqrt(496) = sqrt(16*31) = 4sqrt(31).So, d = [-4 ± 4sqrt(31)] / 6 = [ -4 + 4sqrt(31) ] / 6 or [ -4 - 4sqrt(31) ] / 6.Simplify:Factor out 4 in numerator:[4(-1 + sqrt(31))]/6 = [2(-1 + sqrt(31))]/3.Yes, that's correct.So, d ≈ 3.045.Then, p = 50 - 2d ≈ 50 - 6.09 ≈ 43.91.Yes, that seems correct.So, the optimal price per unit is approximately 43.91.But to express it exactly, it's (154 - 4sqrt(31))/3, which simplifies to approximately 43.91.So, I think that's the answer.Final Answer1. The number of units that maximizes the profit is boxed{1} hundred units.2. The price per unit that maximizes profit is boxed{frac{154 - 4sqrt{31}}{3}} dollars.</think>

🧠:<think>Alright, so I have this problem where I need to help allocate resources to three schools to maximize their test score improvements. The resources are funding and teacher training hours. The first part is with given constants a=2 and b=3, and the second part introduces a penalty based on initial scores. Let me try to break this down step by step.Starting with part 1. The goal is to maximize the total test score improvement across all three schools. The improvement function is given by I(F, H) = 2*ln(F + 1) + 3*sqrt(H). We have a total funding of 100,000, which is 100 in thousands, and 400 hours of training. These need to be distributed among the three schools.So, we need to allocate F1, F2, F3 such that F1 + F2 + F3 = 100, and H1 + H2 + H3 = 400. Each school i has its own Fi and Hi. The total improvement is the sum of I(Fi, Hi) for i=1,2,3.Since the problem is about maximizing the sum, and the function is additive, I think we can approach this using calculus, specifically Lagrange multipliers, because we have constraints on the total funding and training hours.Let me denote the total improvement as:Total I = 2*ln(F1 + 1) + 3*sqrt(H1) + 2*ln(F2 + 1) + 3*sqrt(H2) + 2*ln(F3 + 1) + 3*sqrt(H3)We need to maximize this subject to F1 + F2 + F3 = 100 and H1 + H2 + H3 = 400.To use Lagrange multipliers, we can set up the Lagrangian function:L = 2*ln(F1 + 1) + 3*sqrt(H1) + 2*ln(F2 + 1) + 3*sqrt(H2) + 2*ln(F3 + 1) + 3*sqrt(H3) - λ*(F1 + F2 + F3 - 100) - μ*(H1 + H2 + H3 - 400)Then, we take partial derivatives with respect to each variable and set them equal to zero.Let's compute the partial derivatives for F1:dL/dF1 = 2/(F1 + 1) - λ = 0 => 2/(F1 + 1) = λSimilarly, for F2 and F3:2/(F2 + 1) = λ2/(F3 + 1) = λSo, this implies that 2/(F1 + 1) = 2/(F2 + 1) = 2/(F3 + 1) = λWhich means that F1 + 1 = F2 + 1 = F3 + 1 = 2/λTherefore, F1 = F2 = F3 = 2/λ - 1Similarly, for the H variables:dL/dH1 = (3)/(2*sqrt(H1)) - μ = 0 => 3/(2*sqrt(H1)) = μSame for H2 and H3:3/(2*sqrt(H2)) = μ3/(2*sqrt(H3)) = μSo, sqrt(H1) = sqrt(H2) = sqrt(H3) = 3/(2μ)Therefore, H1 = H2 = H3 = (3/(2μ))^2So, from this, we can see that all schools should receive the same amount of funding and the same number of training hours. Because the marginal improvement per unit funding and per unit training is the same across all schools.So, since F1 = F2 = F3, and total funding is 100, each school gets 100/3 ≈ 33.333 thousand dollars.Similarly, each school gets 400/3 ≈ 133.333 hours of training.Wait, but let me verify if this is correct. Because the functions are concave, right? The ln function is concave, and the sqrt function is also concave. So, the total improvement function is concave, which means that the maximum is achieved when the marginal benefits are equal across all schools.Therefore, equal allocation is optimal.So, the optimal allocation is F1 = F2 = F3 = 100/3 ≈ 33.333, and H1 = H2 = H3 = 400/3 ≈ 133.333.But let me double-check the math.From the partial derivatives, for each Fi, 2/(Fi + 1) = λ, so all Fi + 1 are equal, so Fi are equal.Similarly, for each Hi, 3/(2*sqrt(Hi)) = μ, so sqrt(Hi) are equal, so Hi are