Question:

You are in charge of marketing a new pair of sunglasses: The Ray Bob. For a sales cycle, it costs a base amount of 90,000$, and then 8$ for each pair of glasses produced. When you set the price at 600$ for a pair, you sell 5,000 pairs. If you increase the price by 2$, you sell 12 fewer pairs. Let x be the number of pairs of glasses produced. 

Model the function C(x), the production cost. 
Model the function p(x), the price per pair of glasses.
Model the function R(x), the revenue from production.
Model the function P(x), the profit from production.
Find the value of x that maximizes the profit.
Calculate P(x), the maximum profit. Submit the answer below.
Note: Your answer must be written in decimal form with EXACTLY 2 decimal places of precision.

Your Answer:

Let's go through the process step-by-step to model the functions $ C(x) $, $ p(x) $, $ R(x) $, and $ P(x) $, and to determine the value of $ x $ that maximizes the profit and the maximum profit value.

### 1. **Modeling the Production Cost $ C(x) $:**
The cost function consists of a fixed base cost of $90,000 and a variable cost of $8 per pair of glasses produced.

\[
C(x) = 90000 + 8x
\]
where $ x $ is the number of pairs of glasses produced.

### 2. **Modeling the Price Per Pair $ p(x) $:**
We are told that when the price is set at $600, 5,000 pairs of glasses are sold. Moreover, for every $2 increase in price, 12 fewer pairs of glasses are sold.

Let’s express the price per pair of glasses $ p(x) $ as a linear function of the number of glasses sold $ x $. 

From the given information:
- When $ x = 5000 $, the price is $600.
- For every 12 fewer pairs sold (a decrease in $ x $ by 12), the price increases by $2.

The slope $ m $ of the price function can be calculated as:
\[
m = \frac{2}{-12} = -\frac{1}{6}
\]

Using the point-slope form of a line $ p(x) = mx + b $, and using $ (5000, 600) $, we can plug in the values to find the intercept $ b $:

\[
600 = -\frac{1}{6}(5000) + b
\]
\[
600 = -833.33 + b
\]
\[
b = 1433.33
\]

Thus, the price function is:
\[
p(x) = -\frac{1}{6}x + 1433.33
\]

### 3. **Modeling the Revenue $ R(x) $:**
Revenue is the price per pair multiplied by the number of pairs sold. Thus, the revenue function $ R(x) $ is:

\[
R(x) = p(x) \cdot x = \left(-\frac{1}{6}x + 1433.33\right)x
\]
\[
R(x) = -\frac{1}{6}x^2 + 1433.33x
\]

### 4. **Modeling the Profit $ P(x) $:**
Profit is the difference between revenue and cost. Thus, the profit function $ P(x) $ is:

\[
P(x) = R(x) - C(x)
\]
\[
P(x) = \left(-\frac{1}{6}x^2 + 1433.33x\right) - (90000 + 8x)
\]
\[
P(x) = -\frac{1}{6}x^2 + 1433.33x - 90000 - 8x
\]
\[
P(x) = -\frac{1}{6}x^2 + (1433.33 - 8)x - 90000
\]
\[
P(x) = -\frac{1}{6}x^2 + 1425.33x - 90000
\]

### 5. **Maximizing the Profit:**
To find the value of $ x $ that maximizes profit, we take the derivative of $ P(x) $ and set it equal to 0.

\[
P'(x) = -\frac{1}{3}x + 1425.33
\]
Set the derivative equal to 0:

\[
-\frac{1}{3}x + 1425.33 = 0
\]
\[
\frac{1}{3}x = 1425.33
\]
\[
x = 1425.33 \times 3 = 4275.99
\]

Thus, the value of $ x $ that maximizes the profit is approximately $ x = 4276 $ pairs of glasses.

### 6. **Calculating the Maximum Profit:**
Substitute $ x = 4276 $ into the profit function $ P(x) $:

\[
P(4276) = -\frac{1}{6}(4276)^2 + 1425.33(4276) - 90000
\]

Let’s calculate this value.The maximum profit is approximately **$2,957,348.41** when 4,276 pairs of glasses are produced.

Let me know if you'd like more details or if you have any questions! Here are some related questions to explore:

1. How would the profit function change if the base cost increased?
2. What would happen to the revenue if the price per pair decreased by a fixed amount?
3. How sensitive is the maximum profit to small changes in production cost?
4. Can you model the effect of a price drop on the overall profit?
5. How does adjusting the price elasticity (change in demand for price changes) affect revenue?

**Tip:** Maximizing profit often involves balancing the marginal cost with the marginal revenue!

You are in charge of marketing a new pair of sunglasses: The Ray Bob. For a sales cycle, it costs a base amount of 90,000$, and then 8$ for each pair of glasses produced. When you set the price at 600$ for a pair, you sell 5,000 pairs. If you increase the price by 2$, you sell 12 fewer pairs. Let x be the number of pairs of glasses produced.

Model the function C(x), the production cost. Model the function p(x), the price per pair of glasses. Model the function R(x), the revenue from production. Model the function P(x), the profit from production. Find the value of x that maximizes the profit. Calculate P(x), the maximum profit. Submit the answer below. Note: Your answer must be written in decimal form with EXACTLY 2 decimal places of precision.

Explore how to model and solve a profit maximization problem for The Ray Bob sunglasses. Learn how to derive the production cost, price function, revenue, and profit. Discover how to find the maximum profit and the number of pairs to produce. Suitable for students in Grades 10-12.

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