Problem 66
Question
BUSINESS: Sales The number of bottles of whiskey that a store will sell in a month at a price of \(p\) dollars per bottle is $$ N(p)=\frac{2250}{p+7} \quad(p \geq 5) $$ Find the rate of change of this quantity when the price is \(\$ 8\) and interpret your answer.
Step-by-Step Solution
Verified Answer
The sales decrease by 10 bottles per month for every dollar increase in price when the price is \$8.
1Step 1: Identify the function
The given function for the number of bottles sold is \( N(p) = \frac{2250}{p+7} \). This function gives the number of bottles sold as a function of price \( p \).
2Step 2: Understand the requirement
We need to find the rate of change of the quantity \( N(p) \) with respect to the price \( p \) when \( p = 8 \). This requires finding the derivative of \( N(p) \) with respect to \( p \).
3Step 3: Differentiate N(p) with respect to p
To find \( \frac{dN}{dp} \), use the quotient rule: If \( N(p) = \frac{u(p)}{v(p)} \), then \( \frac{dN}{dp} = \frac{v(p)\frac{du}{dp} - u(p)\frac{dv}{dp}}{[v(p)]^2} \). Here, \( u(p) = 2250 \) and \( v(p) = p+7 \), with \( \frac{du}{dp} = 0 \) and \( \frac{dv}{dp} = 1 \).
4Step 4: Apply the quotient rule
Substitute into the quotient rule formula: \[\frac{dN}{dp} = \frac{(p+7) \cdot 0 - 2250 \cdot 1}{(p+7)^2} = \frac{-2250}{(p+7)^2}\]
5Step 5: Substitute p = 8 into the derivative
Now, substitute \( p = 8 \) into the derivative \[\frac{dN}{dp} = \frac{-2250}{(8+7)^2} = \frac{-2250}{15^2} = \frac{-2250}{225} = -10\]
6Step 6: Interpret the result
The derivative \( \frac{dN}{dp} = -10 \) means that for every additional dollar increase in the price at \( p = \\(8 \), the store expects to sell 10 fewer bottles of whiskey per month. This is the rate of change of sales with respect to price at \( \\)8 \).
Key Concepts
DerivativeRate of ChangeQuotient Rule
Derivative
A derivative represents the rate at which a function changes as its input changes. It gives you instantaneous rate of change at any given point.
In this exercise, we're interested in finding the derivative of the function \( N(p) = \frac{2250}{p+7} \). Here, \( N(p) \) describes how many whiskey bottles are sold based on the bottle price \( p \).
By computing \( \frac{dN}{dp} \), the derivative of \( N \) with respect to \( p \), we determine how the sales numbers change as the price shifts. When you calculate a derivative, you use specific rules depending on the form of the function, like the product rule, chain rule, or in this case, the quotient rule. Derivatives are fundamental in calculus and essential for understanding changes and trends in functions.
In this exercise, we're interested in finding the derivative of the function \( N(p) = \frac{2250}{p+7} \). Here, \( N(p) \) describes how many whiskey bottles are sold based on the bottle price \( p \).
By computing \( \frac{dN}{dp} \), the derivative of \( N \) with respect to \( p \), we determine how the sales numbers change as the price shifts. When you calculate a derivative, you use specific rules depending on the form of the function, like the product rule, chain rule, or in this case, the quotient rule. Derivatives are fundamental in calculus and essential for understanding changes and trends in functions.
Rate of Change
The rate of change is essentially how a quantity responds to variations in another quantity. In practical terms, it shows how something like sales, production, or distance changes as the defining variable, like price or time, changes.
In this problem, we are concerned with how many bottles are sold as the price per bottle changes. The rate of change is given by the derivative \( \frac{dN}{dp} \).
Here, a rate of \(-10\) means that for each dollar increase in the price, the amount of whiskey bottles sold decreases by 10 bottles. Rates of change give businesses and scientists valuable insight into trends, allowing them to optimize prices or other factors to achieve desired outcomes.
In this problem, we are concerned with how many bottles are sold as the price per bottle changes. The rate of change is given by the derivative \( \frac{dN}{dp} \).
Here, a rate of \(-10\) means that for each dollar increase in the price, the amount of whiskey bottles sold decreases by 10 bottles. Rates of change give businesses and scientists valuable insight into trends, allowing them to optimize prices or other factors to achieve desired outcomes.
Quotient Rule
The quotient rule is a method for taking derivatives of functions that are fractions of two other functions. When we have a function given by \( \frac{u(p)}{v(p)} \), the quotient rule helps us find its derivative. The rule is:
\[ \frac{d}{dp}\left(\frac{u(p)}{v(p)}\right) = \frac{v(p)\frac{du}{dp} - u(p)\frac{dv}{dp}}{[v(p)]^2} \]
In this situation, the function \( N(p) = \frac{2250}{p+7} \) can be set up with \( u(p) = 2250 \) and \( v(p) = p+7 \). Since \( u(p) \) is a constant, its derivative \( \frac{du}{dp} \) is 0. The derivative of \( v(p) \) is 1, making the application of the quotient rule straightforward.
\[ \frac{d}{dp}\left(\frac{u(p)}{v(p)}\right) = \frac{v(p)\frac{du}{dp} - u(p)\frac{dv}{dp}}{[v(p)]^2} \]
In this situation, the function \( N(p) = \frac{2250}{p+7} \) can be set up with \( u(p) = 2250 \) and \( v(p) = p+7 \). Since \( u(p) \) is a constant, its derivative \( \frac{du}{dp} \) is 0. The derivative of \( v(p) \) is 1, making the application of the quotient rule straightforward.
- The formula for the derivative becomes: \( \frac{-2250}{(p+7)^2} \).
- This helps us quickly find the rate of change of sales as price changes, crucial in business analysis.
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