Chapter 3.4, Problem 6AYU

(a)

To determine

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

Answer to Problem 6AYU

The expression of the revenue $R$ is $R\left(x\right)=\frac{-1}{20}{x}^2+25x$

Explanation of Solution

Given:

  $x=-20p+500 0<p\le 25$

Explanations:

Here it is given that price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $x=-20p+500$

Solve it for $p$ .

  $\begin{array}{l}x-500=-20p+500-500\\ x-500=-20p\end{array}$

Divide both sides by $-20$ .

  $\begin{array}{l}\frac{x-500}{-20}=\frac{-20p}{-20}\\ p=-\frac{1}{20}x+25\end{array}$

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

So

  $\begin{array}{l}R\left(x\right)=x\times p\\ =x\times \left[\left(\frac{-1}{20}\right)x+25\right]\\ R\left(x\right)=\frac{-1}{20}{x}^2+25x\end{array}$

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

Conclusion:

Hence, the expression of the revenue $R$ is $R\left(x\right)=\frac{-1}{20}{x}^2+25x$

(b)

To determine

To find: the revenue if20units are sold

Answer to Problem 6AYU

So, the required total revenue is $\$480$ .

Explanation of Solution

Given:

  $x=-20p+500 0<p\le 25$

Explanations:

When the quantity sold $x=20$ ,

Total Revenue is,

  $\begin{array}{l}R\left(20\right)=-{\left(20\right)}^2+20\times 25\\ =\frac{-400}{20}+500\\ =-20+500\\ R\left(20\right)=480\end{array}$

So, the required total revenue is $\$480$ .

Conclusion:

So, the required total revenue is $\$480$ .

(c)

To determine

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

Answer to Problem 6AYU

Maximum revenue is $\$3125$

Total $250$ units maximize revenue.

Explanation of Solution

Given:

  $x=-20p+500 0<p\le 25$

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, $a=\frac{-1}{20}$ and $b=25$

So $x$ coordinate of vertex

  $\begin{array}{l}=\frac{-25}{2\times \frac{-1}{20}}\\ =25\times \frac{20}{2}\\ =250\end{array}$

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

That shows that $x=250$ or total $250$ units maximize revenue.

Now the value of $R\left(x\right)$ at $x=250$ would be the maximum revenue now that is attainable at vertex only, so on plug in $x=250$ in given revenue function.

  $\begin{array}{l}R\left(250\right)=\frac{-1}{20}\times {\left(250\right)}^2+25\times 250\\ =\frac{-62500}{5}+6250\\ =-3125+6250\\ R\left(250\right)=3125\end{array}$

Or

Maximum revenue is $\$3125$ that is obtained when maximum $250$ units of product are sold.

Conclusion:

Hence, Maximum revenue is $\$3125$

Total $250$ units maximize revenue.

(d)

To determine

To find: price charged by the company to maximize revenue

Answer to Problem 6AYU

Hence, maximum revenue is achieved when each item is sold at the rate of $\$12.5$ per unit.

Explanation of Solution

Given:

Maximizing Revenue The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation

  $x=-20p+500 0<p\le 25$

Explanations:

As given that $R=3125$ and $x=250$

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

  $3125=250\times p$

Now divide both sides by $250$ ,

  $\begin{array}{l}\frac{250\times p}{250}=\frac{3125}{250}\\ p=12.5\end{array}$

Or, maximum revenue is achieved when each item is sold at the rate of $\$12.5$ per unit.

Conclusion:

Hence, maximum revenue is achieved when each item is sold at the rate of $\$12.5$ per unit.

(e)

To determine

To find: price charged by the company to earn at least $\$3000$

Answer to Problem 6AYU

The company should charge between $\$10$ and $\$15$ to earn at least $\$3000$ in revenue

Explanation of Solution

Given:

  $x=-20p+500 0<p\le 25$

Explanations:

Now again as $Revenue=Price\times totalitemsold$

  $\begin{array}{l}R=xp\\ =\left(-20p+500\right)\times p\\ =-20\times p\times p+500\times p\\ =-20p^2+500p\end{array}$

Now for revenue to be at least $\$480$ or for $R\ge 3000$

  $\begin{array}{l}i.e.-20p^2+500p\ge 3000\\ or,-20\left(p^2-25p\right)\ge 3000\end{array}$

Now divide both sides by $-20$ .

  $\begin{array}{l}\frac{-20\left(p^2-25p\right)}{-20}\le \frac{3000}{-20}\\ p^2-25p\le -150\\ p^2-25p+150\le 0\\ \left(p-15\right)\left(p-10\right)\le 0\\ Either\\ p-15\le 0\\ p\le 15\\ p-10\le 0\\ p\le 10\end{array}$

So $p\le 10$

Further as price is always positive so $p>0$

So, the domain of $R=\left\{p|0<p\le 10\right\}$

Conclusion:

Therefore, the company should charge between $\$10$ and $\$15$ to earn at least $\$3000$ in revenue