Chapter 2.7, Problem 70E

To determine

To find: the number of units that must be sold to obtain a profit of at least $1,650,000 and the price per unit.

Answer to Problem 70E

  $90,000\le x\le 100,000$

  $\$30\le p\le \$32$

Explanation of Solution

Given:

The revenue equation for a product is $R=x\left(50-0.0002x\right)$

The cost equation for that product is $C=12x+150,000$

Here, x represents the number of units sold.

It is known that,

Profit = Revenue − Cost

  $\begin{array}{c}P=R-C\\ =x\left(50-0.0002x\right)-\left(12x+150000\right)\\ =-0.0002x^2+38x-150000\end{array}$

Now, it is required that profit must be at least $1,650,000. That is,

  $\begin{array}{c}-0.0002x^2+38x-150,000\ge 1,650,000\\ 0.0002x^2-38x+1,800,000\le 0\\ x^2-190,000x+9\times {10}^9\le 0\end{array}$

  $\begin{array}{c}{x}^2-190,000x+9\times {10}^9\le 0\\ x^2-90000x-100000x+9\times {10}^9\le 0\\ x\left(x-90000\right)-100000\left(x-90000\right)\le 0\\ \left(x-90000\right)\left(x-100000\right)\le 0\end{array}$

This inequality is satisfied only when $90,000\le x\le 100,000$

Now, the price per unit is found by dividing the revenue by the total number of units sold.

So, when 90,000 units are sold, the price per unit will be:

  $\begin{array}{c}p=\frac{R\left(90000\right)}{90000}\\ =\frac{\left(90000\right)\left(50-0.0002\left(90000\right)\right)}{90000}\\ =\$32\end{array}$

And when 100,000 units are sold, the price per unit will be:

  $\begin{array}{c}p=\frac{R\left(100,000\right)}{100,000}\\ =\frac{\left(100,000\right)\left(50-0.0002\left(100,000\right)\right)}{100,000}\\ =\$30\end{array}$

Conclusion:

To obtain a profit of at least $1,650,000 the number of units sold should be $90,000\le x\le 100,000$ , and the price per unit should be $\$30\le p\le \$32$ .