🧠:<think>Alright, so I have this problem about Mr. Capital and his firm. It's about maximizing profits under two different economic systems. Let me try to break it down step by step.First, the problem says that the firm's profit, Π(q), is a quadratic function given by Π(q) = -2q² + 40q - 150. Okay, so that's a quadratic equation, and since the coefficient of q² is negative (-2), the parabola opens downward. That means the vertex of this parabola will give me the maximum profit. Cool, so I need to find the vertex to determine the output level q that maximizes profit.I remember that for a quadratic function in the form f(q) = aq² + bq + c, the vertex occurs at q = -b/(2a). Let me apply that here. So, in this case, a is -2 and b is 40. Plugging into the formula: q = -40/(2*(-2)) = -40/(-4) = 10. So, the output level q that maximizes profit is 10 units.Now, to find the maximum profit, I need to plug q = 10 back into the profit function Π(q). Let's compute that:Π(10) = -2*(10)² + 40*(10) - 150= -2*100 + 400 - 150= -200 + 400 - 150= (400 - 200) - 150= 200 - 150= 50So, the maximum profit is 50 units. That seems straightforward.Moving on to the second part, after a political shift, communism is introduced, and profits must be equally distributed among the workers. There are 20 employees, so the profit per employee, P_e(q), is Π(q)/20. So, P_e(q) = (-2q² + 40q - 150)/20.Now, Mr. Capital can re-optimize the output level to maximize this new profit per employee. So, I need to find the q that maximizes P_e(q). Since P_e(q) is just Π(q) scaled down by a factor of 20, the shape of the function remains the same. It's still a quadratic function, just with different coefficients.Let me write out P_e(q):P_e(q) = (-2q² + 40q - 150)/20= (-2/20)q² + (40/20)q - (150/20)Simplify each term:= (-1/10)q² + 2q - 7.5So, P_e(q) = -0.1q² + 2q - 7.5Again, this is a quadratic function with a negative coefficient on q², so it opens downward, and the vertex will give the maximum profit per employee.Using the vertex formula again, q = -b/(2a). Here, a = -0.1 and b = 2.So, q = -2/(2*(-0.1)) = -2/(-0.2) = 10.Wait, that's the same output level as before, q = 10. Hmm, so does that mean the optimal output level doesn't change? Let me think about that.Since P_e(q) is just Π(q) divided by 20, which is a constant factor, the shape of the function doesn't change. So, the maximum occurs at the same q value. Therefore, the optimal output level remains 10.But let me verify that. Maybe I made a mistake in the calculation. Let's compute P_e(q) at q = 10.P_e(10) = (-0.1)*(10)^2 + 2*(10) - 7.5= (-0.1)*100 + 20 - 7.5= -10 + 20 - 7.5= 10 - 7.5= 2.5So, the profit per employee is 2.5 units when q = 10.But wait, is there a different q that could give a higher P_e(q)? Let me test another value, say q = 11.P_e(11) = (-0.1)*(121) + 22 - 7.5= -12.1 + 22 - 7.5= 9.9 - 7.5= 2.4Hmm, that's lower than 2.5. How about q = 9?P_e(9) = (-0.1)*(81) + 18 - 7.5= -8.1 + 18 - 7.5= 9.9 - 7.5= 2.4Same as q = 11. So, it seems that q = 10 is indeed the maximum.Wait, but let me think about this differently. Since P_e(q) is just Π(q)/20, maximizing P_e(q) is equivalent to maximizing Π(q). So, naturally, the maximum occurs at the same q. So, it's logical that q = 10 is still the optimal output level.But just to be thorough, let me take the derivative of P_e(q) with respect to q and set it to zero to find the maximum.P_e(q) = -0.1q² + 2q - 7.5dP_e/dq = -0.2q + 2Set derivative equal to zero:-0.2q + 2 = 0-0.2q = -2q = (-2)/(-0.2)q = 10Yep, same result. So, the optimal output level is still 10, and the profit per employee is 2.5.Wait, but let me think about the implications. Under the capitalist system, the total profit was 50, which is then distributed equally among 20 employees, so each gets 2.5. If the output level was changed, say, to 11, the total profit would be:Π(11) = -2*(121) + 40*(11) - 150= -242 + 440 - 150= (440 - 242) - 150= 198 - 150= 48So, total profit is 48, which divided by 20 is 2.4 per employee, which is less than 2.5. Similarly, at q=9:Π(9) = -2*(81) + 40*(9) - 150= -162 + 360 - 150= (360 - 162) - 150= 198 - 150= 48Same as q=11, so profit per employee is 2.4.Therefore, it's confirmed that q=10 gives the maximum profit per employee.So, summarizing:1. Under capitalism, the optimal output is 10, with a maximum profit of 50.2. Under communism, the optimal output remains 10, with a profit per employee of 2.5.Wait, but the problem says "after a significant political shift, a form of communism is introduced, leading to a mandatory equal distribution of profits among the workers." So, does that mean that the firm's profit function changes? Or is it just the distribution?Wait, in the problem, it says the profit per employee becomes Π(q)/20. So, the total profit is still Π(q), but now it's divided equally. So, the firm's total profit is still Π(q), but each employee gets Π(q)/20. So, if the goal is to maximize profit per employee, which is Π(q)/20, then yes, we need to maximize Π(q)/20, which is equivalent to maximizing Π(q). So, the optimal q remains the same.But let me think again. Is there a scenario where the optimal q could change? For example, if the firm's objective changes from maximizing total profit to maximizing profit per employee, which is a different objective. However, in this case, since profit per employee is just a linear transformation of total profit, the optimal q remains the same.But just to be thorough, let me think about if the firm's objective function changes. If the firm was now trying to maximize per employee profit, which is Π(q)/20, then yes, it's the same as maximizing Π(q). So, the optimal q is the same.Alternatively, if the firm had a different objective, like maximizing total profit minus some cost related to distribution, that might change things. But in this case, it's just dividing the profit equally, so the optimization remains the same.Therefore, I think my conclusion is correct.Final Answer1. The output level that maximizes profit is boxed{10}, and the maximum profit is boxed{50}.2. The new optimal output level is boxed{10}, and the corresponding profit per employee is boxed{2.5}.</think>

