a company that sells dirt bikes determines that in order to sell x bikes in a year they should use a unit price of p = 8610 - 0.8x dollars. how many bikes must they sell per year to maximize revenue?

To determine the number of bikes the company must sell to maximize revenue, we need to find the maximum value of the revenue function. 

The revenue \( R \) generated from selling \( x \) bikes can be calculated as the product of the unit price \( p \) and the number of bikes sold \( x \). Given the unit price function:

\[
p = 8610 - 0.8x
\]

the revenue function \( R(x) \) is:

\[
R(x) = x \cdot p = x \cdot (8610 - 0.8x)
\]

Let's expand and simplify this:

\[
R(x) = 8610x - 0.8x^2
\]

This is a quadratic function in the form:

\[
R(x) = -0.8x^2 + 8610x
\]

### Step 1: Determine the Maximum Revenue

A quadratic function \( ax^2 + bx + c \) reaches its maximum at \( x = -\frac{b}{2a} \). Here, \( a = -0.8 \) and \( b = 8610 \).

Substitute these values into the formula:

\[
x = -\frac{8610}{2 \cdot (-0.8)} = \frac{8610}{1.6}
\]

### Step 2: Calculate \( x \)

Now, let's compute the value:

\[
x = \frac{8610}{1.6} = 5381.25
\]

Since the number of bikes \( x \) must be an integer, we round to the nearest whole number. Thus, the company must sell **5381 bikes** per year to maximize revenue.

### Conclusion

To maximize revenue, the company should aim to sell **5381 bikes** per year.

Would you like further details or have any other questions?

### Related Questions:

1. What would the maximum revenue be when selling 5381 bikes?
2. How does changing the unit price formula affect the revenue-maximizing number of bikes?
3. What happens to revenue if the company sells 100 bikes more or fewer than the optimal amount?
4. Can the revenue be maximized if the company changes the price function to a different linear relationship?
5. What is the break-even point for the company if the cost per bike is known?

#### Tip: Always remember that quadratic functions representing revenue or profit maximization problems will open downward (negative \( a \) value) and have a maximum point at the vertex.

A company that sells dirt bikes determines that in order to sell x bikes in a year they should use a unit price of p = 8610 - 0.8x dollars. How many bikes must they sell per year to maximize revenue?

Learn how to maximize the revenue for a company selling dirt bikes with the function p = 8610 - 0.8x. This solution uses quadratic functions and revenue maximization techniques, suitable for high school algebra students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation