3. ( 5 pts ) The demand for bottles of sparkling cider is given by $$ D(p)=749.619\left(0.2^{p}\right) $$ million bottles, where $$ p $$ is the price in dollars per botte, $$ 2 \leq p \leq 4.5 $$. a. Write a fully defined model for the revenue from the sale of sparkling cider, $$ R(p) $$. b. Find $$ R^{\prime}(p) $$ and write a fully defined model for it. c. Find and interpret $$ R^{\prime}(3.25) $$.

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**a. Model for Revenue**

The number of bottles demanded is given by  
$$
D(p)=749.619\,(0.2)^p,
$$  
where $$p$$ is the price in dollars (with $$2\leq p\leq 4.5$$).  
Revenue is the product of the price and the quantity sold. Thus, the revenue function is  
$$
R(p)=p\cdot D(p)=749.619\,p\,(0.2)^p,\quad 2\leq p\leq 4.5.
$$

**b. Finding $$R'(p)$$**

We have  
$$
R(p)=749.619\,p\,(0.2)^p.
$$
To differentiate $$R(p)$$ with respect to $$p$$, we use the product rule. Let  
$$
u(p)=p \quad \text{and} \quad v(p)=(0.2)^p.
$$
Then,  
$$
u'(p)=1
$$
and we differentiate $$v(p)$$ as follows. Recall that  
$$
\frac{d}{dp}\,(a)^p=(a)^p\ln(a)
$$
for any positive constant $$a$$. Thus,  
$$
v'(p)=(0.2)^p\ln(0.2).
$$
By the product rule,  
$$
R'(p)=749.619\left[u'(p)v(p)+u(p)v'(p)\right]=749.619\left[(0.2)^p+ p\,(0.2)^p\ln(0.2)\right].
$$
This can be factored as  
$$
R'(p)=749.619\,(0.2)^p\left[1+p\ln(0.2)\right],\quad 2\leq p\leq 4.5.
$$

**c. Evaluating and Interpreting $$R'(3.25)$$**

To find $$R'(3.25)$$, substitute $$p=3.25$$ into the derivative:
$$
R'(3.25)=749.619\,(0.2)^{3.25}\left[1+3.25\ln(0.2)\right].
$$

First, compute the values step by step:

1. Calculate $$\ln(0.2)$$:  
   $$
   \ln(0.2)\approx -1.60944.
   $$
2. Compute the bracket term:  
   $$
   1+3.25\ln(0.2)\approx 1+3.25(-1.60944)\approx 1-5.23168\approx -4.23168.
   $$
3. Evaluate $$(0.2)^{3.25}$$:  
   $$
   (0.2)^{3.25}=\exp\left(3.25\ln(0.2)\right)=\exp(-5.23168)\approx 0.00536.
   $$
4. Multiply the terms:
   $$
   749.619\times0.00536\approx 4.0219,
   $$
   and then multiply by the bracket term:
   $$
   4.0219\times(-4.23168)\approx -17.023.
   $$

Thus,  
$$
R'(3.25)\approx -17.02.
$$

**Interpretation:**  
The derivative $$R'(3.25)$$ represents the marginal revenue when the price is $$\$3.25$$ per bottle. A value of approximately $$-17.02$$ means that if the price is increased marginally at $$p=3.25$$, the revenue will decrease by about $$17.02$$ million dollars per one dollar increase in the price (since $$R(p)$$ is measured in millions of dollars).
**a. Revenue Model** The revenue from selling sparkling cider is $$ R(p) = 749.619 \times p \times (0.2)^p, $$ where $$ p $$ is the price in dollars per bottle, and $$ 2 \leq p \leq 4.5 $$. **b. Marginal Revenue** The rate of change of revenue with respect to price is $$ R'(p) = 749.619 \times (0.2)^p \times (1 + p \times \ln(0.2)), $$ for $$ 2 \leq p \leq 4.5 $$. **c. Marginal Revenue at $$ p = 3.25 $$** At a price of \$3.25 per bottle, the marginal revenue is approximately $$ R'(3.25) \approx -17.02 \text{ million dollars per dollar increase in price}. $$ This means that increasing the price by \$1 at \$3.25 will decrease revenue by about \$17.02 million.

( 5 pts ) The demand for bottles of sparkling cider is given by million
bottles, where is the price in dollars per botte, .
a. Write a fully defined model for the revenue from the sale of sparkling cider, .
b. Find and write a fully defined model for it.
c. Find and interpret .

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a. Revenue Model

b. Finding

c. Finding and Interpreting

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( 5 pts ) The demand for bottles of sparkling cider is given by million bottles, where is the price in dollars per botte, . a. Write a fully defined model for the revenue from the sale of sparkling cider, . b. Find and write a fully defined model for it. c. Find and interpret .