🧠:<think>Okay, so I have this problem where a chemical engineer is trying to develop an eco-friendly paper production process. They've come up with a new chemical treatment that affects both energy consumption and waste production. The goal is to minimize both of these simultaneously by adjusting the amount of chemical used, denoted as ( C ). First, I need to formulate a multi-objective optimization problem. I remember that multi-objective optimization deals with situations where you have more than one objective to optimize, and they might conflict with each other. So in this case, minimizing energy consumption and minimizing waste production could be conflicting because using more chemical might reduce energy but increase waste, or something like that.The functions given are quadratic in terms of ( C ). So, ( E(C) = aC^2 + bC + c ) and ( W(C) = dC^2 + eC + f ). The constants ( a, b, c, d, e, f ) are determined by experimental data, so they are known values.The first part of the problem is to formulate the optimization problem. So, I need to set up the objectives and the constraints. The objectives are to minimize ( E(C) ) and ( W(C) ). The constraint is that ( C ) must be within a certain range, specifically ( [C_{text{min}}, C_{text{max}}] ). So, mathematically, the problem can be written as:Minimize ( E(C) = aC^2 + bC + c )Minimize ( W(C) = dC^2 + eC + f )Subject to:( C_{text{min}} leq C leq C_{text{max}} )That seems straightforward. I think this is the correct way to formulate the problem. It's a constrained multi-objective optimization problem with two objectives and one variable.Now, moving on to the second part: determining the Pareto optimal set. I remember that in multi-objective optimization, the Pareto optimal set consists of solutions where you can't improve one objective without worsening another. So, these are the non-dominated solutions.Given that both ( E(C) ) and ( W(C) ) are quadratic functions, their shapes depend on the coefficients. Since they are quadratic, they can either open upwards or downwards depending on the sign of the leading coefficient. If ( a ) is positive, ( E(C) ) opens upwards, meaning it has a minimum point. Similarly, if ( d ) is positive, ( W(C) ) opens upwards.Assuming that both ( a ) and ( d ) are positive, which is typical in such contexts because increasing ( C ) beyond a certain point might start increasing energy or waste, so the functions have a minimum. If that's the case, each function will have a single minimum point within the feasible region.But since we're dealing with two objectives, the Pareto optimal set isn't just a single point but a set of points where each point represents a trade-off between energy and waste. So, to find the Pareto optimal set, I need to find all the points where improving one objective doesn't worsen the other.One method to find the Pareto optimal set is to use the concept of non-dominated solutions. For each ( C ) in the feasible range, we check if there's another ( C' ) such that ( E(C') leq E(C) ) and ( W(C') leq W(C) ). If such a ( C' ) exists, then ( C ) is dominated and not part of the Pareto set. Otherwise, it is.Alternatively, since both functions are quadratic, we can find their individual minima and then see how they trade off. Let me think about the steps:1. Find the minimum of ( E(C) ). The derivative of ( E(C) ) with respect to ( C ) is ( 2aC + b ). Setting this equal to zero gives ( C_E = -b/(2a) ).2. Similarly, find the minimum of ( W(C) ). The derivative is ( 2dC + e ). Setting this equal to zero gives ( C_W = -e/(2d) ).Now, these minima might lie within the feasible range ( [C_{text{min}}, C_{text{max}}] ) or not. If they do, those are the points where each function is minimized individually. However, since we're minimizing both, the Pareto optimal set will likely include these points and possibly others where the trade-off between energy and waste is balanced.But since both functions are convex (assuming ( a, d > 0 )), the Pareto front will be a convex curve. The Pareto optimal set in this case can be found by considering all ( C ) values where the trade-off between the two objectives is optimal.Another approach is to use the method of weighted sums. For different weights ( lambda ) and ( 1 - lambda ), we can combine the two objectives into a single function:( lambda E(C) + (1 - lambda) W(C) )Minimizing this for different ( lambda ) in [0,1] will give different points on the Pareto front. However, since both ( E(C) ) and ( W(C) ) are quadratic, the combined function is also quadratic, and its minimum can be found by taking the derivative.But wait, if we use the weighted sum method, each weight will give a different optimal ( C ). So, the Pareto optimal set can be parameterized by ( lambda ), and each ( lambda ) gives a specific ( C ) that minimizes the weighted sum.Alternatively, we can use the concept of marginal rates of substitution. At a Pareto optimal point, the marginal rate of substitution between energy and waste should be equal to the ratio of their gradients. That is:( frac{dE}{dC} = lambda frac{dW}{dC} )Where ( lambda ) is the trade-off rate between energy and waste. This gives us:( 2aC + b = lambda (2dC + e) )Solving for ( C ) in terms of ( lambda ):( 2aC + b = 2lambda d C + lambda e )( (2a - 2lambda d)C = lambda e - b )( C = frac{lambda e - b}{2(a - lambda d)} )This gives us a relationship between ( C ) and ( lambda ). For each ( lambda ) in [0,1], we get a corresponding ( C ) that lies on the Pareto front.However, we also need to ensure that ( C ) lies within ( [C_{text{min}}, C_{text{max}}] ). So, we might have to adjust ( lambda ) such that ( C ) stays within the feasible range.But this seems a bit involved. Maybe another way is to consider that since both functions are convex, the Pareto optimal set will consist of all points between the minima of the two functions, considering the trade-offs.Wait, actually, in two-objective optimization with convex objectives, the Pareto optimal set is the set of all points where the gradient of one objective is a scalar multiple of the gradient of the other. This is similar to what I wrote earlier with the marginal rates of substitution.So, to find the Pareto optimal set, we can set the gradients equal up to a scalar multiple. That is:( nabla E = lambda nabla W )Since these are functions of a single variable ( C ), the gradients are just the derivatives:( E'(C) = lambda W'(C) )Which is the same as:( 2aC + b = lambda (2dC + e) )This equation relates ( C ) and ( lambda ). For each ( lambda ), we can solve for ( C ). However, ( lambda ) is a positive scalar, and typically, in the context of Pareto optimality, ( lambda ) can be considered as the ratio of the weights on the two objectives.But since we are dealing with minimization, we can also think of ( lambda ) as the trade-off rate between the two objectives. So, varying ( lambda ) from 0 to infinity (or 0 to 1, depending on normalization) will give us different points on the Pareto front.However, since ( C ) must be within ( [C_{text{min}}, C_{text{max}}] ), we need to ensure that the solution ( C ) from the above equation lies within this interval. If not, the optimal ( C ) would be at the boundary.So, to summarize, the Pareto optimal set can be found by solving for ( C ) in terms of ( lambda ) from the equation ( 2aC + b = lambda (2dC + e) ), and then ensuring that ( C ) is within the feasible range. Each such ( C ) corresponds to a Pareto optimal solution, and the set of all such ( C ) (and their corresponding ( E(C) ) and ( W(C) )) forms the Pareto optimal set.But wait, actually, in the case of two objectives and one variable, the Pareto optimal set is typically a curve in the objective space, but in terms of the