The revenue derived from the production of x units of a particular commodity is $R ( x ) = \frac { 80 x - x ^ { 2 } } { x ^ { 2 } + 80 }$ million dollars.What level of production results in maximum revenue? What is the maximum revenue? Round numbers to two decimal places,if necessary.
A)Maximum at x = 8 and maximum revenue is R(8)= 4 (million dollars)
B)Maximum at x = 8 and maximum revenue is R(8)= 32 (million dollars)
C)Maximum at x = 9 and maximum revenue is R(9)= 2.67 (million dollars)
D)Maximum at x = 8 and maximum revenue is R(8)= 2.67 (thousand dollars)

Question

Quizplus · Accepted Answer

To find the maximum revenue, we differentiate $R(x)$ with respect to $x$ and set the derivative equal to zero to find critical points. The derivative of $R(x)$ is $R'(x) = \frac{80 - 2x}{x^2 + 80} - \frac{(80x - x^2)(2x)}{(x^2 + 80)^2}$. Setting $R'(x) = 0$ and solving for $x$ will give the value of $x$ that maximizes $R(x)$. Without explicitly solving the derivative equation here, we can verify the correct answer by evaluating $R(x)$ at the given $x$ values. Plugging $x = 8$ into $R(x)$ gives $R(8) = \frac{80(8) - 8^2}{8^2 + 80} = \frac{640 - 64}{64 + 80} = \frac{576}{144} = 4$ million dollars, indicating that the maximum revenue is achieved at $x = 8$ and the maximum revenue is 4 million dollars.

The Revenue Derived from the Production of X Units of a Particular