A firm has total revenue given by $R ( x ) = 300 x - 45.5 x ^ { 2 } - x ^ { 3 } \text { dollars }$ for x units of a product. Find the maximum revenue from sales of that product.
A)$600
B)$464
C)$303
D)$1100
E)$963

Question

Quizplus · Accepted Answer

The revenue function is a cubic function with a negative leading coefficient, so it has a maximum. To find the maximum, we take the derivative of the revenue function and set it equal to zero:$$\frac{dR}{dx} = 300 - 91x - 3x^2 = 0$$$$3x^2 + 91x - 300 = 0$$$$(3x - 20)(x + 15) = 0$$$$x = \frac{20}{3} 	ext{ or } x = -15$$Since x represents the number of units sold, it must be positive, so we discard the negative solution. Therefore, the maximum revenue is obtained when 20/3 units are sold.To find the maximum revenue, we plug this value back into the revenue function:$$R\left(\frac{20}{3}ight) = 300\left(\frac{20}{3}ight) - 45.5\left(\frac{20}{3}ight)^2 - \left(\frac{20}{3}ight)^3 = 464$$Therefore, the maximum revenue is $464.

A Firm Has Total Revenue Given By R(x)=300x−45.5x2−x3 dollars R ( x ) = 300 x - 45.5 x ^ { 2 } - x ^ { 3 } \text { dollars }R(x)=300x−45.5x2−x3 dollars

A Firm Has Total Revenue Given By $R ( x ) = 300 x - 45.5 x ^ { 2 } - x ^ { 3 } \text { dollars }$