91Ó°ÊÓ

Given a equation for the price p (in dollars) and the quantity x sold of a certain product as$p=-\frac{1}{3}x+100$.

As $R=px$which means

$$\begin{gathered} R=x(-\frac{1}{3}x+100) \\ =-\frac{1}{3}{x}^2+100x \end{gathered}$$

The revenue R as a function of x is$R=-\frac{1}{3}{x}^2+100x$.

As x represents the number of quantity sold, we have price $p>0$, so $-\frac{1}{3}x+100>0$.

Solving this linear inequality, we find that

$$\begin{gathered} -\frac{1}{3}x+100>0 \\ 100>\frac{1}{3}x \\ 300>x \\ x<300 \end{gathered}$$

In addition, quantity sold $xâ‰\yen 0$.

Combining these inequalities, we have the domain of R is role="math" localid="1647559214417" $\{x:0≤x<300\}=[0,300)$.

Substitute $x=100$in $R=-\frac{1}{3}{x}^2+100$.

$$\begin{gathered} R=-\frac{1}{3}{\left(100\right)}^2+100\left(100\right) \\ =-\frac{10000}{3}+10000 \\ =\frac{-10000+30000}{3} \\ =\frac{20000}{3} \\ =6666.66 \end{gathered}$$

The revenue is 6666.66 dollars when 100 units are sold.

The function R is a quadratic function with $a=-\frac{1}{3},b=100$, and $c=0$. Because $a<0$, the vertex is the highest point on the parabola.

The revenue R is maximum when x is

$$\begin{gathered} x=-\frac{b}{2a} \\ =\frac{-100}{2(-{\displaystyle \frac{1}{3}})} \\ =50\left(3\right) \\ =150 \end{gathered}$$

The revenue maximizes when 150 units are sold.

The maximum revenue is

$$\begin{gathered} R=-\frac{1}{3}{x}^2+100x \\ =-\frac{1}{3}{\left(150\right)}^2+100\left(150\right) \\ =-7500+15000 \\ =75,000 \end{gathered}$$

The maximum revenue is 75,000 dollars.

The revenue maximizes when 150units are sold.

So to maximize the price, substitute $x=150$in $p=-\frac{1}{3}x+100$.

$$\begin{gathered} p=-\frac{1}{3}\left(150\right)+100 \\ =-50+100 \\ =50 \end{gathered}$$

So the company should charge 50 dollars to maximize the revenue.