Questions: Question 18, 2.7.43-BE Part 1 of 6 The price p (in dollars) and the demand x for a particular clock radio are related by the equation x=3000-10 p. (A) Express the price p in terms of the demand x, and find the domain of this function. (B) Find the revenue R(x) from the sale of x clock radios. What is the domain of R ? (C) Find the marginal revenue at a production level of 1600 clock radios. (D) Interpret R'(1900)=-80.00. (A) p=p(x)=

Question 18, 2.7.43-BE
Part 1 of 6

The price p (in dollars) and the demand x for a particular clock radio are related by the equation x=3000-10 p.
(A) Express the price p in terms of the demand x, and find the domain of this function.
(B) Find the revenue R(x) from the sale of x clock radios. What is the domain of R ?
(C) Find the marginal revenue at a production level of 1600 clock radios.
(D) Interpret R'(1900)=-80.00.
(A) p=p(x)=

Transcript text: Question 18, 2.7.43-BE Part 1 of 6 The price p (in dollars) and the demand x for a particular clock radio are related by the equation $\mathrm{x}=3000-10 \mathrm{p}$. (A) Express the price $p$ in terms of the demand $x$, and find the domain of this function. (B) Find the revenue $R(x)$ from the sale of $x$ clock radios. What is the domain of $R$ ? (C) Find the marginal revenue at a production level of 1600 clock radios. (D) Interpret $R^{\prime}(1900)=-80.00$. (A) $p=p(x)=$ $\square$

Solution

Solution Steps

Solution Approach

(A) To express the price $ p $ in terms of the demand $ x $, solve the equation $ x = 3000 - 10p $ for $ p $. The domain of this function is determined by the realistic values of $ x $, which should be non-negative and less than or equal to 3000.

(B) The revenue $ R(x) $ is calculated as the product of the price per unit and the number of units sold, i.e., $ R(x) = p(x) \times x $. Substitute the expression for $ p(x) $ from part (A) into this equation. The domain of $ R $ is the same as the domain of $ p(x) $.

(C) The marginal revenue is the derivative of the revenue function $ R(x) $. Calculate $ R'(x) $ and evaluate it at $ x = 1600 $.

Step 1: Express the Price $ p $ in Terms of Demand $ x $

To express the price $ p $ in terms of the demand $ x $, solve the equation $ x = 3000 - 10p $ for $ p $:

\[ p = 300 - \frac{x}{10} \]

The domain of this function is determined by the realistic values of $ x $, which should be non-negative and less than or equal to 3000. Therefore, the domain is $ 0 \leq x \leq 3000 $.

Step 2: Find the Revenue Function $ R(x) $

The revenue $ R(x) $ is calculated as the product of the price per unit and the number of units sold:

\[ R(x) = p(x) \times x = \left(300 - \frac{x}{10}\right) \times x = 300x - \frac{x^2}{10} \]

The domain of $ R(x) $ is the same as the domain of $ p(x) $, which is $ 0 \leq x \leq 3000 $.

Step 3: Calculate the Marginal Revenue at $ x = 1600 $

The marginal revenue is the derivative of the revenue function $ R(x) $:

\[ R'(x) = \frac{d}{dx}\left(300x - \frac{x^2}{10}\right) = 300 - \frac{x}{5} \]

Evaluate $ R'(x) $ at $ x = 1600 $:

\[ R'(1600) = 300 - \frac{1600}{5} = -20 \]

Final Answer

The expression for price $ p $ in terms of demand $ x $ is $ p = 300 - \frac{x}{10} $ with domain $ 0 \leq x \leq 3000 $.
The revenue function is $ R(x) = 300x - \frac{x^2}{10} $ with domain $ 0 \leq x \leq 3000 $.
The marginal revenue at a production level of 1600 clock radios is $ R'(1600) = -20 $.

\[ \boxed{p = 300 - \frac{x}{10}}, \quad \boxed{R(x) = 300x - \frac{x^2}{10}}, \quad \boxed{R'(1600) = -20} \]

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