the price p (in dollars) and the demand x for a particular clock - radio are related by the equation x =…

Explanation:

Step1: Rearrange the demand - price equation

Given $x = 5000-50p$, solve for $p$. First, isolate $50p$: $50p=5000 - x$. Then $p=\frac{5000 - x}{50}=100-\frac{x}{50}$. Since $x\geq0$ (demand cannot be negative) and $p\geq0$, when $p = 0$, $x = 5000$. So the domain of $p(x)$ is $0\leq x\leq5000$.

Step2: Find the revenue function

Revenue $R(x)=p\times x$. Substitute $p = 100-\frac{x}{50}$ into the revenue formula: $R(x)=(100-\frac{x}{50})x=100x-\frac{x^{2}}{50}$. The domain of $R(x)$ is the same as the domain of $x$ in the demand - price relationship, so $0\leq x\leq5000$.

Step3: Find the marginal revenue function

The marginal revenue function $R^\prime(x)$ is the derivative of $R(x)$. Since $R(x)=100x-\frac{x^{2}}{50}$, using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $R^\prime(x)=100-\frac{2x}{50}=100-\frac{x}{25}$.

Step4: Evaluate the marginal revenue at $x = 3900$

Substitute $x = 3900$ into $R^\prime(x)$: $R^\prime(3900)=100-\frac{3900}{25}=100 - 156=- 56$.

Step5: Interpret $R^\prime(4400)=-76$

The marginal revenue $R^\prime(4400)=-76$ means that when the production level is 4400 clock - radios, the revenue from selling one additional clock - radio will decrease by $76$.

Answer:

(A) $p(x)=100-\frac{x}{50}$, domain: $0\leq x\leq5000$ (B) $R(x)=100x-\frac{x^{2}}{50}$, domain: $0\leq x\leq5000$ (C) $-56$ (D) When the production level is 4400 clock - radios, the revenue from selling one additional clock - radio will decrease by $76$.