🧠:<think>Alright, so I have this problem about a retired NFL player who wants to organize a concert and needs help with the acoustics. He's modeled his concert hall as an ellipsoid, and he wants to find the focal points to place speakers optimally. Then, he also wants to ensure sound waves don't interfere destructively, so he's using the Helmholtz equation. Hmm, okay, let's break this down step by step.First, part 1 is about finding the focal points of the ellipsoid. The equation given is (frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} = 1), with (a = 30), (b = 20), and (c = 10). So, I remember that for an ellipsoid, the focal points are along the major axis. Since (a > b > c), the major axis is along the x-axis. I think the formula for the distance from the center to each focus is (c = sqrt{a^2 - b^2}) for an ellipse, but wait, this is a triaxial ellipsoid, so it's a bit different. Let me recall. For a triaxial ellipsoid, each pair of axes defines an ellipse, and the foci are located along the major axis of each of these cross-sections.But in this case, since it's a three-dimensional ellipsoid, the foci are along the major axis, which is the x-axis here. The distance from the center to each focus is given by (f = sqrt{a^2 - b^2}) if (a > b > c). Wait, but actually, I think it's (f = sqrt{a^2 - b^2}) if the major axis is along x, but I need to confirm.Wait, no, actually, for an ellipsoid, the foci are located at a distance of (f = sqrt{a^2 - b^2}) along the major axis if it's a prolate spheroid. But in this case, it's a triaxial ellipsoid, so maybe the formula is similar but considering all three axes? Hmm, I might be mixing things up.Let me think. For an ellipsoid, the foci are located along the major axis, and the distance from the center is given by (f = sqrt{a^2 - b^2}) if it's a prolate spheroid (where a > b = c). But in this case, all three axes are different: a=30, b=20, c=10. So, it's a triaxial ellipsoid.I think the formula for the distance from the center to each focus is still (f = sqrt{a^2 - b^2}), but only if the ellipsoid is prolate. Wait, no, actually, for a triaxial ellipsoid, the foci are still along the major axis, and the distance is (f = sqrt{a^2 - b^2}) if a > b > c. Let me check.Wait, actually, I think the formula is (f = sqrt{a^2 - b^2}) for the distance along the major axis, but I'm not entirely sure. Alternatively, maybe it's (f = sqrt{a^2 - c^2}) since c is the smallest. Hmm, this is confusing.Let me look it up in my mind. For an ellipsoid, the foci are located along the major axis, and the distance from the center is given by (f = sqrt{a^2 - b^2}) if it's a prolate spheroid (where a > b = c). But in a triaxial ellipsoid, where all three axes are different, the foci are still along the major axis, and the distance is (f = sqrt{a^2 - b^2}) if a > b > c. Wait, but in this case, a=30, b=20, c=10, so a > b > c, so the major axis is x, and the distance from center to foci is (f = sqrt{a^2 - b^2}).Wait, but actually, I think it's (f = sqrt{a^2 - c^2}) because c is the smallest. Hmm, no, that doesn't make sense because in a prolate spheroid, the foci are along the major axis, and the distance is (f = sqrt{a^2 - b^2}). So, in this case, since a > b > c, the major axis is x, and the distance should be (f = sqrt{a^2 - b^2}).Wait, but let me think again. For an ellipse in 2D, the distance from center to foci is (c = sqrt{a^2 - b^2}), where a > b. So, in 3D, for an ellipsoid, the foci are along the major axis, and the distance is similar. So, in this case, since a is the major axis, the distance should be (f = sqrt{a^2 - b^2}).Let me compute that. So, (a = 30), (b = 20). So, (f = sqrt{30^2 - 20^2} = sqrt{900 - 400} = sqrt{500} = 10sqrt{5}). So, approximately 22.36 units. Therefore, the focal points are at (±10√5, 0, 0). So, the coordinates are (10√5, 0, 0) and (-10√5, 0, 0).Wait, but hold on, in 3D, is it the same? Because in 3D, the ellipsoid has three axes, so maybe the foci are still along the major axis, but the formula is similar to the 2D case. So, yes, I think that's correct.So, for part 1, the focal points are at (±10√5, 0, 0). So, that's the answer.Now, moving on to part 2. He wants to ensure that sound waves from the focal points don't destructively interfere. He's using the Helmholtz equation: (nabla^2 phi + k^2 phi = 0), where (phi) is the sound pressure and (k) is the wave number. Given that the frequency is 440 Hz, and the speed of sound is 343 m/s, we need to find (k) and solve the Helmholtz equation in ellipsoidal coordinates.First, let's find the wave number (k). The wave number is given by (k = frac{omega}{c}), where (omega) is the angular frequency and (c) is the speed of sound. The angular frequency (omega) is (2pi f), where (f) is the frequency.So, let's compute that. (f = 440 text{ Hz}), so (omega = 2pi times 440 approx 2 times 3.1416 times 440 approx 2763.96 text{ rad/s}).Then, (k = frac{omega}{c} = frac{2763.96}{343} approx 8.06 text{ m}^{-1}).So, (k approx 8.06 text{ m}^{-1}).Now, solving the Helmholtz equation in ellipsoidal coordinates. Hmm, this seems more complicated. I remember that solving PDEs in ellipsoidal coordinates can be done using separation of variables, but it's quite involved.First, let's recall that ellipsoidal coordinates are a generalization of elliptic coordinates to three dimensions. They are defined by three parameters, and the coordinate surfaces are ellipsoids, hyperboloids, etc.The Helmholtz equation in ellipsoidal coordinates can be solved using the method of separation of variables. The solutions are products of functions of each coordinate, and they involve associated Legendre functions or spheroidal wave functions, depending on the coordinate system.But since the concert hall is modeled as an ellipsoid, we can use ellipsoidal coordinates to solve the Helmholtz equation. However, the exact solution would be quite complex and might involve special functions.Alternatively, since the problem is about ensuring that sound waves from the foci don't destructively interfere, perhaps we can consider the properties of ellipsoids and their focal points.In an ellipse, a wave emanating from one focus reflects off the ellipse and passes through the other focus. This property is used in whispering galleries. So, in 3D, for an ellipsoid, a sound wave emanating from one focus will reflect off the surface and pass through the other focus. This means that if you place a speaker at one focus, the sound will be concentrated at the other focus, creating a focal point.But in this case, the player wants to place high-quality speakers at the focal points. So, if he places speakers at both foci, the sound waves will emanate from both points and reflect off the ellipsoid surface, potentially creating constructive interference at certain points.However, to ensure that there is no destructive interference, we need to make sure that the sound waves from both speakers reach the audience at the same time or in a way that their phases don't cancel each other out.But solving the Helmholtz equation in ellipsoidal coordinates is non-trivial. I think the general solution involves a series expansion in terms of ellipsoidal harmonics, which are solutions to the Helmholtz equation in ellipsoidal coordinates.The Helmholtz equation in ellipsoidal coordinates can be written as:[frac{1}{sin theta cos phi} left[ frac{partial}{partial theta} left( sin theta cos phi frac{partial phi}{partial theta} right) + frac{partial}{partial phi} left( sin theta cos phi frac{partial phi}{partial phi} right) right] + frac{1}{r^2} frac{partial^2 phi}{partial r^2} + k^2 phi = 0]Wait, no, that's not quite right. The Helmholtz equation in ellipsoidal coordinates is more complex. Let me recall the form.In ellipsoidal coordinates, the Laplacian operator is more complicated. The Helmholtz equation would then involve the Laplacian in ellipsoidal coordinates plus (k^2 phi = 0).The general solution to the Helmholtz equation in ellipsoidal coordinates can be expressed as a sum of products of functions of each coordinate, involving ellipsoidal harmonics. These harmonics are solutions to the angular part of the equation and are related to associated Legendre functions or spheroidal wave functions.However, solving this equation requires a deep understanding of special functions and separation of variables in ellipsoidal coordinates, which is beyond my current knowledge. So, perhaps I can outline the steps involved.1. Express the Helmholtz equation in ellipsoidal coordinates.2. Use separation of variables to reduce the PDE to a set of ODEs.3. Solve the ODEs, which will involve special functions like ellipsoidal harmonics.4. Apply boundary conditions, such as the sound pressure being zero on the surface of the ellipsoid or some radiation condition at infinity.But since the concert hall is a bounded domain (the ellipsoid), we might impose Dirichlet boundary conditions, meaning the sound pressure is zero on the surface.However, given the complexity, I think the problem might be expecting a more conceptual answer rather than an explicit solution. Maybe it's about understanding that the Helmholtz equation governs the sound propagation and that the wave number is calculated as above.Alternatively, since the problem mentions solving the Helmholtz equation in ellipsoidal coordinates, perhaps it's expecting the expression for the wave number and an acknowledgment that the solution involves ellipsoidal harmonics, but without computing the exact solution.So, to summarize, for part 2:- The wave number (k) is approximately 8.06 m⁻¹.- Solving the Helmholtz equation in ellipsoidal coordinates involves separation of variables and results in solutions expressed in terms of ellipsoidal harmonics, which are special functions.But I'm not sure if that's sufficient. Maybe I need to write the general form of the solution.In ellipsoidal coordinates, the solution to the Helmholtz equation can be written as a sum over ellipsoidal harmonics, which are products of functions of each coordinate. The general solution is:[phi(r, theta, phi) = sum_{n=0}^{infty} sum_{m=-n}^{n} A_{nm} Gamma_n(r) Y_{nm}(theta, phi)]Where (Gamma_n(r)) are the radial functions and (Y_{nm}(theta, phi)) are the ellipsoidal harmonics, which generalize the spherical harmonics to ellipsoidal coordinates. The coefficients (A_{nm}) are determined by the boundary conditions.But without specific boundary conditions, we can't determine the exact form of the solution. However, if we consider the boundary condition that the sound pressure is zero on the surface of the ellipsoid, we can set up an eigenvalue problem where the wave number (k) must satisfy certain conditions for the solution to exist.But again, this is quite involved, and I might not be able to provide a complete solution here. So, perhaps the answer is just to compute the wave number and recognize that the solution involves ellipsoidal harmonics.So, to wrap up:1. The focal points are at (±10√5, 0, 0).2. The wave number (k) is approximately 8.06 m⁻¹, and the solution to the Helmholtz equation in ellipsoidal coordinates involves ellipsoidal harmonics, which are special functions used to describe wave propagation in such geometries.I think that's as far as I can go without delving into more advanced mathematical methods.</think>

🧠:<think>Alright, so I have this problem about modeling the growth of a rare medicinal plant influenced by lunar cycles and nutrient diffusion. It's divided into two parts: first, determining the constants in a sinusoidal growth model, and second, solving a diffusion equation for nutrient concentration. Let me tackle each part step by step.Part 1: Lunar Influence on Growth RateThe growth rate ( G(t) ) is given by:[G(t) = A sinleft( frac{2pi}{T} t right) + B]where ( T = 29.53 ) days. The maximum growth rate is 5 cm/day, and the minimum is 1 cm/day. I need to find ( A ) and ( B ).Hmm, okay. Since it's a sinusoidal function, I remember that the sine function oscillates between -1 and 1. So when we multiply by ( A ), it will oscillate between ( -A ) and ( A ). Then adding ( B ) shifts it vertically. Therefore, the maximum value of ( G(t) ) will be ( A + B ) and the minimum will be ( -A + B ).Given that the maximum is 5 and the minimum is 1, I can set up two equations:1. ( A + B = 5 )2. ( -A + B = 1 )Let me write these down:1. ( A + B = 5 )2. ( -A + B = 1 )If I subtract the second equation from the first, I can eliminate ( B ):( (A + B) - (-A + B) = 5 - 1 )Simplify:( A + B + A - B = 4 )( 2A = 4 )So, ( A = 2 ).Now plug ( A = 2 ) back into the first equation:( 2 + B = 5 )Thus, ( B = 3 ).Okay, so ( A = 2 ) and ( B = 3 ). That seems straightforward.Part 2: Nutrient DiffusionThe nutrient concentration ( N(x, t) ) follows the diffusion equation:[frac{partial N}{partial t} = D frac{partial^2 N}{partial x^2}]with ( D = 0.1 , text{cm}^2/text{day} ) and initial condition ( N(x, 0) = N_0 e^{-alpha x} ) where ( alpha = 0.5 , text{cm}^{-1} ). I need to find ( N(2, 10) ), the concentration at ( x = 2 , text{cm} ) after 10 days.Hmm, solving the diffusion equation. I remember that the solution to the diffusion equation with an initial condition can be found using the heat kernel or the fundamental solution. The general solution is a convolution of the initial condition with the heat kernel.The heat kernel (or Green's function) for the diffusion equation is:[G(x, t) = frac{1}{sqrt{4 pi D t}} e^{-x^2 / (4 D t)}]So, the solution ( N(x, t) ) is the convolution of ( N(x, 0) ) with ( G(x, t) ):[N(x, t) = int_{-infty}^{infty} N(x', 0) G(x - x', t) dx']But since our initial condition is ( N_0 e^{-alpha x} ), which is defined for all ( x ), but in reality, the concentration is probably zero for ( x < 0 ). Wait, the problem didn't specify, but since it's a depth, ( x ) is positive. So maybe we can consider ( x ) from 0 to infinity.But the initial condition is given as ( N(x, 0) = N_0 e^{-alpha x} ) for some constants ( N_0 ) and ( alpha ). So, it's an exponentially decaying function with depth.Therefore, the integral becomes:[N(x, t) = int_{0}^{infty} N_0 e^{-alpha x'} cdot frac{1}{sqrt{4 pi D t}} e^{-(x - x')^2 / (4 D t)} dx']Hmm, that seems a bit complicated. Maybe I can simplify this integral.Let me factor out constants:[N(x, t) = frac{N_0}{sqrt{4 pi D t}} int_{0}^{infty} e^{-alpha x'} e^{-(x - x')^2 / (4 D t)} dx']Combine the exponents:[e^{-alpha x'} e^{-(x - x')^2 / (4 D t)} = e^{-alpha x' - (x - x')^2 / (4 D t)}]Let me expand the quadratic term:[(x - x')^2 = x^2 - 2 x x' + x'^2]So, the exponent becomes:[- alpha x' - frac{x^2 - 2 x x' + x'^2}{4 D t}]Let me rewrite it:[- frac{x'^2}{4 D t} + left( frac{2 x}{4 D t} - alpha right) x' - frac{x^2}{4 D t}]Simplify the coefficients:- Coefficient of ( x'^2 ): ( -1/(4 D t) )- Coefficient of ( x' ): ( (x/(2 D t) - alpha) )- Constant term: ( -x^2/(4 D t) )So, the integral becomes:[int_{0}^{infty} e^{- frac{1}{4 D t} x'^2 + left( frac{x}{2 D t} - alpha right) x' - frac{x^2}{4 D t}} dx']This is a Gaussian integral, which can be evaluated using the formula:[int_{-infty}^{infty} e^{a x'^2 + b x' + c} dx' = sqrt{frac{pi}{-a}} e^{c - b^2/(4a)}]But our integral is from 0 to infinity, not the entire real line. Hmm, that complicates things. Maybe we can use the error function or something.Alternatively, perhaps we can complete the square in the exponent.Let me write the exponent as:[- frac{1}{4 D t} x'^2 + left( frac{x}{2 D t} - alpha right) x' - frac{x^2}{4 D t}]Let me factor out the coefficient of ( x'^2 ):[- frac{1}{4 D t} left( x'^2 - 4 D t left( frac{x}{2 D t} - alpha right) x' + 4 D t cdot frac{x^2}{4 D t} right)]Wait, that might not be the best approach. Let me try completing the square.Let me denote:( a = -1/(4 D t) )( b = (x/(2 D t) - alpha) )( c = -x^2/(4 D t) )So, the exponent is ( a x'^2 + b x' + c ).Completing the square:( a x'^2 + b x' = a left( x'^2 + (b/a) x' right) )Complete the square inside:( x'^2 + (b/a) x' = left( x' + frac{b}{2a} right)^2 - frac{b^2}{4a^2} )So, the exponent becomes:( a left( left( x' + frac{b}{2a} right)^2 - frac{b^2}{4a^2} right) + c )Simplify:( a left( x' + frac{b}{2a} right)^2 - frac{b^2}{4a} + c )So, substituting back:( a = -1/(4 D t) )( b = (x/(2 D t) - alpha) )( c = -x^2/(4 D t) )Thus, the exponent is:[- frac{1}{4 D t} left( x' + frac{(x/(2 D t) - alpha)}{2 cdot (-1/(4 D t))} right)^2 - frac{(x/(2 D t) - alpha)^2}{4 cdot (-1/(4 D t))} - frac{x^2}{4 D t}]Wait, this is getting messy. Let me compute each term step by step.First, compute ( frac{b}{2a} ):( b = frac{x}{2 D t} - alpha )( a = -1/(4 D t) )So, ( frac{b}{2a} = frac{ (frac{x}{2 D t} - alpha) }{ 2 cdot (-1/(4 D t)) } = frac{ (frac{x}{2 D t} - alpha) }{ -1/(2 D t) } = - (frac{x}{2 D t} - alpha) cdot 2 D t = -x + 2 D t alpha )Next, compute ( - frac{b^2}{4a} ):( b^2 = left( frac{x}{2 D t} - alpha right)^2 )( 4a = 4 cdot (-1/(4 D t)) = -1/D t )So, ( - frac{b^2}{4a} = - frac{ left( frac{x}{2 D t} - alpha right)^2 }{ -1/D t } = D t left( frac{x}{2 D t} - alpha right)^2 )Therefore, the exponent becomes:[- frac{1}{4 D t} left( x' - x + 2 D t alpha right)^2 + D t left( frac{x}{2 D t} - alpha right)^2 - frac{x^2}{4 D t}]Simplify the constant terms:First, expand ( D t left( frac{x}{2 D t} - alpha right)^2 ):[D t left( frac{x^2}{4 D^2 t^2} - frac{x alpha}{D t} + alpha^2 right ) = frac{x^2}{4 D t} - x alpha + D t alpha^2]So, the exponent is:[- frac{1}{4 D t} left( x' - x + 2 D t alpha right)^2 + frac{x^2}{4 D t} - x alpha + D t alpha^2 - frac{x^2}{4 D t}]Simplify the constants:The ( frac{x^2}{4 D t} ) and ( - frac{x^2}{4 D t} ) cancel out. So we're left with:[- frac{1}{4 D t} left( x' - x + 2 D t alpha right)^2 - x alpha + D t alpha^2]Therefore, the exponent is:[- frac{1}{4 D t} left( x' - x + 2 D t alpha right)^2 - x alpha + D t alpha^2]So, putting it all together, the integral becomes:[int_{0}^{infty} e^{- frac{1}{4 D t} left( x' - x + 2 D t alpha right)^2 - x alpha + D t alpha^2 } dx']Factor out the constants from the exponent:[e^{- x alpha + D t alpha^2} int_{0}^{infty} e^{- frac{1}{4 D t} left( x' - x + 2 D t alpha right)^2 } dx']Let me make a substitution to simplify the integral. Let ( y = x' - x + 2 D t alpha ). Then, when ( x' = 0 ), ( y = -x + 2 D t alpha ), and as ( x' to infty ), ( y to infty ). So, the integral becomes:[e^{- x alpha + D t alpha^2} int_{-x + 2 D t alpha}^{infty} e^{- frac{y^2}{4 D t} } dy]This integral is related to the error function. Recall that:[int_{a}^{infty} e^{- y^2 / (4 D t) } dy = sqrt{pi D t} cdot text{erfc}left( frac{a}{sqrt{4 D t}} right )]Where ( text{erfc} ) is the complementary error function.So, substituting back:[int_{-x + 2 D t alpha}^{infty} e^{- y^2 / (4 D t) } dy = sqrt{pi D t} cdot text{erfc}left( frac{ -x + 2 D t alpha }{ sqrt{4 D t} } right )]Simplify the argument of erfc:[frac{ -x + 2 D t alpha }{ sqrt{4 D t} } = frac{2 D t alpha - x}{2 sqrt{D t}} = sqrt{D t} alpha - frac{x}{2 sqrt{D t}}]So, putting it all together, the integral becomes:[sqrt{pi D t} cdot text{erfc}left( sqrt{D t} alpha - frac{x}{2 sqrt{D t}} right )]Therefore, the concentration ( N(x, t) ) is:[N(x, t) = frac{N_0}{sqrt{4 pi D t}} cdot e^{- x alpha + D t alpha^2} cdot sqrt{pi D t} cdot text{erfc}left( sqrt{D t} alpha - frac{x}{2 sqrt{D t}} right )]Simplify constants:( frac{1}{sqrt{4 pi D t}} cdot sqrt{pi D t} = frac{1}{2} )So,[N(x, t) = frac{N_0}{2} e^{- x alpha + D t alpha^2} cdot text{erfc}left( sqrt{D t} alpha - frac{x}{2 sqrt{D t}} right )]Hmm, that seems manageable. Now, let's plug in the given values:- ( D = 0.1 , text{cm}^2/text{day} )- ( alpha = 0.5 , text{cm}^{-1} )- ( x = 2 , text{cm} )- ( t = 10 , text{days} )First, compute ( sqrt{D t} ):( sqrt{0.1 times 10} = sqrt{1} = 1 )Compute ( sqrt{D t} alpha ):( 1 times 0.5 = 0.5 )Compute ( frac{x}{2 sqrt{D t}} ):( frac{2}{2 times 1} = 1 )So, the argument of erfc is:( 0.5 - 1 = -0.5 )But erfc is defined for all real numbers, and erfc(-z) = 2 - erfc(z). So,( text{erfc}(-0.5) = 2 - text{erfc}(0.5) )I need to compute ( text{erfc}(0.5) ). From tables or calculators, ( text{erfc}(0.5) approx 0.4795 ). Therefore, ( text{erfc}(-0.5) = 2 - 0.4795 = 1.5205 ).Next, compute the exponent:( -x alpha + D t alpha^2 = -2 times 0.5 + 0.1 times 10 times (0.5)^2 )Calculate each term:- ( -2 times 0.5 = -1 )- ( 0.1 times 10 = 1 )- ( (0.5)^2 = 0.25 )- So, ( 1 times 0.25 = 0.25 )Thus, total exponent: ( -1 + 0.25 = -0.75 )So, ( e^{-0.75} approx 0.4724 )Putting it all together:[N(2, 10) = frac{N_0}{2} times 0.4724 times 1.5205]Compute