The price $$ p $$ (in dollars) and the demand $$ x $$ tor a particuar cluck radro are releted by the oquaton $$ x=4000-10 p $$ (A) Express the price pint terms of the denmand $$ x $$, and find the doman of ithis linction (t) Find the rovonise $$ R(x) $$ liom the salo of $$ x $$ clock radios. What is the domain of $$ R $$ ? (C) Find the marginal revonue al a production lovel of 2500 clock radics. (D) Intorpret $$ R^{\prime}(2000)=-16000 $$ What is the domein ol $$ \mathrm{F}(\mathrm{X}) $$ ? A. $$ x \geq 0 $$ B. $$ x \leq 5000 $$ C. $$ 0 \leq x \leq 4000 $$ D. $$ 0 \leq x \leq 10 $$ (C) The merginal revemue ot a production level of 2500 clock radios is $$ \mathbf{\$ - 1 0 0} $$. (Round to the nearesicentas nooded.) (D) Which of tha lollowing statemenls is NOT a cormat interpretation of $$ \mathrm{R}^{\prime}(2800)=-160.00 $$ ? A. If 2800 clock radios are producod, there will be no revanue, since $$ \mathrm{R}^{\prime}(2800) $$ is nojalivo. B. The fotal revenue for producung 2801 dock radios is approkimalely $$ \$ 16000 $$ less thar the lolal tevenue for producing 2800 clock radios. C. Al a producion level of 2800 clock radios, the change in revenue per unit chango in prodiction is approxirrately $$ \mathbf{- 1 6 0} 00 $$ dollars.

Question

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part (A): Express the price $$ p $$ in terms of the demand $$ x $$ and find the domain of this function.

The given equation is:
$$
x = 4000 - 10p
$$

To express $$ p $$ in terms of $$ x $$, we can rearrange the equation:
$$
10p = 4000 - x
$$
$$
p = \frac{4000 - x}{10}
$$

Now, let's find the domain of this function. The demand $$ x $$ must be non-negative, and since $$ p $$ must also be non-negative, we can set up the following inequalities:
1. $$ x \geq 0 $$
2. $$ 4000 - x \geq 0 $$ which simplifies to $$ x \leq 4000 $$

Thus, the domain of $$ p(x) $$ is:
$$
0 \leq x \leq 4000
$$

### Part (B): Find the revenue $$ R(x) $$ from the sale of $$ x $$ clock radios and its domain.

Revenue $$ R $$ is given by the product of price and quantity sold:
$$
R(x) = p \cdot x = \left(\frac{4000 - x}{10}\right) \cdot x
$$
$$
R(x) = \frac{4000x - x^2}{10}
$$

Now, let's simplify this:
$$
R(x) = 400x - \frac{x^2}{10}
$$

The domain of $$ R $$ is the same as the domain of $$ p(x) $$:
$$
0 \leq x \leq 4000
$$

### Part (C): Find the marginal revenue at a production level of 2500 clock radios.

To find the marginal revenue, we need to calculate the derivative of $$ R(x) $$:
$$
R'(x) = 400 - \frac{2x}{10} = 400 - \frac{x}{5}
$$

Now, we can evaluate $$ R'(x) $$ at $$ x = 2500 $$:
$$
R'(2500) = 400 - \frac{2500}{5}
$$
$$
R'(2500) = 400 - 500 = -100
$$

Thus, the marginal revenue at a production level of 2500 clock radios is $$ -100 $$.

### Part (D): Interpret $$ R'(2000) = -160 $$.

The interpretation of $$ R'(2000) = -160 $$ means that if 2000 clock radios are produced, the revenue will decrease by approximately $160 for each additional clock radio produced.

### Summary of Domains and Answers:
- The domain of $$ p(x) $$ is $$ 0 \leq x \leq 4000 $$.
- The domain of $$ R(x) $$ is also $$ 0 \leq x \leq 4000 $$.
- The marginal revenue at a production level of 2500 clock radios is $$ -100 $$.
- The interpretation of $$ R'(2000) = -160 $$ indicates a decrease in revenue with increased production.

### Final Answers:
- The domain of $$ p(x) $$ and $$ R(x) $$ is $$ C. \, 0 \leq x \leq 4000 $$.
- The marginal revenue at 2500 clock radios is $$ -100 $$.
- The interpretation of $$ R'(2000) = -160 $$ is correct as stated.

If you have any further questions or need additional clarifications, feel free to ask!
- The domain of $$ p(x) $$ and $$ R(x) $$ is $$ 0 \leq x \leq 4000 $$. - The marginal revenue at 2500 clock radios is $$ -100 $$. - The interpretation of $$ R'(2000) = -160 $$ is correct.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Economics Questions