Chapter 7.1, Problem 70E

Solution

To calculate: The number of units sold to gives the maximum revenue.

The number units sold for maximum revenue is 6.

Given Information:

The total number of units sold and the relation between $p$ and $x$

are:

  $R=px,p=80-x^2$

Calculation:

Consider the given information,

Substitute $p=80-x^2$ in $R=px$ .

  $\begin{array}{l}R=\left(80-x^2\right)x\\ R=80x-x^3\end{array}$

Differentiate the obtained equation with respect $x$ to

  $\frac{dR}{dx}=80-3x^2$

The maximum value of revenue is obtained by substituting first differentiation is equal to 0 if second differentiation is less than 0.

  $\begin{array}{c}\frac{dR}{dx}=0\\ 80-3x^2=0\\ x=\pm \sqrt{\frac{80}{3}}\\ x=\pm 5.16\end{array}$

Differentiate again with respect to $x$ ,

  $\frac{d^{2}R}{d{x}^2}=-6x$

The positive value of x gives the double differentiate negative. It implies,

  $x=5.16$

The number of units must be integer. it implies,

  $x=6$

Therefore, the number of units sold to gives the maximum revenue is 6.