🧠:<think>Okay, so I have this problem about solving a logistic differential equation and then modifying it with an external factor. Let me try to work through it step by step.Starting with Sub-problem 1: Solving the logistic differential equation. The equation is given as:[ frac{dA}{dt} = r cdot A cdot left(1 - frac{A}{K} right) ]They've given me specific values: ( r = 0.5 ) and ( K = 1 ). So plugging those in, the equation becomes:[ frac{dA}{dt} = 0.5 cdot A cdot (1 - A) ]And the initial condition is ( A(0) = A_0 ). I remember that the logistic equation is a separable equation, so I can try to separate the variables A and t.Let me rewrite the equation:[ frac{dA}{dt} = 0.5 A (1 - A) ]So, separating variables:[ frac{dA}{A (1 - A)} = 0.5 dt ]Now, I need to integrate both sides. The left side looks like it can be integrated using partial fractions. Let me recall how to do that.Let me express ( frac{1}{A(1 - A)} ) as partial fractions. Let me set:[ frac{1}{A(1 - A)} = frac{C}{A} + frac{D}{1 - A} ]Multiplying both sides by ( A(1 - A) ):[ 1 = C(1 - A) + D A ]Expanding:[ 1 = C - C A + D A ]Grouping like terms:[ 1 = C + (D - C) A ]Since this must hold for all A, the coefficients of like terms must be equal on both sides. So:For the constant term: ( C = 1 )For the coefficient of A: ( D - C = 0 ) => ( D = C = 1 )So, the partial fractions decomposition is:[ frac{1}{A(1 - A)} = frac{1}{A} + frac{1}{1 - A} ]Therefore, the integral becomes:[ int left( frac{1}{A} + frac{1}{1 - A} right) dA = int 0.5 dt ]Integrating term by term:Left side:[ int frac{1}{A} dA + int frac{1}{1 - A} dA = ln |A| - ln |1 - A| + C_1 ]Right side:[ 0.5 int dt = 0.5 t + C_2 ]So combining both sides:[ ln |A| - ln |1 - A| = 0.5 t + C ]Where ( C = C_2 - C_1 ) is a constant.Simplify the left side using logarithm properties:[ ln left| frac{A}{1 - A} right| = 0.5 t + C ]Exponentiating both sides to eliminate the logarithm:[ left| frac{A}{1 - A} right| = e^{0.5 t + C} = e^C e^{0.5 t} ]Since ( e^C ) is just another positive constant, let me denote it as ( C' ). So:[ frac{A}{1 - A} = C' e^{0.5 t} ]Now, solving for A:Multiply both sides by ( 1 - A ):[ A = C' e^{0.5 t} (1 - A) ]Expand the right side:[ A = C' e^{0.5 t} - C' e^{0.5 t} A ]Bring the term with A to the left side:[ A + C' e^{0.5 t} A = C' e^{0.5 t} ]Factor out A:[ A (1 + C' e^{0.5 t}) = C' e^{0.5 t} ]Solve for A:[ A = frac{C' e^{0.5 t}}{1 + C' e^{0.5 t}} ]Now, apply the initial condition ( A(0) = A_0 ). Let's plug t = 0 into the equation:[ A_0 = frac{C' e^{0}}{1 + C' e^{0}} = frac{C'}{1 + C'} ]Solving for ( C' ):Multiply both sides by ( 1 + C' ):[ A_0 (1 + C') = C' ]Expand:[ A_0 + A_0 C' = C' ]Bring terms with ( C' ) to one side:[ A_0 = C' - A_0 C' ]Factor out ( C' ):[ A_0 = C' (1 - A_0) ]Therefore:[ C' = frac{A_0}{1 - A_0} ]So, substituting back into the expression for A(t):[ A(t) = frac{ left( frac{A_0}{1 - A_0} right) e^{0.5 t} }{1 + left( frac{A_0}{1 - A_0} right) e^{0.5 t} } ]Simplify numerator and denominator:Multiply numerator and denominator by ( 1 - A_0 ):[ A(t) = frac{ A_0 e^{0.5 t} }{ (1 - A_0) + A_0 e^{0.5 t} } ]Alternatively, we can write this as:[ A(t) = frac{ A_0 e^{0.5 t} }{ A_0 e^{0.5 t} + (1 - A_0) } ]Which is the standard solution to the logistic equation. So that's Sub-problem 1 done.Now, moving on to Sub-problem 2: Modifying the differential equation with an external factor ( E(t) = e^{-beta t} ). The original equation is:[ frac{dA}{dt} = 0.5 A (1 - A) ]We need to include this external factor. The problem says to modify the original equation, but it doesn't specify how. Hmm. So I need to figure out how to incorporate E(t) into the equation.One way to model external factors is to scale the growth rate. So perhaps the growth rate r is now time-dependent, multiplied by E(t). So the modified equation would be:[ frac{dA}{dt} = r(t) cdot A cdot (1 - frac{A}{K}) ]Where ( r(t) = r cdot E(t) = 0.5 e^{-0.1 t} )So substituting, the equation becomes:[ frac{dA}{dt} = 0.5 e^{-0.1 t} cdot A (1 - A) ]Given that ( K = 1 ), so the equation is:[ frac{dA}{dt} = 0.5 e^{-0.1 t} A (1 - A) ]So now, we have a differential equation:[ frac{dA}{dt} = 0.5 e^{-0.1 t} A (1 - A) ]With initial condition ( A(0) = A_0 ). So we need to solve this.This is a Bernoulli equation, or perhaps can be solved by separation of variables. Let me see.Rewrite the equation:[ frac{dA}{dt} = 0.5 e^{-0.1 t} A (1 - A) ]We can separate variables:[ frac{dA}{A (1 - A)} = 0.5 e^{-0.1 t} dt ]So similar to the first problem, but now the right side has an exponential term.Again, let's use partial fractions on the left side as before.We already did that earlier, so:[ frac{1}{A(1 - A)} = frac{1}{A} + frac{1}{1 - A} ]Therefore, integrating both sides:Left side:[ int left( frac{1}{A} + frac{1}{1 - A} right) dA = ln |A| - ln |1 - A| + C_1 ]Right side:[ int 0.5 e^{-0.1 t} dt ]Let me compute that integral.Let me make a substitution: let ( u = -0.1 t ), so ( du = -0.1 dt ), which means ( dt = -10 du ). So:[ int 0.5 e^{u} (-10) du = -5 int e^{u} du = -5 e^{u} + C_2 = -5 e^{-0.1 t} + C_2 ]So putting it all together:[ ln left| frac{A}{1 - A} right| = -5 e^{-0.1 t} + C ]Where ( C = C_2 - C_1 ) is a constant.Exponentiating both sides:[ left| frac{A}{1 - A} right| = e^{-5 e^{-0.1 t} + C} = e^C e^{-5 e^{-0.1 t}} ]Again, ( e^C ) is a positive constant, let's denote it as ( C' ). So:[ frac{A}{1 - A} = C' e^{-5 e^{-0.1 t}} ]Solving for A:Multiply both sides by ( 1 - A ):[ A = C' e^{-5 e^{-0.1 t}} (1 - A) ]Expand:[ A = C' e^{-5 e^{-0.1 t}} - C' e^{-5 e^{-0.1 t}} A ]Bring the A terms to the left:[ A + C' e^{-5 e^{-0.1 t}} A = C' e^{-5 e^{-0.1 t}} ]Factor out A:[ A left( 1 + C' e^{-5 e^{-0.1 t}} right) = C' e^{-5 e^{-0.1 t}} ]Solve for A:[ A = frac{C' e^{-5 e^{-0.1 t}}}{1 + C' e^{-5 e^{-0.1 t}}} ]Now, apply the initial condition ( A(0) = A_0 ). Let's plug t = 0 into the equation:[ A_0 = frac{C' e^{-5 e^{0}}}{1 + C' e^{-5 e^{0}}} = frac{C' e^{-5}}{1 + C' e^{-5}} ]Solve for ( C' ):Let me denote ( e^{-5} ) as a constant, let's compute its approximate value. ( e^{-5} approx 0.006737947 ). But maybe we can keep it symbolic for now.So:[ A_0 = frac{C' e^{-5}}{1 + C' e^{-5}} ]Multiply both sides by denominator:[ A_0 (1 + C' e^{-5}) = C' e^{-5} ]Expand:[ A_0 + A_0 C' e^{-5} = C' e^{-5} ]Bring terms with ( C' ) to one side:[ A_0 = C' e^{-5} - A_0 C' e^{-5} ]Factor out ( C' e^{-5} ):[ A_0 = C' e^{-5} (1 - A_0) ]Solve for ( C' ):[ C' = frac{A_0}{(1 - A_0) e^{-5}} = frac{A_0 e^{5}}{1 - A_0} ]So, substituting back into the expression for A(t):[ A(t) = frac{ left( frac{A_0 e^{5}}{1 - A_0} right) e^{-5 e^{-0.1 t}} }{1 + left( frac{A_0 e^{5}}{1 - A_0} right) e^{-5 e^{-0.1 t}} } ]Simplify numerator and denominator:Multiply numerator and denominator by ( 1 - A_0 ):[ A(t) = frac{ A_0 e^{5} e^{-5 e^{-0.1 t}} }{ (1 - A_0) + A_0 e^{5} e^{-5 e^{-0.1 t}} } ]Simplify the exponents:Note that ( e^{5} e^{-5 e^{-0.1 t}} = e^{5 (1 - e^{-0.1 t})} ). Wait, let me check:Wait, ( e^{5} times e^{-5 e^{-0.1 t}} = e^{5 - 5 e^{-0.1 t}} = e^{5(1 - e^{-0.1 t})} ). Yes, that's correct.So, we can write:[ A(t) = frac{ A_0 e^{5(1 - e^{-0.1 t})} }{ (1 - A_0) + A_0 e^{5(1 - e^{-0.1 t})} } ]Alternatively, factor out ( e^{5(1 - e^{-0.1 t})} ) in the denominator:[ A(t) = frac{ A_0 e^{5(1 - e^{-0.1 t})} }{ A_0 e^{5(1 - e^{-0.1 t})} + (1 - A_0) } ]Which is the solution to the modified logistic equation with the external factor.Let me just recap the steps to make sure I didn't make a mistake.1. For Sub-problem 1, I separated variables, used partial fractions, integrated, applied initial condition, and got the standard logistic solution.2. For Sub-problem 2, I included the external factor by scaling the growth rate. Then, separated variables again, integrated both sides, applied the initial condition, solved for the constant, and expressed A(t) in terms of the initial condition and the external factor.I think that makes sense. Let me check the integration step for the right side again.We had:[ int 0.5 e^{-0.1 t} dt ]Let me compute that again without substitution:The integral of ( e^{kt} ) is ( frac{1}{k} e^{kt} ). So here, k = -0.1, so:[ int 0.5 e^{-0.1 t} dt = 0.5 times frac{e^{-0.1 t}}{-0.1} + C = -5 e^{-0.1 t} + C ]Yes, that's correct. So the integral was done correctly.Then, exponentiating both sides:[ ln left| frac{A}{1 - A} right| = -5 e^{-0.1 t} + C ]So exponentiating gives:[ frac{A}{1 - A} = e^{C} e^{-5 e^{-0.1 t}} ]Which is correct.Then solving for A(t) step by step, applying initial condition, and solving for the constant. That seems correct.So, I think the solutions are correct.Final AnswerSub-problem 1: The solution is ( boxed{A(t) = dfrac{A_0 e^{0.5 t}}{A_0 e^{0.5 t} + (1 - A_0)}} ).Sub-problem 2: The solution is ( boxed{A(t) = dfrac{A_0 e^{5(1 - e^{-0.1 t})}}{A_0 e^{5(1 - e^{-0.1 t})} + (1 - A_0)}} ).</think>

