91Ó°ÊÓ

To find the daily total revenue function, we will multiply the unit price of the deluxe edition by the number of deluxe editions demanded and add this to the product of the unit price of the standard edition and the number of standard editions demanded. The price of a deluxe edition is given by \(p = 20 - 0.005x - 0.001y\), and the price of a standard edition is given by \(q = 15 - 0.001x - 0.003y\). Therefore, the daily total revenue function, R(x, y), can be expressed as: \[R(x, y) = xp + yq\] \[R(x, y) = x(20 - 0.005x - 0.001y) + y(15 - 0.001x - 0.003y)\]

Now we will simplify the revenue function by distributing x and y to the terms inside the brackets: \[R(x, y) = 20x - 0.005x^2 - 0.001xy + 15y - 0.001xy - 0.003y^2\] Combine the similar terms: \[R(x, y) = -0.005x^2 - 0.003y^2 - 0.002xy + 20x + 15y\] So, the daily total revenue function is: \[R(x, y) = -0.005x^2 - 0.003y^2 - 0.002xy + 20x + 15y\]

To find the domain of the function R(x, y), we will determine the values of x and y for which both the price functions are non-negative. The prices are: \[p = 20 - 0.005x - 0.001y\] \[q = 15 - 0.001x - 0.003y\] We want \(p \geq 0\) and \(q \geq 0\), which can be expressed as: \(20 - 0.005x - 0.001y \geq 0\) \(15 - 0.001x - 0.003y \geq 0\) Now, we will find the ranges of x and y that satisfy these inequalities: From the first inequality: \(0.005x + 0.001y \leq 20\) \(x \leq 4000 - 200y\) From the second inequality: \(0.001x + 0.003y \leq 15\) \(x \leq 1500 - 3000y\) Since both x and y represent the number of copies demanded, they must be non-negative, so \(x \geq 0\) and \(y \geq 0\). Putting all these restrictions together, we have the domain for R(x, y): \[0 \leq x \leq 4000 - 200y\] \[0 \leq y \leq 1500 - 3000y\] So, the domain of the function R(x, y) is given by \((x, y) \in [0, 4000 - 200y] \times [0, 1500 - 3000y]\) where x and y are non-negative integers.