Business Applications of Differentiation
1. Optimization
Example 4
An automobile company manufactures and sells $x$ number of cars per week. If the weekly price equation is $P(x)=\frac{1}{3} x^2-x-3$, (where $P(X)$ is in millions of Ghana Cedi)
Find for each week the maximum revenue.
Example 5
A company manufactures and sells $x$ transistor radios per week. If the weekly price equation is $P ( x )=10-\frac{x}{1000}, \quad 0 \leq x \leq 8000$
Find for the week the maximum revenue.
Answer (1)
For Example 4, the revenue function is R ( x ) = 3 1 x 3 − x 2 − 3 x , and its derivative is R ′ ( x ) = x 2 − 2 x − 3 . Setting R ′ ( x ) = 0 gives critical points x = − 1 and x = 3 .
The second derivative is R ′′ ( x ) = 2 x − 2 . At x = 3 , 0"> R ′′ ( 3 ) = 4 > 0 , indicating a local minimum. The price at x = 3 is negative, so there is no maximum revenue.
For Example 5, the revenue function is R ( x ) = 10 x − 1000 x 2 , and its derivative is R ′ ( x ) = 10 − 1000 2 x . Setting R ′ ( x ) = 0 gives x = 5000 .
The second derivative is R ′′ ( x ) = − 500 1 < 0 , indicating a maximum. The maximum revenue is R ( 5000 ) = 25000 .
The maximum revenue for Example 5 is 25000 .
Explanation
Problem Setup We are asked to find the maximum weekly revenue for an automobile company. The weekly price equation is given by P ( x ) = 3 1 x 2 − x − 3 , where x is the number of cars manufactured and sold per week, and P ( x ) is in millions of Ghana Cedi.
Revenue Function The revenue function R ( x ) is the product of the number of cars sold and the price per car, i.e., R ( x ) = x × P ( x ) . Therefore, we have R ( x ) = x ( 3 1 x 2 − x − 3 ) = 3 1 x 3 − x 2 − 3 x .
Finding Critical Points To find the maximum revenue, we need to find the critical points of the revenue function. We take the derivative of R ( x ) with respect to x and set it to zero: R ′ ( x ) = d x d R = x 2 − 2 x − 3 = 0.
Solving for x Now, we solve the quadratic equation x 2 − 2 x − 3 = 0 for x . We can factor the quadratic as ( x − 3 ) ( x + 1 ) = 0 . Thus, the solutions are x = 3 and x = − 1 . Since the number of cars cannot be negative, we discard x = − 1 . So, x = 3 is the only critical point we consider.
Second Derivative Test To determine if x = 3 corresponds to a maximum or minimum, we use the second derivative test. We find the second derivative of R ( x ) :
R ′′ ( x ) = 2 x − 2.
Evaluating the Second Derivative We evaluate R ′′ ( x ) at x = 3 :
R ′′ ( 3 ) = 2 ( 3 ) − 2 = 6 − 2 = 4. Since 0"> R ′′ ( 3 ) = 4 > 0 , the critical point x = 3 corresponds to a local minimum.
Analyzing the Price and Revenue However, since the price equation P ( x ) = 3 1 x 2 − x − 3 is a parabola opening upwards, it is possible that the revenue function does not have a maximum. Let's analyze the price equation at x = 3 : P ( 3 ) = 3 1 ( 3 ) 2 − 3 − 3 = 3 − 3 − 3 = − 3 . Since the price is negative, this does not make sense in the context of the problem. The problem states that P ( x ) is in millions of Ghana Cedi, so a negative price means the company is paying people to take the cars. This implies that the problem is not well-posed, and there is no maximum revenue. However, if we ignore the fact that the price is negative, we can proceed to calculate the revenue at x = 3 : R ( 3 ) = 3 1 ( 3 ) 3 − ( 3 ) 2 − 3 ( 3 ) = 9 − 9 − 9 = − 9 .
Final Answer Since the problem is not well-posed and the price becomes negative, there is no maximum revenue in the realistic sense. However, if we were to proceed with the given equations, we would find a minimum revenue of -9 million Ghana Cedi at x = 3 .
Example 5 Let's consider the second example. The weekly price equation is P ( x ) = 10 − 1000 x , with 0 ≤ x ≤ 8000 . The revenue function is R ( x ) = x P ( x ) = x ( 10 − 1000 x ) = 10 x − 1000 x 2 . To find the maximum revenue, we take the derivative of R ( x ) with respect to x and set it to zero: R ′ ( x ) = 10 − 1000 2 x = 0 . Solving for x , we get 1000 2 x = 10 , so x = 5000 . The second derivative is R ′′ ( x ) = − 1000 2 = − 500 1 . Since R ′′ ( x ) < 0 , we have a maximum at x = 5000 . The maximum revenue is R ( 5000 ) = 10 ( 5000 ) − 1000 ( 5000 ) 2 = 50000 − 1000 25000000 = 50000 − 25000 = 25000 .
Final Answer for Example 5 The maximum revenue for Example 5 is 25000.
Examples
Optimization problems like finding maximum revenue are crucial in business. For instance, a retail store can use this approach to determine the optimal pricing strategy for a product. By analyzing the relationship between price, demand, and production costs, the store can identify the price point that maximizes its profit. This ensures the business operates efficiently and achieves its financial goals.