🧠:<think>Okay, so I have this problem about Alex who has a terminal illness, and his remaining lifespan follows an exponential decay model. The PDF is given as f_T(t) = λe^{-λt} where λ is 0.25 per year. There are two parts to the problem. First, I need to calculate the expected value or mean of Alex's remaining lifespan. Hmm, I remember that for an exponential distribution, the mean is 1/λ. Let me verify that. The exponential distribution has a PDF f(t) = λe^{-λt} for t ≥ 0, and its mean is indeed E[T] = 1/λ. So in this case, λ is 0.25, so the mean should be 1/0.25, which is 4 years. That seems straightforward. Wait, just to make sure, maybe I should derive it from the integral. The expected value E[T] is the integral from 0 to infinity of t*f_T(t) dt. So that would be ∫₀^∞ t*λe^{-λt} dt. I think this integral evaluates to 1/λ. Let me recall integration by parts. Let u = t, dv = λe^{-λt} dt. Then du = dt, and v = -e^{-λt}. So integrating by parts, we get uv|₀^∞ - ∫₀^∞ v du. That is, [-t e^{-λt}]₀^∞ + ∫₀^∞ e^{-λt} dt. The first term, as t approaches infinity, e^{-λt} goes to zero, and at t=0, it's zero as well. So the first term is zero. The second integral is ∫₀^∞ e^{-λt} dt, which is [ -1/λ e^{-λt} ] from 0 to ∞, which is 0 - (-1/λ) = 1/λ. So yes, E[T] is indeed 1/λ, which is 4 years. Got it.Now, moving on to the second part. Alex wants to allocate his remaining time T between two activities: fulfilling personal wishes (X) and spending time with family (Y), with the constraint that X + Y = T. He wants the ratio of X to Y to be 2:3. So, I need to determine the joint PDF of X and Y and find E[X] and E[Y].First, let's parse the ratio. The ratio of X to Y is 2:3, which means X/Y = 2/3, so X = (2/5)T and Y = (3/5)T. Because 2 + 3 = 5 parts, so X is 2/5 of T and Y is 3/5 of T. So, X and Y are linear functions of T. Since T is a random variable, X and Y will also be random variables.Given that X = (2/5)T and Y = (3/5)T, and since X + Y = T, this makes sense. So, to find the joint PDF of X and Y, I need to perform a transformation of variables. Since X and Y are both functions of T, and T is a single random variable, the joint distribution might have some constraints.Wait, but X and Y are perfectly dependent because Y = T - X, so they are not independent. Therefore, the joint PDF will be supported on the line X + Y = T, which complicates things a bit. Alternatively, since both X and Y are linear transformations of T, maybe we can express their joint distribution in terms of T.Alternatively, perhaps it's easier to consider that since X = (2/5)T and Y = (3/5)T, we can express T in terms of X or Y. For example, T = (5/2)X or T = (5/3)Y. Then, since T has an exponential distribution, we can find the distribution of X and Y.But since X and Y are dependent, their joint PDF isn't just the product of their individual PDFs. Instead, we need to find the joint PDF considering their relationship.Let me think about the transformation. Let me denote the transformation as:X = (2/5)TY = (3/5)TSo, we can write this as a linear transformation:X = aTY = bTwhere a = 2/5 and b = 3/5.Since X and Y are both functions of T, we can think of this as a mapping from T to (X, Y). However, since X and Y are dependent, the joint PDF will be a function along the line X + Y = T, but since T itself is a random variable, it's a bit more involved.Alternatively, perhaps we can parameterize the joint distribution in terms of T. Let me consider that for each T, X and Y are determined, so the joint PDF f_{X,Y}(x,y) is non-zero only when x + y = t, and t is a random variable with PDF f_T(t). So, maybe f_{X,Y}(x,y) = f_T(t) * |J|, where J is the Jacobian determinant of the transformation.Wait, but in this case, since X and Y are both functions of T, the transformation is not invertible in the usual sense because we have two variables depending on one. So, the standard Jacobian method might not apply here because the transformation isn't from (X, Y) to (T, something else); it's from T to (X, Y), but we don't have a second variable. So, perhaps we need a different approach.Alternatively, since X = (2/5)T and Y = (3/5)T, we can express T in terms of X or Y, and then express the joint PDF accordingly. Let's try to express T in terms of X. So, T = (5/2)X. Then, since T has PDF f_T(t) = 0.25 e^{-0.25 t}, substituting t = (5/2)x, we get f_T((5/2)x) = 0.25 e^{-0.25*(5/2)x} = 0.25 e^{-(5/8)x}.But since X is a function of T, we also need to account for the change of variable. The PDF of X would be f_X(x) = f_T((5/2)x) * |d/dx (5/2 x)|. Wait, no, more accurately, if T = (5/2)X, then X = (2/5)T, so dX/dT = 2/5. Therefore, f_X(x) = f_T(t) / |dX/dT| = f_T((5/2)x) / (2/5) = (0.25 e^{-0.25*(5/2)x}) / (2/5) = (0.25 / (2/5)) e^{-(5/8)x} = (0.25 * 5/2) e^{-(5/8)x} = (5/8) e^{-(5/8)x}.Similarly, for Y, since Y = (3/5)T, T = (5/3)Y. So, f_Y(y) = f_T((5/3)y) / (3/5) = (0.25 e^{-0.25*(5/3)y}) / (3/5) = (0.25 * 5/3) e^{-(5/12)y} = (5/12) e^{-(5/12)y}.But wait, this gives us the marginal PDFs of X and Y, but the question asks for the joint PDF. Since X and Y are perfectly correlated (because Y = T - X), their joint distribution is singular and lies on the line X + Y = T. Therefore, the joint PDF isn't straightforward; it's actually a distribution along a curve in the X-Y plane.In such cases, the joint PDF can be expressed using a Dirac