/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Revenue A firm has a revenue fun... [FREE SOLUTION] | 91Ó°ÊÓ

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Revenue A firm has a revenue function given by \(R(p)=\) \(10 p-p^{2},\) where \(p\) is the price of a chocolate bar sold in dollars and \(R\) is measured in thousands of dollars per day. Suppose the firm is able to increase the price of each bar by 5 cents each year (without affecting demand). If \(t\) is time measured in years, write an equation for the revenue as a function of \(t\) if the price of a candy bar starts out at 25 cents.

Short Answer

Expert verified
Revenue function: \( R(t) = 2.4375 + 0.475t - 0.0025t^2 \).

Step by step solution

01

Initial Price Setup

Start by understanding the initial conditions. The price of a chocolate bar, denoted by \( p \), begins at $0.25 (or 25 cents). Thus, the initial price can be written as \( p = 0.25 \).
02

Rate of Price Increase

The price increases by 5 cents each year. Thus, each year \( t \), the price increase is \( 0.05t \), where \( t \) represents the number of years.
03

Expression for Price Over Time

Combine the initial price with the annual increase to form a general expression for price. The equation becomes \( p = 0.25 + 0.05t \).
04

Substitute Price into Revenue Function

Substitute \( p = 0.25 + 0.05t \) back into the revenue function \( R(p) = 10p - p^2 \).
05

Expanded Revenue Function

The equation becomes: \( R(t) = 10(0.25 + 0.05t) - (0.25 + 0.05t)^2 \).
06

Simplify Revenue Expression

Calculate each part: \( 10(0.25 + 0.05t) \) and expand \( (0.25 + 0.05t)^2 \). Simplify to obtain the revenue as a function of time \( t \).
07

Final Revenue Expression

After simplification, the revenue function in terms of \( t \) is: \( R(t) = 2.5 + 0.5t - (0.25^2 + 2 \times 0.25 \times 0.05t + (0.05t)^2) \). Simplify further to get the final expression: \( R(t) = 2.5 + 0.5t - 0.0625 - 0.025t - 0.0025t^2 \).
08

Final Simplified Revenue Function

Combine the terms to obtain the final revenue equation: \( R(t) = 2.4375 + 0.475t - 0.0025t^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Price Elasticity
At its core, price elasticity is a measure of how sensitive the demand for a product is to changes in price. In other words, it helps us understand how much the quantity demanded of a product will change as its price changes. This is particularly important for businesses because it impacts revenue and profitability when prices fluctuate. - **Elastic Demand:** If a small change in price leads to a large change in quantity demanded, the product is said to have elastic demand. - **Inelastic Demand:** Conversely, if a substantial price change results in little alteration to the quantity demanded, the demand is inelastic. Calculating price elasticity in this scenario can enhance our understanding of how the gradual increase in price of a chocolate bar affects demand and subsequently, the firm's revenue. In this exercise, we assume that increasing the price by 5 cents annually does not affect demand. This suggests that the demand is perfectly inelastic, meaning that demand remains constant regardless of price.
Calculus Applications
The realm of calculus provides powerful tools to find rates of change and comprehend how functions behave. In problems related to revenue functions, calculus helps us analyze how changing prices impact revenue over time, which is vital for strategic financial planning. One common calculus application is differentiating the revenue function to find the rate of change of revenue concerning price. However, in our exercise, we are directly using algebraic manipulations by plugging in the equation for price into the revenue function. - Once the new expression for revenue as a function of time, \( R(t) \), is obtained, it can further be differentiated to understand how revenue changes over specific time intervals.Ultimately, calculus applications like differentiation equip us to optimize functions to maximize or minimize outcomes, crucial in the business decision-making process.
Mathematical Modeling
Mathematical modeling involves constructing mathematical representations of real-world scenarios in order to predict future behavior. In this exercise, we are modeling how the firm's revenue progresses over time given the price increase of chocolate bars. Key Points:- We define the initial price and predict future prices using a linear model: \( p = 0.25 + 0.05t \). This straightforward model allows us to substitute and find the revenue function as it relates to time, \( R(t) = 2.4375 + 0.475t - 0.0025t^2 \).- Such modeling helps us forecast the revenue, enabling informed decision-making and strategic planning.Through effective mathematical modeling, businesses can identify potential risks, forecast financial outcomes, and experiment with different pricing strategies without real-world consequences. This valuable insight supports robust business strategies for revenue optimization and sustainability.

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