company manufactures and sells x cellphones per week. The weekly​ price-demand and cost equations are given below.
p equals 600 minus 0.1 x and Upper C left parenthesis x right parenthesis equals 15 comma 000 plus 135 x
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Part 1
​(A) What price should the company charge for the​ phones, and how many phones should be produced to maximize the weekly​ revenue? What is the maximum weekly​ revenue?
The company should produce

3000 phones each week at a price of ​$

300.
​(Round to the nearest cent as​ needed.)
Part 2
The maximum weekly revenue is ​$

900000. ​(Round to the nearest cent as​ needed.)
Part 3
​(B) What price should the company charge for the​ phones, and how many phones should be produced to maximize the weekly​ profit? What is the maximum weekly​ profit?
The company should produce

2325 phones each week at a price of ​$

367.50.
​(Round to the nearest cent as​ needed.)
Part 4
The maximum weekly profit is ​$

84562.50. ​(Round to the nearest cent as​ needed.)

To solve these problems, we need to use the given equations:

1. **Price-demand equation:** $ p = 600 - 0.1x $
2. **Cost equation:** $ C(x) = 15,000 + 135x $

### Part 1: Maximizing Revenue

Revenue $ R $ is given by the product of the number of phones sold $ x $ and the price $ p $:
\[ R(x) = x \cdot p = x(600 - 0.1x) = 600x - 0.1x^2 \]

To find the number of phones $ x $ that maximizes revenue, we take the derivative of $ R(x) $ with respect to $ x $ and set it to zero:
\[ R'(x) = 600 - 0.2x \]

Setting the derivative equal to zero:
\[ 600 - 0.2x = 0 \]
\[ 0.2x = 600 \]
\[ x = 3000 \]

To find the price at this number of phones:
\[ p = 600 - 0.1(3000) = 600 - 300 = 300 \]

Thus, the company should produce **3000 phones** each week at a price of **$300**.

To find the maximum revenue:
\[ R(3000) = 3000 \cdot 300 = 900,000 \]

So, the maximum weekly revenue is **$900,000**.

### Part 2: Maximizing Profit

Profit $ P $ is given by the difference between revenue and cost:
\[ P(x) = R(x) - C(x) = (600x - 0.1x^2) - (15,000 + 135x) \]
\[ P(x) = 600x - 0.1x^2 - 15,000 - 135x \]
\[ P(x) = 465x - 0.1x^2 - 15,000 \]

To find the number of phones $ x $ that maximizes profit, we take the derivative of $ P(x) $ with respect to $ x $ and set it to zero:
\[ P'(x) = 465 - 0.2x \]

Setting the derivative equal to zero:
\[ 465 - 0.2x = 0 \]
\[ 0.2x = 465 \]
\[ x = 2325 \]

To find the price at this number of phones:
\[ p = 600 - 0.1(2325) = 600 - 232.5 = 367.5 \]

Thus, the company should produce **2325 phones** each week at a price of **$367.50**.

To find the maximum profit:
\[ P(2325) = 465(2325) - 0.1(2325)^2 - 15,000 \]
\[ P(2325) = 465 \cdot 2325 - 0.1 \cdot 540625 - 15,000 \]
\[ P(2325) = 1,081,125 - 54,062.5 - 15,000 \]
\[ P(2325) = 1,081,125 - 69,062.5 \]
\[ P(2325) = 1,012,062.5 \]

So, the maximum weekly profit is **$1,012,062.50**.

### Summary:
- To maximize revenue, produce 3000 phones at $300 each. Maximum revenue is $900,000.
- To maximize profit, produce 2325 phones at $367.50 each. Maximum profit is $1,012,062.50.

If you have any questions or need further details, feel free to ask!

Here are some additional questions you might consider:
1. How does the cost structure impact profit maximization?
2. What would happen to revenue if the price-demand relationship changed?
3. How sensitive is the optimal production quantity to changes in cost per unit?
4. What is the breakeven point for the company?
5. How would introducing fixed costs change the analysis?

**Tip:** Always check the second derivative to confirm whether a critical point is a maximum or minimum.

Learn how to maximize weekly revenue and profit for a cellphone manufacturing company using price-demand and cost equations. Discover the optimal production quantities and prices to achieve maximum revenue and profit. Solve real-world optimization problems with step-by-step solutions.

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