91Ó°ÊÓ

Given that the price p (in dollars) and the quantity x sold of a certain product obey the demand equation :

$x=-5p+100;0<p≤20$

We get

$$\begin{gathered} x=-5p+100 \\ \frac{x}{-5}=p+\frac{100}{-5} \\ -\frac{1}{5}x=p-20 \\ -\frac{1}{5}x+20=p \\ p=-\frac{1}{5}x+20 \end{gathered}$$

As we know that $R=px$then

$$\begin{gathered} R=x(-\frac{1}{5}x+20) \\ =-\frac{1}{5}{x}^2+20x \end{gathered}$$

We get

$$\begin{gathered} R=-\frac{1}{5}{\left(15\right)}^2+20\left(15\right) \\ =-45+300 \\ =255 \end{gathered}$$

So the revenue is 255 dollars when 15 units are sold.

The function R is a quadratic function with $a=-\frac{1}{5},b=20$, and $c=0$. Because $a<0$, the vertex is the highest point on the parabola.

The revenue R is a maximum when x is

$$\begin{gathered} x=-\frac{b}{2a} \\ =-\frac{20}{2(-{\displaystyle \frac{1}{5}})} \\ =50 \end{gathered}$$

The maximum revenue is

$$\begin{gathered} R=-\frac{1}{5}{\left(50\right)}^2+20\left(50\right) \\ =-500+1000 \\ =500 \end{gathered}$$

$R=500$dollars is maximum.

We get

$$\begin{gathered} p=-\frac{1}{5}\left(50\right)+20 \\ =-10+20 \\ =10 \end{gathered}$$

So the company should charge 10 dollars per unit to maximize revenue.

As $R=xp$then

$$\begin{gathered} R=p(-5p+100) \\ =-5p^2+100p \end{gathered}$$

Now for revenue at least 480 dollars,

$$\begin{gathered} -5p^2+100p=480 \\ -p^2+20p=96 \\ -p^2+20p-96=0 \\ {p}^2-20p+96=0 \end{gathered}$$

By quadratic formula,

localid="1647563123280" $$\begin{gathered} p=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{-(-20)Â\pm \sqrt{{(-20)}^2-4\left(1\right)\left(96\right)}}{2\left(1\right)} \\ =\frac{20Â\pm \sqrt{400-384}}{2} \\ =\frac{20Â\pm 4}{2} \end{gathered}$$

we have

$$\begin{gathered} p=\frac{20+4}{2}orp=\frac{20-4}{2} \\ p=12orp=8 \end{gathered}$$

So the company should charge between 8 and 12 dollars to earn at least 480 dollars revenue.