decision variable ( C ), it's a set of points. However, since ( C ) is one-dimensional, the Pareto optimal set in the decision space is also one-dimensional, meaning it's a range of ( C ) values.But I think more accurately, for each ( C ) in the feasible range, we can evaluate ( E(C) ) and ( W(C) ), and then determine which of these points are non-dominated. However, since both functions are convex, the Pareto optimal set will likely consist of a single interval where each point represents a trade-off between the two objectives.Alternatively, considering the gradients, if the minima of both functions are within the feasible region, the Pareto optimal set might be the interval from the minimum of ( E(C) ) to the minimum of ( W(C) ), but I'm not entirely sure.Wait, let's think about this. If both ( E(C) ) and ( W(C) ) have their minima within ( [C_{text{min}}, C_{text{max}}] ), then the Pareto optimal set would include all points between these two minima where increasing ( C ) beyond the energy minimum might start increasing energy but decreasing waste, and vice versa.But actually, no. Because both functions are convex, the Pareto optimal set would be the set of points where the trade-off between the two objectives is optimal. This can be found by solving the equation ( E'(C) = lambda W'(C) ) for ( C ) in terms of ( lambda ), and then considering all such ( C ) that lie within the feasible range.So, solving ( 2aC + b = lambda (2dC + e) ) for ( C ):( 2aC + b = 2lambda d C + lambda e )Rearranging terms:( 2aC - 2lambda d C = lambda e - b )( C(2a - 2lambda d) = lambda e - b )( C = frac{lambda e - b}{2(a - lambda d)} )This gives ( C ) as a function of ( lambda ). Now, ( lambda ) can be any positive real number, but in the context of optimization, it's often considered between 0 and 1 when using weights. However, in this case, since we're dealing with gradients, ( lambda ) can be any positive value, representing the trade-off rate.But we also need to ensure that ( C ) is within ( [C_{text{min}}, C_{text{max}}] ). So, for each ( lambda ), we calculate ( C ) and check if it's within the feasible range. If it is, then that ( C ) is part of the Pareto optimal set. If not, the optimal ( C ) would be at the boundary.However, this approach might not capture all Pareto optimal points because the equation ( E'(C) = lambda W'(C) ) gives a specific ( C ) for each ( lambda ), but the Pareto optimal set is the set of all such ( C ) where this condition holds, considering the feasible range.Alternatively, another method is to consider that the Pareto optimal set is the set of all ( C ) where there is no other ( C' ) such that ( E(C') leq E(C) ) and ( W(C') leq W(C) ) with at least one inequality strict. So, to find this set, we can plot ( E(C) ) and ( W(C) ) against each other and identify the non-dominated region.But since both functions are quadratic, their relationship can be analyzed analytically. Let's consider the trade-off between ( E(C) ) and ( W(C) ). If increasing ( C ) beyond the energy minimum reduces waste but increases energy, and decreasing ( C ) beyond the waste minimum reduces energy but increases waste, then the Pareto optimal set would be the set of ( C ) values where the trade-off is optimal.So, to find the Pareto optimal set, we can consider the following steps:1. Find the minima of ( E(C) ) and ( W(C) ) within the feasible range.2. If both minima are within the feasible range, then the Pareto optimal set will be the interval between these two minima, where each point represents a trade-off between energy and waste.3. If one or both minima are outside the feasible range, then the Pareto optimal set will be at the boundaries or around the feasible minima.But wait, actually, that might not be accurate. Because even if both minima are within the feasible range, the Pareto optimal set isn't necessarily the interval between them. Instead, it's the set of points where the trade-off between the two objectives is such that you can't improve one without worsening the other.So, perhaps a better approach is to consider the equation ( E'(C) = lambda W'(C) ) and solve for ( C ) as a function of ( lambda ), then express ( E(C) ) and ( W(C) ) in terms of ( lambda ), and plot them to get the Pareto front.But since we're dealing with a single variable, the Pareto optimal set in the decision space (i.e., the values of ( C )) can be found by solving for ( C ) in terms of ( lambda ) and ensuring it's within the feasible range.So, to formalize this, the Pareto optimal set ( S ) is given by:( S = left{ C in [C_{text{min}}, C_{text{max}}] bigg| exists lambda > 0 text{ such that } 2aC + b = lambda (2dC + e) right} )This means that for each ( C ) in ( S ), there exists a ( lambda ) such that the marginal increase in energy is proportional to the marginal increase in waste, with the proportionality constant ( lambda ) representing the trade-off rate.Therefore, the Pareto optimal set is the set of all ( C ) in the feasible range that satisfy the above condition for some ( lambda > 0 ).But to express this more explicitly, we can solve for ( C ) in terms of ( lambda ):( C = frac{lambda e - b}{2(a - lambda d)} )Now, we need to ensure that ( C ) is within ( [C_{text{min}}, C_{text{max}}] ). So, for each ( lambda ), we can compute ( C ) and check if it's within the feasible range. If it is, then that ( C ) is part of the Pareto optimal set.However, ( lambda ) can vary over all positive real numbers, so we can parameterize the Pareto optimal set by ( lambda ). Each ( lambda ) gives a specific ( C ), and thus a specific point on the Pareto front.But in practice, we might want to express the Pareto optimal set without the parameter ( lambda ). To do this, we can eliminate ( lambda ) from the equations.From the equation ( 2aC + b = lambda (2dC + e) ), we can solve for ( lambda ):( lambda = frac{2aC + b}{2dC + e} )Now, substituting this into the expressions for ( E(C) ) and ( W(C) ), we can express one objective in terms of the other.But this might not be straightforward. Alternatively, we can express the relationship between ( E ) and ( W ) by eliminating ( C ).Let me try that. From ( C = frac{lambda e - b}{2(a - lambda d)} ), we can express ( lambda ) in terms of ( C ):( lambda = frac{2aC + b}{2dC + e} )Now, substituting this into the expressions for ( E(C) ) and ( W(C) ):( E(C) = aC^2 + bC + c )( W(C) = dC^2 + eC + f )But I don't see an immediate way to eliminate ( C ) and get a direct relationship between ( E ) and ( W ). However, since both are quadratic, their relationship might be a quadratic curve in the ( E )-( W ) plane.But perhaps it's more useful to consider that the Pareto optimal set in the decision space (i.e., the values of ( C )) is given by the solutions to ( 2aC + b = lambda (2dC + e) ) for some ( lambda > 0 ), within the feasible range.Therefore, the Pareto optimal set is the set of all ( C ) in ( [C_{text{min}}, C_{text{max}}] ) such that ( C = frac{lambda