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 bottle, $$ 2 \leq p \leq 4.5 $$. 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 $$ R^{\prime}(3.25) $$.

Question

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**a. Revenue Model**

The revenue from the sparkling cider is the price per bottle times the number of bottles sold. Since the demand is given by  
$$
D(p)=749.619\,(0.2)^p,
$$  
the revenue function is  
$$
R(p)=p\cdot D(p)=749.619\,p\,(0.2)^p,
$$  
with the domain $$ 2\le p\le 4.5 $$.

**b. Derivative of the Revenue Model**

We want to find $$ R'(p) $$ where  
$$
R(p)=749.619\,p\,(0.2)^p.
$$  
Differentiate using the product rule. Let  
$$
f(p)=p \quad \text{and} \quad g(p)=(0.2)^p.
$$
Then  
$$
f'(p)=1 \quad \text{and} \quad g'(p)=(0.2)^p\ln(0.2).
$$
Thus,  
$$
R'(p)=749.619\,[f'(p)g(p)+f(p)g'(p)]=749.619\Big[(0.2)^p+p\,(0.2)^p\ln(0.2)\Big].
$$
Factor $$ (0.2)^p $$ out:  
$$
R'(p)=749.619\,(0.2)^p\Big[1+p\ln(0.2)\Big],
$$
with the domain $$ 2\le p\le 4.5 $$.

**c. Interpretation of $$ R'(3.25) $$**

Substitute $$ p=3.25 $$ into the derivative:
$$
R'(3.25)=749.619\,(0.2)^{3.25}\Big[1+3.25\ln(0.2)\Big].
$$
This value represents the instantaneous rate of change of revenue (in millions of dollars, since numbers are in millions) with respect to the price at $$ p=3.25 $$ dollars per bottle.

A negative value of $$ R'(3.25) $$ would mean that a small increase in the price at $$ p=3.25 $$ would lead to a decrease in revenue. Conversely, a positive value would indicate an increase. Calculating approximately:

1. Compute $$ \ln(0.2) $$. Since $$ \ln(0.2)\approx -1.60944 $$, then  
   $$
   1+3.25\ln(0.2)\approx 1+3.25(-1.60944) \approx 1-5.2300\approx -4.23.
   $$
2. Compute $$ (0.2)^{3.25} $$. Since  
   $$
   (0.2)^{3.25}=e^{3.25\ln(0.2)}\approx e^{-5.230}\approx 0.00536,
   $$
3. Thus,  
   $$
   R'(3.25)\approx 749.619\times 0.00536\times (-4.23).
   $$
4. Calculating the product,  
   $$
   749.619\times 0.00536\approx 4.02,
   $$
   and so  
   $$
   4.02\times(-4.23)\approx -17.0.
   $$

So,  
$$
R'(3.25)\approx -17.0.
$$

This means that at a price of $$ \$3.25 $$ per bottle, a small increase in the price would decrease the revenue by approximately $$ 17 $$ million dollars per additional dollar in price.
**a. Revenue Model** The revenue from selling sparkling cider is calculated by multiplying the price per bottle by the number of bottles sold. Given the demand function $$ D(p)=749.619\,(0.2)^p, $$ the revenue function is $$ R(p)=p\cdot D(p)=749.619\,p\,(0.2)^p, $$ where $$ p $$ is the price in dollars per bottle, and $$ 2 \leq p \leq 4.5 $$. **b. Derivative of the Revenue Model** To find the rate at which revenue changes with respect to price, we take the derivative of $$ R(p) $$. Using the product rule: $$ R'(p)=749.619\,(0.2)^p\Big[1+p\ln(0.2)\Big], $$ with $$ 2 \leq p \leq 4.5 $$. **c. Interpretation of $$ R'(3.25) $$** At a price of $$ \$3.25 $$ per bottle, the rate of change of revenue is approximately $$ -17.0 $$ million dollars per dollar increase in price. This means that increasing the price slightly at this point would lead to a decrease in total revenue by about $$ 17 $$ million dollars.

( 5 pts ) The demand for bottles of sparkling cider is given by million
bottles, where is the price in dollars per bottle, .
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 .

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

a. Revenue Model

b. Derivative of Revenue Model

c. Compute

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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 bottle, . 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 .