Question

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!

Answer 1

Answer:

When the price of a ticket = $0 or $40, there will be no revenue.

The revenue will be $500 if each ticket cost $37.3 or $2.68

Step-by-step explanation:

Let R(p) = p(200-5p)

when R(p) = 0

p(200-5p) = 0

p = 0 or 200-5p = 0

When the price of a ticket = $0 or $40, there will be no revenue.

when R(p) = 500

200p - 5p² = 500

5p² - 200p + 500 = 0

p = $37.3 or $2.68 (3 sig. fig.)

The required prices are $37.3 and $2.68

Answer 2

Answer:

A) 0, 40

B) 2.68, 37.32

Step-by-step explanation:

Part A:

Let R= revenue

Let = price

Use the factored form p(200-5p) to solve the first part.  The question asks what price will revenue be zero.  Therefore, we should let R= 0 and solve for p.

r= p(200-5p)

0=  p(200-5p)

The equation will equal zero when either factor is zero, therefore solve for p when either factor equals zero.

p= 0

200- 5p= 0

p= 40

Rearrange and find that p in the second factor and find that p= 40

Therefore, the school will make no revenue when the price is either 0$ or 40$.

Part B;

Let r= $500

r= 200p- 5p^2

500= 200p- 5p^2

Set equation to zero and solve for p by using quadratic formula where a= -5, b= 200, c= -500.

0= -5p^2 + 200p - 500

After plugging in values to quadratic formula, you get p= $2.68 or p= $37.32.

Therefore, there will be $500 revenue when ticket price is either p= $2.68 or p= $37.32.