delta function. Specifically, since X + Y = T, we can write the joint PDF as f_{X,Y}(x,y) = f_T(t) * δ(x + y - t), but since t is a random variable, this might not be directly applicable. Alternatively, perhaps we can express it as f_{X,Y}(x,y) = f_T(x + y) * δ(y - (3/5)(x + y)) or something similar.Wait, maybe a better approach is to consider that since X = (2/5)T and Y = (3/5)T, we can express T in terms of X and Y, but since X and Y are dependent, we can write the joint PDF as f_{X,Y}(x,y) = f_T((x + y)) * |J|, where J is the Jacobian determinant of the transformation from (X, Y) to (T, something). But since we only have one variable T, this might not be straightforward.Alternatively, perhaps we can think of this as a change of variables from T to (X, Y), but since we have two variables and only one equation, we need to introduce another variable, say Z, which is arbitrary. But that complicates things.Wait, maybe I'm overcomplicating it. Since X and Y are both linear functions of T, and T is exponential, perhaps the joint PDF can be expressed as f_{X,Y}(x,y) = f_T(t) * δ(x - (2/5)t) * δ(y - (3/5)t). But I'm not sure if that's the right way to express it.Alternatively, since X = (2/5)T and Y = (3/5)T, we can write T = (5/2)X = (5/3)Y. Therefore, for any given X and Y, they must satisfy Y = (3/2)X. So, the joint PDF is non-zero only along the line Y = (3/2)X. Therefore, we can express the joint PDF as f_{X,Y}(x,y) = f_T((5/2)x) * (5/2) if y = (3/2)x, and zero otherwise.Wait, let me think about that. If we have X = (2/5)T, then T = (5/2)X. So, for a given X, T is determined, and Y is determined as Y = (3/5)T = (3/5)(5/2)X = (3/2)X. So, the joint PDF is only non-zero when y = (3/2)x. Therefore, we can express the joint PDF as f_{X,Y}(x,y) = f_T((5/2)x) * (5/2) * δ(y - (3/2)x). But I'm not sure if that's correct. Alternatively, since X and Y are both functions of T, perhaps we can express the joint PDF as f_{X,Y}(x,y) = f_T(t) * |J|, where J is the Jacobian determinant of the transformation from (X, Y) to T. But since we have two variables and one equation, the Jacobian would be a matrix, and its determinant would be zero, which doesn't help.Wait, maybe a better approach is to use the method of transformation of variables. Let me consider that we have a transformation from T to (X, Y). Since X = (2/5)T and Y = (3/5)T, we can write this as a linear transformation:X = (2/5)TY = (3/5)TSo, the inverse transformation is T = (5/2)X = (5/3)Y. But since both expressions must hold, we have (5/2)X = (5/3)Y, which simplifies to 3X = 2Y, or Y = (3/2)X. So, the joint PDF is supported on the line Y = (3/2)X.To find the joint PDF, we can use the fact that for transformations where the variables are dependent, the joint PDF can be expressed as f_{X,Y}(x,y) = f_T(t) * |J|, where J is the Jacobian determinant of the transformation from (X, Y) to T. But since we have two variables and only one equation, the Jacobian is a bit tricky.Alternatively, perhaps we can parameterize the joint PDF in terms of T. Since X and Y are both functions of T, we can express the joint PDF as f_{X,Y}(x,y) = f_T(t) * δ(x - (2/5)t) * δ(y - (3/5)t). But integrating this over x and y would give us f_T(t), which might not be the right approach.Wait, maybe a better way is to recognize that since X and Y are linear functions of T, and T is exponential, the joint distribution of X and Y is a singular distribution along the line X + Y = T, but since T itself is random, it's a bit more involved.Alternatively, perhaps we can consider that since X = (2/5)T and Y = (3/5)T, we can express the joint PDF as f_{X,Y}(x,y) = f_T((5/2)x) * (5/2) * δ(y - (3/2)x). This is because for a given X, Y must be (3/2)X, and the PDF is scaled by the derivative of T with respect to X, which is 5/2.So, putting it all together, the joint PDF would be:f_{X,Y}(x,y) = (5/2) * f_T((5/2)x) * δ(y - (3/2)x)Substituting f_T(t) = 0.25 e^{-0.25 t}, we get:f_{X,Y}(x,y) = (5/2) * 0.25 e^{-0.25*(5/2)x} * δ(y - (3/2)x)Simplifying, 5/2 * 0.25 = 5/8, and 0.25*(5/2) = 5/8, so:f_{X,Y}(x,y) = (5/8) e^{-(5/8)x} * δ(y - (3/2)x)This makes sense because the joint PDF is non-zero only along the line y = (3/2)x, and the density is scaled appropriately.Now, to find E[X] and E[Y], we can use the linearity of expectation. Since X = (2/5)T and Y = (3/5)T, we have:E[X] = E[(2/5)T] = (2/5) E[T] = (2/5)*4 = 8/5 = 1.6 yearsSimilarly,E[Y] = E[(3/5)T] = (3/5)*4 = 12/5 = 2.4 yearsAlternatively, we can compute E[X] and E[Y] using the joint PDF. For E[X], we can integrate x * f_{X,Y}(x,y) over all x and y. But since the PDF is non-zero only along y = (3/2)x, we can express the integral as:E[X] = ∫₀^∞ ∫_{y=(3/2)x} x * f_{X,Y}(x,y) dy dxBut since