🧠:<think>Okay, so I have this problem where Ahmed is organizing an interfaith dialogue event in a hall that's 30 meters by 40 meters. He wants to place circular tables with a radius of 1.5 meters each. The key thing is that the distance between the centers of any two tables has to be at least 3 meters. I need to figure out the maximum number of tables he can fit in the hall while maintaining that distance. Then, there's a second part about the cost of carpeting the entire hall at 45 Turkish Lira per square meter.Let me start with the first question. I think this is a circle packing problem in a rectangle. The hall is a rectangle, and we need to place as many circles as possible without overlapping, with the centers at least 3 meters apart. Each table has a radius of 1.5 meters, so the diameter is 3 meters. Wait, that's interesting because the required distance between centers is also 3 meters. Hmm, so if the centers are 3 meters apart, that means the edges of the tables just touch each other, right? Because each table has a radius of 1.5 meters, so 1.5 + 1.5 = 3 meters. So, the tables will be tangent to each other but not overlapping.So, if the tables are arranged in a grid pattern, each table will take up a space of 3 meters in both length and width. That is, each table effectively occupies a square of 3x3 meters, but since the hall is a rectangle, we can calculate how many such squares fit into the hall's dimensions.The hall is 30 meters by 40 meters. If each table requires 3 meters in both directions, then the number of tables along the 30-meter side would be 30 divided by 3, which is 10. Similarly, along the 40-meter side, it would be 40 divided by 3, which is approximately 13.333. But since we can't have a fraction of a table, we take the integer part, which is 13. So, 10 tables along the width and 13 along the length.Wait, but hold on. Is that the most efficient way to arrange them? Because sometimes, arranging circles in a hexagonal pattern can allow for more density. But in this case, since the distance between centers is exactly equal to the diameter, maybe the square packing is the most efficient here. Because in hexagonal packing, the centers are closer in one direction, but in this case, the required distance is fixed at 3 meters, so maybe it doesn't matter.Alternatively, maybe we can think of it as a grid where each table is spaced 3 meters apart from each other both horizontally and vertically. So, the number of tables would be (30 / 3) * (40 / 3). But since 30 divided by 3 is 10, and 40 divided by 3 is about 13.333, so we take 10 * 13 = 130 tables. But wait, let me double-check that.If we have 10 tables along the 30-meter side, each taking up 3 meters, that's 10 * 3 = 30 meters, which fits exactly. Along the 40-meter side, 13 tables would take up 13 * 3 = 39 meters, leaving 1 meter unused. So, that seems okay.But is there a way to fit more tables by adjusting the arrangement? For example, maybe staggering the rows so that each row is offset by 1.5 meters, allowing the tables to fit into the gaps of the previous row. In that case, the vertical distance between rows would be less, but the horizontal distance would remain the same.Wait, but the required distance between centers is 3 meters. If we stagger the rows, the vertical distance between centers would be sqrt(3^2 - 1.5^2) = sqrt(9 - 2.25) = sqrt(6.75) ≈ 2.598 meters. But that's less than 3 meters, so the distance between centers in the vertical direction would be less than 3 meters, which violates the requirement. Therefore, we can't do that because the distance between centers must be at least 3 meters in all directions.So, the hexagonal packing isn't applicable here because it would reduce the distance between some centers below the required 3 meters. Therefore, the most efficient packing is the square grid where each table is