Question

The price p​ (in dollars) and the demand $x$ for a particular clock radio are related by the equation \[x\text{ }=2000-2020p.\] ?
a).How to express the price $p$ in terms of the demand​ $x$, and find the domain of this function?
b).How to find the revenue​ \[R\left( x \right)\] from the sale of $x$ clock radios. What is the domain of​ \[R\]?
c).How to find the marginal revenue at a production level of \[1400\] clock radios.
d).How to interpret \[R'\left( 1900 \right)=-90.00\]?

Answer 1

Hint: As we can see that there are sub-questions to this question, so in order to find a solution to this question, we have to solve the problem sub-question wise and solve accordingly. Also, there is a typographical error in the question, so on correction, we have \[x=2000-20p\]

Complete step by step solution:
Starting with our question 1, we have:
a).How to express the price $p$ in terms of the demand​ $x$, and find the domain of this function?
This means we have to solve so that we have $p$ on the left and an expression involving $x$ on the right.
That is, we get:
$p=\dfrac{2000-x}{20}$
$p=\dfrac{2000-x}{20}=100-\dfrac{x}{20}$
Since, Demand $x$ must be non-negative and price $p$ must also be non-negative.
Therefore, to solve this we have to make it greater than zero.
That is, we get:
$\Rightarrow \dfrac{2000-x}{20}\ge 0$
Therefore, with this we can get the domain of the function as:
\[0\le x\le 2000\]

Now question 2:
b).How to find the revenue​ \[R\left( x \right)\] from the sale of $x$ clock radios. What is the domain of​ \[R\]?
In this we have to apply some formula to find revenue \[R\left( x \right)\] that is:
\[Revenue=Quantity~\times ~Price\]
We have Quantity as $x$ and Price as $\dfrac{2000-x}{20}$
So by applying this formula we get:
$R\left( x \right)=x\times \left( \dfrac{2000-x}{20} \right)$
On simplifying, we get:
$R\left( x \right)=x\left( \dfrac{2000-x}{20} \right)=100x-\dfrac{{{x}^{2}}}{20}$
Therefore, Revenue \[R\left( x \right)\] is:
$R\left( x \right)=100x-\dfrac{{{x}^{2}}}{20}$
The domain is the same as the previous domain that is,
\[0\le x\le 2000\]

Now question 3:
c).How to find the marginal revenue at a production level of \[1400\] clock radios.
The marginal revenue is the derivative of revenue.
With this, we get:
$R\left( x \right)=100x-\dfrac{{{x}^{2}}}{20}$
So now differentiate on both the sides, we get:
${R}'\left( x \right)=100-\dfrac{x}{10}$
Since, we have a production level of \[1400\] clock radios that is $x=1400$.
So by substituting $x=1400$, we get:
${R}'\left( 1400 \right)=100-\dfrac{1400}{10}$
On simplifying, we have:
${R}'\left( 1400 \right)=-40$
So the marginal revenue at a production level of \[1400\] clock radios is,
${R}'\left( 1400 \right)=-40$

Now coming to question 4:
d).How to interpret \[R'\left( 1900 \right)=-90.00\]?
There are a couple of ways to interpret the marginal revenue. (by using the given value of \[-90.00\])
When selling \[1900\] units, the revenue is decreasing at a rate of \[90.00~\] per additional unit sold.
If sales are \[1900\] units, a $1$ unit increase in sales will result in approximately a \[90.00~\] decrease in revenue.

Note:
On finding marginal revenue, the word "approximate" is important when writing the answer.
The Marginal Revenue gives us the change along the tangent line, not along the curve.
If we calculated the actual change along the curve from \[x=1900~\] to \[~x=1901\] we would get a change in revenue of (exactly)
\[R\left( 1901 \right)-R\left( 1900 \right)=-90.05\]
The marginal revenue has other uses and it is easier to calculate than the exact change in revenue.