f_{X,Y}(x,y) is expressed with a delta function, this integral simplifies to:E[X] = ∫₀^∞ x * (5/8) e^{-(5/8)x} * (3/2) dxWait, no, actually, when integrating with the delta function, the inner integral over y would just pick up the value at y = (3/2)x, so the integral becomes:E[X] = ∫₀^∞ x * (5/8) e^{-(5/8)x} dxWhich is the same as E[X] = (5/8) ∫₀^∞ x e^{-(5/8)x} dxThis integral is the expectation of an exponential distribution with λ = 5/8, so E[X] = 1/(5/8) = 8/5 = 1.6, which matches our earlier result.Similarly, for E[Y], since Y = (3/2)X, we can compute E[Y] = (3/2) E[X] = (3/2)*(8/5) = 24/10 = 12/5 = 2.4, which again matches our earlier result.So, to summarize, the joint PDF of X and Y is f_{X,Y}(x,y) = (5/8) e^{-(5/8)x} δ(y - (3/2)x), and the expected values are E[X] = 1.6 years and E[Y] = 2.4 years.</think>

🧠:<think>Okay, so I have this problem about a commercial real estate agent analyzing a land lease deal. The land is shaped like a parabola, and there are two sub-problems to solve. Let me start with Sub-problem 1.Sub-problem 1: Finding the Area of the Parabolic PlotFirst, the parabola is defined by the equation ( y = ax^2 + bx + c ). We know a few things:1. The vertex is at (3, -9).2. The parabola passes through the origin (0,0).3. We need to find the area between x = 0 and x = 6.Alright, so I need to find the equation of the parabola first. Since we know the vertex, it might be easier to use the vertex form of a parabola. The vertex form is ( y = a(x - h)^2 + k ), where (h, k) is the vertex.Given the vertex is (3, -9), plugging that in:( y = a(x - 3)^2 - 9 )Now, we also know that the parabola passes through (0,0). Let's plug that point into the equation to find 'a':( 0 = a(0 - 3)^2 - 9 )( 0 = a(9) - 9 )( 9a = 9 )( a = 1 )So, the equation in vertex form is ( y = (x - 3)^2 - 9 ). Let me expand this to standard form to make it easier for integration later.Expanding ( (x - 3)^2 ):( (x - 3)^2 = x^2 - 6x + 9 )So, substituting back:( y = x^2 - 6x + 9 - 9 )( y = x^2 - 6x )So, the equation is ( y = x^2 - 6x ).Now, to find the area between x = 0 and x = 6, we need to integrate the function from 0 to 6. But wait, since it's a parabola, I should check if it's above or below the x-axis in this interval.Let me find the roots of the parabola to understand where it crosses the x-axis. Setting y = 0:( 0 = x^2 - 6x )( x(x - 6) = 0 )So, the roots are at x = 0 and x = 6. That means the parabola crosses the x-axis at these points. Since the vertex is at (3, -9), which is below the x-axis, the parabola opens upwards because the coefficient of ( x^2 ) is positive (a = 1). So, between x = 0 and x = 6, the parabola is below the x-axis, meaning the area will be the integral of the absolute value of the function.But since we're calculating the area, we can take the integral from 0 to 6 of |y| dx. However, since y is negative in this interval, the area will be the integral of -y dx from 0 to 6.So, the area A is:( A = int_{0}^{6} |y| dx = int_{0}^{6} |x^2 - 6x| dx )But since ( x^2 - 6x ) is negative between 0 and 6 (except at the endpoints), we can write:( A = int_{0}^{6} (6x - x^2) dx )Let me compute this integral step by step.First, find the antiderivative of ( 6x - x^2 ):The antiderivative of 6x is ( 3x^2 ).The antiderivative of ( -x^2 ) is ( -frac{1}{3}x^3 ).So, the antiderivative F(x) is:( F(x) = 3x^2 - frac{1}{3}x^3 )Now, evaluate F(x) from 0 to 6:( F(6) = 3*(6)^2 - (1/3)*(6)^3 )( F(6) = 3*36 - (1/3)*216 )( F(6) = 108 - 72 )( F(6) = 36 )( F(0) = 3*(0)^2 - (1/3)*(0)^3 = 0 )So, the area A is:( A = F(6) - F(0) = 36 - 0 = 36 )Therefore, the area of the parabolic plot is 36 square units.Wait, let me double-check my calculations because sometimes signs can be tricky.We had y = x^2 - 6x, which is negative between 0 and 6. So, integrating (6x - x^2) from 0 to 6 gives the area above the curve, which is positive. So, 36 seems correct.Alternatively, if I didn't take the absolute value, integrating y from 0 to 6 would give a negative area, but since we're talking about physical area, we take the positive value.So, yes, 36 is correct.Sub-problem 2: Maximizing the Area of a Rectangular PatchNow, the agent wants to renovate a rectangular section within the plot. The rectangle has its base along the x-axis, with one vertex at the origin (0,0) and the opposite vertex lying on the parabola. We need to find the maximum possible area of this rectangle and the coordinates of the opposite vertex.Let me visualize this. The rectangle has its base from (0,0) to (x,0) along the x-axis, and its top side from (x,0) to (x,y), where (x,y) is on the parabola. So, the opposite vertex is (x,y), and since it's on the parabola, y = x^2 - 6x.But wait, the parabola is y = x^2 - 6x, which is negative between 0 and 6. However, the rectangle's height can't be negative because it's a physical area. So, does that mean the rectangle can only exist where y is positive? But in our case, the parabola is below the x-axis between 0 and 6, so y is negative. Hmm, that seems contradictory.Wait, maybe I made a mistake here. If the parabola is below the x-axis, then the y-coordinate is negative. But a rectangle can't have negative height. So, perhaps the rectangle is actually above the x-axis, but since the parabola is below, maybe the rectangle is between the x-axis and the parabola? But that would imply the rectangle is above the parabola, but the parabola is below the x-axis. Hmm, this is confusing.Wait, no. Let me think again. The rectangle has its base on the x-axis, from (0,0) to (x,0), and then goes up to (x,y), where y is on the parabola. But since the parabola is below the x-axis, y is negative. So, the rectangle would actually be below the x-axis, but that doesn't make sense because the area can't be negative. Maybe the rectangle is constructed such that its top side is on the parabola, but since the parabola is below the x-axis, the rectangle would have negative height. That doesn't make physical sense.Wait, perhaps I misinterpreted the problem. Maybe the rectangle is constructed such that its top side is on the parabola, but the parabola is below the x-axis, so the rectangle's height is the distance from the x-axis to the parabola. But in that case, the height would be |y|, which is positive.So, perhaps the area of the rectangle is x * |y|, where y is the y-coordinate on the parabola. Since y is negative, |y| = -y.Given that, the area A is:( A = x * |y| = x * (-y) = x * (6x - x^2) )Because y = x^2 - 6x, so |y| = -(x^2 - 6x) = 6x - x^2.Therefore, the area is:( A = x*(6x - x^2) = 6x^2 - x^3 )Now, to find the maximum area, we need to find the value of x that maximizes A.So, let's consider A as a function of x:( A(x) = 6x^2 - x^3 )To find the maximum, take the derivative of A with respect to x, set it equal to zero, and solve for x.First, compute the derivative:( A'(x) = d/dx [6x^2 - x^3] = 12x - 3x^2 )Set A'(x) = 0:( 12x - 3x^2 = 0 )( 3x(4 - x) = 0 )So, the critical points are at x = 0 and x = 4.Now, we need to determine which of these gives a maximum. Since x = 0 would give A = 0, which is a minimum, the maximum must be at x = 4.To confirm, we can take the second derivative test.Compute the second derivative:( A''(x) = d/dx [12x - 3x^2] = 12 - 6x )Evaluate at x = 4:( A''(4) = 12 - 6*4 = 12 - 24 = -12 )Since A''(4) is negative, the function is concave down at x = 4, indicating a local maximum.Therefore, the maximum area occurs at x = 4.Now, let's find the corresponding y-coordinate on the parabola:( y = x^2 - 6x )( y = (4)^2 - 6*(4) = 16 - 24 = -8 )But since the area is x * |y|, the height is |y| = 8.So, the opposite vertex is at (4, -8), but since we're talking about the rectangle's height, it's 8 units above the x-axis, but the actual point is (4, -8). Wait, that seems contradictory.Wait, no. The rectangle has its base on the x-axis from (0,0) to (4,0), and then goes up to (4, y), but y is negative. So, the rectangle would actually extend below the x-axis, but since we're talking about area, it's the absolute value. So, the height is 8 units, but the point is (4, -8). So, the rectangle's top side is from (0,0) to (4, -8), but that would make it a rectangle below the x-axis. However, in reality, the area is positive, so we can consider the height as 8 units regardless of the direction.But in terms of coordinates, the opposite vertex is (4, -8). So, that's the point on the parabola.Therefore, the maximum area is:( A = 4 * 8 = 32 ) square units.Wait, let me compute A(x) at x = 4:( A(4) = 6*(4)^2 - (4)^3 = 6*16 - 64 = 96 - 64 = 32 )Yes, that's correct.So, the maximum area is 32, and the opposite vertex is at (4, -8).Wait, but let me double-check if this makes sense. If x = 4, then the rectangle goes from (0,0) to (4,0) along the x-axis, and then up to (4, -8). But since the rectangle is below the x-axis, the height is 8 units downward. However, in terms of area, it's still 32 square units.Alternatively, if we consider the rectangle above the x-axis, but since the parabola is below, that's not possible. So, the rectangle must be below the x-axis, but the area is positive.Therefore, the maximum area is 32, and the opposite vertex is at (4, -8).Wait, but let me think again. If the rectangle is constructed with one vertex at (0,0) and the opposite vertex on the parabola, then the rectangle's top side is from (x,0) to (x,y), where y is on the parabola. Since y is negative, the rectangle is below the x-axis, but the area is positive.Alternatively, perhaps the rectangle is constructed such that its top side is on the parabola, but since the parabola is below the x-axis, the rectangle is below the x-axis. So, the height is |y|, which is positive, but the y-coordinate is negative.So, yes, the area is 32, and the opposite vertex is at (4, -8).Therefore, the maximum area is 32, and the opposite vertex is at (4, -8).Wait, but let me check if there's another way to interpret the problem. Maybe the rectangle is constructed such that its top side is on the parabola, but the parabola is below the x-axis, so the rectangle is below the x-axis, but the area is still positive. So, yes, that makes sense.Therefore, I think my solution is correct.Summary of Solutions:Sub-problem 1: The area of the parabolic plot is 36 square units.Sub-problem 2: The maximum area of the rectangle is 32 square units, with the opposite vertex at (4, -8).Wait, but let me just make sure I didn't make any calculation errors.For Sub-problem 1:Equation of parabola: y = x^2 - 6x.Integral from 0 to 6 of (6x - x^2) dx.Antiderivative: 3x^2 - (1/3)x^3.At x=6: 3*(36) - (1/3)*(216) = 108 - 72 = 36.At x=0: 0.So, area is 36. Correct.For Sub-problem 2:Area function: A = x*(6x - x^2) = 6x^2 - x^3.Derivative: 12x - 3x^2.Set to zero: 12x - 3x^2 = 0 => x=0 or x=4.Second derivative: 12 - 6x. At x=4, it's -12, so concave down, maximum.A(4) = 6*(16) - 64 = 96 - 64 = 32.Coordinates: (4, y) where y = 4^2 - 6*4 = 16 - 24 = -8.So, yes, correct.I think that's solid.</think>

🧠:<think>Alright, so I have this problem where I need to figure out the minimum number of days required to train 400 employees. Each training session can have up to 25 employees, and each session is led by one of 5 trainers. Each trainer can only do 3 sessions a day. The training hours are from 9 AM to 5 PM, which is 8 hours. Each session is 1 hour long, so that means in one day, a trainer can conduct up to 3 sessions, each taking an hour, right?First, I need to figure out how many sessions are required in total. Since each session can have 25 employees, I can divide the total number of employees by the session capacity. So, 400 divided by 25 is 16. That means I need 16 training sessions in total.Now, each day, each trainer can conduct up to 3 sessions. There are 5 trainers, so the maximum number of sessions that can be conducted in a day is 5 trainers multiplied by 3 sessions each, which is 15 sessions per day.Wait, hold on. If I need 16 sessions and can only do 15 per day, that means I need at least 2 days. Because on the first day, I can do 15 sessions, and on the second day, I can do the remaining 1 session. So, the minimum number of days required is 2.But let me double-check that. Each day, 5 trainers can do 3 sessions each, so 15 sessions. 15 sessions can cover 15*25 = 375 employees. But we have 400 employees, so 375 isn't enough. So, on the second day, we need to cover the remaining 25 employees, which is 1 session. So yes, 2 days are needed.Wait, but each day is 8 hours, and each session is 1 hour. So, in one day, a trainer can do 3 sessions, but each session is 1 hour, so 3 hours per trainer. That leaves 5 hours unused per trainer each day. But since we have multiple trainers, we can stagger their sessions. So, in one day, the maximum number of sessions is 15, as I thought.So, 16 sessions needed, 15 per day, so 2 days.Now, moving on to the second part. Each employee needs a 30-minute follow-up session exactly one week after their initial training. So, each employee will have two sessions: the initial 1-hour training and a 30-minute follow-up.First, let's figure out how many follow-up sessions there are. Since each of the 400 employees needs one, that's 400 follow-up sessions. Each follow-up session is 30 minutes, which is half an hour. So, in terms of session hours, each follow-up is 0.5 hours.But the follow-up sessions are also conducted from 9 AM to 5 PM, which is 8 hours. So, each day, each trainer can conduct multiple follow-up sessions, depending on the duration.Wait, the problem says each session is 30 minutes, so each trainer can conduct up to how many follow-up sessions per day? Since each session is 0.5 hours, and the trainer can work up to 8 hours a day, but the constraint is that each trainer can conduct a maximum of 3 sessions per day, regardless of the session length? Or is it based on time?Wait, the original problem says each trainer can conduct a maximum of 3 sessions per day, and no two sessions can overlap. So, for the follow-up sessions, each session is 30 minutes, but the constraint is still 3 sessions per trainer per day, regardless of the session length.But wait, actually, the original problem says each