spaced 3 meters apart both horizontally and vertically.Therefore, the number of tables along the 30-meter side is 30 / 3 = 10, and along the 40-meter side, it's 40 / 3 ≈ 13.333, so 13 tables. So, total tables would be 10 * 13 = 130.Wait, but let me visualize this. If each table is 3 meters apart, and the hall is 30x40, then starting from one corner, the first table is at (0,0), the next at (3,0), then (6,0), etc., up to (27,0), which is the 10th table. Similarly, the next row would be at (0,3), (3,3), ..., (27,3), and so on, up to (0,39), (3,39), etc. So, the last table in the 40-meter direction would be at 39 meters, leaving 1 meter unused. So, yes, 13 rows of 10 tables each, totaling 130 tables.But wait, is that the maximum? Let me think again. Maybe we can fit more tables if we adjust the starting position. For example, if we shift the starting point so that we can fit an extra table in the 40-meter direction.Wait, the total length is 40 meters. If each table is spaced 3 meters apart, starting from 0, the positions would be 0, 3, 6, ..., 39, which is 13 tables. If we shift the starting point by 1.5 meters, then the first table would be at 1.5 meters, and the last table would be at 1.5 + 12*3 = 1.5 + 36 = 37.5 meters. Then, the next table would be at 40.5 meters, which is beyond the hall's length. So, that doesn't help us fit an extra table. Similarly, shifting by other amounts won't allow us to fit more than 13 tables along the 40-meter side.Therefore, 13 tables along the length and 10 along the width, totaling 130 tables.Wait, but let me check the area to see if this makes sense. Each table has an area of πr² = π*(1.5)^2 ≈ 7.0686 square meters. 130 tables would occupy 130 * 7.0686 ≈ 918.918 square meters. The hall's area is 30*40=1200 square meters. So, the tables take up about 76.5% of the hall's area. That seems reasonable, but I wonder if we can fit more tables by optimizing the arrangement.Alternatively, maybe we can fit more tables by considering that the distance between centers is 3 meters, but the tables themselves have a radius of 1.5 meters. So, the distance from the wall to the center of the first table must be at least 1.5 meters to prevent the table from extending beyond the wall. Similarly, the last table must be at least 1.5 meters away from the opposite wall.Therefore, the effective length available for placing tables along the 30-meter side is 30 - 2*1.5 = 27 meters. Similarly, along the 40-meter side, it's 40 - 2*1.5 = 37 meters.Wait, that's a different approach. So, if we have to leave 1.5 meters on each end, then the available space for the centers is 27 meters along the 30-meter side and 37 meters along the 40-meter side.Then, the number of tables along the 30-meter side would be 27 / 3 = 9, and along the 40-meter side, 37 / 3 ≈ 12.333, so 12 tables.Therefore, total tables would be 9 * 12 = 108 tables.Wait, that's significantly less than 130. So, which approach is correct?I think the key here is whether the tables can be placed right up to the walls or not. If the tables must be at least 1.5 meters away from the walls, then the available space is reduced by 3 meters in each dimension (1.5 meters on each side). But if the tables can be placed with their edges touching the walls, then the centers can be as close as 1.5 meters from the walls.Wait, the problem says the distance between the centers of any two tables is at least 3 meters. It doesn't specify the distance from the walls. So, perhaps the tables can be placed with their edges touching the walls, meaning the centers can be 1.5 meters away from the walls.Therefore, the available space for centers is 30 meters along the width and 40 meters along the length, but the centers must be at least 1.5 meters away from the walls. So, the effective area for placing centers is (30 - 2*1.5) = 27 meters in width and (40 - 2*1.5) = 37 meters in length.Therefore, the number of tables along the width would be 27 / 3 = 9, and along the length, 37 / 3 ≈ 12.333, so 12 tables.Thus, total tables would be 9 * 12 = 108.But wait, this contradicts the earlier approach where we didn't subtract the 1.5 meters from the walls. So, which is correct?I think the correct approach is to consider that the tables must be placed such that their edges do not extend beyond the walls. Therefore, the centers must be at least 1.5 meters away from each wall. Therefore, the effective area for placing centers is 27x37 meters.Therefore, the number of tables along the width is 27 / 3 = 9, and along the length, 37 / 3 ≈ 12.333, so 12 tables.Thus, total tables would be 9 * 12 = 108.But wait, let me think again. If we don't subtract the 1.5 meters from the walls, then the first table's center is at 1.5 meters from the wall, and the last table's center is at 30 - 1.5 = 28.5 meters along the width. So, the distance between the first and last centers is 28.5 - 1.5 = 27 meters. So, the number of intervals between centers is 27 / 3 = 9, which means 10 tables along the width.Wait, that's different. So, if the first center is at 1.5 meters, and each subsequent center is 3 meters apart, then the positions are 1.5, 4.5, 7.5, ..., up to 1.5 + 9*3 = 1.5 + 27 = 28.5 meters. So, that's 10 tables along the width.Similarly, along the length, starting at 1.5 meters, the positions would be 1.5, 4.5, ..., up to 1.5 + n*3 meters, where 1.5 + n*3 ≤ 40 - 1.5 = 38.5 meters.So, n*3 ≤ 37 meters, so n ≤ 37 / 3 ≈ 12.333, so n = 12 intervals, meaning 13 tables.Wait, so that would give us 10 tables along the width and 13 along the length, totaling 130 tables.But wait, this seems conflicting with the earlier calculation where we subtracted 3 meters from each dimension.I think the confusion arises from whether we subtract the 1.5 meters from each end or not. Let me clarify.If the tables must be placed such that their edges do not extend beyond the walls, then the centers must be at least 1.5 meters away from each wall. Therefore, the first center is at 1.5 meters from the wall, and the last center is at 30 - 1.5 = 28.5 meters along the width. The distance between the first and last centers is 28.5 - 1.5 = 27 meters. If each interval between centers is 3 meters, then the number of intervals is 27 / 3 = 9, which means 10 tables along the width.Similarly, along the length, the first center is at 1.5 meters, and the last center is at 40 - 1.5 = 38.5 meters. The distance between them is 38.5 - 1.5 = 37 meters. Divided by 3 meters per interval, that's 37 / 3 ≈ 12.333 intervals, so 12 full intervals, meaning 13 tables along the length.Therefore, the total number of tables is 10 * 13 = 130.But wait, earlier I thought that if we subtract 3 meters from each dimension, we get 27x37, which would allow 9x12 tables, but that seems incorrect because we can actually fit 10 tables along the width and 13 along the length by starting the first center at 1.5 meters and ending at 28.5 meters (for width) and 38.5 meters (for length).Therefore, the correct number is 130 tables.But let me verify this with an alternative approach. Let's calculate how many tables can fit along each dimension considering the required spacing and the wall constraints.For the width (30 meters):- Each table requires a space of 3 meters (diameter) plus 1.5 meters on each side for the walls. Wait, no, the 1.5 meters is just the radius, so the center must be 1.5 meters away from the wall. Therefore, the first center is at 1.5 meters, and each subsequent center is 3 meters apart.So, the positions along the width are: 1.5, 4.5, 7.5, ..., up to the maximum position ≤ 30 - 1.5 = 28.5 meters.The number of tables is given by the number of terms in the sequence starting at 1.5, with a common difference of 3, up to 28.5.The nth term of an arithmetic sequence is a_n = a_1 + (n-1)d.So, 28.5 = 1.5 + (n-1)*328.5 - 1.5 = (n-1)*327 = (n-1)*3n-1 = 9n = 10So, 10 tables along the width.Similarly, along the length (40 meters):First center at 1.5 meters, last center at 40 - 1.5 = 38.5 meters.Number of tables: a_n = 1.5 + (n-1)*3 ≤ 38.538.5 - 1.5 = (n-1)*337 = (n-1)*3n-1 = 37 / 3 ≈ 12.333So, n-1 = 12, n = 13Therefore, 13 tables along the length.Thus, total tables: 10 * 13 = 130.Therefore, the maximum number of tables is 130.Now, for the second question: If Ahmed decides to cover the floor of the hall with a special carpet that costs 45 Turkish Lira per square meter, what will be the total cost?The hall is 30 meters by 40 meters, so the area is 30 * 40 = 1200 square meters.The cost per square meter is 45 TL, so total cost is 1200 * 45.Let me calculate that:1200 * 45 = (1000 * 45) + (200 * 45) = 45,000 + 9,000 = 54,000 TL.Therefore, the total cost is 54,000 Turkish Lira.But wait, let me make sure I didn't make a calculation error.1200 * 45:Break it down:45 * 1000 = 45,00045 * 200 = 9,000Adding them together: 45,000 + 9,000 = 54,000.Yes, that's correct.So, the answers are 130 tables and 54,000 TL.</think>

🧠:<think>Okay, so I've got this problem about modeling the viewership and ratings of a Marvel series episode. It's divided into two parts. Let me take them one by one.First, the differential equation for the number of viewers is given as:[ frac{dV}{dt} = kVleft(1 - frac{V}{M}right) ]Hmm, this looks familiar. It seems like a logistic growth model. I remember that logistic equations have the form:[ frac{dN}{dt} = rNleft(1 - frac{N}{K}right) ]where ( N ) is the population, ( r ) is the growth rate, and ( K ) is the carrying capacity. So in this case, ( V ) is analogous to the population, ( k ) is the growth rate, and ( M ) is the maximum potential viewership, which is like the carrying capacity.So, to solve this differential equation, I need to find the general solution for ( V(t) ). Since it's a logistic equation, the solution should be similar to the logistic function.The standard solution for the logistic equation is:[ N(t) = frac{K}{1 + left(frac{K - N_0}{N_0}right)e^{-rt}} ]Applying this to our problem, replacing ( N ) with ( V ), ( K ) with ( M ), and ( r ) with ( k ), and assuming ( V_0 ) is the initial number of viewers at ( t = 0 ), the solution should be:[ V(t) = frac{M}{1 + left(frac{M - V_0}{V_0}right)e^{-kt}} ]Wait, let me verify that. Let me try solving the differential equation step by step.The equation is:[ frac{dV}{dt} = kVleft(1 - frac{V}{M}right) ]This is a separable equation, so I can rewrite it as:[ frac{dV}{Vleft(1 - frac{V}{M}right)} = k dt ]Now, I need to integrate both sides. The left side can be integrated using partial fractions. Let me set up the partial fractions decomposition.Let me write:[ frac{1}{Vleft(1 - frac{V}{M}right)} = frac{A}{V} + frac{B}{1 - frac{V}{M}} ]Multiplying both sides by ( Vleft(1 - frac{V}{M}right) ):[ 1 = Aleft(1 - frac{V}{M}right) + B V ]Let me solve for A and B. Let's plug in ( V = 0 ):[ 1 = A(1 - 0) + B(0) implies A = 1 ]Now, plug in ( V = M ):[ 1 = A(1 - 1) + B M implies 1 = 0 + B M implies B = frac{1}{M} ]So, the partial fractions decomposition is:[ frac{1}{Vleft(1 - frac{V}{M}right)} = frac{1}{V} + frac{1}{Mleft(1 - frac{V}{M}right)} ]Therefore, the integral becomes:[ int left( frac{1}{V} + frac{1}{Mleft(1 - frac{V}{M}right)} right) dV = int k dt ]Let me compute the left integral term by term.First term:[ int frac{1}{V} dV = ln|V| + C_1 ]Second term:Let me make a substitution for the second integral. Let ( u = 1 - frac{V}{M} ), then ( du = -frac{1}{M} dV implies -M du = dV ).So,[ int frac{1}{Mleft(1 - frac{V}{M}right)} dV = int frac{1}{M u} (-M du) = - int frac{1}{u} du = -ln|u| + C_2 = -lnleft|1 - frac{V}{M}right| + C_2 ]Putting it all together:[ ln|V| - lnleft|1 - frac{V}{M}right| = kt + C ]Where ( C = C_1 + C_2 ) is the constant of integration.Simplify the left side using logarithm properties:[ lnleft| frac{V}{1 - frac{V}{M}} right| = kt + C ]Exponentiate both sides to eliminate the logarithm:[ frac{V}{1 - frac{V}{M}} = e^{kt + C} = e^{kt} cdot e^{C} ]Let me denote ( e^{C} ) as another constant, say ( C' ), so:[ frac{V}{1 - frac{V}{M}} = C' e^{kt} ]Now, solve for ( V ):Multiply both sides by ( 1 - frac{V}{M} ):[ V = C' e^{kt} left(1 - frac{V}{M}right) ]Expand the right side:[ V = C' e^{kt} - frac{C'}{M} e^{kt} V ]Bring the term with ( V ) to the left:[ V + frac{C'}{M} e^{kt} V = C' e^{kt} ]Factor out ( V ):[ V left(1 + frac{C'}{M} e^{kt}right) = C' e^{kt} ]Solve for ( V ):[ V = frac{C' e^{kt}}{1 + frac{C'}{M} e^{kt}} ]Let me simplify this expression. Multiply numerator and denominator by ( M ):[ V = frac{M C' e^{kt}}{M + C' e^{kt}} ]Now, let's apply the initial condition to find ( C' ). At ( t = 0 ), ( V = V_0 ).So,[ V_0 = frac{M C' e^{0}}{M + C' e^{0}} = frac{M C'}{M + C'} ]Solve for ( C' ):Multiply both sides by ( M + C' ):[ V_0 (M + C') = M C' ]Expand:[ V_0 M + V_0 C' = M C' ]Bring terms with ( C' ) to one side:[ V_0 M = M C' - V_0 C' ]Factor out ( C' ):[ V_0 M = C' (M - V_0) ]Solve for ( C' ):[ C' = frac{V_0 M}{M - V_0} ]So, plug this back into the expression for ( V(t) ):[ V(t) = frac{M cdot frac{V_0 M}{M - V_0} e^{kt}}{M + frac{V_0 M}{M - V_0} e^{kt}} ]Simplify numerator and denominator:Numerator:[ M cdot frac{V_0 M}{M - V_0} e^{kt} = frac{M^2 V_0}{M - V_0} e^{kt} ]Denominator:[ M + frac{V_0 M}{M - V_0} e^{kt} = M left(1 + frac{V_0}{M - V_0} e^{kt}right) ]So,[ V(t) = frac{frac{M^2 V_0}{M - V_0} e^{kt}}{M left(1 + frac{V_0}{M - V_0} e^{kt}right)} ]Simplify by canceling an ( M ):[ V(t) = frac{frac{M V_0}{M - V_0} e^{kt}}{1 + frac{V_0}{M - V_0} e^{kt}} ]Let me factor out ( frac{V_0}{M - V_0} e^{kt} ) from the denominator:Wait, actually, let me write it as:[ V(t) = frac{M V_0 e^{kt}}{(M - V_0) + V_0 e^{kt}} ]Yes, that's a cleaner way. So,[ V(t) = frac{M V_0 e^{kt}}{(M - V_0) + V_0 e^{kt}} ]Alternatively, this can be written as:[ V(t) = frac{M}{1 + left( frac{M - V_0}{V_0} right) e^{-kt}} ]Which is the standard form of the logistic function. So, that's the general solution.Alright, so that's part 1 done. Now, moving on to part 2.We have the average rating ( R(V) ) given by:[ R(V) = R_0 e^{-lambda V} ]We need to find the time ( t ) at which the rating drops to half of its initial value, i.e., ( R(V) = frac{R_0}{2} ).So, let's set ( R(V) = frac{R_0}{2} ):[ frac{R_0}{2} = R_0 e^{-lambda V} ]Divide both sides by ( R_0 ):[ frac{1}{2} = e^{-lambda V} ]Take the natural logarithm of both sides:[ lnleft( frac{1}{2} right) = -lambda V ]Simplify the left side:[ -ln(2) = -lambda V ]Multiply both sides by -1:[ ln(2) = lambda V ]So,[ V = frac{ln(2)}{lambda} ]Okay, so we have ( V ) when the rating is half. But we need to find the time ( t ) when this happens. So, we need to relate ( V ) to ( t ) using the solution from part 1.From part 1, we have:[ V(t) = frac{M V_0 e^{kt}}{(M - V_0) + V_0 e^{kt}} ]We can set this equal to ( frac{ln(2)}{lambda} ) and solve for ( t ).So,[ frac{M V_0 e^{kt}}{(M - V_0) + V_0 e^{kt}} = frac{ln(2)}{lambda} ]Let me denote ( V(t) = V ) for simplicity:[ frac{M V_0 e^{kt}}{(M - V_0) + V_0 e^{kt}} = frac{ln(2)}{lambda} ]Let me cross-multiply:[ M V_0 e^{kt} cdot lambda = (M - V_0) ln(2) + V_0 e^{kt} ln(2) ]Bring all terms to one side:[ M V_0 lambda e^{kt} - V_0 ln(2) e^{kt} - (M - V_0) ln(2) = 0 ]Factor out ( e^{kt} ) from the first two terms:[ e^{kt} (M V_0 lambda - V_0 ln(2)) - (M - V_0) ln(2) = 0 ]Factor ( V_0 ) from the first term:[ e^{kt} V_0 (M lambda - ln(2)) - (M - V_0) ln(2) = 0 ]Let me solve for ( e^{kt} ):[ e^{kt} V_0 (M lambda - ln(2)) = (M - V_0) ln(2) ]So,[ e^{kt} = frac{(M - V_0) ln(2)}{V_0 (M lambda - ln(2))} ]Take the natural logarithm of both sides:[ kt = lnleft( frac{(M - V_0) ln(2)}{V_0 (M lambda - ln(2))} right) ]Therefore,[ t = frac{1}{k} lnleft( frac{(M - V_0) ln(2)}{V_0 (M lambda - ln(2))} right) ]Hmm, let me check if this makes sense. Let me make sure the denominator inside the logarithm isn't zero or negative because ( ln(2) ) is positive, and ( M lambda ) is positive since ( M ) and ( lambda ) are positive constants. So, as long as ( M lambda > ln(2) ), the argument of the logarithm is positive, which it should be because otherwise, we wouldn't have a real solution.So, assuming ( M lambda > ln(2) ), which is reasonable because if ( M lambda ) were less than ( ln(2) ), the rating might not even reach half of its initial value, or it might happen at a negative time, which doesn't make sense.So, that's the expression for ( t ).Wait, let me recap:We started with ( R(V) = R_0 e^{-lambda V} ), set it equal to ( R_0 / 2 ), solved for ( V ), which gave ( V = ln(2)/lambda ). Then, using the logistic growth solution for ( V(t) ), set that equal to ( ln(2)/lambda ) and solved for ( t ).Yes, that seems correct.Let me just write the final expression clearly:[ t = frac{1}{k} lnleft( frac{(M - V_0) ln(2)}{V_0 (M lambda - ln(2))} right) ]So, that's the time when the rating drops to half its initial value.Wait, just to make sure, let me think about the case when ( V_0 ) is very small compared to ( M ). Then, the expression inside the log would be approximately ( (M ln(2))/(V_0 M lambda) ), which simplifies to ( ln(2)/(V_0 lambda) ). So, as ( V_0 ) approaches zero, the time ( t ) would approach infinity, which makes sense because if you start with almost no viewers, it would take a long time to reach the required ( V ).Alternatively, if ( V_0 ) is close to ( M ), then the numerator ( (M - V_0) ) becomes small, so the time ( t ) would be negative, which doesn't make sense, meaning that if the initial viewership is already high, the rating might have already dropped below half before ( t = 0 ).But since ( V_0 ) is the initial number of viewers, it's reasonable to assume ( V_0 < M ), so ( M - V_0 > 0 ). Also, as discussed, ( M lambda > ln(2) ) is necessary for the argument of the logarithm to be positive.Alright, I think that's solid.Final Answer1. The general solution for the number of viewers is (boxed{V(t) = dfrac{M}{1 + left( dfrac{M - V_0}{V_0} right) e^{-kt}}}).2. The time at which the rating drops to half of its initial value is (boxed{t = dfrac{1}{k} lnleft( dfrac{(M - V_0) ln 2}{V_0 (M lambda - ln 2)} right)}).</think>

🧠:<think>Okay, so I need to help Alex optimize their online portfolio by finding the critical points of the engagement score function E(x, y). The function is given by:[ E(x, y) = frac{k cdot e^{-x^2 - y^2}}{sqrt{x^2 + y^2 + 1}} ]First, I remember that to find critical points, I need to compute the partial derivatives of E with respect to x and y, set them equal to zero, and solve the resulting system of equations. Let me start by computing the partial derivative with respect to x. Since E is a function of both x and y, I'll treat y as a constant when taking the derivative with respect to x.The function E can be written as:[ E(x, y) = k cdot e^{-x^2 - y^2} cdot (x^2 + y^2 + 1)^{-1/2} ]This looks like a product of two functions: the exponential part and the power function. So, I think I'll need to use the product rule for differentiation. Alternatively, maybe logarithmic differentiation would be easier? Hmm, not sure. Let me try the product rule.Let me denote:[ u = e^{-x^2 - y^2} ][ v = (x^2 + y^2 + 1)^{-1/2} ]So, E = k * u * v.First, compute the partial derivative of u with respect to x:[ frac{partial u}{partial x} = e^{-x^2 - y^2} cdot (-2x) = -2x e^{-x^2 - y^2} ]Next, compute the partial derivative of v with respect to x:[ frac{partial v}{partial x} = (-1/2) cdot (x^2 + y^2 + 1)^{-3/2} cdot 2x = -x (x^2 + y^2 + 1)^{-3/2} ]Now, using the product