e - b}{2(a - lambda d)} ) for some ( lambda > 0 ).But to express this without ( lambda ), we can consider that for each ( C ) in the feasible range, if the ratio ( frac{2aC + b}{2dC + e} ) is positive, then ( C ) is part of the Pareto optimal set.Wait, that might not be correct. Because ( lambda ) must be positive, so ( frac{2aC + b}{2dC + e} > 0 ). Therefore, ( 2aC + b ) and ( 2dC + e ) must have the same sign.So, the condition for ( C ) to be in the Pareto optimal set is:1. ( C in [C_{text{min}}, C_{text{max}}] )2. ( (2aC + b)(2dC + e) > 0 )This ensures that ( lambda ) is positive.Therefore, the Pareto optimal set is the subset of ( [C_{text{min}}, C_{text{max}}] ) where ( (2aC + b)(2dC + e) > 0 ).This makes sense because if both derivatives are positive or both are negative, then the trade-off rate ( lambda ) is positive, indicating a valid trade-off between the two objectives.So, to find the Pareto optimal set, we need to solve the inequality ( (2aC + b)(2dC + e) > 0 ) within the interval ( [C_{text{min}}, C_{text{max}}] ).This inequality can be solved by finding the roots of the equation ( (2aC + b)(2dC + e) = 0 ), which are ( C = -b/(2a) ) and ( C = -e/(2d) ). These are the points where each derivative is zero, i.e., the minima of ( E(C) ) and ( W(C) ).The sign of the product ( (2aC + b)(2dC + e) ) will depend on the intervals determined by these roots. So, we can analyze the sign of the product in each interval:1. For ( C < min(-b/(2a), -e/(2d)) ): Depending on the signs of ( a ) and ( d ), the product could be positive or negative.2. For ( min(-b/(2a), -e/(2d)) < C < max(-b/(2a), -e/(2d)) ): The product will have the opposite sign.3. For ( C > max(-b/(2a), -e/(2d)) ): The product will have the same sign as the leading terms.But since ( a ) and ( d ) are positive (assuming convex functions), the derivatives ( 2aC + b ) and ( 2dC + e ) will increase as ( C ) increases. Therefore, if ( -b/(2a) < -e/(2d) ), then for ( C < -b/(2a) ), both derivatives are negative, so their product is positive. Between ( -b/(2a) ) and ( -e/(2d) ), one derivative is positive and the other is negative, so the product is negative. For ( C > -e/(2d) ), both derivatives are positive, so the product is positive.Similarly, if ( -e/(2d) < -b/(2a) ), the intervals would be similar but reversed.Therefore, the product ( (2aC + b)(2dC + e) ) is positive in the intervals ( C < min(-b/(2a), -e/(2d)) ) and ( C > max(-b/(2a), -e/(2d)) ), and negative in between.But since we are considering ( C ) within ( [C_{text{min}}, C_{text{max}}] ), we need to see where this interval overlaps with the regions where the product is positive.So, the Pareto optimal set will be the union of the intervals within ( [C_{text{min}}, C_{text{max}}] ) where ( (2aC + b)(2dC + e) > 0 ).This means that if the feasible region ( [C_{text{min}}, C_{text{max}}] ) overlaps with the regions where the product is positive, those overlapping intervals are part of the Pareto optimal set.However, if the entire feasible region lies within the interval where the product is negative, then there are no Pareto optimal points, which doesn't make sense because we can always find a trade-off.Wait, actually, no. Because even if the product is negative, it just means that the trade-off rate ( lambda ) would be negative, which isn't meaningful in this context since we're dealing with minimization. Therefore, only the regions where the product is positive contribute to the Pareto optimal set.Therefore, the Pareto optimal set is the set of all ( C ) in ( [C_{text{min}}, C_{text{max}}] ) such that ( (2aC + b)(2dC + e) > 0 ).So, to summarize, the Pareto optimal set is determined by solving the inequality ( (2aC + b)(2dC + e) > 0 ) within the feasible range ( [C_{text{min}}, C_{text{max}}] ). The solutions to this inequality give the values of ( C ) that are Pareto optimal.Therefore, the Pareto optimal set is:( S = left{ C in [C_{text{min}}, C_{text{max}}] bigg| (2aC + b)(2dC + e) > 0 right} )This set represents all the chemical amounts ( C ) where the trade-off between energy consumption and waste production is optimal, meaning you can't reduce one without increasing the other.So, putting it all together, the multi-objective optimization problem is formulated with two objectives to minimize and a constraint on ( C ). The Pareto optimal set is found by identifying the regions where the product of the derivatives is positive within the feasible range.</think>

🧠:<think>Alright, so I'm trying to figure out this problem about the speed of sound in the ocean. It's a bit complex, but I'll take it step by step. First, the problem gives me an equation for the speed of sound ( c(d) ) which depends on depth ( d ). The formula is:[ c(d) = 1449.2 + 4.6T(d) - 0.055T(d)^2 + 0.00029T(d)^3 + (1.34 - 0.01T(d))(S(d) - 35) + 0.016d ]And they tell me that both temperature ( T(d) ) and salinity ( S(d) ) vary linearly with depth. Specifically:- ( T(d) = 20 - 0.01d )- ( S(d) = 35 + 0.002d )So, my first task is to substitute these expressions into the formula for ( c(d) ) and then find the depth ( d ) where ( c(d) ) is maximized between 0 and 1000 meters.Alright, let's start by substituting ( T(d) ) and ( S(d) ) into the equation.First, let me write down all the components:1. ( T(d) = 20 - 0.01d )2. ( S(d) = 35 + 0.002d )So, I'll substitute these into each part of the equation for ( c(d) ).Let me break down the equation term by term:1. The constant term: 1449.22. The term with ( T(d) ): 4.6T(d)3. The quadratic term with ( T(d) ): -0.055T(d)^24. The cubic term with ( T(d) ): 0.00029T(d)^35. The term involving both ( T(d) ) and ( S(d) ): (1.34 - 0.01T(d))(S(d) - 35)6. The term with depth ( d ): 0.016dSo, substituting ( T(d) ) and ( S(d) ) into each term:1. 1449.2 remains as is.2. 4.6*(20 - 0.01d)3. -0.055*(20 - 0.01d)^24. 0.00029*(20 - 0.01d)^35. (1.34 - 0.01*(20 - 0.01d))*( (35 + 0.002d) - 35 )6. 0.016dNow, let's compute each term step by step.Starting with term 2:2. 4.6*(20 - 0.01d) = 4.6*20 - 4.6*0.01d = 92 - 0.046dTerm 3:3. -0.055*(20 - 0.01d)^2First, compute (20 - 0.01d)^2:= 20^2 - 2*20*0.01d + (0.01d)^2= 400 - 0.4d + 0.0001d^2So, term 3 becomes:-0.055*(400 - 0.4d + 0.0001d^2)= -0.055*400 + 0.055*0.4d - 0.055*0.0001d^2= -22 + 0.022d - 0.0000055d^2Term 4:4. 0.00029*(20 - 0.01d)^3First, compute (20 - 0.01d)^3:= 20^3 - 3*20^2*0.01d + 3*20*(0.01d)^2 - (0.01d)^3= 8000 - 3*400*0.01d + 3*20*0.0001d^2 - 0.000001d^3= 8000 - 12d + 0.006d^2 - 0.000001d^3So, term 4 becomes:0.00029*(8000 - 12d + 0.006d^2 - 0.000001d^3)= 0.00029*8000 - 0.00029*12d + 0.00029*0.006d^2 - 0.00029*0.000001d^3= 2.32 - 0.00348d + 0.00000174d^2 - 0.00000000029d^3Term 5:5. (1.34 - 0.01*(20 - 0.01d))*( (35 + 0.002d) - 35 )Simplify the second part first: (35 + 0.002d - 35) = 0.002dSo, term 5 becomes:(1.34 - 0.01*(20 - 0.01d)) * 0.002dCompute inside the first parenthesis:1.34 - 0.01*(20 - 0.01d) = 1.34 - 0.2 + 0.0001d = 1.14 + 0.0001dSo, term 5 is:(1.14 + 0.0001d) * 0.002d = 1.14*0.002d + 0.0001*0.002d^2= 0.00228d + 0.0000002d^2Term 6:6. 