Chapter 7.1, Problem 69E

Solution

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

The obtained value is 13.

Given Information:

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

  $R=px,p=100-4x$

Calculation:

Consider the given information,

Substitute $p=100-4x$ in $R=px$

  $\begin{array}{l}R=\left(100-4x\right)x\\ R=100x-4x^2\end{array}$

Differentiate the obtained equation w.r.t x

  $\frac{dR}{dx}=100-8x$

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

  $\begin{array}{c}\frac{dR}{dx}=0\\ 100-8x=0\\ x=\frac{100}{8}\\ x=\frac{25}{2}\end{array}$

Differentiate again with respect to $x$ .

  $\frac{d^{2}R}{d{x}^2}=-8$

Thus, the value obtained by substituting first differentiation is equal to 0 is always gives the required number of units to maximize the Revenue.

The number of units must be integer. it is implies that,

  $\begin{array}{c}x=12.5\\ \approx 13\text{ units}\end{array}$

Therefore, the number of units sold to gives the maximum revenue is 13.