the constants:( 0.4724 times 1.5205 approx 0.4724 times 1.5205 approx 0.718 )So,( N(2, 10) approx frac{N_0}{2} times 0.718 = N_0 times 0.359 )Wait, but I don't know the value of ( N_0 ). The problem didn't specify it. Hmm, maybe I missed something.Looking back at the problem statement: "the initial concentration ( N(x, 0) = N_0 e^{-alpha x} ) for some constants ( N_0 ) and ( alpha )." It doesn't give a specific value for ( N_0 ), so perhaps we can express ( N(2, 10) ) in terms of ( N_0 ).Alternatively, maybe ( N_0 ) is a normalization constant such that the total initial concentration is finite. But without more information, I think we can only express ( N(2, 10) ) as ( 0.359 N_0 ).Wait, but maybe I made a mistake in the calculation. Let me double-check.Compute ( text{erfc}(-0.5) approx 1.5205 ) as before.Compute ( e^{-0.75} approx 0.4724 ).Multiply them: ( 0.4724 times 1.5205 approx 0.4724 * 1.5 ≈ 0.7086, plus 0.4724 * 0.0205 ≈ 0.0097, total ≈ 0.7183 ).Then, ( N(2, 10) = frac{N_0}{2} times 0.7183 ≈ N_0 times 0.35915 ).So, approximately ( 0.359 N_0 ).But since ( N_0 ) is a constant, unless it's given, we can't compute a numerical value. Maybe the problem expects an expression in terms of ( N_0 ), or perhaps ( N_0 ) is 1? The problem doesn't specify, so perhaps I should leave it as ( 0.359 N_0 ).Alternatively, maybe I should express it as ( frac{N_0}{2} e^{-0.75} text{erfc}(0.5 - 1) ), but that might not be necessary.Wait, actually, let me think again. The initial condition is ( N(x, 0) = N_0 e^{-alpha x} ). If we integrate this over all space, it would be ( N_0 int_{0}^{infty} e^{-alpha x} dx = N_0 / alpha ). So, unless ( N_0 ) is given, we can't find a numerical value. Therefore, I think the answer should be expressed in terms of ( N_0 ).So, summarizing:[N(2, 10) = frac{N_0}{2} e^{-0.75} text{erfc}(-0.5) approx 0.359 N_0]But since the problem might expect a numerical value, perhaps ( N_0 ) is 1? Or maybe it's a different approach.Wait, another thought: maybe the problem assumes that the initial condition is normalized such that ( N(0, 0) = N_0 ). So, if ( N_0 ) is the concentration at ( x=0 ) at ( t=0 ), then perhaps we can express the answer in terms of ( N_0 ). Since the problem doesn't specify ( N_0 ), I think we have to leave it as ( 0.359 N_0 ).Alternatively, maybe I made a mistake in the integral limits. Let me check.Wait, when I did the substitution ( y = x' - x + 2 D t alpha ), the lower limit became ( y = -x + 2 D t alpha ). For ( x = 2 ), ( D t = 1 ), so ( y = -2 + 2*1*0.5 = -2 + 1 = -1 ). So, the integral is from ( y = -1 ) to ( infty ).But in the expression, I used ( text{erfc}(-0.5) ). Wait, actually, the argument was ( sqrt{D t} alpha - frac{x}{2 sqrt{D t}} = 0.5 - 1 = -0.5 ). So, that's correct.Therefore, I think my calculation is correct, and the result is ( N(2, 10) approx 0.359 N_0 ).But since the problem didn't specify ( N_0 ), maybe I should leave it in terms of ( N_0 ). Alternatively, perhaps ( N_0 ) is 1, making ( N(2, 10) approx 0.359 ).Wait, let me check the problem statement again: "the initial concentration ( N(x, 0) = N_0 e^{-alpha x} ) for some constants ( N_0 ) and ( alpha )." It says "some constants," so unless more information is given, we can't determine ( N_0 ). Therefore, the answer should be expressed as ( N(2, 10) = frac{N_0}{2} e^{-0.75} text{erfc}(-0.5) ), which is approximately ( 0.359 N_0 ).Alternatively, perhaps the problem expects a different approach, like using separation of variables or eigenfunction expansion. But given the initial condition is ( N_0 e^{-alpha x} ), which is not a simple exponential decay, the solution I derived using the heat kernel seems appropriate.Wait, another thought: maybe the problem assumes that the concentration is only a function of time, not space? But no, the problem states it's a function of both ( x ) and ( t ).Alternatively, perhaps I can use Laplace transforms. Let me try that approach.Taking Laplace transform with respect to ( x ). Wait, but the equation is in ( x ) and ( t ). Maybe Fourier transform? Hmm, but I think the approach I took earlier is correct.Alternatively, perhaps the solution can be expressed as a product of functions, but given the initial condition is not separable in a simple way, I think the convolution approach is necessary.Therefore, I think my earlier result is correct, and the concentration at ( x=2 ) cm after 10 days is approximately ( 0.359 N_0 ).But since the problem didn't specify ( N_0 ), maybe I should express it in terms of ( N_0 ). Alternatively, perhaps ( N_0 ) is 1, making it approximately 0.359.Wait, let me think again. The initial condition is ( N(x, 0) = N_0 e^{-0.5 x} ). If we consider the total initial concentration, it's ( int_{0}^{infty} N(x, 0) dx = N_0 int_{0}^{infty} e^{-0.5 x} dx = N_0 times 2 ). So, unless the total concentration is given, we can't find ( N_0 ). Therefore, I think the answer must be expressed in terms of ( N_0 ).So, to summarize Part 2, ( N(2, 10) approx 0.359 N_0 ).Combined Effects DiscussionNow, combining both models, the growth rate ( G(t) ) varies sinusoidally with peaks at full moon, while the nutrient concentration ( N(x, t) ) diffuses over time, decreasing with depth and time. The plant's growth would be highest when both the lunar influence is at its peak and the nutrient concentration is sufficient at the root level.Over one lunar cycle (29.53 days), the growth rate would oscillate, with maximum growth around the full moon. Meanwhile, the nutrient concentration at a given depth (like 2 cm) would decrease over time due to diffusion, but the rate of decrease depends on the initial concentration and the diffusion coefficient.Therefore, the plant's growth would be most optimal around the full moon when the growth rate is highest, provided that the nutrient concentration is still sufficient. As time progresses, the nutrient concentration diminishes, which might limit the growth even during the high-growth lunar phases.So, the combined effect is that the plant experiences periodic growth spurts synchronized with the lunar cycle, but the overall growth is also constrained by the availability of nutrients, which diminish over time due to diffusion.</think>

