Gymnast Clothing manufactures expensive hockey jerseys for sale to college bookstores in runs of up to 300. Its cost (in dollars) for a run of x hockey jerseys is
C(x) = 3000 + 10x + 0.2x2    (0 ≤ x ≤ 300).
Gymnast Clothing sells the jerseys at $100 each. Find the revenue function.
R(x) =

Find the profit function.
P(x) =

What is the minimum number of jerseys that Gymnast Clothing should manufacture to make a profit? (Round your answer up to the nearest whole number.)
jerseys

Let's break down each part of the problem:

### 1. **Revenue Function $ R(x) $**:
Revenue is calculated as the number of jerseys sold multiplied by the selling price per jersey. Since each jersey is sold at $100, the revenue function $ R(x) $ will be:

\[
R(x) = 100x
\]

### 2. **Profit Function $ P(x) $**:
Profit is the difference between revenue and cost. The cost function is given by:

\[
C(x) = 3000 + 10x + 0.2x^2
\]

Therefore, the profit function $ P(x) $ will be:

\[
P(x) = R(x) - C(x)
\]

Substituting the expressions for $ R(x) $ and $ C(x) $:

\[
P(x) = 100x - (3000 + 10x + 0.2x^2)
\]

Simplifying the expression:

\[
P(x) = 100x - 3000 - 10x - 0.2x^2
\]
\[
P(x) = 90x - 3000 - 0.2x^2
\]
\[
P(x) = -0.2x^2 + 90x - 3000
\]

### 3. **Minimum Number of Jerseys to Make a Profit**:
To find the minimum number of jerseys that will make a profit, we need to solve for when $ P(x) > 0 $, i.e., when the profit is positive. First, set $ P(x) = 0 $ to find the break-even point:

\[
-0.2x^2 + 90x - 3000 = 0
\]

This is a quadratic equation. Let's solve it using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here:
- $ a = -0.2 $
- $ b = 90 $
- $ c = -3000 $

Let's calculate the solution for $ x $.The solutions to the quadratic equation are approximately $ x_1 = 36.25 $ and $ x_2 = 413.75 $. Since the maximum number of jerseys that can be produced is 300, the valid solution is $ x_1 = 36.25 $.

To make a profit, Gymnast Clothing needs to produce more than 36.25 jerseys. Rounding up to the nearest whole number, the minimum number of jerseys that should be manufactured to make a profit is **37 jerseys**.

Would you like further details or clarifications?

Here are five related questions for further exploration:
1. What would happen to the profit function if the cost per jersey increased?
2. How would the revenue change if the selling price per jersey was $120 instead of $100?
3. Can we determine the number of jerseys that would maximize the profit?
4. What if Gymnast Clothing had a different cost function, like $ C(x) = 2000 + 15x + 0.3x^2 $?
5. How does the break-even point change with a higher or lower fixed cost?

**Tip**: When solving quadratic equations, always check whether both solutions fit within the problem's constraints!

Gymnast Clothing manufactures expensive hockey jerseys for sale to college bookstores in runs of up to 300. Its cost (in dollars) for a run of x hockey jerseys is C(x) = 3000 + 10x + 0.2x² (0 ≤ x ≤ 300). Gymnast Clothing sells the jerseys at $100 each. Find the revenue function R(x). Find the profit function P(x). What is the minimum number of jerseys that Gymnast Clothing should manufacture to make a profit? (Round your answer up to the nearest whole number.)

Explore how to determine the revenue function, profit function, and the minimum number of hockey jerseys Gymnast Clothing must manufacture to make a profit. Learn key concepts such as quadratic equations, revenue analysis, and cost functions. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation