🧠:<think>Okay, so I have this problem about bacterial growth models. It's divided into two parts. Let me try to tackle them one by one.Starting with part 1: Strain A follows a logistic growth model. I remember that the logistic equation is a common model for population growth where the growth rate decreases as the population approaches the carrying capacity. The differential equation given is:[frac{dP_A}{dt} = r_A P_A left(1 - frac{P_A}{K_A}right)]And the initial condition is ( P_A(0) = P_{A0} ). I need to find the general solution for ( P_A(t) ).Hmm, I think the logistic equation has a standard solution. Let me recall. The general solution for the logistic equation is:[P(t) = frac{K}{1 + left(frac{K}{P_0} - 1right) e^{-rt}}]Where ( K ) is the carrying capacity, ( r ) is the growth rate, and ( P_0 ) is the initial population. So, applying this to Strain A, substituting ( K ) with ( K_A ), ( r ) with ( r_A ), and ( P_0 ) with ( P_{A0} ), the solution should be:[P_A(t) = frac{K_A}{1 + left(frac{K_A}{P_{A0}} - 1right) e^{-r_A t}}]Wait, let me verify that. If I plug in ( t = 0 ), does it give me ( P_{A0} )?At ( t = 0 ), the exponential term becomes 1, so:[P_A(0) = frac{K_A}{1 + left(frac{K_A}{P_{A0}} - 1right) times 1} = frac{K_A}{frac{K_A}{P_{A0}}}]Simplifying that:[P_A(0) = frac{K_A times P_{A0}}{K_A} = P_{A0}]Yes, that checks out. So, I think that's correct.Moving on to part 2: Strain B's growth is modeled by a modified logistic equation. The differential equation is:[frac{dP_B}{dt} = r_B P_B left(1 - frac{P_B}{K_B}right) + alpha P_A P_B]And the initial conditions are ( P_A(0) = P_{A0} ) and ( P_B(0) = P_{B0} ). I need to derive the coupled differential equations for ( P_A(t) ) and ( P_B(t) ), and discuss the steady-state conditions.Wait, the problem says "derive the coupled differential equations," but aren't they already given? Strain A's equation is the logistic model, and Strain B's equation is the modified logistic with an additional term involving ( P_A ). So, perhaps the coupled system is just the two differential equations together.So, the coupled system is:[frac{dP_A}{dt} = r_A P_A left(1 - frac{P_A}{K_A}right)][frac{dP_B}{dt} = r_B P_B left(1 - frac{P_B}{K_B}right) + alpha P_A P_B]Yes, that makes sense. So, these two equations are coupled because ( P_B ) depends on ( P_A ), but ( P_A ) doesn't depend on ( P_B ). So, it's a system where Strain A grows logistically, and Strain B's growth is influenced by both its own logistic term and a mutualistic interaction with Strain A.Now, I need to discuss the conditions under which both populations reach a steady state. A steady state occurs when the derivatives are zero, meaning ( frac{dP_A}{dt} = 0 ) and ( frac{dP_B}{dt} = 0 ).So, setting ( frac{dP_A}{dt} = 0 ):[r_A P_A left(1 - frac{P_A}{K_A}right) = 0]This gives two possibilities: either ( P_A = 0 ) or ( 1 - frac{P_A}{K_A} = 0 ), which implies ( P_A = K_A ).Similarly, for ( frac{dP_B}{dt} = 0 ):[r_B P_B left(1 - frac{P_B}{K_B}right) + alpha P_A P_B = 0]Factor out ( P_B ):[P_B left[ r_B left(1 - frac{P_B}{K_B}right) + alpha P_A right] = 0]So, either ( P_B = 0 ) or ( r_B left(1 - frac{P_B}{K_B}right) + alpha P_A = 0 ).Now, let's analyze the possible steady states.1. Extinction of both strains: ( P_A = 0 ) and ( P_B = 0 ). This is a trivial steady state, but it's unlikely unless initial populations are zero or conditions are very harsh.2. Strain A at carrying capacity, Strain B extinct: ( P_A = K_A ) and ( P_B = 0 ). Let's check if this is a steady state. For Strain A, ( P_A = K_A ) satisfies ( frac{dP_A}{dt} = 0 ). For Strain B, ( P_B = 0 ) satisfies ( frac{dP_B}{dt} = 0 ). So, yes, this is a steady state. However, whether this is stable depends on the interaction term.3. Strain A extinct, Strain B at carrying capacity: ( P_A = 0 ) and ( P_B = K_B ). For Strain A, ( P_A = 0 ) is a steady state. For Strain B, substituting ( P_A = 0 ): [ r_B P_B left(1 - frac{P_B}{K_B}right) = 0 ] So, ( P_B = 0 ) or ( P_B = K_B ). So, ( P_B = K_B ) is another steady state. But in this case, Strain A is extinct, so Strain B can reach its carrying capacity on its own.4. Both strains at non-zero steady states: ( P_A neq 0 ), ( P_B neq 0 ). For this, we need both ( frac{dP_A}{dt} = 0 ) and ( frac{dP_B}{dt} = 0 ). From Strain A's equation, ( P_A = K_A ) is the only non-zero steady state. Wait, no. If ( P_A ) is not at ( K_A ), then ( r_A P_A (1 - P_A/K_A) = 0 ) implies ( P_A = 0 ) or ( P_A = K_A ). So, Strain A can't have a steady state at any other population except 0 or ( K_A ). Therefore, for both strains to be non-zero, Strain A must be at ( K_A ), and then we can find ( P_B ). So, substituting ( P_A = K_A ) into Strain B's steady state equation: [ r_B left(1 - frac{P_B}{K_B}right) + alpha K_A = 0 ] Solving for ( P_B ): [ r_B left(1 - frac{P_B}{K_B}right) = -alpha K_A ] [ 1 - frac{P_B}{K_B} = -frac{alpha K_A}{r_B} ] [ frac{P_B}{K_B} = 1 + frac{alpha K_A}{r_B} ] [ P_B = K_B left(1 + frac{alpha K_A}{r_B}right) ] Wait, but ( P_B ) can't exceed the carrying capacity unless the interaction term allows it. But in reality, carrying capacity is a limit, so if ( 1 + frac{alpha K_A}{r_B} > 1 ), then ( P_B ) would be greater than ( K_B ), which isn't possible. Hmm, that suggests that this steady state is only possible if ( 1 + frac{alpha K_A}{r_B} leq 1 ), which would require ( alpha leq 0 ). But ( alpha ) is a mutualistic interaction coefficient, which should be positive because mutualism benefits both species. Wait, that seems contradictory. Let me double-check. The equation for Strain B's steady state when ( P_A = K_A ): [ r_B left(1 - frac{P_B}{K_B}right) + alpha K_A = 0 ] Rearranged: [ r_B left(1 - frac{P_B}{K_B}right) = -alpha K_A ] Since ( r_B ) and ( K_A ) are positive, and ( alpha ) is positive (mutualism), the right side is negative. Therefore, ( 1 - frac{P_B}{K_B} ) must be negative, which implies ( P_B > K_B ). But that's impossible because ( K_B ) is the carrying capacity. So, this suggests that when Strain A is at ( K_A ), Strain B cannot reach a steady state unless ( P_B > K_B ), which isn't feasible. Therefore, the only possible steady states are when one of the strains is extinct. Wait, but that doesn't make sense because mutualism should allow both to coexist. Maybe I made a mistake in the analysis. Let me think again. If both strains are present, then Strain A is at ( K_A ), but Strain B's equation requires ( P_B ) to be greater than ( K_B ), which isn't possible. Therefore, perhaps mutualism in this model doesn't lead to a stable coexistence, but rather one strain outcompetes the other. Alternatively, maybe I need to consider that Strain A's growth affects Strain B's carrying capacity. Let me see. Alternatively, perhaps the steady state for Strain B when Strain A is at ( K_A ) is: [ r_B left(1 - frac{P_B}{K_B}right) + alpha K_A = 0 ] So, [ 1 - frac{P_B}{K_B} = -frac{alpha K_A}{r_B} ] [ frac{P_B}{K_B} = 1 + frac{alpha K_A}{r_B} ] [ P_B = K_B left(1 + frac{alpha K_A}{r_B}right) ] But as I said, this would require ( P_B > K_B ), which isn't possible. Therefore, this suggests that when Strain A is at ( K_A ), Strain B cannot reach a steady state unless ( alpha ) is negative, which would be a competitive interaction, not mutualistic. Therefore, perhaps the only steady states are when one strain is extinct. So, either both are extinct, Strain A is at ( K_A ) and Strain B is extinct, or Strain B is at ( K_B ) and Strain A is extinct. But that seems counterintuitive because mutualism should allow both to coexist. Maybe the model is set up in a way that mutualism doesn't allow coexistence unless certain conditions are met. Alternatively, perhaps I need to consider that Strain A's growth affects Strain B's carrying capacity. Let me see. Wait, in the equation for Strain B, the term ( alpha P_A P_B ) is added, which is positive because ( alpha ) is positive. So, when Strain A is present, it increases Strain B's growth rate. Therefore, Strain B can grow even when its own logistic term is negative, as long as the mutualistic term is positive enough. So, perhaps the steady state for Strain B is when the two terms balance each other. Let me try to find the steady state where both are non-zero. So, setting ( frac{dP_B}{dt} = 0 ): [ r_B P_B left(1 - frac{P_B}{K_B}right) + alpha P_A P_B = 0 ] Factor out ( P_B ): [ P_B left[ r_B left(1 - frac{P_B}{K_B}right) + alpha P_A right] = 0 ] So, either ( P_B = 0 ) or: [ r_B left(1 - frac{P_B}{K_B}right) + alpha P_A = 0 ] Rearranged: [ r_B left(1 - frac{P_B}{K_B}right) = -alpha P_A ] Since ( r_B ) and ( alpha ) are positive, the left side must be negative, which implies ( 1 - frac{P_B}{K_B} < 0 ), so ( P_B > K_B ). But again, this is impossible because ( P_B ) can't exceed ( K_B ) in the logistic model. Therefore, the only solution is ( P_B = 0 ). Wait, that can't be right. If mutualism is present, Strain B should be able to coexist with Strain A. Maybe I'm missing something. Alternatively, perhaps the model is such that Strain B's carrying capacity is effectively increased by the presence of Strain A. Let me think about it differently. Suppose Strain A is at ( K_A ), then the equation for Strain B becomes: [ frac{dP_B}{dt} = r_B P_B left(1 - frac{P_B}{K_B}right) + alpha K_A P_B ] Let me rewrite this: [ frac{dP_B}{dt} = P_B left[ r_B left(1 - frac{P_B}{K_B}right) + alpha K_A right] ] For this to reach a steady state, the term in brackets must be zero: [ r_B left(1 - frac{P_B}{K_B}right) + alpha K_A = 0 ] Solving for ( P_B ): [ 1 - frac{P_B}{K_B} = -frac{alpha K_A}{r_B} ] [ frac{P_B}{K_B} = 1 + frac{alpha K_A}{r_B} ] [ P_B = K_B left(1 + frac{alpha K_A}{r_B}right) ] But as before, this implies ( P_B > K_B ), which isn't possible. Therefore, in this model, mutualism doesn't allow both strains to coexist at a steady state. Instead, it leads to Strain B exceeding its carrying capacity, which isn't feasible. Therefore, the only steady states are when one strain is extinct. Alternatively, perhaps the model allows for a steady state where Strain A is not at ( K_A ). Wait, but from Strain A's equation, the only steady states are ( P_A = 0 ) or ( P_A = K_A ). So, if Strain A is not at ( K_A ), it must be at 0. Therefore, the only possible steady states are: - Both extinct: ( P_A = 0 ), ( P_B = 0 ) - Strain A at ( K_A ), Strain B extinct: ( P_A = K_A ), ( P_B = 0 ) - Strain B at ( K_B ), Strain A extinct: ( P_A = 0 ), ( P_B = K_B ) But wait, if Strain A is extinct, then Strain B's equation becomes: [ frac{dP_B}{dt} = r_B P_B left(1 - frac{P_B}{K_B}right) ] So, Strain B can reach ( K_B ) on its own. Therefore, the steady states are as above. However, mutualism usually allows for coexistence, so perhaps this model isn't capturing that. Maybe the interaction term is too strong, forcing one strain to outcompete the other. Alternatively, perhaps the steady state where both are present requires that Strain A is not at ( K_A ). But from Strain A's equation, it can't be at any other steady state except 0 or ( K_A ). Therefore, in this model, mutualism doesn't lead to coexistence. Instead, it's a competition where one strain outcompetes the other. Therefore, the conditions for steady state are: - Both strains extinct. - Strain A at ( K_A ), Strain B extinct. - Strain B at ( K_B ), Strain A extinct. But wait, if Strain A is at ( K_A ), Strain B can't reach a steady state unless ( P_B > K_B ), which isn't possible. Therefore, the only feasible steady states are when one strain is extinct. So, in conclusion, the coupled system has steady states where either both are extinct, or one is at its carrying capacity and the other is extinct. Mutualism in this model doesn't allow for both to coexist at a steady state. Therefore, the conditions for steady state are: - ( P_A = 0 ) and ( P_B = 0 ) - ( P_A = K_A ) and ( P_B = 0 ) - ( P_A = 0 ) and ( P_B = K_B ) However, whether these are stable depends on the parameters. For example, if Strain A is at ( K_A ), Strain B's population would tend to grow beyond ( K_B ), which isn't possible, so Strain B would either die out or the model would break down. Therefore, the only stable steady states are when one strain is extinct. Alternatively, perhaps I need to consider that when Strain A is present, it can support a higher population of Strain B, effectively increasing its carrying capacity. But in the model, the carrying capacity ( K_B ) is fixed, so the mutualistic term can't increase ( K_B ). Therefore, Strain B can't exceed ( K_B ), so the mutualistic term can only help Strain B reach ( K_B ) faster, but not exceed it. Wait, let me think about this again. If Strain A is present, the term ( alpha P_A P_B ) adds to Strain B's growth rate. So, even if Strain B's own logistic term is negative (i.e., ( P_B > K_B )), the mutualistic term could still make the overall growth rate positive, allowing Strain B to grow beyond ( K_B ). But in reality, ( K_B ) is the maximum population due to resource limitations, so Strain B can't exceed ( K_B ). Therefore, the model might not be realistic in that aspect. Alternatively, perhaps the mutualistic interaction increases Strain B's carrying capacity. If that's the case, the model should have ( K_B ) replaced by ( K_B + alpha P_A ), but that's not how it's written. The equation is: [ frac{dP_B}{dt} = r_B P_B left(1 - frac{P_B}{K_B}right) + alpha P_A P_B ] So, the mutualistic term is additive, not multiplicative on the carrying capacity. Therefore, it's possible for Strain B to grow beyond ( K_B ) if ( alpha P_A ) is large enough, which isn't realistic. Therefore, perhaps the model is flawed in that aspect, or it's assuming that the mutualistic interaction allows Strain B to exceed its carrying capacity, which might not be the case in reality. Given that, perhaps the only feasible steady states are when one strain is extinct. Therefore, the conditions for steady state are when either both are extinct, or one is at its carrying capacity and the other is extinct. So, to summarize: - The coupled differential equations are the logistic equation for Strain A and the modified logistic equation for Strain B. - The steady states occur when either both populations are zero, or one population is at its carrying capacity and the other is zero. - Mutualism in this model doesn't allow for both populations to coexist at a steady state because the interaction term would require one population to exceed its carrying capacity, which isn't feasible. Therefore, the conditions for steady state are when one of the populations is extinct, and the other is at its carrying capacity, or both are extinct. I think that's the conclusion I can draw from this model.</think>

🧠:<think>Okay, so I have these two problems to solve, both related to drug interactions. Let me take them one at a time.Starting with the first problem: There's this function ( f(x, y) = x^2 + y^2 - 2xycos(theta) ). We're given that ( theta = frac{pi}{4} ) and ( C = 10 ). We need to find the values of ( x ) and ( y ) such that ( f(x, y) = 10 ).Hmm, let me think. The function looks a bit like the law of cosines. In trigonometry, the law of cosines is ( c^2 = a^2 + b^2 - 2abcos(C) ), where ( C ) is the angle opposite side ( c ). So, if I compare that to our function ( f(x, y) ), it seems similar. So, maybe ( f(x, y) ) represents the square of the length of a side in a triangle where ( x ) and ( y ) are the other two sides and ( theta ) is the angle between them.If that's the case, then ( f(x, y) = C ) would mean that the length of the side opposite the angle ( theta ) is ( sqrt{C} ). But wait, in our case, ( C = 10 ), so the length would be ( sqrt{10} ). But I'm not sure if that helps directly. Maybe I need to think differently.Alternatively, perhaps I can rewrite the function ( f(x, y) ). Let me try expanding it:( f(x, y) = x^2 + y^2 - 2xycos(theta) ).Given ( theta = frac{pi}{4} ), so ( cos(frac{pi}{4}) = frac{sqrt{2}}{2} ). Plugging that in:( f(x, y) = x^2 + y^2 - 2xy cdot frac{sqrt{2}}{2} ).Simplify that:( f(x, y) = x^2 + y^2 - sqrt{2}xy ).So, the equation becomes:( x^2 + y^2 - sqrt{2}xy = 10 ).Hmm, this is a quadratic equation in two variables. I need to find all pairs ( (x, y) ) that satisfy this equation. It might be helpful to think of this as a quadratic form or perhaps complete the square.Alternatively, maybe I can use substitution. Let me try expressing ( y ) in terms of ( x ) or vice versa.Let me rearrange the equation:( x^2 - sqrt{2}xy + y^2 = 10 ).This looks like a quadratic in ( x ):( x^2 - sqrt{2}y x + y^2 - 10 = 0 ).Using the quadratic formula for ( x ):( x = frac{ sqrt{2}y pm sqrt{ ( sqrt{2}y )^2 - 4 cdot 1 cdot (y^2 - 10) } }{ 2 } ).Calculating the discriminant:( ( sqrt{2}y )^2 - 4(y^2 - 10) = 2y^2 - 4y^2 + 40 = -2y^2 + 40 ).So, the discriminant is ( -2y^2 + 40 ). For real solutions, the discriminant must be non-negative:( -2y^2 + 40 geq 0 )Multiply both sides by (-1) (remember to reverse inequality):( 2y^2 - 40 leq 0 )Divide by 2:( y^2 - 20 leq 0 )So,( y^2 leq 20 )Which means:( |y| leq sqrt{20} = 2sqrt{5} approx 4.472 ).So, ( y ) must be between ( -2sqrt{5} ) and ( 2sqrt{5} ).Now, plugging back into the expression for ( x ):( x = frac{ sqrt{2}y pm sqrt{ -2y^2 + 40 } }{ 2 } ).Simplify the square root:( sqrt{ -2y^2 + 40 } = sqrt{ 40 - 2y^2 } = sqrt{2(20 - y^2)} = sqrt{2} sqrt{20 - y^2} ).So, substituting back:( x = frac{ sqrt{2}y pm sqrt{2} sqrt{20 - y^2} }{ 2 } = frac{ sqrt{2} ( y pm sqrt{20 - y^2} ) }{ 2 } ).Factor out ( sqrt{2}/2 ):( x = frac{ sqrt{2} }{ 2 } ( y pm sqrt{20 - y^2} ) ).Simplify ( sqrt{2}/2 ) as ( 1/sqrt{2} ):( x = frac{ y pm sqrt{20 - y^2} }{ sqrt{2} } ).So, that's an expression for ( x ) in terms of ( y ). Similarly, if I had solved for ( y ) in terms of ( x ), I would get a similar expression.Alternatively, maybe I can parametrize this equation. Let me think about it as a quadratic form.The equation is:( x^2 - sqrt{2}xy + y^2 = 10 ).This is a quadratic equation, and it represents a conic section. Let me find its type.The general form is ( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 ). In our case, ( A = 1 ), ( B = -sqrt{2} ), ( C = 1 ), and ( D = E = 0 ), ( F = -10 ).To determine the type of conic, we can compute the discriminant ( B^2 - 4AC ).Here, ( B^2 - 4AC = ( -sqrt{2} )^2 - 4(1)(1) = 2 - 4 = -2 ).Since the discriminant is negative, it's an ellipse.So, the equation represents an ellipse. Therefore, there are infinitely many solutions ( (x, y) ) lying on this ellipse.But the problem asks to "find the values of ( x ) and ( y )" such that ( f(x, y) = 10 ). It doesn't specify any additional constraints, so I think the answer is all pairs ( (x, y) ) lying on this ellipse.But maybe the question is expecting specific solutions, perhaps in terms of parametric equations or something else.Alternatively, perhaps I can write this equation in matrix form:( [x quad y] begin{bmatrix} 1 & -sqrt{2}/2 -sqrt{2}/2 & 1 end{bmatrix} begin{bmatrix} x y end{bmatrix} = 10 ).But I'm not sure if that helps directly.Alternatively, maybe I can diagonalize the quadratic form by rotating the axes to eliminate the cross term.Yes, that might be a good approach. Since it's an ellipse, we can rotate the coordinate system to remove the ( xy ) term.The angle of rotation ( phi ) needed to eliminate the cross term is given by:( tan(2phi) = frac{B}{A - C} ).In our case, ( B = -sqrt{2} ), ( A = 1 ), ( C = 1 ).So,( tan(2phi) = frac{ -sqrt{2} }{ 1 - 1 } = frac{ -sqrt{2} }{ 0 } ).Hmm, division by zero, which implies that ( 2phi = frac{pi}{2} ), so ( phi = frac{pi}{4} ).So, we need to rotate the axes by ( phi = frac{pi}{4} ).Let me recall the rotation formulas:( x = x'cosphi - y'sinphi )( y = x'sinphi + y'cosphi )Since ( phi = frac{pi}{4} ), ( cosphi = sinphi = frac{sqrt{2}}{2} ).So,( x = frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y' )( y = frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y' )Now, substitute these into the original equation:( x^2 + y^2 - sqrt{2}xy = 10 ).Let me compute each term:First, ( x^2 ):( x^2 = left( frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y' right)^2 = frac{2}{4}(x')^2 - 2 cdot frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2}x'y' + frac{2}{4}(y')^2 = frac{1}{2}(x')^2 - frac{1}{2}x'y' + frac{1}{2}(y')^2 ).Similarly, ( y^2 ):( y^2 = left( frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y' right)^2 = frac{2}{4}(x')^2 + 2 cdot frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2}x'y' + frac{2}{4}(y')^2 = frac{1}{2}(x')^2 + frac{1}{2}x'y' + frac{1}{2}(y')^2 ).Now, ( xy ):( xy = left( frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y' right)left( frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y' right) ).This is of the form ( (a - b)(a + b) = a^2 - b^2 ).So,( xy = left( frac{sqrt{2}}{2}x' right)^2 - left( frac{sqrt{2}}{2}y' right)^2 = frac{2}{4}(x')^2 - frac{2}{4}(y')^2 = frac{1}{2}(x')^2 - frac{1}{2}(y')^2 ).Now, putting it all together:( x^2 + y^2 - sqrt{2}xy = [ frac{1}{2}(x')^2 - frac{1}{2}x'y' + frac{1}{2}(y')^2 ] + [ frac{1}{2}(x')^2 + frac{1}{2}x'y' + frac{1}{2}(y')^2 ] - sqrt{2} [ frac{1}{2}(x')^2 - frac{1}{2}(y')^2 ] ).Simplify term by term:First, combine ( x^2 + y^2 ):( frac{1}{2}(x')^2 - frac{1}{2}x'y' + frac{1}{2}(y')^2 + frac{1}{2}(x')^2 + frac{1}{2}x'y' + frac{1}{2}(y')^2 )The ( x'y' ) terms cancel out:( (-frac{1}{2}x'y' + frac{1}{2}x'y') = 0 )So, we have:( frac{1}{2}(x')^2 + frac{1}{2}(y')^2 + frac{1}{2}(x')^2 + frac{1}{2}(y')^2 = ( frac{1}{2} + frac{1}{2} )(x')^2 + ( frac{1}{2} + frac{1}{2} )(y')^2 = (x')^2 + (y')^2 ).Now, subtract ( sqrt{2}xy ):( (x')^2 + (y')^2 - sqrt{2} left( frac{1}{2}(x')^2 - frac{1}{2}(y')^2 right ) ).Compute the subtraction:( (x')^2 + (y')^2 - frac{sqrt{2}}{2}(x')^2 + frac{sqrt{2}}{2}(y')^2 ).Factor terms:( left( 1 - frac{sqrt{2}}{2} right)(x')^2 + left( 1 + frac{sqrt{2}}{2} right)(y')^2 ).So, the equation becomes:( left( 1 - frac{sqrt{2}}{2} right)(x')^2 + left( 1 + frac{sqrt{2}}{2} right)(y')^2 = 10 ).Let me compute the coefficients:( 1 - frac{sqrt{2}}{2} approx 1 - 0.7071 approx 0.2929 )( 1 + frac{sqrt{2}}{2} approx 1 + 0.7071 approx 1.7071 )So, the equation is:( 0.2929(x')^2 + 1.7071(y')^2 = 10 ).To write it in standard ellipse form, divide both sides by 10:( frac{(x')^2}{10 / 0.2929} + frac{(y')^2}{10 / 1.7071} = 1 ).Compute the denominators:( 10 / 0.2929 approx 34.14 )( 10 / 1.7071 approx 5.86 )So, approximately:( frac{(x')^2}{34.14} + frac{(y')^2}{5.86} = 1 ).This is the standard form of an ellipse centered at the origin in the rotated coordinate system.Therefore, the solutions ( (x, y) ) correspond to all points on this ellipse, rotated back by ( phi = frac{pi}{4} ).But since the problem doesn't specify any particular constraints, like specific concentrations or additional equations, I think the answer is that all pairs ( (x, y) ) lying on this ellipse satisfy ( f(x, y) = 10 ).Alternatively, if we need to express ( x ) and ( y ) in terms of parameters, we could parameterize the ellipse.Let me denote:( a^2 = 34.14 ) so ( a approx 5.84 )( b^2 = 5.86 ) so ( b approx 2.42 )Then, the parametric equations in the rotated system are:( x' = a cos theta' )( y' = b sin theta' )Where ( theta' ) is the parameter varying from 0 to ( 2pi ).Then, converting back to the original coordinates:( x = frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y' = frac{sqrt{2}}{2}(a cos theta' - b sin theta') )( y = frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y' = frac{sqrt{2}}{2}(a cos theta' + b sin theta') )So, substituting ( a approx 5.84 ) and ( b approx 2.42 ):( x approx frac{sqrt{2}}{2}(5.84 cos theta' - 2.42 sin theta') )( y approx frac{sqrt{2}}{2}(5.84 cos theta' + 2.42 sin theta') )But since the problem doesn't specify a particular method or form, I think the answer is that all pairs ( (x, y) ) satisfying the ellipse equation ( x^2 + y^2 - sqrt{2}xy = 10 ) are solutions.Alternatively, if they expect specific values, perhaps assuming symmetry or something else, but without more information, I can't determine specific numerical values. So, I think the answer is the set of all ( (x, y) ) such that ( x^2 + y^2 - sqrt{2}xy = 10 ).Moving on to the second problem: Dr. Smith mentions the half-life of drug B is affected by the concentration of drug A. The half-life ( T ) is given by ( T = T_0 cdot e^{-alpha x} ), where ( T_0 = 8 ) hours, ( alpha = 0.1 ), and we need to find ( x ) such that ( T = 4 ) hours.So, let's plug in the known values:( 4 = 8 cdot e^{-0.1 x} ).Divide both sides by 8:( frac{4}{8} = e^{-0.1 x} )Simplify:( frac{1}{2} = e^{-0.1 x} )Take the natural logarithm of both sides:( lnleft( frac{1}{2} right) = -0.1 x )We know that ( lnleft( frac{1}{2} right) = -ln(2) approx -0.6931 ).So,( -0.6931 = -0.1 x )Multiply both sides by (-1):( 0.6931 = 0.1 x )Divide both sides by 0.1:( x = frac{0.6931}{0.1} = 6.931 ).So, ( x approx 6.931 ). Depending on the required precision, we can round it to, say, two decimal places: 6.93.But let me verify the calculation:Given ( T = T_0 e^{-alpha x} )We have ( T = 4 ), ( T_0 = 8 ), ( alpha = 0.1 ).So,( 4 = 8 e^{-0.1 x} )Divide both sides by 8:( 0.5 = e^{-0.1 x} )Take natural log:( ln(0.5) = -0.1 x )( ln(0.5) approx -0.6931 )So,( -0.6931 = -0.1 x )Multiply both sides by (-1):( 0.6931 = 0.1 x )Divide by 0.1:( x = 6.931 ).Yes, that seems correct. So, the concentration ( x ) of drug A needed is approximately 6.931.But let me express it more precisely. Since ( ln(2) approx 0.69314718056 ), so:( x = frac{ln(2)}{0.1} = 10 ln(2) approx 10 times 0.69314718056 approx 6.9314718056 ).So, ( x approx 6.931 ). Depending on the context, maybe we can write it as ( 10 ln(2) ), which is exact.But since the problem gives numerical values, probably expects a numerical answer. So, approximately 6.93.So, summarizing:1. The values of ( x ) and ( y ) lie on the ellipse defined by ( x^2 + y^2 - sqrt{2}xy = 10 ).2. The concentration ( x ) of drug A needed is approximately 6.93.Final Answer1. The concentrations ( x ) and ( y ) satisfy the equation ( x^2 + y^2 - sqrt{2}xy = 10 ). Thus, the solution is (boxed{x^2 + y^2 - sqrt{2}xy = 10}).2. The concentration ( x ) of drug A is (boxed{6.93}).</think>

🧠:<think>Okay, so I have this problem where an expert foodie is planning gourmet tours, and they want to maximize the satisfaction score based on the allocation of vegetables, fruits, and herbs. The satisfaction function is given by S(v, f, h) = 3v² + 4f² + 2h² + 6vf + 5fh - 4vh. The total available quantities are V=100, F=80, H=60, and the constraint is v + f + h ≤ 200. First, I need to figure out how to maximize S(v, f, h) under these constraints. Since it's a quadratic function, maybe I can use calculus to find the maximum. But before that, I should check if the function is concave or convex because that affects whether the critical point is a maximum or minimum.Looking at the function S(v, f, h), the quadratic terms have coefficients 3, 4, and 2 for v², f², and h² respectively. The cross terms are 6vf, 5fh, and -4vh. To determine concavity, I can look at the Hessian matrix. The Hessian will be a 3x3 matrix of second derivatives.The second derivatives for v² is 6, for f² is 8, and for h² is 4. The mixed partial derivatives: for vf, it's 6 (since the first derivative with respect to v is 6f + 6v -4h, so the second derivative with respect to f is 6). Similarly, for fh, it's 5, and for vh, it's -4.So the Hessian matrix H is:[6, 6, -4][6, 8, 5][-4, 5, 4]To check if it's positive definite or negative definite, we can look at the leading principal minors. If all leading principal minors are positive, it's positive definite; if they alternate starting with negative, it's negative definite.First minor: 6 > 0.Second minor: determinant of top-left 2x2 matrix:|6 6||6 8| = 6*8 - 6*6 = 48 - 36 = 12 > 0.Third minor: determinant of the full Hessian. Let's compute that.Compute determinant:6*(8*4 - 5*5) - 6*(6*4 - 5*(-4)) + (-4)*(6*5 - 8*(-4))First term: 6*(32 - 25) = 6*7 = 42Second term: -6*(24 + 20) = -6*44 = -264Third term: -4*(30 + 32) = -4*62 = -248So total determinant: 42 - 264 - 248 = 42 - 512 = -470 < 0Since the third leading principal minor is negative, the Hessian is indefinite. That means the function is neither concave nor convex, so there might be a saddle point. Hmm, that complicates things because if it's indefinite, the critical point might not be a maximum or minimum.But maybe we can still find a critical point and see if it's a maximum. Alternatively, since the problem is constrained, maybe we can use Lagrange multipliers.So, let's set up the Lagrangian. The function to maximize is S(v, f, h) = 3v² + 4f² + 2h² + 6vf + 5fh - 4vh, subject to the constraint v + f + h ≤ 200. Also, we have the availability constraints: v ≤ 100, f ≤ 80, h ≤ 60.But since the total available is V=100, F=80, H=60, and the total allocation is v + f + h ≤ 200, which is more than the sum of available (100+80+60=240). Wait, actually, 200 is less than 240. So, the constraint is v + f + h ≤ 200, but each variable is also bounded by their own maximums.So, perhaps the maximum allocation is 200, but each variable can't exceed their own limits. So, the feasible region is defined by v ≤ 100, f ≤80, h ≤60, and v + f + h ≤200.But since 100 +80 +60=240>200, the sum constraint is binding.So, we can set up the Lagrangian with the sum constraint and the individual constraints.But since the Hessian is indefinite, the critical point might not be a maximum, so maybe the maximum occurs at the boundary of the feasible region.Alternatively, perhaps we can assume that the maximum occurs at the interior point where the gradient of S is proportional to the gradient of the constraint.So, let's set up the Lagrangian:L = 3v² + 4f² + 2h² + 6vf + 5fh - 4vh - λ(v + f + h - 200)Take partial derivatives:∂L/∂v = 6v + 6f -4h - λ = 0∂L/∂f = 8f + 6v +5h - λ = 0∂L/∂h = 4h +5f -4v - λ = 0∂L/∂λ = v + f + h -200 =0So, we have four equations:1. 6v +6f -4h = λ2. 6v +8f +5h = λ3. -4v +5f +4h = λ4. v + f + h =200Now, let's subtract equation 1 from equation 2:(6v +8f +5h) - (6v +6f -4h) = 0This simplifies to 2f +9h =0But f and h are quantities, so they can't be negative. 2f +9h=0 implies f=0 and h=0, which contradicts the availability. So, this suggests that the critical point is not in the interior, but on the boundary.Therefore, the maximum must occur at the boundary of the feasible region.So, we need to check the boundaries. The boundaries are defined by either v=100, f=80, h=60, or v + f + h=200.But since the sum constraint is v + f + h ≤200, and the maximum sum is 200, the maximum might occur at the sum constraint.But also, individual constraints may limit the variables.So, perhaps the maximum occurs when v=100, f=80, h=20 (since 100+80+20=200). But h can't exceed 60, so 20 is fine.Alternatively, maybe v=100, f=80, h=20 is the allocation.But let's check if that's the case.Alternatively, maybe the maximum occurs when v=100, f=80, h=20, but we need to check if that's the optimal.Alternatively, maybe we can set up the problem with v, f, h as variables, with v ≤100, f ≤80, h ≤60, and v + f + h=200.So, let's assume that the maximum occurs at the sum constraint, so v + f + h=200.Now, we can express h=200 -v -f, and substitute into S(v, f, h).So, S(v, f, h) = 3v² +4f² +2h² +6vf +5fh -4vhSubstitute h=200 -v -f:S(v, f) =3v² +4f² +2(200 -v -f)² +6vf +5f(200 -v -f) -4v(200 -v -f)Let me expand this step by step.First, expand 2(200 -v -f)²:=2*(40000 -400v -400f +v² +2vf +f²)=80000 -800v -800f +2v² +4vf +2f²Next, expand 5f(200 -v -f):=1000f -5vf -5f²Next, expand -4v(200 -v -f):=-800v +4v² +4vfNow, put all together:S(v, f) =3v² +4f² + [80000 -800v -800f +2v² +4vf +2f²] +6vf + [1000f -5vf -5f²] + [-800v +4v² +4vf]Now, let's combine like terms.First, constants: 80000v terms: -800v -800v = -1600vf terms: -800f +1000f = 200fv² terms: 3v² +2v² +4v² =9v²f² terms:4f² +2f² -5f²=1f²vf terms:4vf +6vf -5vf +4vf=9vfSo, S(v, f)=9v² +f² +9vf -1600v +200f +80000Now, we can take partial derivatives with respect to v and f to find the critical points.∂S/∂v =18v +9f -1600 =0∂S/∂f =2f +9v +200 =0So, we have the system:18v +9f =16009v +2f =-200Wait, that can't be right because the second equation would imply negative values, which is impossible since v and f are positive.Wait, let me check the partial derivatives again.Wait, S(v, f)=9v² +f² +9vf -1600v +200f +80000So, ∂S/∂v=18v +9f -1600=0∂S/∂f=2f +9v +200=0So, the second equation is 2f +9v +200=0But 2f +9v +200=0 implies 2f= -9v -200, which would make f negative, which is impossible because f ≥0.Therefore, the critical point is not in the feasible region, so the maximum must occur on the boundary.Therefore, we need to check the boundaries where either v=100, f=80, or h=60.But since v + f + h=200, if we set one variable to its maximum, the others are adjusted accordingly.Let's consider the cases:Case 1: v=100Then, f + h=100But f ≤80, h ≤60So, f can be up to 80, h=20Or f=80, h=20Alternatively, f can be less than 80, h more than 20, but h can't exceed 60.So, let's set v=100, f=80, h=20Compute S(100,80,20):=3*(100)^2 +4*(80)^2 +2*(20)^2 +6*100*80 +5*80*20 -4*100*20=3*10000 +4*6400 +2*400 +6*8000 +5*1600 -4*2000=30000 +25600 +800 +48000 +8000 -8000Compute step by step:30000 +25600=5560055600 +800=5640056400 +48000=104400104400 +8000=112400112400 -8000=104400So, S=104400Case 2: f=80Then, v + h=120But v ≤100, h ≤60So, if h=60, then v=60So, v=60, f=80, h=60Compute S(60,80,60):=3*60² +4*80² +2*60² +6*60*80 +5*80*60 -4*60*60=3*3600 +4*6400 +2*3600 +6*4800 +5*4800 -4*3600=10800 +25600 +7200 +28800 +24000 -14400Compute step by step:10800 +25600=3640036400 +7200=4360043600 +28800=7240072400 +24000=9640096400 -14400=82000So, S=82000Case 3: h=60Then, v + f=140But v ≤100, f ≤80So, if v=100, f=40Compute S(100,40,60):=3*100² +4*40² +2*60² +6*100*40 +5*40*60 -4*100*60=3*10000 +4*1600 +2*3600 +6*4000 +5*2400 -4*6000=30000 +6400 +7200 +24000 +12000 -24000Compute step by step:30000 +6400=3640036400 +7200=4360043600 +24000=6760067600 +12000=7960079600 -24000=55600So, S=55600Now, compare the three cases:Case1: S=104400Case2: S=82000Case3: S=55600So, the maximum is in Case1: v=100, f=80, h=20, with S=104400But wait, let's check if there are other boundary cases where two variables are at their maximums.For example, v=100, f=80, h=20 is already considered.Alternatively, v=100, h=60, then f=40, which is Case3.Alternatively, f=80, h=60, then v=60, which is Case2.So, the maximum is indeed in Case1.But let's also check if the sum constraint is binding. Since v=100, f=80, h=20, sum=200, which is exactly the constraint. So, that's the optimal allocation.Now, for part 2, the cost function is given by C(v, f, h)=2v +3f +1h.We need to find the maximum allowable cost C_max such that S(v, f, h)≥1500 and C(v, f, h)≤500.Wait, but in part1, we found that the maximum S is 104400, which is way above 1500. So, perhaps the question is to find the maximum C_max such that S≥1500 and C≤C_max.But the wording is: "determine the maximum allowable cost C_max that ensures a high satisfaction score S≥1500 while maintaining the total cost below 500."Wait, so the total cost must be below 500, and we need to find the maximum C_max such that S≥1500.Wait, but if the cost is constrained to be below 500, then C_max is 500, but we need to ensure that S≥1500.But in part1, the maximum S is 104400, which is much higher than 1500, so perhaps the question is to find the maximum C such that S≥1500, but C≤500.Wait, but the wording is: "formulate the cost function C(v, f, h) and determine the maximum allowable cost C_max that ensures a high satisfaction score S(v, f, h)≥1500 while maintaining the total cost below 500."Hmm, perhaps it's to find the maximum C_max such that S≥1500 and C≤500. But since S can be as high as 104400, which is way above 1500, the maximum C_max is 500, but we need to check if there's a feasible allocation where C=500 and S≥1500.Alternatively, perhaps the question is to find the maximum C_max such that S≥1500 and C≤C_max, but the total cost must be below 500, so C_max is 500.But maybe I need to find the maximum C such that S≥1500 and C≤500.Wait, perhaps the problem is to find the maximum C_max such that there exists an allocation where S≥1500 and C≤C_max, but C_max is as large as possible without exceeding 500.Wait, but since C is constrained to be below 500, the maximum C_max is 500, but we need to ensure that there's an allocation where S≥1500 and C≤500.Alternatively, perhaps the question is to find the maximum C such that S≥1500, and C is as large as possible without exceeding 500.Wait, maybe I need to set up an optimization problem where we maximize C(v, f, h)=2v +3f +h, subject to S(v, f, h)≥1500 and v + f + h ≤200, v≤100, f≤80, h≤60, and C(v, f, h)≤500.But the question says: "determine the maximum allowable cost C_max that ensures a high satisfaction score S(v, f, h)≥1500 while maintaining the total cost below 500."So, perhaps we need to find the maximum C_max such that there exists an allocation where S≥1500 and C≤C_max, with C_max as large as possible, but not exceeding 500.But since 500 is the upper limit, perhaps C_max=500, but we need to check if there's an allocation with C=500 and S≥1500.Alternatively, maybe the question is to find the maximum C such that S≥1500 and C≤500.Wait, perhaps I need to set up the problem as maximizing C=2v +3f +h, subject to S(v, f, h)≥1500, v + f + h ≤200, v≤100, f≤80, h≤60, and C≤500.But since we need to find C_max, perhaps it's the maximum C such that S≥1500, but C≤500.Alternatively, maybe the question is to find the maximum C such that S≥1500, and C is as large as possible without exceeding 500.Wait, perhaps I need to use Lagrange multipliers again, but with the constraints S≥1500 and C≤500.But this is getting complicated. Maybe a better approach is to find the minimum cost required to achieve S=1500, and then set C_max as the minimum between that cost and 500.Alternatively, perhaps we can find the allocation that gives S=1500 with the minimum cost, and then if that cost is less than 500, then C_max=500, otherwise, C_max is the cost needed to achieve S=1500.But this is getting a bit involved. Maybe I can set up the problem as minimizing C=2v +3f +h, subject to S(v, f, h)=1500, and the other constraints.But since S is a quadratic function, it's not linear, so it's a nonlinear optimization problem.Alternatively, perhaps we can use the method of Lagrange multipliers with the equality constraint S=1500.But this might be too time-consuming.Alternatively, since in part1, the maximum S is 104400, which is way above 1500, so achieving S=1500 is trivial, and the cost can be as high as 500.Therefore, perhaps the maximum allowable cost is 500, as long as the satisfaction is at least 1500.But to be precise, we need to check if there's an allocation where C=500 and S≥1500.So, let's try to find such an allocation.We need to solve for v, f, h such that:2v +3f +h=500v + f + h ≤200v ≤100, f ≤80, h ≤60And S(v, f, h)≥1500But wait, 2v +3f +h=500 and v + f + h ≤200.Subtracting the second equation from the first:(2v +3f +h) - (v + f + h)=500 - (v + f + h)Which simplifies to v +2f=500 - (v + f + h)But since v + f + h ≤200, then 500 - (v + f + h) ≥300So, v +2f ≥300But v ≤100, f ≤80So, v +2f ≤100 +160=260But 260 <300, which is a contradiction.Therefore, there is no solution where 2v +3f +h=500 and v + f + h ≤200.Therefore, the maximum possible cost C_max is less than 500.So, we need to find the maximum C such that 2v +3f +h=C, v + f + h ≤200, v ≤100, f ≤80, h ≤60, and S(v, f, h)≥1500.This is a constrained optimization problem.Alternatively, perhaps we can find the maximum C such that S≥1500, and C is as large as possible.But this is getting too involved. Maybe a better approach is to use the method of Lagrange multipliers with the constraints.But given the time, perhaps I can assume that the maximum C_max is 500, but since it's impossible to reach 500 without violating the sum constraint, the maximum C_max is the maximum possible C under the sum constraint.So, the maximum C is when v=100, f=80, h=20, which gives C=2*100 +3*80 +1*20=200 +240 +20=460.So, C=460, which is less than 500, and S=104400≥1500.Therefore, the maximum allowable cost is 460, but since the question says "maintaining the total cost below 500", so 460 is below 500, so C_max=460.But wait, maybe we can increase C beyond 460 without violating the sum constraint.Wait, the sum constraint is v + f + h ≤200.If we set v + f + h=200, then the maximum C is achieved by maximizing 2v +3f +h.To maximize 2v +3f +h, given v + f + h=200, and v ≤100, f ≤80, h ≤60.We can set up the problem as maximizing 2v +3f +h, subject to v + f + h=200, v ≤100, f ≤80, h ≤60.This is a linear optimization problem.We can use the method of corners.The feasible region is defined by v + f + h=200, v ≤100, f ≤80, h ≤60.We can express h=200 -v -f, and substitute into the cost function:C=2v +3f + (200 -v -f)=v +2f +200We need to maximize C=v +2f +200, subject to:v ≤100f ≤80200 -v -f ≤60 => v + f ≥140v ≥0, f ≥0So, the constraints are:v ≤100f ≤80v + f ≥140v ≥0, f ≥0We can plot this in the v-f plane.The feasible region is a polygon with vertices at:1. v=100, f=80 (but v + f=180 ≥140)2. v=100, f=40 (since v + f=140)3. v=60, f=80 (since v + f=140)4. v=0, f=140 (but f=140>80, so not feasible)So, the feasible vertices are:(100,80), (100,40), (60,80)Compute C at these points:At (100,80): C=100 +2*80 +200=100+160+200=460At (100,40): C=100 +2*40 +200=100+80+200=380At (60,80): C=60 +2*80 +200=60+160+200=420So, the maximum C is 460 at (100,80), which is the same as in part1.Therefore, the maximum allowable cost C_max is 460, which is below 500, and S=104400≥1500.Therefore, the answer is C_max=460.</think>

🧠:<think>Okay, so I have this problem about population growth and resource allocation. Let me try to figure it out step by step. First, part 1 asks for the general solution of the logistic growth model. I remember that the logistic equation is a differential equation that models population growth considering a carrying capacity. The equation is given as:[ frac{dP}{dt} = rP left(1 - frac{P}{K}right) ]Where:- ( P(t) ) is the population at time ( t )- ( r ) is the intrinsic growth rate- ( K ) is the carrying capacityGiven values:- Initial population ( P(0) = 1000 )- Carrying capacity ( K = 10,000 )- Growth rate ( r = 0.1 )I need to find the general solution ( P(t) ). I recall that the logistic equation can be solved using separation of variables. Let me write down the equation again:[ frac{dP}{dt} = 0.1 P left(1 - frac{P}{10000}right) ]So, to solve this, I should separate the variables ( P ) and ( t ). Let me rewrite the equation:[ frac{dP}{P left(1 - frac{P}{10000}right)} = 0.1 dt ]Now, I need to integrate both sides. The left side looks a bit complicated, so I think partial fractions might be useful here. Let me set up the integral:[ int frac{1}{P left(1 - frac{P}{10000}right)} dP = int 0.1 dt ]Let me simplify the denominator on the left side. Let's make a substitution to make it easier. Let me denote ( u = frac{P}{10000} ), so ( P = 10000u ) and ( dP = 10000 du ). Substituting, the integral becomes:[ int frac{10000}{10000u (1 - u)} du = int 0.1 dt ]Simplify the constants:[ int frac{1}{u(1 - u)} du = int 0.0001 dt ]Wait, actually, let me check that substitution again. If ( P = 10000u ), then ( dP = 10000 du ). So substituting into the integral:Left side becomes:[ int frac{10000}{10000u (1 - u)} du = int frac{1}{u(1 - u)} du ]Right side:[ int 0.1 dt ]So, the equation is:[ int frac{1}{u(1 - u)} du = int 0.1 dt ]Now, I can use partial fractions on the left integral. Let me express ( frac{1}{u(1 - u)} ) as ( frac{A}{u} + frac{B}{1 - u} ).So:[ 1 = A(1 - u) + B u ]To find A and B, let me set ( u = 0 ):[ 1 = A(1 - 0) + B(0) implies A = 1 ]Set ( u = 1 ):[ 1 = A(1 - 1) + B(1) implies B = 1 ]So, the partial fractions decomposition is:[ frac{1}{u(1 - u)} = frac{1}{u} + frac{1}{1 - u} ]Therefore, the integral becomes:[ int left( frac{1}{u} + frac{1}{1 - u} right) du = int 0.1 dt ]Integrate term by term:Left side:[ int frac{1}{u} du + int frac{1}{1 - u} du = ln|u| - ln|1 - u| + C ]Right side:[ 0.1 t + C ]So, combining both sides:[ ln|u| - ln|1 - u| = 0.1 t + C ]Substitute back ( u = frac{P}{10000} ):[ lnleft|frac{P}{10000}right| - lnleft|1 - frac{P}{10000}right| = 0.1 t + C ]Simplify the left side using logarithm properties:[ lnleft( frac{P}{10000(1 - frac{P}{10000})} right) = 0.1 t + C ]Which simplifies to:[ lnleft( frac{P}{10000 - P} right) = 0.1 t + C ]Exponentiate both sides to eliminate the natural log:[ frac{P}{10000 - P} = e^{0.1 t + C} = e^{0.1 t} cdot e^{C} ]Let me denote ( e^{C} ) as a constant ( C' ) for simplicity:[ frac{P}{10000 - P} = C' e^{0.1 t} ]Now, solve for ( P ):Multiply both sides by ( 10000 - P ):[ P = C' e^{0.1 t} (10000 - P) ]Expand the right side:[ P = 10000 C' e^{0.1 t} - C' e^{0.1 t} P ]Bring all terms with ( P ) to the left:[ P + C' e^{0.1 t} P = 10000 C' e^{0.1 t} ]Factor out ( P ):[ P (1 + C' e^{0.1 t}) = 10000 C' e^{0.1 t} ]Solve for ( P ):[ P = frac{10000 C' e^{0.1 t}}{1 + C' e^{0.1 t}} ]Now, apply the initial condition ( P(0) = 1000 ) to find ( C' ). At ( t = 0 ):[ 1000 = frac{10000 C' e^{0}}{1 + C' e^{0}} implies 1000 = frac{10000 C'}{1 + C'} ]Multiply both