0.016d remains as is.Now, let's collect all the terms together:1. 1449.22. 92 - 0.046d3. -22 + 0.022d - 0.0000055d^24. 2.32 - 0.00348d + 0.00000174d^2 - 0.00000000029d^35. 0.00228d + 0.0000002d^26. 0.016dNow, let's combine all these terms.First, let's list all constants, linear terms, quadratic terms, and cubic terms.Constants:1449.2 + 92 - 22 + 2.32Linear terms:-0.046d + 0.022d - 0.00348d + 0.00228d + 0.016dQuadratic terms:-0.0000055d^2 + 0.00000174d^2 + 0.0000002d^2Cubic terms:-0.00000000029d^3Let's compute each category.Constants:1449.2 + 92 = 1541.21541.2 - 22 = 1519.21519.2 + 2.32 = 1521.52So, constants sum to 1521.52.Linear terms:-0.046d + 0.022d = (-0.046 + 0.022)d = -0.024d-0.024d - 0.00348d = (-0.024 - 0.00348)d = -0.02748d-0.02748d + 0.00228d = (-0.02748 + 0.00228)d = -0.0252d-0.0252d + 0.016d = (-0.0252 + 0.016)d = -0.0092dSo, linear terms sum to -0.0092d.Quadratic terms:-0.0000055d^2 + 0.00000174d^2 = (-0.0000055 + 0.00000174)d^2 = -0.00000376d^2-0.00000376d^2 + 0.0000002d^2 = (-0.00000376 + 0.0000002)d^2 = -0.00000356d^2So, quadratic terms sum to -0.00000356d^2.Cubic terms:Only one term: -0.00000000029d^3So, putting it all together, the equation for ( c(d) ) becomes:[ c(d) = 1521.52 - 0.0092d - 0.00000356d^2 - 0.00000000029d^3 ]Hmm, that seems a bit odd. Let me double-check my calculations because the coefficients are very small for quadratic and cubic terms, which might be correct, but I want to make sure I didn't make a mistake.Wait, when I computed term 3, I had:-0.055*(400 - 0.4d + 0.0001d^2) = -22 + 0.022d - 0.0000055d^2That seems correct.Term 4:0.00029*(8000 - 12d + 0.006d^2 - 0.000001d^3) = 2.32 - 0.00348d + 0.00000174d^2 - 0.00000000029d^3Yes, that looks correct.Term 5:(1.14 + 0.0001d)*0.002d = 0.00228d + 0.0000002d^2Yes, that's correct.So, when I added up the constants, linear, quadratic, and cubic terms, I think I did it right.So, the final expression for ( c(d) ) is:[ c(d) = 1521.52 - 0.0092d - 0.00000356d^2 - 0.00000000029d^3 ]Wait, but this seems like a very small coefficient for the quadratic term. Let me see, maybe I made a mistake in the signs or the coefficients.Wait, in term 3, the quadratic term was -0.0000055d^2, and in term 4, it was +0.00000174d^2, and in term 5, +0.0000002d^2. So, adding them up:-0.0000055 + 0.00000174 + 0.0000002 = (-0.0000055 + 0.00000194) = -0.00000356Yes, that's correct.Similarly, the cubic term is only from term 4: -0.00000000029d^3So, seems correct.Now, so the function is:[ c(d) = 1521.52 - 0.0092d - 0.00000356d^2 - 0.00000000029d^3 ]Wait, but this is a cubic function, and we're looking for its maximum between 0 and 1000 meters.To find the maximum, we can take the derivative of ( c(d) ) with respect to ( d ), set it equal to zero, and solve for ( d ).So, let's compute the derivative ( c'(d) ):[ c'(d) = frac{d}{dd} [1521.52 - 0.0092d - 0.00000356d^2 - 0.00000000029d^3] ]Which is:[ c'(d) = -0.0092 - 2*0.00000356d - 3*0.00000000029d^2 ][ c'(d) = -0.0092 - 0.00000712d - 0.00000000087d^2 ]We set this equal to zero to find critical points:[ -0.0092 - 0.00000712d - 0.00000000087d^2 = 0 ]Multiply both sides by -1 to make it easier:[ 0.0092 + 0.00000712d + 0.00000000087d^2 = 0 ]This is a quadratic equation in terms of ( d ):[ 0.00000000087d^2 + 0.00000712d + 0.0092 = 0 ]Let me write it as:[ a d^2 + b d + c = 0 ]Where:- ( a = 0.00000000087 )- ( b = 0.00000712 )- ( c = 0.0092 )To solve for ( d ), we can use the quadratic formula:[ d = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]Plugging in the values:First, compute the discriminant ( D = b^2 - 4ac ):( D = (0.00000712)^2 - 4 * 0.00000000087 * 0.0092 )Compute each part:( (0.00000712)^2 = 5.06944 * 10^{-11} )( 4 * 0.00000000087 * 0.0092 = 4 * 0.00000000087 * 0.0092 )First, 0.00000000087 * 0.0092 = 7.956 * 10^{-12}Then, multiply by 4: 3.1824 * 10^{-11}So, ( D = 5.06944 * 10^{-11} - 3.1824 * 10^{-11} = 1.88704 * 10^{-11} )Now, square root of D:( sqrt{D} = sqrt{1.88704 * 10^{-11}} approx 1.3737 * 10^{-5.5} )Wait, let me compute it more accurately.First, 1.88704 * 10^{-11} is 1.88704e-11.The square root of 1.88704e-11 is sqrt(1.88704)*10^{-5.5}.Wait, sqrt(1.88704) is approximately 1.3737, and 10^{-5.5} is 10^{-5} * 10^{-0.5} ≈ 1e-5 * 0.3162 ≈ 3.162e-6.So, sqrt(D) ≈ 1.3737 * 3.162e-6 ≈ 4.343e-6Wait, let me compute it step by step:sqrt(1.88704e-11) = sqrt(1.88704) * sqrt(1e-11) = 1.3737 * 1e-5.5But 1e-5.5 is 1e-5 * 1e-0.5 ≈ 1e-5 * 0.3162 ≈ 3.162e-6So, 1.3737 * 3.162e-6 ≈ 4.343e-6So, sqrt(D) ≈ 4.343e-6Now, plug into the quadratic formula:[ d = frac{-b pm sqrt{D}}{2a} ]So,[ d = frac{-0.00000712 pm 4.343e-6}{2 * 0.00000000087} ]Compute denominator:2 * 0.00000000087 = 0.00000000174So,First, compute the two possible solutions:1. ( d = frac{-0.00000712 + 4.343e-6}{0.00000000174} )2. ( d = frac{-0.00000712 - 4.343e-6}{0.00000000174} )Compute numerator for first solution:-0.00000712 + 0.000004343 = (-7.12e-6 + 4.343e-6) = -2.777e-6So,d = (-2.777e-6) / 0.00000000174 ≈ (-2.777e-6) / (1.74e-9) ≈ -1596.0Second solution:Numerator: -0.00000712 - 0.000004343 = -0.000011463 = -1.1463e-5So,d = (-1.1463e-5) / 0.00000000174 ≈ (-1.1463e-5) / (1.74e-9) ≈ -6588.0Wait, both solutions are negative, which is outside our depth range of 0 to 1000 meters.Hmm, that suggests that the derivative ( c'(d) ) never crosses zero within the interval [0, 1000]. So, the maximum must occur at one of the endpoints.But wait, let's check the derivative at d=0 and d=1000 to see if it's increasing or decreasing.At d=0:c'(0) = -0.0092 - 0 - 0 = -0.0092 < 0So, the function is decreasing at d=0.At d=1000:Compute c'(1000):= -0.0092 - 0.00000712*1000 - 0.00000000087*(1000)^2= -0.0092 - 0.00712 - 0.00000087= (-0.0092 - 0.00712) - 0.00000087 ≈ -0.01632 - 0.00000087 ≈ -0.01632087 < 0So, the derivative is negative at both ends, meaning the function is decreasing throughout the interval. Therefore, the maximum occurs at the left endpoint, which is d=0.But wait, that seems counterintuitive because usually, the speed of sound in the ocean can have a maximum at a certain depth due to the thermocline and halocline. But according to this calculation, the speed is decreasing throughout the depth from 0 to 1000 meters. So, the maximum speed is at d=0.But let me double-check my calculations because this seems odd.Wait, perhaps I made a mistake in substituting the terms or in the derivative.Let me go back to the expression for ( c(d) ):After substitution, I had:[ c(d) = 1521.52 - 0.0092d - 0.00000356d^2 - 0.00000000029d^3 ]Is this correct?Wait, let me recompute the constants and coefficients step by step.Original equation:c(d) = 1449.2 + 4.6T - 0.055T^2 + 0.00029T^3 + (1.34 - 0.01T)(S - 35) + 0.016dWith T = 20 - 0.01d and S = 35 + 0.002d.So, let's compute each term again carefully.Compute 4.6T:4.6*(20 - 0.01d) = 92 - 0.046dCompute -0.055T^2:First, T^2 = (20 - 0.01d)^2 = 400 - 0.4d + 0.0001d^2So, -0.055*(400 - 0.4d + 0.0001d^2) = -22 + 0.022d - 0.0000055d^2Compute 0.00029T^3:First, T^3 = (20 - 0.01d)^3 = 8000 - 12d + 0.006d^2 - 0.000001d^3So, 0.00029*(8000 - 12d + 0.006d^2 - 0.000001d^3) = 2.32 - 0.00348d + 0.00000174d^2 - 0.00000000029d^3Compute (1.34 - 0.01T)(S - 35):First, S - 35 = 0.002dThen, 1.34 - 0.01T = 1.34 - 0.01*(20 - 0.01d) = 1.34 - 0.2 + 0.0001d = 1.14 + 0.0001dSo, (1.14 + 0.0001d)*0.002d = 0.00228d + 0.0000002d^2So, adding all terms:1449.2 + 92 - 0.046d -22 + 0.022d -0.0000055d^2 + 2.32 -0.00348d +0.00000174d^2 -0.00000000029d^3 +0.00228d +0.0000002d^2 +0.016dNow, let's compute constants:1449.2 + 92 = 1541.21541.2 -22 = 1519.21519.2 + 2.32 = 1521.52Linear terms:-0.046d + 0.022d = -0.024d-0.024d -0.00348d = -0.02748d-0.02748d +0.00228d = -0.0252d-0.0252d +0.016d = -0.0092dQuadratic terms:-0.0000055d^2 +0.00000174d^2 = -0.00000376d^2-0.00000376d^2 +0.0000002d^2 = -0.00000356d^2Cubic term:-0.00000000029d^3So, yes, the expression is correct.Therefore, the derivative is:c'(d) = -0.0092 -0.00000712d -0.00000000087d^2Which is always negative in the interval [0,1000], as we saw.Therefore, the function is decreasing throughout the interval, so the maximum occurs at d=0.But wait, in reality, the speed of sound in the ocean typically has a maximum at a certain depth because of the temperature and salinity profiles. So, maybe the given temperature and salinity profiles are such that the speed decreases with depth.Given that T(d) = 20 - 0.01d, so temperature decreases with depth, which is typical. Salinity S(d) = 35 + 0.002d, so salinity increases with depth, which is also typical in some regions.But the speed of sound depends on both temperature and salinity. The effect of temperature is more significant in the thermocline, and salinity in the halocline.But according to our calculations, the speed is decreasing with depth. So, perhaps in this specific case, the combined effect of decreasing temperature and increasing salinity leads to an overall decrease in sound speed.Therefore, the maximum speed is at the surface, d=0.So, the answer to part 1 is d=0 meters.But let me check the speed at d=0 and d=1000 to confirm.At d=0:c(0) = 1449.2 + 4.6*20 -0.055*(20)^2 +0.00029*(20)^3 + (1.34 -0.01*20)*(35 -35) +0.016*0Compute each term:1449.24.6*20 = 92-0.055*(400) = -220.00029*(8000) = 2.32(1.34 -0.2)*(0) = 00.016*0 = 0So, c(0) = 1449.2 +92 -22 +2.32 = 1449.2 +92 = 1541.2; 1541.2 -22=1519.2; 1519.2 +2.32=1521.52 m/sAt d=1000:Compute c(1000):First, T(1000)=20 -0.01*1000=10°CS(1000)=35 +0.002*1000=37 pptSo,c(1000)=1449.2 +4.6*10 -0.055*(10)^2 +0.00029*(10)^3 + (1.34 -0.01*10)*(37 -35) +0.016*1000Compute each term:1449.24.6*10=46-0.055*100= -5.50.00029*1000=0.29(1.34 -0.1)*(2)= (1.24)*2=2.480.016*1000=16So, summing up:1449.2 +46=1495.21495.2 -5.5=1489.71489.7 +0.29=1490.01490.0 +2.48=1492.481492.48 +16=1508.48 m/sSo, c(0)=1521.52 m/s and c(1000)=1508.48 m/sTherefore, the speed decreases from 1521.52 m/s at the surface to 1508.48 m/s at 1000m, confirming that the maximum is at d=0.So, the answer to part 1 is d=0 meters.Now, part 2: calculate the corresponding speed of sound c(d) at this depth (which is d=0, so c=1521.52 m/s) and analyze how a 1% increase in salinity S(d) at this depth would affect the speed of sound, assuming other parameters remain unchanged.First, the current salinity at d=0 is S(0)=35 +0.002*0=35 ppt.A 1% increase in salinity would make it 35 *1.01=35.35 ppt.But wait, actually, the formula uses S(d) -35, so let's see how the term involving salinity is structured.In the original equation:c(d) = ... + (1.34 -0.01T(d))(S(d) -35) + ...So, the term is (1.34 -0.01T(d))*(S -35)At d=0, T=20°C, so 1.34 -0.01*20=1.34 -0.2=1.14So, the term is 1.14*(S -35)Currently, S=35, so this term is 0.If we increase S by 1%, S becomes 35*1.01=35.35So, the new term is 1.14*(35.35 -35)=1.14*0.35=0.399So, the change in c(d) is +0.399 m/sTherefore, the new speed of sound would be 1521.52 +0.399≈1521.919 m/sSo, the increase is approximately 0.399 m/s, which is a small increase.Alternatively, to find the sensitivity, we can compute the derivative of c with respect to S at d=0.From the original equation:c = ... + (1.34 -0.01T)(S -35) + ...So, dc/dS = (1.34 -0.01T)At d=0, T=20, so dc/dS=1.34 -0.2=1.14 m/s per ppt.Therefore, a 1% increase in S from 35 to 35.35 is an increase of 0.35 ppt.So, the change in c is 1.14 *0.35≈0.399 m/s, same as before.So, the speed increases by approximately 0.4 m/s.Therefore, the effect is a small increase in sound speed due to the 1% increase in salinity.So, summarizing:1. The depth where speed is maximum is d=0 meters.2. The speed at d=0 is 1521.52 m/s, and a 1% increase in salinity increases the speed by approximately 0.4 m/s.Final Answer1. The depth at which the speed of sound reaches a maximum is boxed{0} meters.2. The corresponding speed of sound is boxed{1521.52} meters per second, and a 1% increase in salinity would increase the speed by approximately boxed{0.4} meters per second.</think>

🧠:<think>Alright, so I have this problem about an artist who's created a grid painting with some specific requirements. Let me try to break it down step by step.First, part 1 is about coloring the grid such that no two adjacent squares have the same color. I need to find the minimum number of colors required for this. Hmm, okay. This sounds familiar, like a graph coloring problem. Each square is a vertex, and edges connect adjacent squares. So, we need to find the chromatic number of this grid graph.I remember that for a grid, which is a bipartite graph, the chromatic number is 2. Wait, is that right? Because in a bipartite graph, you can color the vertices with two colors such that no two adjacent vertices share the same color. So, for a chessboard, which is an 8x8 grid, you only need two colors, right? Black and white, alternating.But wait, is a grid always bipartite? Let me think. A bipartite graph is a graph whose vertices can be divided into two disjoint sets such that no two graph vertices within the same set are adjacent. In a grid, each square can be colored like a checkerboard, alternating between two colors. So, yes, that should work. So, regardless of the size of the grid, as long as it's a grid graph, it's bipartite, so 2 colors suffice.But hold on, the problem says m and n are greater than 1. So, it's at least a 2x2 grid. In a 2x2 grid, you can color it with two colors, alternating. So, yeah, I think 2 is the minimum number of colors needed.Let me try to visualize a 3x3 grid. If I color the top-left corner with color A, then the adjacent squares must be color B. Then the next row would start with color B, then A, and so on. This way, no two adjacent squares share the same color. So, it seems consistent.But wait, what if the grid is odd by odd? Like 3x3. Does it still work? Let me check. The center square is surrounded by four squares. If the center is color A, then all four surrounding it must be color B. Then, the squares adjacent to those would be color A, and so on. So, yes, it still works. So, regardless of whether m and n are even or odd, as long as it's a grid, two colors suffice.Therefore, the minimum number of colors required is 2. I think that's the answer for part 1.Now, moving on to part 2. The artist wants to paint each square with a gradient that changes linearly. The gradient is represented by a function f(x, y) that varies between 0 and 1, where (x, y) are the coordinates of the center of each square. We need to find a possible function f(x, y) such that the gradient changes smoothly across the entire canvas and the integral of f(x, y) over the entire grid is equal to mn/2.Okay, so first, the function f(x, y) should vary smoothly. Smoothly usually implies that the function is continuous and differentiable, but since we're dealing with a grid of squares, each square has a center point (x, y). So, maybe f(x, y) is a function defined on these discrete points, but the artist wants it to change smoothly, so perhaps the function should vary linearly across the entire grid.Also, the integral over the entire grid should be mn/2. Since each square is a unit square, I assume, or maybe not? Wait, the problem doesn't specify the size of each square, just that it's an m x n grid. So, perhaps each square is of unit length, so the entire grid is m units wide and n units tall.But the function f(x, y) is defined on the centers of each square. So, the centers would be at positions (i + 0.5, j + 0.5) for i from 0 to m-1 and j from 0 to n-1, assuming the grid starts at (0,0).Wait, but the integral over the entire grid would be the sum of f(x, y) over all squares, right? Because each square contributes f(x, y) * area, but if each square is a unit square, then the area is 1, so the integral is just the sum of f(x, y) over all m x n squares.So, the integral of f(x, y) over the entire grid is equal to the sum of f(x, y) for all squares, which should be mn/2. So, the average value of f(x, y) over the grid should be 1/2.So, we need a function f(x, y) that varies smoothly from 0 to 1 across the grid, such that the average value is 1/2.Hmm, a linear gradient could work. For example, if we have a gradient that goes from 0 at the bottom-left corner to 1 at the top-right corner, that would be a linear function.But wait, the function is defined at the centers of each square. So, perhaps f(x, y) = (x + y)/ (m + n - 1). Wait, but if we do that, the maximum value would be at (m, n), but since the centers are at (i + 0.5, j + 0.5), the maximum x would be m - 0.5 and maximum y would be n - 0.5.Alternatively, maybe f(x, y) = (x + y)/(m + n). Hmm, but let's think about the integral.Wait, if we define f(x, y) as a linear function, say f(x, y) = ax + by + c, then the integral over the grid would be the sum over all squares of f(x, y). Since each square is a unit square, the integral is the sum.But we need the sum to be mn/2. So, if we can define f(x, y) such that the average value is 1/2, that would satisfy the integral condition.Alternatively, maybe f(x, y) is a function that increases linearly from 0 to 1 across the grid. For example, f(x, y) = (x)/(m) or f(x, y) = (y)/(n). But that would make the function vary along one axis only.Wait, but the problem says the gradient changes smoothly across the entire canvas. So, maybe a linear gradient in both x and y directions. For example, f(x, y) = (x + y)/(m + n). But let's test this.Wait, the centers are at (i + 0.5, j + 0.5) for i from 0 to m-1 and j from 0 to n-1. So, the minimum x is 0.5, maximum x is m - 0.5, similarly for y.If we define f(x, y) = (x + y)/(m + n), then the minimum value would be (0.5 + 0.5)/(m + n) = 1/(m + n), and the maximum would be (m - 0.5 + n - 0.5)/(m + n) = (m + n - 1)/(m + n). So, the range is from 1/(m + n) to (m + n - 1)/(m + n). That's not exactly 0 to 1, but close.Alternatively, maybe f(x, y) = (x - 0.5)/(m - 1) + (y - 0.5)/(n - 1). Wait, that might not be linear. Alternatively, f(x, y) = (x)/(m) + (y)/(n). Let's see.Wait, if we define f(x, y) = (x + y)/(m + n), then the average value would be the sum over all squares of (x + y)/(m + n). Let's compute that.The sum over all squares of x is the sum over i from 0 to m-1 of (i + 0.5) times n, because for each x-coordinate, there are n y-coordinates. Similarly, the sum over all squares of y is the sum over j from 0 to n-1 of (j + 0.5) times m.So, the total sum of x over all squares is n * sum_{i=0}^{m-1} (i + 0.5) = n * [sum_{i=0}^{m-1} i + sum_{i=0}^{m-1} 0.5] = n * [(m(m-1)/2) + 0.5m] = n * [ (m^2 - m)/2 + m/2 ] = n * [ (m^2 - m + m)/2 ] = n * (m^2)/2.Similarly, the total sum of y over all squares is m * sum_{j=0}^{n-1} (j + 0.5) = m * [sum_{j=0}^{n-1} j + sum_{j=0}^{n-1} 0.5] = m * [ (n(n-1)/2) + 0.5n ] = m * [ (n^2 - n)/2 + n/2 ] = m * (n^2)/2.So, the total sum of (x + y) over all squares is n*(m^2)/2 + m*(n^2)/2 = (m^2 n + m n^2)/2 = mn(m + n)/2.Therefore, the sum of f(x, y) over all squares is [mn(m + n)/2] / (m + n) ) = mn/2. Perfect! So, the integral (which is the sum in this case) is mn/2 as required.Therefore, the function f(x, y) = (x + y)/(m + n) satisfies the condition that the integral over the entire grid is mn/2, and it's a linear function, so it changes smoothly across the canvas.Wait, but does this function vary between 0 and 1? Let's check the minimum and maximum values.The minimum x is 0.5, minimum y is 0.5, so f_min = (0.5 + 0.5)/(m + n) = 1/(m + n). The maximum x is m - 0.5, maximum y is n - 0.5, so f_max = (m - 0.5 + n - 0.5)/(m + n) = (m + n - 1)/(m + n) = 1 - 1/(m + n).So, the function varies from 1/(m + n) to 1 - 1/(m + n). It doesn't exactly reach 0 or 1, but it's close. If we want it to vary exactly from 0 to 1, maybe we can adjust the function.Alternatively, we can define f(x, y) = (x - 0.5)/(m - 1) + (y - 0.5)/(n - 1). Wait, let's see.Wait, if we define f(x, y) = (x - 0.5)/(m - 1) + (y - 0.5)/(n - 1), then at the bottom-left corner, x=0.5, y=0.5, so f=0 + 0=0. At the top-right corner, x=m - 0.5, y=n - 0.5, so f=(m - 1)/(m - 1) + (n - 1)/(n - 1)=1 +1=2. That's too much, it goes beyond 1.Alternatively, maybe f(x, y) = (x - 0.5)/(m) + (y - 0.5)/(n). Then at (0.5, 0.5), f=0 + 0=0. At (m - 0.5, n - 0.5), f=(m - 1)/m + (n - 1)/n ≈ 1 + 1=2. Still too much.Wait, maybe f(x, y) = (x)/(m + 1) + (y)/(n + 1). Then at (0.5, 0.5), f≈0.5/(m+1) + 0.5/(n+1). At (m - 0.5, n - 0.5), f≈(m - 0.5)/(m + 1) + (n - 0.5)/(n + 1). Not sure if this reaches 1.Alternatively, maybe f(x, y) = (x + y - 1)/(m + n - 1). Let's test this.At (0.5, 0.5), f=(0.5 + 0.5 -1)/(m + n -1)=0/(m + n -1)=0. At (m - 0.5, n - 0.5), f=(m - 0.5 + n - 0.5 -1)/(m + n -1)= (m + n - 2)/(m + n -1)= (m + n -1 -1)/(m + n -1)=1 - 1/(m + n -1). Close to 1 but not exactly.Hmm, maybe it's acceptable since the function needs to vary between 0 and 1, but perhaps the artist is okay with it approaching 1 asymptotically. Alternatively, maybe a different function.Wait, perhaps f(x, y) = (x + y -1)/(m + n -1). Then, as above, it