The expressions p(200-5p)and 200p-5p2define the same function. The function models the revenue a school would earn from selling raffle tickets at p dollars each.

At what price or prices would the school collect $0 revenue from raffle sales? Explain or show your reasoning.

The school staff noticed that there are two ticket prices that would both result in a revenue of $500. How would you find out what those two prices are?

I really have no idea. If anyone could help that would be great!

Answers

Answer 1
a. Prices at $0 revenue

The prices which the school would collect $0 revenue are $0 and $ 40 respectively.

Since the revenue R(p) = p(200 - 5p).

The price at which the school earn s $0 revenue from the raffle sales is when R(p) = 0.

So, p(200 - 5p) = 0

p = 0 or 200 - 5p = 0

p = 0 or 200 = 5p

p = 0 or p = 200/5 = 4

So, the prices which the school would collect $0 revenue are $0 and $ 40 respectively.

b. Prices at $500 revenue

The two ticket prices that would produce a revenue of $500 are $2.68 and $ 37.32

We need to find the two ticket prices that would result in a revenue of $500. So, R(p) = 500.

So, p(200 - 5p) = 500

200p - 5p² = 500

Re-arranging, we have

5p² - 200p + 500 = 0

Dividing through by 5, we have

p² - 40p + 100 = 0

Using the quadratic formula, we find the value of p.

So,

[tex]p = \frac{-(-40) +/- \sqrt{(-40)^{2} - 4 X 1 X 100} }{2 X 1} \\p = \frac{40 +/- \sqrt{1600 - 400} }{2} \\p = \frac{40)+/- \sqrt{1200} }{2} \\p = \frac{40 +/- 34.64 }{2} \\p = \frac{40 - 34.64 }{2} or p = \frac{40 + 34.64 }{2} \\p = \frac{5.36}{2} or p = \frac{74.64 }{2} \\p = 2.68 or 37.32[/tex]

So, the two ticket prices that would produce a revenue of $500 are $2.68 and $ 37.32

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