Question

The revenue for a certain product is given by the equation $R\left( x \right) = 100 – \dfrac{{400}}{{x + 5}} – x$ , where x is the number of produced items. Find the value of x that results in maximum revenue.

Answer 1

Hint: When we need to find the maximum of an equation, the equation has to be differentiated and equated with 0.

Complete step-by-step answer:
Differentiate the function R(x) so as to determine the critical point.
By differentiating,
${R^\prime\left( x \right) }={ \left( {100 – \dfrac{{400}}{{x + 5}} – x} \right)^\prime }$
Simplify the equation,
${ \dfrac{{400}}{{{{\left( {x + 5} \right)}^2}}} – 1}$
Equate the expression with 0,
${R^\prime\left( x \right) = 0}$
 $\Rightarrow {\dfrac{{400}}{{{{\left( {x + 5} \right)}^2}}} – 1 = 0}$
Simplify the equation and separating the variables,
$400 = \left( {x + 5} \right)^2 $
Find the square root,
 x + 5 = 20
Find x,
$ \Rightarrow x = 15 $

Using the First Derivative Test, it can be verified that x = 15 is a point of maximum.
Thus, the maximum revenue occurs when x = 15.


Note: In these types of questions first find the differentiation of equation and after finding the maximum value use the First Derivative Test to verify.