The price p and the quantity x sold of a certain product obey the demand equation
p equals negative one eighth x plus 50 comma 0 less than or equals x less than or equals 400
​(a)
Express the revenue R as a function of x.
​(b)
What is the revenue if 352 units are​ sold?
​(c)
What quantity x maximizes​ revenue? What is the maximum​ revenue?
​(d)
What price should the company charge to maximize​ revenue?
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Part 1
​(a)  Express the revenue R as a function of x.


To solve the problem, we start with the given demand equation:

\[
p = -\frac{1}{8}x + 50
\]

### (a) Express the revenue $ R $ as a function of $ x $.

Revenue $ R $ is calculated as the product of the price $ p $ and the quantity sold $ x $:

\[
R = p \cdot x
\]

Substituting the demand equation into this formula gives:

\[
R = \left(-\frac{1}{8}x + 50\right) \cdot x
\]

This simplifies to:

\[
R = -\frac{1}{8}x^2 + 50x
\]

Thus, the revenue function $ R(x) $ is:

\[
R(x) = -\frac{1}{8}x^2 + 50x
\]

### (b) What is the revenue if 352 units are sold?

Now, we substitute $ x = 352 $ into the revenue function:

\[
R(352) = -\frac{1}{8}(352)^2 + 50(352)
\]

Calculating this:

1. Calculate $ (352)^2 = 123904 $.
2. Calculate $ -\frac{1}{8}(123904) = -15488 $.
3. Calculate $ 50(352) = 17600 $.
4. Combine these: 

\[
R(352) = -15488 + 17600 = 2112
\]

So, the revenue if 352 units are sold is **$2112**.

### (c) What quantity $ x $ maximizes revenue? What is the maximum revenue?

The revenue function $ R(x) = -\frac{1}{8}x^2 + 50x $ is a quadratic equation that opens downward (since the coefficient of $ x^2 $ is negative). The maximum revenue occurs at the vertex of the parabola, which can be found using the formula:

\[
x = -\frac{b}{2a}
\]

Here, $ a = -\frac{1}{8} $ and $ b = 50 $:

\[
x = -\frac{50}{2 \cdot -\frac{1}{8}} = \frac{50 \cdot 8}{2} = 200
\]

To find the maximum revenue, substitute $ x = 200 $ back into the revenue function:

\[
R(200) = -\frac{1}{8}(200)^2 + 50(200)
\]

Calculating this:

1. $ (200)^2 = 40000 $.
2. $ -\frac{1}{8}(40000) = -5000 $.
3. $ 50(200) = 10000 $.
4. Combine:

\[
R(200) = -5000 + 10000 = 5000
\]

So, the quantity that maximizes revenue is **200 units**, and the maximum revenue is **$5000**.

### (d) What price should the company charge to maximize revenue?

To find the price at which this maximum revenue occurs, substitute $ x = 200 $ back into the demand equation:

\[
p = -\frac{1}{8}(200) + 50
\]

Calculating this:

1. $ -\frac{1}{8}(200) = -25 $.
2. $ p = -25 + 50 = 25 $.

Thus, the price to maximize revenue is **$25**.

### Summary of Answers:
- (a) $ R(x) = -\frac{1}{8}x^2 + 50x $
- (b) Revenue for 352 units: **$2112**
- (c) Quantity to maximize revenue: **200 units**, Maximum revenue: **$5000**
- (d) Price to maximize revenue: **$25**

Do you want further details on any of these steps? Here are some questions for exploration:

1. How does the shape of the revenue function impact pricing strategy?
2. What happens to revenue if the demand equation changes?
3. Can we derive other important economic metrics from the revenue function?
4. How would fixed costs influence the overall profit calculation?
5. What are the implications of selling beyond the optimal quantity?

The price p and the quantity x sold of a certain product obey the demand equation p = -1/8 x + 50, 0 <= x <= 400. Express the revenue R as a function of x, find the revenue if 352 units are sold, determine what quantity x maximizes revenue, and what price should the company charge to maximize revenue.

Explore how to express revenue as a function of quantity sold using the demand equation p = -1/8 x + 50. Learn to calculate revenue for 352 units sold, find the optimal quantity for maximum revenue (200 units), and determine the price ($25) that maximizes revenue.

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