Chapter ISG, Problem 8.2.6P

a.

To determine

To find: The expression of the expected revenue.

Answer to Problem 8.2.6P

Expression for revenue is, $56.25x-6.25x^2$ .

Explanation of Solution

Given information:

A company is expected to sell $45-5x$ and price of each is $1.25x$ .

Calculation:

The general formula to calculate the revenue is,

  $R=\frac{1}{2}\times \left(\text{number of units}\right)\times \left(\text{price per unit}\right)$

Revenue generation is,

  $\begin{array}{c}R=\left(45-5x\right)\times 1.25x\\ =56.25x-6.25x^2\end{array}$

Expression for revenue is $56.25x-6.25x^2$ .

b.

To determine

To find: The maximum revenue and the price of each piece.

Answer to Problem 8.2.6P

The maximum revenue is $126.5625\text{dollar}$ and price per rupees is $5.625\text{dollar}$ .

Explanation of Solution

Given information:

Expression for revenue is $56.25x-6.25x^2$ .

Calculation:

The given revenue is,

  $R=56.25x-6.25x^2$ …..(1)

Differentiate equation 1 with respect to x .

  $\begin{array}{c}\frac{dR}{dx}=\frac{d}{dx}\left(56.25x-6.25x^2\right)\\ =56.25-12.5x\end{array}$ …..(2)

Equate equation 2 with respect to x $\begin{array}{c}\frac{dR}{dx}=0\\ 56.25-12.5x=0\\ x=\frac{56.25}{12.5}\\ x=4.5\end{array}$

Again differentiate equation 2 with respect to h .

  $\begin{array}{c}\frac{d^{2}R}{d{x}^2}=\frac{d}{dx}\left(56.25-12.5x\right)\\ =-12.5<0\end{array}$

Since second derivative is less than 0, the maximum revenue will be at $x=4.5$ .

Maximum Area is,

  $\begin{array}{c}R=56.25\left(4.5\right)-6.25{\left(4.5\right)}^2\\ =253.125-126.5625\\ =126.5625\end{array}$

Price per unit will be,

  $\begin{array}{c}\text{Price}=1.25\left(4.5\right)\\ =5.625\end{array}$ .

Hence, the maximum revenue is $126.5625\text{dollar}$ and price per rupees is $5.625\text{dollar}$ .