🧠:<think>Alright, so I have this problem where an entrepreneur wants to optimize their crop yields on a 500-acre farm while also considering environmental impact. They grow wheat, corn, and soybeans, each with their own yield functions. The goal is to figure out how many acres to allocate to each crop to maximize total yield without exceeding 500 acres. Then, in part two, we have to add an environmental impact constraint and solve again.Starting with part 1. Let me outline what I know:- The farm has 500 acres.- Three crops: wheat, corn, soybeans.- Yield per acre for each crop is a function of the number of acres allocated to it. So, the more acres you allocate to a crop, the lower the yield per acre because of diminishing returns, I suppose.- The yield functions are: - Wheat: Y_w(x) = 100 - 0.5x - Corn: Y_c(x) = 150 - 0.8x - Soybeans: Y_s(x) = 120 - 0.6x- We need to maximize the total yield, which would be the sum of (yield per acre * acres allocated) for each crop.Wait, hold on. The yield functions are given as Y_w(x) = 100 - 0.5x, but is x the number of acres allocated to that specific crop? So, for wheat, x is just the acres for wheat, not the total acres. That makes sense because each crop's yield depends only on its own allocation.So, if we let x_w be the acres for wheat, x_c for corn, and x_s for soybeans, then the total yield would be:Total Yield = (100 - 0.5x_w) * x_w + (150 - 0.8x_c) * x_c + (120 - 0.6x_s) * x_sAnd the total acres constraint is x_w + x_c + x_s <= 500.We need to maximize Total Yield subject to x_w + x_c + x_s <= 500 and x_w, x_c, x_s >= 0.Hmm, okay, so this is a quadratic optimization problem because each term in the total yield is quadratic in x.To solve this, I can set up the problem with variables x_w, x_c, x_s.Let me write the total yield function:Total Yield = (100x_w - 0.5x_w^2) + (150x_c - 0.8x_c^2) + (120x_s - 0.6x_s^2)So, TY = -0.5x_w^2 + 100x_w - 0.8x_c^2 + 150x_c - 0.6x_s^2 + 120x_sWe need to maximize TY with respect to x_w, x_c, x_s, subject to x_w + x_c + x_s <= 500 and x_w, x_c, x_s >= 0.Since each crop's yield function is concave (the coefficients of x^2 are negative), the total yield function is also concave. Therefore, the maximum occurs at the critical point or at the boundaries.To find the critical point, we can take partial derivatives of TY with respect to each variable, set them equal to zero, and solve the system of equations.Let's compute the partial derivatives.Partial derivative with respect to x_w:d(TY)/dx_w = -1x_w + 100Similarly, partial derivative with respect to x_c:d(TY)/dx_c = -1.6x_c + 150Partial derivative with respect to x_s:d(TY)/dx_s = -1.2x_s + 120Set each of these equal to zero to find the critical points.So,-1x_w + 100 = 0 => x_w = 100-1.6x_c + 150 = 0 => x_c = 150 / 1.6 = 93.75-1.2x_s + 120 = 0 => x_s = 120 / 1.2 = 100So, the critical point is at x_w = 100, x_c = 93.75, x_s = 100.Now, let's check the total acres: 100 + 93.75 + 100 = 293.75 acres. That's well below the 500-acre limit.So, in this case, the maximum occurs at this critical point because it's within the feasible region.Therefore, the optimal allocation is 100 acres for wheat, 93.75 acres for corn, and 100 acres for soybeans.Wait, but 93.75 is a fraction. Since we're dealing with acres, it's okay to have fractional acres, right? I mean, in reality, you can't have a fraction of an acre, but in an optimization problem, unless specified otherwise, fractional solutions are acceptable.So, moving on.Let me compute the total yield at this point.Compute TY:For wheat: (100 - 0.5*100)*100 = (100 - 50)*100 = 50*100 = 5000 bushelsFor corn: (150 - 0.8*93.75)*93.75First, 0.8*93.75 = 75, so 150 - 75 = 75. Then, 75*93.75 = let's compute that.75 * 93.75: 75*90=6750, 75*3.75=281.25, so total is 6750 + 281.25 = 7031.25 bushelsFor soybeans: (120 - 0.6*100)*100 = (120 - 60)*100 = 60*100 = 6000 bushelsTotal TY = 5000 + 7031.25 + 6000 = 18031.25 bushelsSo, that's the maximum yield.But wait, just to be thorough, what if we tried to allocate more acres? Since the critical point is within the 500-acre limit, but what if we tried to allocate more to the crops with higher marginal yields?Wait, but the marginal yield is given by the derivative, which is 100 - x_w for wheat, 150 - 1.6x_c for corn, and 120 - 1.2x_s for soybeans.At the critical point, all marginal yields are zero, meaning that any reallocation would decrease the total yield.But just to check, suppose we take an acre from wheat and give it to corn. The change in total yield would be:Marginal yield of corn at x_c = 93.75 + 1 = 94.75: 150 - 1.6*94.75 = 150 - 151.6 = -1.6Marginal yield of wheat at x_w = 100 - 1 = 99: 100 - 99 = 1So, the net change is -1.6 + 1 = -0.6, which is negative. So, total yield decreases.Similarly, moving an acre from soybeans to corn:Marginal yield of corn at 94.75: -1.6 as aboveMarginal yield of soybeans at 99: 120 - 1.2*99 = 120 - 118.8 = 1.2Net change: -1.6 + 1.2 = -0.4, still negative.So, indeed, the critical point is the maximum.Therefore, the optimal allocation is 100 acres for wheat, 93.75 acres for corn, and 100 acres for soybeans, totaling 293.75 acres, with the remaining 206.25 acres left fallow or used for other purposes.But wait, the problem says "the total number of acres used does not exceed 500 acres." So, it's okay if we don't use all 500 acres. We can leave some unused.So, that's part 1.Now, moving on to part 2. We have to incorporate the environmental impact constraint.The environmental impact index is given by I(x_w, x_c, x_s) = 0.1x_w + 0.2x_c + 0.15x_s <= 300.So, we need to maximize the total yield subject to:1. x_w + x_c + x_s <= 5002. 0.1x_w + 0.2x_c + 0.15x_s <= 3003. x_w, x_c, x_s >= 0So, now we have two constraints.This is a constrained optimization problem with two inequality constraints.To solve this, we can use the method of Lagrange multipliers with multiple constraints.Alternatively, since it's a quadratic optimization problem with linear constraints, we can set up the Lagrangian with two multipliers.Let me denote the Lagrangian as:L = -0.5x_w^2 + 100x_w - 0.8x_c^2 + 150x_c - 0.6x_s^2 + 120x_s - λ1(x_w + x_c + x_s - 500) - λ2(0.1x_w + 0.2x_c + 0.15x_s - 300)Wait, but actually, since the constraints are <=, we need to consider whether they are binding or not.In the original problem without the environmental constraint, the optimal solution used only 293.75 acres, so the environmental impact would be:I = 0.1*100 + 0.2*93.75 + 0.15*100 = 10 + 18.75 + 15 = 43.75, which is way below 300. So, in the original problem, the environmental constraint wasn't binding.But now, we have to include it, so perhaps the optimal solution will change.Wait, but actually, the environmental impact is a separate constraint. So, even if the original solution didn't hit the environmental limit, now we have to ensure that the environmental impact doesn't exceed 300. So, we might have to reallocate acres to reduce the environmental impact.But 300 is a much higher threshold than the original 43.75, so maybe the environmental constraint isn't binding either. But let's check.Wait, no, the environmental impact is 0.1x_w + 0.2x_c + 0.15x_s. So, if we increase x_c, which has the highest coefficient (0.2), the environmental impact increases the most.But in our original solution, x_c was 93.75, which is not that high. So, perhaps even if we allocate more acres, the environmental impact might not reach 300.Wait, let's compute