Chapter 3.4, Problem 5AYU

(a)

To determine

To express: the revenue $R$ as a function of $x$ .

Answer to Problem 5AYU

The expression of the revenue $R$ is $R\left(x\right)=\frac{-1}{5}{x}^2+20x$

Explanation of Solution

Given:

  $x=-5p+100 0<p\le 20$

Explanations:

Here it is given that price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $x=-5p+100$

Solve it for $p$ .

  $\begin{array}{l}x-100=-5p+100-100\\ x-100=-5p\end{array}$

Divide both sides by $-5$ .

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

Now as it is given that Revenue $R$ is the product of price $p$ and quantity $x$ .

So

  $\begin{array}{l}R\left(x\right)=x\times p\\ =x\times \left[\left(\frac{-1}{5}\right)x+20\right]\\ R\left(x\right)=\frac{-1}{5}{x}^2+20x\end{array}$

This is the model that express the Revenue $R$ as a function of $x$ .

Conclusion:

Hence the expression of the revenue $R$ is $R\left(x\right)=\frac{-1}{5}{x}^2+20x$

(b)

To determine

To find: the revenue if15 units are sold

Answer to Problem 5AYU

So, the required total revenue is $\$255$ .

Explanation of Solution

Given:

  $x=-5p+100 0<p\le 20$

Explanations:

When the quantity sold $x=15$ ,

Total Revenue is,

  $\begin{array}{l}R\left(15\right)=-{\left(15\right)}^2+20\times 15\\ =\frac{-225}{5}+300\\ =-45+300\\ R\left(15\right)=255\end{array}$

So, the required total revenue is $\$255$ .

Conclusion:

So, the required total revenue is $\$255$ .

(c)

To determine

To find: the maximum revenue and the quantity $x$ that maximizes the revenue

Answer to Problem 5AYU

  $50$ units maximize revenue.

Maximum revenue is $\$500$

Explanation of Solution

Given:

  $x=-5p+100 0<p\le 20$

Explanations:

As the square term in the given function is negative that shows that its graph opens down or it has a maximum value at its vertex only. Now for the quadratic function $f\left(x\right)=a{x}^2+bx+c$ .

As $x$ coordinate of vertex $=\frac{-b}{2a}$

As in given function of Revenue.

  $R\left(x\right)=\frac{-1}{5}{x}^2+20x$

Here, $a=\frac{-1}{5}$ and $b=20$ .

So, $x$ coordinate of vertex

  $\begin{array}{l}=\frac{-20}{2\times \frac{-1}{5}}\\ =20\times \frac{5}{2}\\ =50\end{array}$

That shows that $x=50$ or total $50$ units maximize revenue.

Now the value of $R\left(x\right)$ at $x=50$ would be the maximum revenue now that is attainable at vertex only, so on plug in $x=50$ in given revenue function.

  $\begin{array}{l}R\left(50\right)=\frac{-1}{5}\times {\left(50\right)}^2+20\times 50\\ =\frac{-2500}{5}+1000\\ =-500+1000\\ R\left(50\right)=500\end{array}$

Or

Maximum revenue is $\$500$ which is obtained when maximum $50$ units of product are sold.

Conclusion:

Hence, $50$ units maximize revenue and Maximumrevenue is $\$500$

(d)

To determine

To find:price charged by the company to maximize revenue?

Answer to Problem 5AYU

Hence, maximum revenue is achieved when each item is sold at the rate of $\$10$ per unit.

Explanation of Solution

Given:

  $x=-5p+100 0<p\le 20$

Explanations:

As given that $R=500$ and $x=50$

Now as formula for calculating revenue is $R=xp$

  $500=50\times p$

Now divide both sides by $50$ .

  $\begin{array}{l}\frac{50\times p}{50}=\frac{500}{50}\\ p=10\end{array}$

Or, maximum revenue is achieved when each item is sold at the rate of $\$10$ per unit.

Conclusion:

Hence, maximum revenue is achieved when each item is sold at the rate of $\$10$ per unit.

(e)

To determine

To find: price charged by the company to earn at least $\$480$

Answer to Problem 5AYU

Thus, the company should charge between $\$8$ and $\$12$ to earn at least $\$480$ in revenue.

Explanation of Solution

Given:

  $x=-5p+100 0<p\le 20$

Explanations:

Now again as $Revenue=Price\times totalitemsold$

  $\begin{array}{l}R=xp\\ =\left(-5p+100\right)\times p\\ =-5p\times p+100\times p\\ =-5p^2+100p\end{array}$

Now for revenue to be at least $\$480$ ,or for

  $\begin{array}{l}R=480\\ -5p^2+100p=480\\ -5\left(p^2-20p\right)=480\end{array}$

Now divide both sides by $-5$ .

  $\begin{array}{l}\frac{-5\left(p^2-20p\right)}{-5}=\frac{480}{-5}\\ p^2-20p=-96\\ p^2-20p+96=0\\ \left(p-12\right)\left(p-8\right)=0\\ p=8,12\end{array}$

Thus, the company should charge between $\$8$ and $\$12$ to earn at least $\$480$ in revenue.

Conclusion:

Thus, the company should charge between $\$8$ and $\$12$ to earn at least $\$480$ in revenue.