https://file.aiquickdraw.com/mathbot/file/1727375403202gfygspbt.jpg

The maximum profit occurs at $ x = 77.5 $ bikes (rounded to 78 bikes since fractional values are not practical in this context). The second derivative test confirms that the profit function has a maximum at this point since the second derivative is negative $-4$.

The maximum profit, when $ x = 78 $ bikes are sold, is $ \frac{16025}{2} = 8012.5 $ dollars. Rounded to the nearest dollar, the maximum profit is **$8013**.

Do you have any further questions or need additional details?

Here are 5 related questions to explore:
1. How would the result change if the cost function $C(x)$ included a variable cost per unit that increases with production?
2. What is the impact of changes in the quadratic coefficient of the revenue function on the maximum profit?
3. How can you graphically interpret the profit function and its maximum?
4. How would taxes or additional fixed costs alter the maximum profit calculation?
5. How does the second derivative test help in confirming a maximum versus a minimum?

**Tip:** When optimizing profit, always check whether the critical point yields a maximum or minimum using the second derivative test.

If a bike company has a revenue of R(x) = 350x - 2x^2 and their costs are C(x) = 4000 + 40x for x bikes sold, find the maximum profit. Remember P(x) = R(x) - C(x). Round to the nearest dollar.

Find the maximum profit for a bike company using the revenue function R(x) = 350x - 2x^2 and the cost function C(x) = 4000 + 40x. Learn how to apply profit maximization concepts, including the first and second derivative tests, to determine the optimal number of bikes to produce.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation