Chapter 1, Problem 4TP

Solution

To find: The price of product when revenue maximum.

The price of product is equal to $\$10$ .

Given information:

The given table is below:

  

Calculation:

Make a table for total price:

Price	$\$15$	$\$14$	$\$13$	$\$12$
Pairs sold	$50$	$60$	$70$	$80$
Total price	$\$750$	$\$840$	$\$910$	$\$960$

The general equation of the quadratic equation is $y=a{x}^2+bx+c$ .

Substitute $x=15,y=750$ in the above equation:

  $\begin{array}{c}750=a{\left(15\right)}^2+b\left(15\right)+c\\ 750=a\left(225\right)+15b+c\\ 225a+15b+c=750\mathrm{...}\left(1\right)\end{array}$

Substitute $x=14,y=840$ in the above equation:

  $\begin{array}{c}840=a{\left(14\right)}^2+b\left(14\right)+c\\ 840=a\left(196\right)+14b+c\\ 196a+14b+c=840\mathrm{...}\left(2\right)\end{array}$

Substitute $x=13,y=910$ in the above equation:

  $\begin{array}{c}910=a{\left(13\right)}^2+b\left(13\right)+c\\ 910=a\left(169\right)+13b+c\\ 169a+13b+c=910\mathrm{...}\left(3\right)\end{array}$

Solve the equations (1), (2), and (3):

  $a=-10,b=200,c=0$

Therefore, the quadratic equation is

  $\begin{array}{l}y=-10x^2+200x+0\\ y=-10x^2+200x\end{array}$ .

Thus, the price of product when revenue maximum is below:

  $\begin{array}{c}x=-\frac{b}{2a}\\ x=-\frac{200}{2\left(-10\right)}\\ x=10\end{array}$

Hence, the price of product when revenue maximum is $\$10$ .