Chapter 9: Problem 18
Find the amount s of advertising that maximizes the profit \(P\). (s and \(P\) are measured in thousands of dollars.) Find the point of diminishing returns. \(P=-0.1 s^{3}+6 s^{2}+400\)
Short Answer
Expert verified
The amount of advertising that maximizes the profit is $40,000 and the point of diminishing returns is at $20,000.
Step by step solution
01
Finding the derivative of the profit function
You start by finding the derivative of the profit function \(P\). Using the power rule, which states that the derivative of \(x^n\) is \(n*x^{n-1}\), the derivative of \(P\), which will be denoted by \(P'\), is calculated in the following way: \(P'=-0.3*s^{2}+12s\).
02
Finding the critical points
After having found the derivative, you find the critical points of the function by setting the derivative equal to zero and solving for \(s\). This gives you the equation: \(-0.3*s^{2}+12s=0\). Solving this equation gives you \(s=0\) and \(s=40\). These are the critical points.
03
Deciding which critical point maximizes the profit
The profit is maximized at the highest critical point, so you plug in the critical points into the profit function to see which \(s\) maximizes \(P\). You find that \(P(0) = 400\) and \(P(40) = 560\). Therefore, \(s=40\) maximizes the profit.
04
Finding the point of diminishing returns
The point of diminishing returns is the point after which additional input in advertising starts to yield progressively smaller increases in profit. This can be found by looking for the inflection point, which is where the second derivative of the profit function changes sign. The second derivative of \(P\), denoted by \(P''\), is found to be: \(P''=-0.6s+12\). Setting this equal to zero and solving for \(s\) gives \(s=20\). So the point of diminishing returns is when \(s=20\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Profit Maximization
In economics, profit maximization involves finding the ideal point where a company can have the utmost return from its investments and operations. This is typically realized by finding the level of input, such as advertising, that results in the highest profit. For a function like the profit function provided, maximizing profit means identifying the value of advertising spend, denoted by \(s\), that makes the profit \(P(s)\) as large as possible.
In the given exercise, we observe that the profit is represented by a cubic function:
To conclude, it's essential to know that profit maximization may depend on various factors like market conditions and costs, but mathematically, the critical points and evaluation help in attaining the highest feasible profit.
In the given exercise, we observe that the profit is represented by a cubic function:
- \(P(s) = -0.1s^{3} + 6s^{2} + 400\).
To conclude, it's essential to know that profit maximization may depend on various factors like market conditions and costs, but mathematically, the critical points and evaluation help in attaining the highest feasible profit.
Critical Points
Critical points of a function are values of \(s\) where the derivative, \(P'\), equals zero or is undefined. These points are significant because they can indicate potential local maxima, minima, or inflection points.
To find these in the given problem, we first derive the profit function to get the rate of change of profit concerning advertising spend:
Finding these helps determine where the most significant changes in profit occur, assisting businesses in making informed decisions on resource allocation.
To find these in the given problem, we first derive the profit function to get the rate of change of profit concerning advertising spend:
- \(P' = -0.3s^{2} + 12s\).
- \(-0.3s^{2} + 12s = 0\).
Finding these helps determine where the most significant changes in profit occur, assisting businesses in making informed decisions on resource allocation.
Derivative Applications
Derivatives are powerful tools in economic analysis, allowing us to measure how functions change. For profit functions, derivatives indicate how profits vary with different levels of input, like advertising expenditures.
The first derivative, \(P'\), reveals the rate of change of profit with respect to advertising, helping us locate critical points. However, derivatives offer further insights with the second derivative, \(P''\), which helps ascertain the nature of these points (whether they are maxima, minima, or points of inflection). In the context of the problem,
The first derivative, \(P'\), reveals the rate of change of profit with respect to advertising, helping us locate critical points. However, derivatives offer further insights with the second derivative, \(P''\), which helps ascertain the nature of these points (whether they are maxima, minima, or points of inflection). In the context of the problem,
- The first derivative \(P' = -0.3s^{2} + 12s\) helps find critical points.
- The second derivative \(P'' = -0.6s + 12\) helps determine the point of diminishing returns.
Point of Diminishing Returns
The point of diminishing returns is a critical concept that describes a level of input where each additional unit contributes less to the output than the previous one. In the context of advertising, it signifies the point beyond which increasing the advertising spend yields progressively smaller increases in profit.
To find this point, we use the second derivative \(P''\). An inflection point in this derivative indicates where the curvature of the profit graph changes, excellent for pinpointing diminishing returns. In the exercise,
To find this point, we use the second derivative \(P''\). An inflection point in this derivative indicates where the curvature of the profit graph changes, excellent for pinpointing diminishing returns. In the exercise,
- We have the second derivative \(P'' = -0.6s + 12\).
- Setting \(P'' = 0\) results in \(s = 20\), which is the point of diminishing returns.