rule, the partial derivative of E with respect to x is:[ frac{partial E}{partial x} = k left( frac{partial u}{partial x} cdot v + u cdot frac{partial v}{partial x} right) ]Plugging in the derivatives:[ frac{partial E}{partial x} = k left( (-2x e^{-x^2 - y^2}) cdot (x^2 + y^2 + 1)^{-1/2} + e^{-x^2 - y^2} cdot (-x (x^2 + y^2 + 1)^{-3/2}) right) ]Let me factor out common terms:First, both terms have -x e^{-x^2 - y^2} and (x^2 + y^2 + 1)^{-3/2}.Wait, let me see:First term: -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2}Second term: -x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2}So, factor out -x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2}:So,[ frac{partial E}{partial x} = k cdot (-x e^{-x^2 - y^2}) (x^2 + y^2 + 1)^{-3/2} [2(x^2 + y^2 + 1) + 1] ]Wait, let me verify that.Let me factor:First term: -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2} = -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2} * 1Second term: -x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2} = -x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2} * 1So, to factor, I can write:= -x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2} [2(x^2 + y^2 + 1) + 1]Wait, because:First term: -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2} = -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2} * (x^2 + y^2 + 1)^0But to factor out (x^2 + y^2 + 1)^{-3/2}, we need to write:First term: -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2} = -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2} * (x^2 + y^2 + 1)^{-1} * (x^2 + y^2 + 1)^1Wait, maybe it's better to factor out (x^2 + y^2 + 1)^{-3/2}:So, (x^2 + y^2 + 1)^{-1/2} = (x^2 + y^2 + 1)^{-3/2} * (x^2 + y^2 + 1)^1Similarly, so:First term: -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-1/2} = -2x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2} * (x^2 + y^2 + 1)Second term: -x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2}So, combining both terms:= -x e^{-x^2 - y^2} (x^2 + y^2 + 1)^{-3/2} [2(x^2 + y^2 + 1) + 1]Simplify inside the brackets:2(x^2 + y^2 + 1) + 1 = 2x^2 + 2y^2 + 2 + 1 = 2x^2 + 2y^2 + 3So, putting it all together:[ frac{partial E}{partial x} = k cdot (-x e^{-x^2 - y^2}) (x^2 + y^2 + 1)^{-3/2} (2x^2 + 2y^2 + 3) ]Similarly, the partial derivative with respect to y will be the same, but with x replaced by y:[ frac{partial E}{partial y} = k cdot (-y e^{-x^2 - y^2}) (x^2 + y^2 + 1)^{-3/2} (2x^2 + 2y^2 + 3) ]So, to find critical points, we set both partial derivatives equal to zero.So, set:1. ( frac{partial E}{partial x} = 0 )2. ( frac{partial E}{partial y} = 0 )Looking at the expressions for the partial derivatives, each is a product of several terms. For the product to be zero, at least one of the factors must be zero.Let me analyze the factors:1. k: This is a constant, so it's non-zero (since k is dependent on contributions, which is 50, so k = sqrt(50 + 1) = sqrt(51) ≈ 7.141, which is non-zero).2. e^{-x^2 - y^2}: The exponential function is always positive, so it's never zero.3. (x^2 + y^2 + 1)^{-3/2}: This is also always positive, so never zero.4. (2x^2 + 2y^2 + 3): This is a sum of squares multiplied by 2 and added to 3, so it's always positive as well.So, the only factors that can be zero are the linear terms in x and y:For ( frac{partial E}{partial x} = 0 ), the factor is -x. So, -x = 0 => x = 0.Similarly, for ( frac{partial E}{partial y} = 0 ), the factor is -y. So, -y = 0 => y = 0.Therefore, the only critical point is at (0, 0).Wait, that seems too straightforward. Let me double-check.Looking back at the partial derivatives:[ frac{partial E}{partial x} = k cdot (-x e^{-x^2 - y^2}) (x^2 + y^2 + 1)^{-3/2} (2x^2 + 2y^2 + 3) ]All the other factors except -x are positive, so the only way this derivative is zero is if x = 0. Similarly, for y. So, the only critical point is at (0, 0).Hmm, that seems correct. Let me think if there could be any other critical points. For example, if the other factors could somehow be zero, but as we saw, they are all positive, so no. So, only (0, 0) is the critical point.Okay, so that answers the first part. The critical point is at (0, 0).Now, moving on to the second part. We need to find the maximum engagement score E at this critical point, given that k = sqrt(C + 1), where C is the number of contributions. Alex has made 50 contributions, so C = 50.First, compute k:k = sqrt(50 + 1) = sqrt(51) ≈ 7.1414Now, evaluate E at (0, 0):E(0, 0) = [k * e^{-0^2 - 0^2}] / sqrt(0^2 + 0^2 + 1) = [k * e^{0}] / sqrt(1) = [k * 1] / 1 = kSo, E(0, 0) = k ≈ 7.1414Wait, that seems too simple. Let me verify.Yes, substituting x=0 and y=0 into E(x, y):E(0, 0) = [k * e^{0}] / sqrt(0 + 0 + 1) = k * 1 / 1 = kSo, E(0, 0) = k.Therefore, with k = sqrt(51), the maximum engagement score is sqrt(51), approximately 7.1414.But wait, is (0, 0) a maximum? We found the critical point, but we need to confirm if it's a maximum.To do that, we can use the second derivative test for functions of two variables. Compute the second partial derivatives and use the discriminant D.Compute the second partial derivatives:First, let's compute f_xx, f_xy, f_yy.But this might get complicated. Alternatively, since the function E(x, y) is a product of a Gaussian (which has a maximum at 0) and a function that decreases as x and y increase, it's likely that (0, 0) is indeed a maximum.But let me try to compute the second derivatives.First, recall that:f_x = k * (-x e^{-x^2 - y^2}) (x^2 + y^2 + 1)^{-3/2} (2x^2 + 2y^2 + 3)Similarly, f_y is similar with y.To compute f_xx, we need to differentiate f_x with respect to x.This seems quite involved. Maybe instead, we can consider the behavior of E(x, y) as x and y increase. As x and y become large, the exponential term decays rapidly, while the denominator grows, so E tends to zero. Therefore, the function has a single critical point at (0, 0), which is likely a maximum.Alternatively, consider the function in polar coordinates. Let me try that.Let x = r cosθ, y = r sinθ. Then, x^2 + y^2 = r^2.So, E(r, θ) = [k e^{-r^2}] / sqrt(r^2 + 1)Since E does not depend on θ, the function is radially symmetric. Therefore, the maximum must occur at r=0, which is (0, 0). So, yes, (0, 0) is the maximum.Therefore, the maximum engagement score is E(0, 0) = k = sqrt(51) ≈ 7.1414.So, summarizing:1. The only critical point is at (0, 0).2. The maximum engagement score is sqrt(51), approximately 7.1414.But wait, let me compute sqrt(51) more accurately.sqrt(49) = 7, sqrt(64)=8, so sqrt(51) is between 7 and 8. 7.1^2 = 50.41, 7.2^2=51.84. So, sqrt(51) is approximately 7.1414.Yes, that's correct.So, the approximate maximum engagement score is 7.1414.But the problem says "approximate", so maybe we can write it as sqrt(51) or approximate to two decimal places as 7.14.Alternatively, the exact value is sqrt(51), but since they asked for approximate, 7.14 is fine.So, to answer the questions:1. Critical point is (0, 0).2. Maximum E is sqrt(51) ≈ 7.14.I think that's it.</think>

🧠:<think>First, I need to calculate the total amount John will have in his savings account after 5 years. He contributes 500 at the end of each month, and the account earns an annual interest rate of 4%, compounded monthly. To find the future value of his monthly contributions, I'll use the future value of an ordinary annuity formula.Next, after the 5-year period, John plans to use 12,000 for his family vacation. I'll subtract this amount from the total savings to determine how much he can invest in the mutual fund. The remaining amount will then be invested in a mutual fund that offers an annual return of 6%, compounded quarterly. For this part, I'll use the compound interest formula to calculate the future value after 3 years.</think>