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.
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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