equal.Therefore, yes, equal allocation is optimal.So, that's part 1.Moving on to part 2. Now, the improvement function is modified to include a penalty based on initial scores: I(F, H, S) = 2*ln(F + 1) + 3*sqrt(H) - 0.1*S.Given S1=60, S2=70, S3=80, and c=0.1.So, the total improvement is sum over i of [2*ln(Fi + 1) + 3*sqrt(Hi) - 0.1*Si].So, the total improvement is:[2*ln(F1 + 1) + 3*sqrt(H1) - 6] + [2*ln(F2 + 1) + 3*sqrt(H2) - 7] + [2*ln(F3 + 1) + 3*sqrt(H3) - 8]Simplify: 2*ln(F1 + 1) + 3*sqrt(H1) + 2*ln(F2 + 1) + 3*sqrt(H2) + 2*ln(F3 + 1) + 3*sqrt(H3) - 21So, the total improvement is the same as before minus 21. But since we're maximizing, the -21 is a constant, so the allocation of resources should be the same as in part 1, right?Wait, no. Because the penalty is per school, it's subtracted based on their initial scores. So, the total improvement is the sum of individual improvements, each of which has a different penalty.Wait, but in the Lagrangian, the penalty is subtracted, so when taking partial derivatives with respect to Fi and Hi, the penalty doesn't affect the derivatives because it's a constant with respect to Fi and Hi.So, the partial derivatives for Fi and Hi remain the same as in part 1, meaning that the optimal allocation is still equal funding and equal training hours across all schools.But wait, let me think again. The penalty is subtracted based on the school's initial score, so schools with higher initial scores have a larger penalty. But since the penalty doesn't depend on the resources allocated, it's just a fixed subtraction.Therefore, when maximizing the total improvement, the penalties are just constants, so they don't affect the allocation of resources. The allocation should still be equal across all schools.But that seems counterintuitive. Maybe I'm missing something.Wait, no. The penalty is subtracted for each school, but it's a fixed amount based on their initial score. So, for school 1, the penalty is 0.1*60=6, school 2 is 7, school 3 is 8. So, the total penalty is 6+7+8=21, which is a constant. So, when maximizing the total improvement, it's equivalent to maximizing the sum of the positive terms minus 21.Therefore, the allocation of resources (funding and training) doesn't depend on the penalties because the penalties are constants. So, the optimal allocation remains the same as in part 1: equal funding and equal training hours.But wait, that doesn't seem right. Because the penalties are different for each school, maybe we should allocate more resources to schools with higher penalties to offset them? But no, because the penalty is fixed and doesn't depend on resources. The resources only affect the positive terms, so to maximize the total, we should allocate resources to maximize the sum of the positive terms, regardless of the penalties.Therefore, the allocation strategy doesn't change. We still allocate equal funding and equal training hours to each school.But let me think again. Suppose we have two schools, one with a high penalty and one with a low penalty. If we allocate more resources to the high penalty school, does that help? No, because the penalty is fixed. The resources only add to the improvement, so to maximize the total, we should allocate resources where the marginal gain is highest, which is the same across all schools because the functions are identical except for the penalties, which are constants.Therefore, the optimal allocation remains equal across all schools.Wait, but in part 2, the improvement function for each school is I(F, H, S) = 2*ln(F + 1) + 3*sqrt(H) - 0.1*S. So, for each school, the improvement is their own function minus their own penalty. So, when summing up, the total improvement is the sum of the positive terms minus the sum of the