p(x).
Given that the price-demand function is p(x) = 57 - 6ln(x), we know that the price is influenced by the natural logarithm of the quantity of games sold, which can be any non-zero and non-negative value.
The total revenue function, R(x), is calculated as the price a consumer pays for the unit product times the quantity sold. Therefore, substituting the price p(x) into the revenue function yields R(x) = x * (57 - 6ln(x)).
Now, to find the value of x that maximizes the revenue, we first need to find the derivative of the revenue function with respect to x. On differentiation, we get R'(x) = 51 - 6ln(x).
Next, we find the critical values where the derivative of the revenue function equals zero. For this, we solve R'(x) = 0 for x, yielding the value x = e^(17/2).
To verify if the critical value we obtained yields a maximum or a minimum, we perform the second derivative test. We differentiate R'(x) once more with respect to x, obtaining R''(x) = -6 * e^(-17/2).
Since the second derivative at the critical value is negative, it means that the critical value of x yields a maximum for R(x). Therefore, the maximum revenue is obtained at x = e^(17/2).
Finally, substituting x = e^(17/2) into the revenue function yields the maximum revenue, R(e^(17/2)) = 6 * e^(17/2) dollars.