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Magazine Chapter 4: Problem 2
The revenue from selling \(q\) items is \(R(q)=500 q-q^{2}\) and the total cost is \(C(q)=150+10 q .\) Write a function that gives the total profit earned, and find the quantity which maximizes the profit.
Short Answer
Expert verified
The maximum profit is achieved at a quantity of 245 items.
Step by step solution
01
Define the Profit Function
The profit function, denoted as \(P(q)\), is the difference between the revenue function \(R(q)\) and the cost function \(C(q)\). To find \(P(q)\), subtract \(C(q)\) from \(R(q)\). Thus, the profit function is given by: \[ P(q) = R(q) - C(q) = (500q - q^2) - (150 + 10q) \] Simplify this to: \[ P(q) = 500q - q^2 - 150 - 10q \] This simplifies further to: \[ P(q) = -q^2 + 490q - 150 \]
02
Find the Critical Points
To find the quantity \(q\) that maximizes the profit, take the derivative of \(P(q)\) with respect to \(q\) and set it equal to zero. Compute the derivative: \[ P'(q) = \frac{d}{dq} \left( -q^2 + 490q - 150 \right) = -2q + 490 \] Set the derivative equal to zero to find the critical points: \[ -2q + 490 = 0 \]
03
Solve for the Critical Value
Solving the equation \(-2q + 490 = 0\) gives:\[ -2q = -490 \] \[ q = \frac{-490}{-2} = 245 \] Thus, the profit-maximizing quantity is \(q = 245\).
04
Confirm Maximum with Second Derivative Test
To confirm that \(q = 245\) is a maximum, use the second derivative test. The second derivative of \(P(q)\) is: \[ P''(q) = \frac{d^2}{dq^2} (-q^2 + 490q - 150) = -2 \] Since \(P''(q) = -2\) is negative, \(q = 245\) is indeed a point of maximum profit.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Calculus Optimization
In economics and business, optimization is the process used to find the best solution to a problem from among a set of feasible options. Calculus optimization specifically involves the use of calculus techniques to solve such problems. In the case of profit maximization, we look for the quantity of products that yields the highest profit.
- The objective function specifies the goal of the optimization, such as maximizing profit or minimizing cost.
- Constraints might exist, limiting the possible solutions to the optimization problem.
- Using calculus, we differentiate the objective function to find critical points, which are potential maxima or minima.
- These critical points help us determine the optimal solution.
Derivative Test
The derivative test is an essential tool in calculus used to find the maxima or minima of a function. In the context of profit maximization, we use it to find the quantity of production that will maximize the profit.
- First Derivative Test: By taking the first derivative of the profit function, we can determine the rate of change of profit with respect to the quantity of goods produced. Setting this derivative equal to zero allows us to find the critical points.
- Second Derivative Test: After finding critical points, we use the second derivative test to confirm whether these points are maxima or minima. If the second derivative is negative at a critical point, it indicates a local maximum; if positive, a local minimum.
Revenue Function
The revenue function, denoted as \( R(q) \), represents the total income from selling \( q \) units of a product. In our exercise, the revenue function is given by \( R(q) = 500q - q^2 \). This equation shows how the revenue depends on the number of items sold.
Take note that revenue functions often include terms that reflect both linear and nonlinear behaviors:
Take note that revenue functions often include terms that reflect both linear and nonlinear behaviors:
- The linear part (\( 500q \)) shows the increase in revenue per unit sold.
- The quadratic term (\( -q^2 \)) indicates a diminishing return as \( q \) increases, accounting for factors like market saturation or reduced demand at higher quantities.
Profit Function
Profit functions are key tools for businesses, as they reveal the net gain obtained after deducting costs from revenue. In mathematical terms, the profit function \( P(q) \) is expressed as the revenue function minus the cost function.
For our example, this is represented by:\[ P(q) = R(q) - C(q) = (500q - q^2) - (150 + 10q) \]Simplifying this, we derive:\[ P(q) = -q^2 + 490q - 150 \]This function helps determine the optimal quantity of goods to maximize profit.
For our example, this is represented by:\[ P(q) = R(q) - C(q) = (500q - q^2) - (150 + 10q) \]Simplifying this, we derive:\[ P(q) = -q^2 + 490q - 150 \]This function helps determine the optimal quantity of goods to maximize profit.
- The coefficient in front of \( q^2 \) often exhibits a negative sign, indicating a downward-opening parabola in a quadratic profit function. Such a shape graphically represents how profit might initially increase and eventually decrease as quantity rises.
- The process involves solving \( P'(q) = 0 \) and using the second derivative test to find the point where profit is maximized.