goes from 0 to almost 1. But if we want exactly 0 to 1, maybe f(x, y) = (x + y -1)/(m + n -1) * 1. So, it's scaled to reach 1 at the top-right corner.Wait, but then the integral would be different. Let me check.If f(x, y) = (x + y -1)/(m + n -1), then the sum over all squares is sum_{i=0}^{m-1} sum_{j=0}^{n-1} [(i + 0.5 + j + 0.5 -1)/(m + n -1)] = sum_{i,j} [(i + j)/(m + n -1)].So, the sum is [sum_{i=0}^{m-1} sum_{j=0}^{n-1} (i + j)] / (m + n -1).Compute sum_{i,j} (i + j) = sum_{i} sum_{j} i + sum_{i} sum_{j} j = m sum_{i} i + n sum_{j} j.Sum_{i} i from 0 to m-1 is m(m-1)/2. Similarly, sum_{j} j from 0 to n-1 is n(n-1)/2.So, total sum is m*(m(m-1)/2) + n*(n(n-1)/2) = (m^2(m-1) + n^2(n-1))/2.Wait, that seems complicated. Let me compute it step by step.sum_{i=0}^{m-1} sum_{j=0}^{n-1} (i + j) = sum_{i=0}^{m-1} [sum_{j=0}^{n-1} i + sum_{j=0}^{n-1} j] = sum_{i=0}^{m-1} [n*i + (n(n-1)/2)].So, this is n * sum_{i=0}^{m-1} i + (n(n-1)/2)*m.Sum_{i=0}^{m-1} i = m(m-1)/2.So, total sum = n*(m(m-1)/2) + (n(n-1)/2)*m = (m n (m -1))/2 + (m n (n -1))/2 = (m n (m -1 + n -1))/2 = (m n (m + n -2))/2.Therefore, the sum of f(x, y) is [ (m n (m + n -2))/2 ] / (m + n -1 ) = (m n (m + n -2))/(2(m + n -1)).We need this to be equal to mn/2. So,(m n (m + n -2))/(2(m + n -1)) = mn/2Multiply both sides by 2:(m n (m + n -2))/(m + n -1) = mnDivide both sides by mn (assuming mn ≠0, which it isn't):(m + n -2)/(m + n -1) =1But (m + n -2)/(m + n -1) =1 - 1/(m + n -1) ≠1. So, this function doesn't satisfy the integral condition.So, maybe my initial function f(x, y) = (x + y)/(m + n) is better because it does satisfy the integral condition, even though it doesn't reach exactly 0 and 1, but varies between 1/(m + n) and 1 - 1/(m + n). Since the problem says the gradient varies between 0 and 1, maybe it's acceptable as long as it's within that range, even if it doesn't reach the endpoints exactly.Alternatively, maybe we can adjust the function to reach exactly 0 and 1. Let's think.Suppose we define f(x, y) = (x + y -1)/(m + n -1). Then, as before, the minimum is 0 at (0.5, 0.5), and maximum is (m + n -2)/(m + n -1) ≈1. But the integral doesn't match.Wait, maybe we can scale it differently. Suppose f(x, y) = (x + y)/(m + n). As before, the integral is mn/2, which is what we need. So, even though it doesn't reach exactly 0 and 1, it's within the range [1/(m + n), 1 - 1/(m + n)], which is close to 0 and 1 for large m and n.Alternatively, if we want it to reach exactly 0 and 1, maybe we can adjust the function. Let's consider f(x, y) = (x + y -1)/(m + n -1). Then, at (0.5, 0.5), f=0, and at (m -0.5, n -0.5), f=(m + n -1 -1)/(m + n -1)= (m + n -2)/(m + n -1)=1 - 1/(m + n -1). So, it approaches 1 as m and n increase.But the integral is (m n (m + n -2))/(2(m + n -1)) which is less than mn/2. So, to make the integral mn/2, maybe we need a different function.Alternatively, perhaps a function that is linear in x and y but scaled appropriately. Let's suppose f(x, y) = a x + b y + c. We need to find a, b, c such that the integral over the grid is mn/2, and the function varies smoothly.But since the function is linear, the integral is the sum over all squares of (a x + b y + c). Let's compute this sum.Sum_{i,j} (a x + b y + c) = a Sum_{i,j} x + b Sum_{i,j} y + c Sum_{i,j} 1.We already computed Sum x and Sum y earlier.Sum x = n * (m^2)/2, Sum y = m * (n^2)/2, Sum 1 = mn.So, the total sum is a*(n m^2 /2) + b*(m n^2 /2) + c*mn.We need this to equal mn/2.So,a*(n m^2 /2) + b*(m n^2 /2) + c*mn = mn/2.Divide both sides by mn:a*(m/2) + b*(n/2) + c = 1/2.So, we have the equation:(a m + b n)/2 + c = 1/2.Additionally, we want f(x, y) to vary between 0 and 1. So, the minimum value of f(x, y) should be 0, and the maximum should be 1.The minimum occurs at the bottom-left corner (0.5, 0.5), so f(0.5, 0.5) = a*0.5 + b*0.5 + c =0.The maximum occurs at the top-right corner (m -0.5, n -0.5), so f(m -0.5, n -0.5)=a*(m -0.5) + b*(n -0.5) + c=1.So, we have two equations:1. 0.5a + 0.5b + c =0.2. a(m -0.5) + b(n -0.5) + c =1.And from the integral condition:3. (a m + b n)/2 + c = 1/2.So, we have three equations:Equation 1: 0.5a + 0.5b + c =0.Equation 2: a(m -0.5) + b(n -0.5) + c =1.Equation 3: 0.5a m + 0.5b n + c =0.5.Let me write these equations more clearly.Equation 1: 0.5a + 0.5b + c =0.Equation 2: (m -0.5)a + (n -0.5)b + c =1.Equation 3: 0.5 m a + 0.5 n b + c =0.5.Now, let's solve this system.From Equation 1: c = -0.5a -0.5b.Substitute c into Equation 3:0.5 m a + 0.5 n b + (-0.5a -0.5b) =0.5.Simplify:0.5 m a + 0.5 n b -0.5a -0.5b =0.5.Factor:0.5a(m -1) + 0.5b(n -1) =0.5.Multiply both sides by 2:a(m -1) + b(n -1) =1.Similarly, substitute c into Equation 2:(m -0.5)a + (n -0.5)b + (-0.5a -0.5b) =1.Simplify:(m -0.5 -0.5)a + (n -0.5 -0.5)b =1.Which is:(m -1)a + (n -1)b =1.Wait, that's the same as the equation from Equation 3 after substitution. So, we have:a(m -1) + b(n -1) =1.But we also have from Equation 3 substitution:a(m -1) + b(n -1) =1.So, both lead to the same equation. So, we have two equations:1. a(m -1) + b(n -1) =1.2. c = -0.5a -0.5b.So, we need another equation to solve for a and b. But we only have one equation. So, we can choose a parameter, say, set a = k, then solve for b.From Equation 1:k(m -1) + b(n -1) =1 => b = (1 -k(m -1))/(n -1).Then, c = -0.5k -0.5*(1 -k(m -1))/(n -1).But we need another condition. Perhaps we can set a = b to make the function symmetric in x and y.Let me try that. Let a = b.Then, from Equation 1:a(m -1) + a(n -1) =1 => a(m + n -2)=1 => a=1/(m + n -2).So, a = b =1/(m + n -2).Then, c = -0.5a -0.5b = -0.5*(2a) = -a = -1/(m + n -2).So, the function is f(x, y) = a x + a y + c = (x + y)/(m + n -2) -1/(m + n -2) = (x + y -1)/(m + n -2).Wait, but let's check the minimum and maximum.At (0.5, 0.5), f= (0.5 +0.5 -1)/(m + n -2)=0/(m +n -2)=0.At (m -0.5, n -0.5), f=(m -0.5 +n -0.5 -1)/(m +n -2)= (m +n -2)/(m +n -2)=1.Perfect! So, the function f(x, y)=(x + y -1)/(m +n -2) varies from 0 to 1, and satisfies the integral condition.Wait, let's verify the integral.Sum over all squares of f(x, y) = sum_{i,j} [(x + y -1)/(m +n -2)].Which is [sum_{i,j} (x + y -1)] / (m +n -2).Sum_{i,j} (x + y -1) = sum_{i,j} x + sum_{i,j} y - sum_{i,j}1.We already computed sum x = n m^2 /2, sum y = m n^2 /2, sum 1= mn.So, total sum = (n m^2 /2 + m n^2 /2 - mn).Factor mn:mn( m/2 + n/2 -1 ) = mn( (m +n)/2 -1 ).Therefore, sum f(x, y) = [mn( (m +n)/2 -1 ) ] / (m +n -2).Simplify numerator:(m +n)/2 -1 = (m +n -2)/2.So, sum f(x, y) = [mn * (m +n -2)/2 ] / (m +n -2 )= mn/2.Perfect! So, this function satisfies the integral condition and varies smoothly from 0 to 1.Therefore, f(x, y) = (x + y -1)/(m +n -2) is a suitable function.Alternatively, if we don't assume a = b, we could have other solutions, but this symmetric solution seems elegant.So, to summarize:1. The minimum number of colors required is 2, as the grid is a bipartite graph.2. A possible function is f(x, y) = (x + y -1)/(m +n -2).I think that's it.</think>