the environmental impact at the original optimal allocation:I = 0.1*100 + 0.2*93.75 + 0.15*100 = 10 + 18.75 + 15 = 43.75, as before.So, 43.75 is much less than 300. Therefore, the environmental constraint is not binding at the original optimal solution. So, perhaps the optimal solution remains the same.But wait, maybe we can allocate more acres to increase the total yield without violating the environmental constraint.Because the environmental constraint is 300, which is much higher than 43.75, so we can potentially use more acres.Wait, but the total acres are limited to 500. So, perhaps we can increase the allocation beyond the critical point, but subject to the environmental constraint.Wait, but the critical point is where the marginal yields are zero. If we go beyond that, the marginal yields become negative, so total yield would decrease.But perhaps, by reallocating acres, we can increase the total yield beyond the critical point without violating the environmental constraint.Wait, no, because the critical point is the maximum of the total yield function. So, any movement away from it would decrease the total yield.But if we have additional constraints, maybe we have to move away from the critical point, but in a way that the constraints are satisfied.Wait, but in this case, the environmental constraint is not binding at the critical point, so the optimal solution remains the same.But let me think again.Suppose we try to allocate more acres, but in such a way that the environmental impact doesn't exceed 300.But since the environmental impact is much higher than the original, we can actually allocate more acres.Wait, but the total yield function is concave, so the maximum is at the critical point. So, if we can allocate more acres without violating the environmental constraint, we can potentially increase the total yield beyond the critical point.Wait, but the critical point is where the marginal yields are zero. So, if we can reallocate acres beyond that, but with positive marginal yields, we can increase the total yield.Wait, but the marginal yields at the critical point are zero. So, moving beyond that would result in negative marginal yields, which would decrease the total yield.Therefore, even if we have more acres available and a higher environmental impact threshold, the maximum total yield is still at the critical point.Therefore, the optimal solution remains the same, with x_w = 100, x_c = 93.75, x_s = 100, and the environmental impact is 43.75, which is below 300.Therefore, the optimal allocation doesn't change.But wait, let me test this.Suppose we try to allocate more acres to corn, which has the highest yield function.Wait, but the yield function for corn is Y_c(x) = 150 - 0.8x. So, as x increases, the yield per acre decreases.But if we increase x_c beyond 93.75, the marginal yield becomes negative, so total yield would decrease.Similarly, for wheat and soybeans.Therefore, even if we have more acres available and a higher environmental impact threshold, the optimal allocation remains at the critical point.Therefore, the solution to part 2 is the same as part 1.But wait, let me double-check.Suppose we try to allocate more acres to corn, which has the highest yield per acre at x=0 (150 bushels/acre). But as we increase x_c, the yield per acre decreases.But if we can increase x_c beyond 93.75, even though the marginal yield becomes negative, maybe the environmental impact allows it.Wait, but the environmental impact is 0.2 per acre for corn, which is the highest among the three crops.So, if we increase x_c, we increase the environmental impact more than the others.But since the environmental impact threshold is 300, which is much higher than our original 43.75, we can potentially increase x_c a lot.But let's see.Suppose we set x_w and x_s to zero, and allocate all 500 acres to corn.Then, environmental impact would be 0.2*500 = 100, which is still below 300.Wait, but that's not correct. Wait, 0.2*500 = 100, which is below 300. So, we can even allocate more than 500 acres? Wait, no, the total acres can't exceed 500.Wait, but if we set x_w and x_s to zero, and x_c = 500, then environmental impact is 100, which is still below 300. So, we could potentially allocate even more to corn, but we are limited by the 500-acre constraint.Wait, but in that case, the total yield would be:Y_c(500) = 150 - 0.8*500 = 150 - 400 = negative, which doesn't make sense. So, the yield per acre can't be negative, so perhaps the yield function is only valid up to a certain point.Wait, the yield functions are given as Y_w(x) = 100 - 0.5x, Y_c(x) = 150 - 0.8x, Y_s(x) = 120 - 0.6x.So, these functions are linear and will eventually become negative if x is large enough. But in reality, yield can't be negative, so perhaps the functions are only valid up to the point where Y=0.For wheat: 100 - 0.5x = 0 => x = 200For corn: 150 - 0.8x = 0 => x = 187.5For soybeans: 120 - 0.6x = 0 => x = 200So, the maximum acres we can allocate to each crop without getting negative yields are 200 for wheat, 187.5 for corn, and 200 for soybeans.But in our case, the total acres are limited to 500, so we can't exceed that.But going back, if we set x_c = 500, the yield per acre would be negative, which isn't practical. So, the maximum x_c we can have is 187.5 acres.But even then, the environmental impact would be 0.2*187.5 = 37.5, which is still below 300.So, in that case, the environmental constraint is not binding.Therefore, the optimal solution is still at the critical point, where x_w = 100, x_c = 93.75, x_s = 100, with total yield 18031.25 bushels and environmental impact 43.75.Therefore, the optimal allocation doesn't change when adding the environmental constraint because the constraint isn't binding.But wait, let me consider another approach. Maybe the environmental constraint can be binding if we try to maximize the environmental impact while keeping the yield as high as possible.But no, the problem is to maximize yield while ensuring that the environmental impact doesn't exceed 300.Since the original optimal solution already has an environmental impact of 43.75, which is way below 300, the constraint doesn't affect the solution.Therefore, the optimal allocation remains the same.So, in conclusion, for both parts 1 and 2, the optimal allocation is 100 acres for wheat, 93.75 acres for corn, and 100 acres for soybeans.But wait, let me think again. Maybe I'm missing something.Suppose we try to use more acres, but in a way that the environmental impact is still below 300. For example, if we allocate more acres to soybeans, which have a lower environmental impact coefficient (0.15) compared to corn (0.2) and wheat (0.1).So, maybe we can increase x_s beyond 100, which would allow us to use more acres without increasing the environmental impact as much.But wait, the total yield function is concave, so the maximum is at the critical point. So, even if we can use more acres, the total yield would decrease because the marginal yields are negative beyond the critical point.Therefore, even if we can use more acres without violating the environmental constraint, it's not beneficial because the total yield would decrease.Therefore, the optimal solution remains at the critical point.So, to summarize:1. The optimal allocation is 100 acres for wheat, 93.75 acres for corn, and 100 acres for soybeans, with a total yield of 18,031.25 bushels.2. Incorporating the environmental impact constraint doesn't change the optimal allocation because the environmental impact at the optimal solution is already well below the threshold of 300.Therefore, the answers are:1. x_w = 100, x_c = 93.75, x_s = 1002. Same as above, since the environmental constraint isn't binding.But wait, let me check if the environmental impact can be increased without decreasing the total yield.Suppose we try to reallocate some acres from wheat or soybeans to corn, which has a higher environmental impact coefficient but also a higher yield per acre.Wait, but at the critical point, the marginal yields are zero. So, reallocating would decrease the total yield.But let's see.Suppose we take 1 acre from wheat and give it to corn.The change in total yield would be:Marginal yield of corn at x_c = 93.75 + 1 = 94.75: 150 - 1.6*94.75 = 150 - 151.6 = -1.6Marginal yield of wheat at x_w = 100 - 1 = 99: 100 - 99 = 1So, net change: -1.6 + 1 = -0.6, which is negative. So, total yield decreases.Similarly, taking 1 acre from soybeans and giving it to corn:Marginal yield of corn at 94.75: -1.6Marginal yield of soybeans at 99: 120 - 1.2*99 = 120 - 118.8 = 1.2Net change: -1.6 + 1.2 = -0.4, still negative.Therefore, any reallocation to corn decreases the total yield.Similarly, reallocating to soybeans:Taking 1 acre from wheat and giving it to soybeans:Marginal yield of soybeans at x_s = 100 + 1 = 101: 120 - 0.6*101 = 120 - 60.6 = 59.4Marginal yield of wheat at 99: 1Net change: 59.4 - 1 = 58.4, which is positive.Wait, that's a positive change. So, total yield would increase.Wait, but that can't be, because the critical point is supposed to be the maximum.Wait, no, because when we take from wheat and give to soybeans, we are moving away from the critical point, but the marginal yield of soybeans is still positive.Wait, but at the critical point, the marginal yields are zero.Wait, no, the marginal yield for soybeans at x_s = 100 is 120 - 1.2*100 = 0.So, if we increase x_s beyond 100, the marginal yield becomes negative.Wait, but if we take 1 acre from wheat and give it to soybeans, the marginal yield of soybeans at 101 is 120 - 1.2*101 = 120 - 121.2 = -1.2Wait, that's negative. So, the net change is -1.2 (from soybeans) + 1 (from wheat) = -0.2, which is negative.Wait, so I made a mistake earlier.Because when we take 1 acre from wheat and give it to soybeans, the marginal yield of soybeans at 101 is negative, so the net change is negative.Therefore, any reallocation away from the critical point would decrease the total yield.Therefore, the critical point is indeed the maximum.Therefore, even if we can allocate more acres without violating the environmental constraint, it's not beneficial because the total yield would decrease.Therefore, the optimal allocation remains the same.So, in conclusion, the optimal allocation is 100 acres for wheat, 93.75 acres for corn, and 100 acres for soybeans, with a total yield of 18,031.25 bushels, and an environmental impact of 43.75, which is well below the threshold of 300.Therefore, the answers are:1. x_w = 100, x_c = 93.75, x_s = 1002. Same as above, since the environmental constraint isn't binding.But wait, let me check if the environmental impact can be increased without decreasing the total yield.Wait, if we can increase x_c beyond 93.75, even though the marginal yield is negative, but the environmental impact is still below 300.But as we saw earlier, increasing x_c beyond 93.75 would decrease the total yield because the marginal yield becomes negative.Therefore, even if the environmental impact is below 300, it's not beneficial to increase x_c because it would decrease the total yield.Therefore, the optimal solution remains the same.So, final answer:1. Allocate 100 acres to wheat, 93.75 acres to corn, and 100 acres to soybeans.2. The same allocation satisfies the environmental constraint, so no change is needed.</think>

🧠:<think>Okay, so I have this problem about two production lines, A and B, and I need to figure out the total units produced and the total cost for each. Then, I have to find the cost per unit for each line. Let me break this down step by step.First, for production line A. It operates for 8 hours a day, and the production rate is given by the function ( P_A(t) = 50 + 3t^2 ) units per hour. To find the total units produced, I think I need to integrate this function over the 8-hour period. Integration makes sense here because it's the accumulation of production over time.So, the total production for line A, let's call it ( T_A ), should be the integral of ( P_A(t) ) from t=0 to t=8. That would be:[T_A = int_{0}^{8} (50 + 3t^2) , dt]Let me compute that. The integral of 50 is 50t, and the integral of ( 3t^2 ) is ( t^3 ) because ( int t^2 dt = frac{t^3}{3} ), so multiplied by 3 gives ( t^3 ). So putting it together:[T_A = [50t + t^3]_{0}^{8}]Now, plugging in 8:50*8 = 400, and 8^3 = 512. So, 400 + 512 = 912.Then, subtracting the value at 0, which is 0 + 0 = 0. So, total units for A is 912.Wait, that seems straightforward. Let me double-check. The integral of 50 is 50t, correct. The integral of 3t^2 is t^3, yes. Evaluated from 0 to 8, so 50*8 is 400, 8^3 is 512, so 400 + 512 is indeed 912. Okay, that seems right.Now, moving on to the cost for line A. The cost function is ( C_A(t) = 100 + 2t + t^2 ) dollars per hour. To find the total cost, I need to integrate this over the 8-hour period as well.So, total cost for A, ( C_{total A} ), is:[C_{total A} = int_{0}^{8} (100 + 2t + t^2) , dt]Let me compute this integral. The integral of 100 is 100t. The integral of 2t is ( t^2 ). The integral of ( t^2 ) is ( frac{t^3}{3} ). So putting it all together:[C_{total A} = [100t + t^2 + frac{t^3}{3}]_{0}^{8}]Evaluating at 8:100*8 = 800, 8^2 = 64, and ( frac{8^3}{3} = frac{512}{3} approx 170.6667 ). Adding these up: 800 + 64 = 864, plus 170.6667 is approximately 1034.6667.Subtracting the value at 0, which is 0, so total cost is approximately 1034.67.Wait, let me write it as an exact fraction instead of a decimal. ( frac{512}{3} ) is exact, so total cost is 800 + 64 + 512/3. Let me compute 800 + 64 first, which is 864. Then, 864 is equal to 2592/3, so adding 512/3 gives (2592 + 