Chapter ISG, Problem 8.3.3P

a.

To determine

To Write: A quadratic expression for P in terms of p.

Answer to Problem 8.3.3P

  $P=-40p^2+p\left(5000+40U\right)-5000U-F$

Explanation of Solution

Given:

Given expressions are

  $n=5000-40p$

  $P=n\left(p-U\right)-F$

Calculation:

  $P=n\left(p-U\right)-F$

  $P=\left(5000-40p\right)\left(p-U\right)-F$

  $P=5000p-5000U-40p^2+40pU-F$

  $P=-40p^2+p\left(5000+40U\right)-5000U-F$

Thus, $P=-40p^2+p\left(5000+40U\right)-5000U-F$

b.

To determine

To Calculate:The best price to charge for the service, and the monthly profit at this price.

The best price to charge for the service is $124.2038542 and the monthly profit at this price is $ 0.000114606.

Given :

U is the unit cost per sale and F is a fixed cost that does not depend on the number of sales. For an online business advice service, the unit cost is $\$30$ per hour-long session and the monthly fixed cost is $\$3000$ .

  $U=\$30$

  $F=\$3000$

Calculation:

Price per session , i.e. p will be such that profit is non negative

P>0

  $-40p^2+p\left(5000+40U\right)-5000U-F>0$

  i.e. $-40p^2+p\left(5000+\left(40\times 30\right)\right)-\left(5000\times 30\right)-3000>0$

  i.e. $-40p^2+6200p-150000-3000>0$

  i.e. $-40p^2+6200p-153000>0$

Applying Sridhara’s method to solve a quadratic equation :

  $x=a{x}^2+bx+c$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The solution for p as follows :

  $p=\frac{-6200\pm \sqrt{{6200}^2-4\left(40\right)153000}}{-2\times 40}$

  i.e. $p=\frac{-6200\pm \sqrt{13960000}}{-80}$

  i.e. $p=\frac{-6200\pm 3736.308338}{-80}$

  either $p=\frac{-6200+3736.308338}{-80}=30.79614578$

OR

  $p=\frac{-6200-3736.308338}{-80}==124.2038542$

Profit is found by using the equation

  $P=-40p^2+p\left(5000+40U\right)-5000U-F$

When

  $p=30.79614578$

  $U=\$30$

  $F=\$3000$

Profit ,

  $\begin{array}{c}P=-(40\times {30.79614578}^2)+(30.79614578\times \left(5000+\left(40\times 30\right)\right))-\left(5000\times 30\right)-3000\\ =0.000039879\end{array}$

  i.e. the monthly profit at price, $p=30.79614578$

  is $\$0.000039879$

When price , $p=124.2038542$

Profit , $P=-(40\times {124.2038542}^2)+(124.2038542\times \left(5000+\left(40\times 30\right)\right))-\left(5000\times 30\right)-3000=0.000114606$ i.e. the monthly profit at price, $p=\$124.2038542$

  is $ $\$124.2038542$

The price will be chosen where the profit is maximum.

Profit at price $\$124.2038542$ is maximum/

Hence, the best price to charge for the service is $\$124.2038542$ and the monthly profit at this price is $\$0.000114606$ .

Answer to Problem 8.3.3P

The best price to charge for the service is $124.2038542 and the monthly profit at this price is $ 0.000114606.

Explanation of Solution

Given :

U is the unit cost per sale and F is a fixed cost that does not depend on the number of sales. For an online business advice service, the unit cost is $\$30$ per hour-long session and the monthly fixed cost is $\$3000$ .

  $U=\$30$

  $F=\$3000$

Calculation:

Price per session , i.e. p will be such that profit is non negative

P>0

  $-40p^2+p\left(5000+40U\right)-5000U-F>0$

  i.e. $-40p^2+p\left(5000+\left(40\times 30\right)\right)-\left(5000\times 30\right)-3000>0$

  i.e. $-40p^2+6200p-150000-3000>0$

  i.e. $-40p^2+6200p-153000>0$

Applying Sridhara’s method to solve a quadratic equation :

  $x=a{x}^2+bx+c$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The solution for p as follows :

  $p=\frac{-6200\pm \sqrt{{6200}^2-4\left(40\right)153000}}{-2\times 40}$

  i.e. $p=\frac{-6200\pm \sqrt{13960000}}{-80}$

  i.e. $p=\frac{-6200\pm 3736.308338}{-80}$

  either $p=\frac{-6200+3736.308338}{-80}=30.79614578$

OR

  $p=\frac{-6200-3736.308338}{-80}==124.2038542$

Profit is found by using the equation

  $P=-40p^2+p\left(5000+40U\right)-5000U-F$

When

  $p=30.79614578$

  $U=\$30$

  $F=\$3000$

Profit ,

  $\begin{array}{c}P=-(40\times {30.79614578}^2)+(30.79614578\times \left(5000+\left(40\times 30\right)\right))-\left(5000\times 30\right)-3000\\ =0.000039879\end{array}$

  i.e. the monthly profit at price, $p=30.79614578$

  is $\$0.000039879$

When price , $p=124.2038542$

Profit , $P=-(40\times {124.2038542}^2)+(124.2038542\times \left(5000+\left(40\times 30\right)\right))-\left(5000\times 30\right)-3000=0.000114606$ i.e. the monthly profit at price, $p=\$124.2038542$

  is $ $\$124.2038542$

The price will be chosen where the profit is maximum.

Profit at price $\$124.2038542$ is maximum/

Hence, the best price to charge for the service is $\$124.2038542$ and the monthly profit at this price is $\$0.000114606$ .