session is 1 hour, but for follow-up, it's 30 minutes. So, does the 3 sessions per day limit apply to the number of sessions, regardless of their duration? I think so, because it's about the number of sessions a trainer can conduct, not the total time.So, each trainer can do up to 3 follow-up sessions per day, each lasting 30 minutes. So, in terms of time, each trainer would use 1.5 hours per day for follow-ups.But the total available time is 8 hours, so in theory, a trainer could do more, but the constraint is 3 sessions per day.So, similar to the initial training, each day, each trainer can do 3 follow-up sessions. So, with 5 trainers, the maximum number of follow-up sessions per day is 5*3=15.But wait, each follow-up session is 30 minutes, so in 8 hours, a trainer could potentially do more, but the constraint is 3 sessions per day.So, regardless of the session length, each trainer is limited to 3 sessions per day. So, 15 follow-up sessions per day.But each follow-up session is 30 minutes, so in 8 hours, a trainer could do 16 follow-up sessions (since 8 hours / 0.5 hours per session = 16). But the constraint is 3 sessions per day, so only 3 per trainer.Therefore, the number of follow-up sessions per day is 15.But wait, we have 400 follow-up sessions. So, 400 divided by 15 per day is approximately 26.666 days. So, we would need 27 days.But wait, that seems too high. Let me think again.Wait, no. The follow-up sessions are conducted one week after the initial training. So, the initial training is done in 2 days, and then the follow-up is one week later. So, the follow-up sessions don't interfere with the initial training schedule.But the question is, how many additional days are needed for the follow-up sessions, considering the same constraints.So, each follow-up session is 30 minutes, but the constraint is 3 sessions per trainer per day, regardless of the duration. So, each day, 5 trainers can do 15 follow-up sessions.Each follow-up session is 30 minutes, so in terms of time, each day, the total time used is 15 sessions * 0.5 hours = 7.5 hours, which is within the 8-hour window.So, the number of days needed is 400 / 15 ≈ 26.666, so 27 days.But wait, that seems like a lot. Let me check again.Wait, each follow-up session is 30 minutes, so in one day, a trainer can do 3 sessions, which is 1.5 hours. So, the rest of the time, 6.5 hours, is unused. But since we have 5 trainers, the total number of follow-up sessions per day is 15, as before.So, 400 / 15 ≈ 26.666, so 27 days.But that seems like a lot. Maybe I'm misunderstanding the constraints.Wait, the problem says "follow-up sessions are also conducted from 9 AM to 5 PM", so the same time constraints as the initial training. But the initial training sessions were 1 hour, and follow-up is 30 minutes. So, perhaps the number of sessions per day can be higher because each session is shorter.Wait, but the constraint is that each trainer can conduct a maximum of 3 sessions per day, regardless of the session length. So, even if the sessions are shorter, they can't do more than 3 per day.Therefore, the number of follow-up sessions per day is still limited to 15, as each trainer can only do 3, regardless of the session duration.Therefore, 400 / 15 ≈ 26.666, so 27 days.But wait, 27 days seems excessive. Maybe I'm missing something.Wait, perhaps the follow-up sessions can be scheduled more efficiently because they are shorter. For example, in the initial training, each session is 1 hour, so in 8 hours, a trainer can do 3 sessions, each taking an hour, with some downtime. But for follow-up sessions, which are 30 minutes, a trainer could potentially do more, but the constraint is still 3 sessions per day.So, regardless of the session length, the maximum number of sessions per trainer per day is 3. Therefore, the number of follow-up sessions per day is 15.So, 400 / 15 ≈ 26.666, so 27 days.But wait, 27 days is almost a month. Is there a way to do it in fewer days?Wait, perhaps the follow-up sessions can be scheduled on the same days as the initial training, but that's not possible because they are one week later. So, the follow-up sessions are separate and need to be scheduled after the initial training.Therefore, the minimum number of additional days required is 27.Wait, but let me think again. Each follow-up session is 30 minutes, so in 8 hours, a trainer could do 16 follow-up sessions. But the constraint is 3 sessions per day, so only 3 per trainer.Therefore, 5 trainers * 3 sessions = 15 per day.So, 400 / 15 ≈ 26.666, so 27 days.Yes, that seems correct.So, the answers are:1. Minimum number of days for initial training: 2 days.2. Additional number of days for follow-up sessions: 27 days.</think>