what is the number of units sold at 500. x=-5p+100, R(p)= -5p^2+100p

To find the number of units sold at a price of \( p = 500 \), we’ll substitute \( p = 500 \) into the demand equation \( x = -5p + 100 \).

Given:
\[
x = -5p + 100
\]
1. Substitute \( p = 500 \):
   \[
   x = -5(500) + 100
   \]
2. Calculate:
   \[
   x = -2500 + 100 = -2400
   \]

The result \( x = -2400 \) indicates that there would be no units sold at a price of 500, as the quantity sold \( x \) cannot be negative. This suggests that a price of \( p = 500 \) is too high for any demand in this model. 

Let me know if you want more details or have any questions. 

Here are some related questions to explore:

1. What is the maximum price at which units are sold in this model?
2. How would you find the price that maximizes revenue?
3. Can we set a range of prices to ensure positive demand?
4. How is revenue affected if the price increases or decreases slightly?
5. What happens to the demand if the price is decreased below a certain threshold?

**Tip:** Negative demand can signal that the price is beyond what customers are willing to pay.

What is the number of units sold at 500. x=-5p+100, R(p)= -5p^2+100p

Understand how to calculate the number of units sold at a price of 500 using demand and revenue functions: x = -5p + 100 and R(p) = -5p^2 + 100p. Learn the effects of price on demand and revenue with a step-by-step solution. Suitable for Grades 9-11.

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