Chapter 2.7, Problem 69E

To determine

To find: the number of units that must be sold to obtain a profit of at least $750,000 and the price per unit.

Answer to Problem 69E

  $40,000\le x\le 50,000$

  $\$50\le p\le \$55$

Explanation of Solution

Given:

The revenue equation for a product is $R=x\left(75-0.0005x\right)$

The cost equation for that product is $C=30x+250,000$

Here, x represents the number of units sold.

It is known that,

Profit = Revenue − Cost

  $\begin{array}{c}P=R-C\\ =x\left(75-0.0005x\right)-\left(30x+250000\right)\\ =-0.0005x^2+45x-250000\end{array}$

Now, it is required that profit must be at least $750,000. That is,

  $\begin{array}{c}-0.0005x^2+45x-250000\ge 750000\\ 0.0005x^2-45x+{10}^6\le 0\\ x^2-90000x+2\times {10}^9\le 0\end{array}$

  $\begin{array}{c}{x}^2-90000x+2\times {10}^9\le 0\\ x^2-50000x-40000x+2\times {10}^9\le 0\\ x\left(x-50000\right)-40000\left(x-50000\right)\le 0\\ \left(x-40000\right)\left(x-50000\right)\le 0\end{array}$

This inequality is satisfied only when $40,000\le x\le 50,000$

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

So, when 40,000 units are sold, the price per unit will be:

  $\begin{array}{c}p=\frac{R\left(40000\right)}{40000}\\ =\frac{\left(40000\right)\left(75-0.0005\left(40000\right)\right)}{40000}\\ =\$55\end{array}$

And when 50,000 units are sold, the price per unit will be:

  $\begin{array}{c}p=\frac{R\left(50000\right)}{50000}\\ =\frac{\left(50000\right)\left(75-0.0005\left(50000\right)\right)}{50000}\\ =\$50\end{array}$

Conclusion:

To obtain a profit of at least $750,000 the number of units sold should be $40,000\le x\le 50,000$ , and the price per unit should be $\$50\le p\le \$55$ .