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Magazine Chapter 2: Problem 5
A one-product firm estimates that its daily total cost function (in suitable units) is \(C(x)=x^{3}-6 x^{2}+13 x+15\) and its total revenue function is \(R(x)=28 x .\) Find the value of \(x\) that maximizes the daily profit.
Short Answer
Expert verified
The value of \ \( x \) that maximizes the daily profit is \ \( 5 \).
Step by step solution
01
Define the profit function
The profit function is the difference between the total revenue function and the total cost function. Define the profit function as \ \( P(x) = R(x) - C(x) \). Substituting the given functions, \ \( P(x) = 28x - (x^3 - 6x^2 + 13x + 15) \). Simplify this to obtain: \ \( P(x) = -x^3 + 6x^2 + 15x - 15 \).
02
Find the derivative of the profit function
To find the critical points which could indicate a maximum profit, calculate the first derivative of the profit function, \ \( P'(x) \). Thus, \ \( P'(x) = \frac{d}{dx} (-x^3 + 6x^2 + 15x - 15) = -3x^2 + 12x + 15 \).
03
Set the derivative equal to zero
To find the critical points, solve \ \( P'(x) = 0 \). This gives us the quadratic equation: \ \( -3x^2 + 12x + 15 = 0 \).
04
Solve the quadratic equation
Solve \ \( -3x^2 + 12x + 15 = 0 \) using the quadratic formula \ \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \ \( a = -3 \), \ \( b = 12 \), and \ \( c = 15 \). This results in: \ \( x = \frac{-12 \pm \sqrt{144 + 180}}{-6} = \frac{-12 \pm \sqrt{324}}{-6} = \frac{-12 \pm 18}{-6} \). The solutions are: \ \( x = -1 \) and \ \( x = 5 \).
05
Determine the maximum profit
To determine which value of \ \( x \) maximizes the profit, evaluate the second derivative, \ \( P''(x) \). So \ \( P''(x) = \frac{d}{dx} (-3x^2 + 12x + 15) = -6x + 12 \). Evaluate \ \( P''(x) \) at both critical points: \ \( P''(-1) = -6(-1) + 12 = 18 \), which is positive, indicating a local minimum. \ \( P''(5) = -6(5) + 12 = -18 \), which is negative, indicating a local maximum. Therefore, the value of \ \( x \) that maximizes profit is \ \( x = 5 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Total Cost Function
The total cost function is a fundamental concept in profit maximization. It represents the total costs a firm incurs to produce a certain number of units, denoted by the variable \(x\). In the given exercise, the total cost function is: \[ C(x) = x^{3} - 6 x^{2} + 13 x + 15 \] To understand this function better:
- \( x^3 \) represents increased costs due to higher production volumes.
- \( -6 x^2 \) might represent cost savings from economies of scale at lower levels of production.
- \( 13 x \) is the linear component indicating constant marginal cost.
- \( 15 \) is the fixed cost incurred regardless of the production level.
Total Revenue Function
The total revenue function tells us how much revenue the firm generates from selling its product. It is a function of the quantity of goods sold, denoted as \(x\). In this exercise, the revenue function is: \[ R(x) = 28 x \] This implies the firm makes 28 units of currency per unit sold. The total revenue function is crucial because it allows us to determine the revenue at various production levels, essential for maximizing profit.
Profit Function
The profit function is the heart of profit maximization problems. It represents the difference between total revenue and total cost: \[ P(x) = R(x) - C(x) \] Substituting the given functions, the profit function becomes: \[ P(x) = 28 x - (x^3 - 6 x^2 + 13 x + 15) = -x^3 + 6 x^2 + 15 x - 15 \] Understanding the profit function allows businesses to see the profit or loss at various production amounts. It's the basis for finding the production level that leads to the highest profit.
Derivative
In calculus, the derivative helps us find the rate of change of a function. For profit maximization, we use the derivative to identify critical points where profit could be maximized or minimized. The first derivative of the profit function \( P(x) = -x^3 + 6x^2 + 15x - 15 \) is: \[ P'(x) = \frac{d}{dx} (-x^3 + 6x^2 + 15x - 15) = -3x^2 + 12x + 15 \] This derivative function, \( P'(x) \), shows the rate of change of profit with respect to production quantity \( x \). Setting this equation to zero helps us find the critical points.
Critical Points
Critical points occur where the derivative of a function equals zero or does not exist. These points are where we might find local maxima or minima. To find critical points, we set the first derivative of the profit function to zero and solve: \[ -3x^2 + 12x + 15 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we calculate: \[ x = \frac{-12 \pm \sqrt{144 + 180}}{-6} = \frac{-12 \pm 18}{-6} \] The solutions to this are \( x = -1 \) and \( x = 5 \). These values represent critical points that need further evaluation to determine if they are maxima or minima.
Second Derivative Test
The second derivative gives us information about the concavity of the profit function, helping us determine if the critical points are maxima or minima. For the profit function \( P(x) = -x^3 + 6 x^2 + 15 x - 15 \), the second derivative is: \[ P''(x) = \frac{d}{dx} (-3x^2 + 12x + 15) = -6x + 12 \] By evaluating \( P''(x) \) at the critical points:
- \( P''(-1) = -6(-1) + 12 = 18 \) (positive, indicating a local minimum)
- \( P''(5) = -6(5) + 12 = -18 \) (negative, indicating a local maximum)
