Problem 79
Question
Business profit. French's Electronics is selling laptop computers. It determines that its total profit, in dollars, is given by $$ P(x)=0.08 x^{2}+80 x $$ where \(x\) is the number of units produced and sold. Suppose that \(x\) is a function of time, in months, where \(x=5 t+1\) a) Find the total profit as a function of time \(t\). b) Find the rate of change of total profit when \(t=48\) months.
Step-by-Step Solution
Verified Answer
a) \( P(t) = 2t^2 + 404t + 80.08 \). b) \( 596 \) dollars per month.
1Step 1: Substitute x in the profit function
Given that the profit function is \( P(x) = 0.08 x^2 + 80x \) and \( x = 5t + 1 \), substitute \( x \) in the profit function to express it in terms of \( t \). This gives us \( P(t) = 0.08(5t + 1)^2 + 80(5t + 1) \).
2Step 2: Expand and simplify the profit function
Expand \( (5t + 1)^2 \) and simplify the expression to obtain \( P(t) = 0.08(25t^2 + 10t + 1) + 400t + 80 \). This simplifies to \( 2t^2 + 4t + 0.08 + 400t + 80 \). Collect like terms to get \( P(t) = 2t^2 + 404t + 80.08 \).
3Step 3: Find the rate of change of profit by differentiating
To find the rate of change of the profit with respect to time, differentiate the profit function \( P(t) = 2t^2 + 404t + 80.08 \) with respect to \( t \). This gives \( \frac{dP}{dt} = 4t + 404 \).
4Step 4: Evaluate the derivative at t = 48
Substitute \( t = 48 \) into \( \frac{dP}{dt} = 4t + 404 \). This yields \( 4(48) + 404 = 192 + 404 = 596 \). The rate of change of profit when \( t = 48 \) months is \( 596 \) dollars per month.
Key Concepts
Differentiation in CalculusProfit MaximizationRate of Change Calculus
Differentiation in Calculus
Differentiation is a key concept in calculus focused on determining the rate at which a quantity changes. In essence, it helps us understand how one variable affects another by looking at their instantaneous rates of change. When we differentiate a function, we're calculating its derivative, which provides valuable insights into the function's behavior over small intervals.
In our problem, we start with the profit function given as \( P(x) = 0.08x^2 + 80x \). When \( x \) is expressed in terms of time \( t \) as \( x = 5t + 1 \), we need to reformulate \( P(x) \) as a function of \( t \). This involves substituting \( x = 5t + 1 \) into \( P(x) \), leading to the new function \( P(t) = 2t^2 + 404t + 80.08 \).
By differentiating this function with respect to time, we find the rate of change of profit according to time variation, as represented by the derivative \( \frac{dP}{dt} = 4t + 404 \). This derivative gives us an understanding of how the profit changes per unit time, helping businesses make decisions based on temporal factors.
In our problem, we start with the profit function given as \( P(x) = 0.08x^2 + 80x \). When \( x \) is expressed in terms of time \( t \) as \( x = 5t + 1 \), we need to reformulate \( P(x) \) as a function of \( t \). This involves substituting \( x = 5t + 1 \) into \( P(x) \), leading to the new function \( P(t) = 2t^2 + 404t + 80.08 \).
By differentiating this function with respect to time, we find the rate of change of profit according to time variation, as represented by the derivative \( \frac{dP}{dt} = 4t + 404 \). This derivative gives us an understanding of how the profit changes per unit time, helping businesses make decisions based on temporal factors.
Profit Maximization
Profit maximization is a central goal in business decision-making, where the objective is to maximize the difference between total revenue and total costs. Understanding how profit behaves with respect to various factors allows businesses to adjust strategies effectively. In mathematical terms, it often involves deriving functions and finding maximum or minimum points using differentiation.
In this example, profit is represented by the function \( P(x) = 0.08x^2 + 80x \), which depends on the number of laptops produced and sold. By transforming this into a time-based function, \( P(t) = 2t^2 + 404t + 80.08 \), one can analyze how changing production over time affects profit.
By calculating the derivative \( \frac{dP}{dt} \), we understand the slope of the profit curve over time. Though this problem does not require finding a maximum or minimum, it sets the groundwork for analyzing scenarios where optimizing production or sales is necessary. Understanding and employing these techniques tap into the broader strategies for achieving profit maximization.
In this example, profit is represented by the function \( P(x) = 0.08x^2 + 80x \), which depends on the number of laptops produced and sold. By transforming this into a time-based function, \( P(t) = 2t^2 + 404t + 80.08 \), one can analyze how changing production over time affects profit.
By calculating the derivative \( \frac{dP}{dt} \), we understand the slope of the profit curve over time. Though this problem does not require finding a maximum or minimum, it sets the groundwork for analyzing scenarios where optimizing production or sales is necessary. Understanding and employing these techniques tap into the broader strategies for achieving profit maximization.
Rate of Change Calculus
Rate of change calculus is crucial in understanding how variables evolve over time or in response to other variables. Calculating the rate of change gives insight into the speed and direction of change, facilitating better prediction and planning in various fields, from physics to economics.
Here, we investigate the rate of change of a business's profit function with respect to time. By differentiating \( P(t) = 2t^2 + 404t + 80.08 \), we obtain \( \frac{dP}{dt} = 4t + 404 \), describing how profit changes as time advances.
To understand the specific rate at a certain time, \( t = 48 \), substituting this into the derivative gives \( 596 \). This indicates that at 48 months, the profit is increasing by 596 dollars per month. Such analyses are vital in strategic planning, allowing companies to respond effectively to trends and temporal changes.
Here, we investigate the rate of change of a business's profit function with respect to time. By differentiating \( P(t) = 2t^2 + 404t + 80.08 \), we obtain \( \frac{dP}{dt} = 4t + 404 \), describing how profit changes as time advances.
To understand the specific rate at a certain time, \( t = 48 \), substituting this into the derivative gives \( 596 \). This indicates that at 48 months, the profit is increasing by 596 dollars per month. Such analyses are vital in strategic planning, allowing companies to respond effectively to trends and temporal changes.
Other exercises in this chapter
Problem 79
For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=6 x-x^{2} $$
View solution Problem 79
Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \t
View solution Problem 79
Find each limit. Use TABLE and start with \(\Delta\) Tbl \(=0.1\). Then use \(0.01,0.001,\) and \(0.0001 .\) When you think you know the limit, graph and use TR
View solution Problem 80
For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=20 x-x^{2} $$
View solution