penalties.Therefore, the allocation of resources should focus on maximizing the sum of the positive terms, which is the same as in part 1. The penalties are just subtracted, so they don't affect the allocation.Therefore, the resource allocation strategy doesn't change. We still allocate equal funding and equal training hours to each school.But wait, maybe I'm wrong. Let me consider the Lagrangian again. The Lagrangian would be:L = sum [2*ln(Fi + 1) + 3*sqrt(Hi) - 0.1*Si] - λ*(sum Fi - 100) - μ*(sum Hi - 400)Taking partial derivatives with respect to Fi:dL/dFi = 2/(Fi + 1) - λ = 0Similarly for Hi:dL/dHi = 3/(2*sqrt(Hi)) - μ = 0The partial derivatives with respect to Si are -0.1, but since Si are constants (given), they don't affect the optimization. Therefore, the optimal allocation is the same as in part 1.So, the resource allocation doesn't change. We still allocate F1=F2=F3=100/3 and H1=H2=H3=400/3.But wait, let me think about the intuition. If a school has a higher initial score, it has a higher penalty, but the resources allocated don't affect the penalty. So, to maximize the total improvement, we should focus on maximizing the sum of the positive terms, which is the same as before. Therefore, equal allocation is still optimal.So, in conclusion, for part 2, the resource allocation remains the same as in part 1.But wait, maybe I'm missing something. Let me consider if the penalty affects the marginal gain. Suppose we have two schools, one with a high penalty and one with a low penalty. If we allocate more resources to the high penalty school, does that help? No, because the penalty is fixed. The resources only add to the improvement, so to maximize the total, we should allocate resources where the marginal gain is highest, which is the same across all schools because the functions are identical except for the penalties, which are constants.Therefore, the optimal allocation remains equal across all schools.So, to summarize:1. For part 1, allocate F1=F2=F3=100/3≈33.333 and H1=H2=H3=400/3≈133.333.2. For part 2, the allocation remains the same because the penalties are constants and don't affect the resource allocation.But wait, let me check the math again. The Lagrangian in part 2 includes the penalties, but when taking derivatives with respect to Fi and Hi, the penalties disappear because they are constants. Therefore, the optimal allocation is the same as in part 1.Yes, that seems correct.</think>

🧠:<think>Alright, so I have two problems to solve here related to a hair care brand's promotional campaign. Let me tackle them one by one.Starting with the first problem: The brand representative noticed that for every 1,000 views, there's a 5% increase in sales, up to a maximum of 60%. The influencer's post is expected to follow a Poisson distribution with a mean of 20,000 views. I need to calculate the expected percentage increase in sales.Hmm, okay. So, first, let's understand the relationship between views and sales. For every 1,000 views, it's a 5% increase. So, if the influencer gets, say, 2,000 views, that would be a 10% increase, right? And this goes on up to 60%. So, the maximum number of views that would give the 60% increase would be 60% / 5% = 12, so 12,000 views. Wait, no, that's not quite right. Because it's 5% per 1,000 views, so 1,000 views = 5%, so 12,000 views would be 60%. But the mean number of views is 20,000, which is higher than 12,000. So, does that mean that beyond 12,000 views, the sales increase doesn't go beyond 60%? So, the sales increase is capped at 60%.Therefore, the percentage increase in sales is a function of the number of views, but it's capped. So, if the number of views is less than or equal to 12,000, the percentage increase is (views / 1,000) * 5%. If it's more than 12,000, it's