sides by ( 1 + C' ):[ 1000 (1 + C') = 10000 C' ]Expand:[ 1000 + 1000 C' = 10000 C' ]Subtract ( 1000 C' ) from both sides:[ 1000 = 9000 C' implies C' = frac{1000}{9000} = frac{1}{9} ]So, ( C' = frac{1}{9} ). Substitute back into the equation for ( P ):[ P(t) = frac{10000 cdot frac{1}{9} e^{0.1 t}}{1 + frac{1}{9} e^{0.1 t}} ]Simplify numerator and denominator:Multiply numerator and denominator by 9 to eliminate the fraction:[ P(t) = frac{10000 e^{0.1 t}}{9 + e^{0.1 t}} ]Alternatively, we can factor out ( e^{0.1 t} ) in the denominator:[ P(t) = frac{10000 e^{0.1 t}}{e^{0.1 t} + 9} ]But both forms are correct. I think the first form is fine.So, that's the general solution for part 1.Moving on to part 2. Amina discovers that the community's resources are allocated based on the population size. The total available resources ( R(t) ) are modeled by ( R(t) = 10,000 ln(1 + t) ). The question is: How long will it take for the population to reach half of the carrying capacity, and what will be the corresponding amount of resources available at that time?First, half of the carrying capacity is ( frac{K}{2} = frac{10,000}{2} = 5,000 ). So, we need to find the time ( t ) when ( P(t) = 5,000 ).From part 1, we have the solution:[ P(t) = frac{10000 e^{0.1 t}}{9 + e^{0.1 t}} ]Set ( P(t) = 5000 ):[ 5000 = frac{10000 e^{0.1 t}}{9 + e^{0.1 t}} ]Let me solve for ( t ). Multiply both sides by ( 9 + e^{0.1 t} ):[ 5000 (9 + e^{0.1 t}) = 10000 e^{0.1 t} ]Expand the left side:[ 45,000 + 5000 e^{0.1 t} = 10,000 e^{0.1 t} ]Subtract ( 5000 e^{0.1 t} ) from both sides:[ 45,000 = 5,000 e^{0.1 t} ]Divide both sides by 5,000:[ 9 = e^{0.1 t} ]Take natural logarithm of both sides:[ ln 9 = 0.1 t ]Solve for ( t ):[ t = frac{ln 9}{0.1} ]Calculate ( ln 9 ). Since ( 9 = 3^2 ), ( ln 9 = 2 ln 3 approx 2 times 1.0986 = 2.1972 ).So,[ t approx frac{2.1972}{0.1} = 21.972 ]So, approximately 21.972 units of time. Depending on the context, if time is in years, it's about 22 years.Now, find the corresponding resources ( R(t) ) at this time. The resource function is:[ R(t) = 10,000 ln(1 + t) ]Substitute ( t approx 21.972 ):[ R(21.972) = 10,000 ln(1 + 21.972) = 10,000 ln(22.972) ]Calculate ( ln(22.972) ). Let me approximate this. I know that ( ln(20) approx 2.9957 ), ( ln(22) approx 3.0910 ), ( ln(23) approx 3.1355 ). Since 22.972 is just slightly less than 23, maybe around 3.135.But let me compute it more accurately. Let me use a calculator approximation:( ln(22.972) approx 3.135 ). Let me verify:( e^{3.135} approx e^{3} times e^{0.135} approx 20.0855 times 1.1447 approx 23.0 ). So, yes, ( ln(23) approx 3.135 ). Since 22.972 is just slightly less, maybe 3.134.So, ( R(t) approx 10,000 times 3.134 = 31,340 ).Alternatively, if I compute it more precisely, using a calculator:Compute ( ln(22.972) ):22.972 is approximately 23, so as above, it's about 3.135. Let me use a calculator:Using natural logarithm:ln(22.972) ≈ 3.135.So, ( R(t) ≈ 10,000 * 3.135 = 31,350 ).But let me check with a calculator:Compute 1 + t = 22.972ln(22.972) ≈ 3.135So, 10,000 * 3.135 = 31,350.But to be precise, let me compute ln(22.972):Using a calculator:ln(22.972) ≈ 3.135Yes, so 31,350.Alternatively, if I use more exact value:Let me compute 22.972:Compute ln(22.972):We can write 22.972 = 23 - 0.028So, ln(23 - 0.028) ≈ ln(23) - (0.028)/23 ≈ 3.1355 - 0.0012 ≈ 3.1343So, ln(22.972) ≈ 3.1343Therefore, R(t) ≈ 10,000 * 3.1343 ≈ 31,343So, approximately 31,343 resources.But since the question says "the corresponding amount of resources", so we can write it as approximately 31,343.Alternatively, if we use the exact value of t, which is ( t = frac{ln 9}{0.1} ), then:Compute R(t) = 10,000 ln(1 + t) = 10,000 ln(1 + (ln 9)/0.1 )But that might not simplify much. Alternatively, we can express it in terms of ln(9):Wait, let me see:t = (ln 9)/0.1 = 10 ln 9So, 1 + t = 1 + 10 ln 9Therefore, R(t) = 10,000 ln(1 + 10 ln 9 )But that's more complicated. Alternatively, compute numerically:ln(9) ≈ 2.1972So, t ≈ 21.972Thus, 1 + t ≈ 22.972ln(22.972) ≈ 3.135So, R(t) ≈ 31,350.So, summarizing:Time to reach half the carrying capacity is approximately 21.972 units, and resources at that time are approximately 31,350.But let me check if I did everything correctly.Wait, in part 1, I had:[ P(t) = frac{10000 e^{0.1 t}}{9 + e^{0.1 t}} ]Set P(t) = 5000:[ 5000 = frac{10000 e^{0.1 t}}{9 + e^{0.1 t}} ]Multiply both sides by denominator:5000*(9 + e^{0.1 t}) = 10000 e^{0.1 t}Which gives:45,000 + 5000 e^{0.1 t} = 10,000 e^{0.1 t}Subtract 5000 e^{0.1 t}:45,000 = 5,000 e^{0.1 t}Divide by 5,000:9 = e^{0.1 t}Take ln:ln 9 = 0.1 tSo, t = (ln 9)/0.1 ≈ 2.1972 / 0.1 ≈ 21.972Yes, that's correct.Then, R(t) = 10,000 ln(1 + t) ≈ 10,000 ln(22.972) ≈ 10,000 * 3.135 ≈ 31,350.So, that seems correct.Alternatively, if I compute ln(22.972) more accurately:Let me use a calculator for ln(22.972):Using a calculator, ln(22.972) ≈ 3.135Yes, so 31,350 is accurate.Therefore, the time is approximately 21.972 units, and resources are approximately 31,350.But since the problem doesn't specify the units of time, I think it's safe to assume it's in years, so approximately 22 years, and resources around 31,350.But to be precise, let me compute ln(22.972) more accurately.Using a calculator:22.972Compute ln(22.972):We know that ln(20) ≈ 2.9957ln(22) ≈ 3.0910ln(23) ≈ 3.135522.972 is 23 - 0.028So, using the approximation ln(a - b) ≈ ln(a) - b/aSo, ln(23 - 0.028) ≈ ln(23) - 0.028/23 ≈ 3.1355 - 0.001217 ≈ 3.1343Therefore, ln(22.972) ≈ 3.1343Thus, R(t) = 10,000 * 3.1343 ≈ 31,343So, approximately 31,343 resources.So, to sum up:Time to reach half the carrying capacity: approximately 21.972 time units (let's say years), and resources available at that time: approximately 31,343.Alternatively, if we want to express t exactly, it's ( t = frac{ln 9}{0.1} = 10 ln 9 ). So, exact form is ( t = 10 ln 9 ), and R(t) = 10,000 ln(1 + 10 ln 9). But that's more complicated, so probably better to give the approximate numerical values.So, final answers:1. The general solution is ( P(t) = frac{10000 e^{0.1 t}}{9 + e^{0.1 t}} ).2. It takes approximately 21.972 years (or time units) for the population to reach 5,000, and the resources available at that time are approximately 31,343.Final Answer1. The general solution is (boxed{P(t) = dfrac{10000 e^{0.1 t}}{9 + e^{0.1 t}}}).2. The population reaches half the carrying capacity in approximately (boxed{21.97}) time units, with resources amounting to approximately (boxed{31343}).</think>