Question

The revenue for a certain product is given by the equation

$R\left(x\right)=100-\frac{400}{x+5}-x$,

where x is the number of produced items. Find the value of x that results in maximum revenue.

Answer 1

We differentiate the function R(x) to determine its critical point.

$R^{\prime }\left(x\right)=(100-\frac{400}{x+5}-x{)}^{\prime }=\frac{400}{(x+5{)}^2}-1$

$R^{\prime }\left(x\right)=0⟹\frac{400}{(x+5{)}^2}-1=0$

$x+5=20$

$x=15$

Using the first derivative test, we can verify that $x=15$ is a point of maximum.

Hence, the maximum revenue occurs when $x=15$.