just 60%.But the views follow a Poisson distribution with a mean of 20,000. So, I need to calculate the expected value of the percentage increase in sales, considering that beyond 12,000 views, the increase is capped at 60%.So, the expected percentage increase E[P] is equal to the expected value of min(5% * (views / 1,000), 60%). Since views are Poisson distributed with λ = 20,000.Wait, but Poisson distributions are typically for counts, and here we're dealing with views, which is a count, so that makes sense. The mean is 20,000, so the variance is also 20,000.But calculating the expectation of a function of a Poisson random variable can be tricky. The expectation of min(5% * (X / 1,000), 60%) where X ~ Poisson(20,000). Hmm.Alternatively, maybe it's easier to model this as a piecewise function. Let me define the percentage increase P as:P = 0.05 * (X / 1000) if X <= 12,000P = 0.60 if X > 12,000So, E[P] = E[0.05 * (X / 1000) | X <= 12,000] * P(X <= 12,000) + E[0.60 | X > 12,000] * P(X > 12,000)But since 0.60 is a constant, the second term simplifies to 0.60 * P(X > 12,000).So, E[P] = E[0.00005 * X | X <= 12,000] * P(X <= 12,000) + 0.60 * P(X > 12,000)Wait, 0.05 / 1000 is 0.00005? Wait, no. 5% is 0.05, and per 1,000 views, so 0.05 * (X / 1000) = 0.00005 * X. Yeah, that's correct.But calculating E[0.00005 * X | X <= 12,000] is the same as 0.00005 * E[X | X <= 12,000].So, E[P] = 0.00005 * E[X | X <= 12,000] * P(X <= 12,000) + 0.60 * P(X > 12,000)But calculating E[X | X <= 12,000] for a Poisson distribution with λ = 20,000 is going to be complicated because 12,000 is much less than the mean of 20,000. So, the probability that X <= 12,000 is going to be very small, and the expectation E[X | X <= 12,000] is going to be less than 12,000.But wait, maybe I can approximate the Poisson distribution with a normal distribution since λ is large (20,000). The Poisson distribution can be approximated by a normal distribution with mean μ = λ = 20,000 and variance σ² = λ = 20,000, so σ = sqrt(20,000) ≈ 141.421.So, let's use the normal approximation. Then, X ~ N(20,000, 141.421²).We can standardize the variable:Z = (X - 20,000) / 141.421We need to find P(X <= 12,000) = P(Z <= (12,000 - 20,000)/141.421) = P(Z <= -5.65685)Looking at standard normal tables, P(Z <= -5.65685) is practically 0. It's way beyond the typical table values. So, essentially, P(X <= 12,000) ≈ 0.Similarly, E[X | X <= 12,000] is going to be a very small value, but since the probability is almost 0, the contribution of the first term to E[P] is negligible.Therefore, E[P] ≈ 0.60 * 1 = 0.60, or 60%.Wait, but that seems counterintuitive. Because the mean is 20,000, which is way above 12,000, so almost all the probability mass is beyond 12,000, so the expected percentage increase is just 60%.But let me verify. If the views are Poisson distributed with mean 20,000, the probability that views are less than or equal to 12,000 is practically zero because 12,000 is more than 5 standard deviations below the mean. So, yes, the probability is negligible.Therefore, the expected percentage increase in sales is 60%.Wait, but let me think again. Is the relationship linear up to 12,000 views, and then capped? So, if the views are 20,000, which is above 12,000, then the sales increase is 60%. So, since the mean is 20,000, which is above 12,000, the expected percentage increase is 60%.But wait, is that the case? Because the views could be less than 12,000, but with such a low probability. So, the expectation would be 60% times the probability that views are above 12,000 plus a lower value times the probability that views are below 12,000. But since the probability that views are below 12,000 is practically zero, the expectation is approximately 60%.Therefore, the expected percentage increase in sales is 60%.Wait, but let me think again. Maybe I'm oversimplifying. Because even though the probability that X <= 12,000 is practically zero, the expectation E[P] is not necessarily 60%. Because E[P] = E[min(0.05*(X/1000), 0.60)]. So, if X is a random variable, then E[min(aX, b)] is not necessarily min(aE[X], b). So, I can't just say E[P] = min(0.05*(E[X]/1000), 0.60). Because E[min(aX, b)] <= min(aE[X], b). So, in this case, since aE[X]/1000 = 0.05*(20,000)/1000 = 1, which is 100%, but the cap is 60%, so E[P] <= 60%. But in reality, since the probability that X <= 12,000 is practically zero, E[P] is approximately 60%.Wait, but let me think about it differently. If X is Poisson(20,000), then P(X <= 12,000) is practically zero, so P(X > 12,000) is practically 1. Therefore, E[P] = E[0.05*(X/1000) | X > 12,000] * P(X > 12,000) + E[0.05*(X/1000) | X <= 12,000] * P(X <= 12,000). But since P(X <= 12,000) is practically zero, E[P] ≈ E[0.05*(X/1000) | X > 12,000]. But wait, no, because for X > 12,000, P is capped at 60%, so E[P] = 60% * P(X > 12,000) + E[0.05*(X/1000) | X <= 12,000] * P(X <= 12,000). Since P(X > 12,000) ≈ 1, E[P] ≈ 60%.Wait, but is that correct? Because for X > 12,000, P is 60%, so E[P] = 60% * P(X > 12,000) + E[P | X <= 12,000] * P(X <= 12,000). Since P(X <= 12,000) is practically zero, E[P] ≈ 60%.Yes, that seems correct. So, the expected percentage increase in sales is 60%.Wait, but let me check with a simpler case. Suppose the mean was 10,000 views. Then, the expected percentage increase would be 0.05*(10,000/1000) = 50%, right? Because 10,000 views would give a 50% increase. But if the mean is 20,000, which is above 12,000, then the expected increase is 60%.So, yes, that makes sense.Therefore, the answer to the first problem is 60%.Now, moving on to the second problem: Using a logistic growth model to predict the saturation point of product sales over time. The logistic growth function is given by S(t) = L / (1 + A e^{-kt}), where S(t) is sales at time t, L is the carrying capacity, A is a constant, e is the base of the natural logarithm, and k is the growth rate.Given that initial sales S(0) are 10% of L, and sales increase to 50% of L in 6 months. We need to find the values of A and k.Alright, let's start by plugging in the initial condition. At t = 0, S(0) = 0.10 L.So, S(0) = L / (1 + A e^{0}) = L / (1 + A) = 0.10 L.Therefore, 1 / (1 + A) = 0.10So, 1 + A = 1 / 0.10 = 10Therefore, A = 10 - 1 = 9.So, A is 9.Now, we need to find k. We know that at t = 6 months, S(6) = 0.50 L.So, plugging into the logistic equation:0.50 L = L / (1 + 9 e^{-6k})Divide both sides by L:0.50 = 1 / (1 + 9 e^{-6k})Take reciprocals:2 = 1 + 9 e^{-6k}Subtract 1:1 = 9 e^{-6k}Divide both sides by 9:1/9 = e^{-6k}Take natural logarithm of both sides:ln(1/9) = -6kSimplify ln(1/9) = -ln(9)So, -ln(9) = -6kMultiply both sides by -1:ln(9) = 6kTherefore, k = ln(9) / 6We can compute ln(9). Since 9 = 3^2, ln(9) = 2 ln(3) ≈ 2 * 1.0986 ≈ 2.1972So, k ≈ 2.1972 / 6 ≈ 0.3662 per month.But let's keep it exact. Since ln(9) = 2 ln(3), so k = (2 ln 3)/6 = (ln 3)/3.So, k = (ln 3)/3.Alternatively, we can write it as k = ln(3^(1/3)) = ln(∛3).But I think expressing it as (ln 3)/3 is fine.So, to summarize:A = 9k = (ln 3)/3Let me double-check the calculations.At t = 0:S(0) = L / (1 + 9 e^{0}) = L / 10 = 0.10 L. Correct.At t = 6:S(6) = L / (1 + 9 e^{-6k}) = L / (1 + 9 e^{-6*(ln3)/3}) = L / (1 + 9 e^{-2 ln3}) = L / (1 + 9*(e^{ln3})^{-2}) = L / (1 + 9*(3)^{-2}) = L / (1 + 9*(1/9)) = L / (1 + 1) = L / 2 = 0.50 L. Correct.Yes, that checks out.So, the values are A = 9 and k = (ln 3)/3.Final Answer1. The expected percentage increase in sales is boxed{60%}.2. The values are ( A = boxed{9} ) and ( k = boxed{dfrac{ln 3}{3}} ).</think>