512)/3 = 3104/3. So, 3104 divided by 3 is approximately 1034.6667, which is 1034.67 when rounded to the nearest cent.Okay, so total cost for A is 1034.67.Now, for production line B. It operates for 10 hours a day, with a production rate of ( P_B(t) = 30 + 4t^2 ) units per hour. So, similar to line A, I need to integrate this over 10 hours.Total production for B, ( T_B ):[T_B = int_{0}^{10} (30 + 4t^2) , dt]Computing the integral. The integral of 30 is 30t. The integral of ( 4t^2 ) is ( frac{4}{3}t^3 ). So,[T_B = [30t + frac{4}{3}t^3]_{0}^{10}]Plugging in 10:30*10 = 300, and ( frac{4}{3}*10^3 = frac{4}{3}*1000 = frac{4000}{3} approx 1333.3333 ). Adding these together: 300 + 1333.3333 ≈ 1633.3333.Subtracting the value at 0, which is 0, so total units for B is approximately 1633.33. But let me keep it exact: 300 is 900/3, so 900/3 + 4000/3 = 4900/3 ≈ 1633.3333. So, 4900/3 is the exact value.Now, moving on to the cost for line B. The cost function is ( C_B(t) = 150 + 3t + 2t^2 ) dollars per hour. So, total cost for B, ( C_{total B} ), is:[C_{total B} = int_{0}^{10} (150 + 3t + 2t^2) , dt]Let me compute this integral. The integral of 150 is 150t. The integral of 3t is ( frac{3}{2}t^2 ). The integral of ( 2t^2 ) is ( frac{2}{3}t^3 ). So,[C_{total B} = [150t + frac{3}{2}t^2 + frac{2}{3}t^3]_{0}^{10}]Evaluating at 10:150*10 = 1500, ( frac{3}{2}*10^2 = frac{3}{2}*100 = 150 ), and ( frac{2}{3}*10^3 = frac{2}{3}*1000 = frac{2000}{3} approx 666.6667 ).Adding these up: 1500 + 150 = 1650, plus 666.6667 is approximately 2316.6667.Subtracting the value at 0, which is 0, so total cost is approximately 2316.67.Again, let me express this as an exact fraction. 1500 is 4500/3, 150 is 450/3, and 2000/3 is already in thirds. So, adding them together: 4500/3 + 450/3 + 2000/3 = (4500 + 450 + 2000)/3 = 7050/3 = 2350. Wait, hold on, that doesn't add up. Wait, 4500 + 450 is 4950, plus 2000 is 6950. So, 6950/3 is approximately 2316.6667, which is 2316.67.Wait, I think I made a mistake in adding. Let me check again:150t evaluated at 10 is 1500.(3/2)t^2 evaluated at 10 is (3/2)*100 = 150.(2/3)t^3 evaluated at 10 is (2/3)*1000 = 2000/3 ≈ 666.6667.So, 1500 + 150 = 1650, plus 666.6667 is 2316.6667, which is 7050/3. Wait, 2316.6667 * 3 = 6950? Wait, 2316.6667 * 3 is 6950? Let me compute 2316.6667 * 3:2316 * 3 = 6948, 0.6667 * 3 ≈ 2, so total ≈ 6950. So, 6950/3 is approximately 2316.6667. So, exact value is 6950/3.Wait, but 150t is 1500, which is 4500/3. (3/2)t^2 is 150, which is 450/3. (2/3)t^3 is 2000/3. So, adding them: 4500/3 + 450/3 + 2000/3 = (4500 + 450 + 2000)/3 = 6950/3. So, correct.So, total cost for B is 6950/3 ≈ 2316.67.Okay, so now I have total units and total costs for both lines. Now, I need to find the cost per unit for each line.For line A, total cost is approximately 1034.67 and total units are 912. So, cost per unit is total cost divided by total units.Similarly, for line B, total cost is approximately 2316.67 and total units are approximately 1633.33. So, cost per unit is total cost divided by total units.Let me compute these.Starting with line A:Cost per unit for A = Total cost A / Total units A = 1034.67 / 912.Let me compute this division. Let me see, 912 goes into 1034.67 how many times?912 * 1 = 912, subtract that from 1034.67, we get 122.67.So, 122.67 / 912 ≈ 0.1345.So, total cost per unit is approximately 1 + 0.1345 ≈ 1.1345 dollars per unit.Wait, that seems low. Let me check my calculations.Wait, 1034.67 divided by 912.Let me compute 1034.67 / 912.First, 912 * 1 = 912, subtract from 1034.67, remainder is 122.67.122.67 / 912 ≈ 0.1345.So, total is approximately 1.1345 dollars per unit.Wait, that's about 1.13 per unit. Hmm, that seems plausible.Alternatively, let me do it as fractions to be exact.Total cost A is 3104/3 dollars, and total units A is 912.So, cost per unit is (3104/3) / 912 = (3104)/(3*912).Simplify 3104 / 912.Let me see, 912 * 3 = 2736, which is less than 3104. 3104 - 2736 = 368.So, 3104 / 912 = 3 + 368/912.Simplify 368/912: divide numerator and denominator by 16: 23/57.So, 3104/912 = 3 + 23/57 ≈ 3.4035.Wait, wait, that can't be. Wait, 3104 divided by 912 is approximately 3.4035.But wait, that contradicts my earlier decimal calculation. Wait, hold on, perhaps I made a mistake in the fraction.Wait, total cost A is 3104/3, which is approximately 1034.67, and total units is 912.So, 3104/3 divided by 912 is equal to (3104)/(3*912) = 3104/2736.Simplify 3104/2736.Divide numerator and denominator by 16: 3104 ÷16=194, 2736 ÷16=171.So, 194/171 ≈ 1.1345.Ah, okay, so 194 divided by 171 is approximately 1.1345. So, that's correct. So, cost per unit is approximately 1.13.Wait, so 194/171 is approximately 1.1345.Yes, because 171*1.1345 ≈ 171 + 171*0.1345 ≈ 171 + 23 ≈ 194.So, that's correct.Now, for line B.Total cost is 6950/3 ≈ 2316.67, and total units is 4900/3 ≈ 1633.33.So, cost per unit is (6950/3) / (4900/3) = 6950/4900.Simplify this fraction.Divide numerator and denominator by 50: 6950 ÷50=139, 4900 ÷50=98.So, 139/98 ≈ 1.418367.So, approximately 1.42 per unit.Alternatively, 139 divided by 98 is approximately 1.4184.So, approximately 1.42 per unit.Wait, let me confirm:98*1.4184 ≈ 98 + 98*0.4184 ≈ 98 + 40.8032 ≈ 138.8032, which is close to 139, so that's correct.So, summarizing:Total units for A: 912Total cost for A: ~1034.67Cost per unit for A: ~1.13Total units for B: ~1633.33Total cost for B: ~2316.67Cost per unit for B: ~1.42So, in conclusion, line A is more cost-efficient per unit than line B.Wait, but let me just make sure I didn't make any calculation errors, especially in the integrals.For line A:Production integral: 50t + t^3 from 0 to 8.At t=8: 50*8=400, 8^3=512, total 912. Correct.Cost integral: 100t + t^2 + (t^3)/3 from 0 to 8.At t=8: 100*8=800, 8^2=64, 8^3/3=512/3≈170.6667. Total: 800+64=864 +170.6667≈1034.6667. Correct.For line B:Production integral: 30t + (4/3)t^3 from 0 to10.At t=10: 30*10=300, (4/3)*1000=1333.3333. Total≈1633.3333. Correct.Cost integral: 150t + (3/2)t^2 + (2/3)t^3 from 0 to10.At t=10: 150*10=1500, (3/2)*100=150, (2/3)*1000≈666.6667. Total≈1500+150=1650+666.6667≈2316.6667. Correct.So, all integrals seem correct.Then, cost per unit:For A: 1034.67 /912≈1.1345≈1.13For B: 2316.67 /1633.33≈1.418≈1.42Yes, that seems correct.Therefore, the answers are:1. Total units produced:Line A: 912 unitsLine B: 1633.33 units (or 4900/3 units)2. Total cost:Line A: 1034.67Line B: 2316.67Cost per unit:Line A: ~1.13Line B: ~1.42So, I think that's all.Final Answer1. Total units produced by line A: boxed{912} units. Total units produced by line B: boxed{dfrac{4900}{3}} units.2. Total cost for line A: boxed{dfrac{3104}{3}} dollars. Total cost for line B: boxed{dfrac{6950}{3}} dollars. Cost per unit for line A: boxed{dfrac{194}{171}} dollars. Cost per unit for line B: boxed{dfrac{139}{98}} dollars.</think>