Jamila sells knitted hats to her friends at school and she is determined that the revenue formula is R(p)=-p²+20p, where R is the revenue in dollars and p is the selling price. she wants to make $200 during her soohmore year. can she achieve this goal and what price wood she have to charge? Use quadratic formula and explain your reasonig. And how would you rewrite the equation with 200 as revenue in factored form?

Question

Jamila sells knitted hats to her friends at school and she is determined that the revenue formula is R(p)=-p²+20p, where R is the revenue in dollars and p is the selling price. she wants to make $200 during her soohmore year. can she achieve this goal and what price wood she have to charge? Use quadratic formula and explain your reasonig. And how would you rewrite the equation with 200 as revenue in factored form?

Answer 1

really? You have the formula, so just solve

p^2 + 2p = 200
p^2 + 20p - 200 = 0
p = (-20±√(20^2+800))/2 = -10 ±10√3

Answer 2

oops. Missed that minus sign

-p^2 + 20p = 200
p^2 - 20p + 200 = 0
p = 10±10i
This is even worse, as it still does not factor, and there are no real solutions!

R(p) = -p^2+20p has a maximum of 100. at p=10.
So, I found my typo -- what is yours?

Answer 3

Yes really. Thanks for helping. I tried 200p=-p²+20p

Answer 4

To determine if Jamila can make $200 during her sophomore year and what price she would have to charge for her knitted hats, we will use the revenue formula that Jamila provided.

The revenue formula given is:

R(p) = -p² + 20p

Here, R(p) represents the revenue in dollars generated from selling the hats, and p represents the selling price.

To find out if Jamila can make $200 in revenue, we need to set up the equation and solve it using the quadratic formula.

So, let's replace R(p) with 200 in the equation:

200 = -p² + 20p

To solve this equation, we rearrange it to a standard quadratic form:

p² - 20p + 200 = 0

Now, we can apply the quadratic formula to find the solutions for p:

p = (-b ± √(b² - 4ac)) / (2a)

Here, a = 1, b = -20, and c = 200.

Plugging these values into the formula:

p = (-(−20) ± √((−20)² - 4(1)(200))) / (2(1))

Simplifying further:

p = (20 ± √(400 - 800)) / 2

p = (20 ± √(-400)) / 2

Since we cannot have a negative square root in this context (as we are dealing with prices), we conclude that Jamila cannot make $200 in revenue during her sophomore year.

Now, let's rewrite the equation with 200 as revenue in factored form:

R(p) = -p² + 20p - 200

To factorize this quadratic equation, we need to find two numbers that add up to 20 (coefficient of 'p') and multiply to -200 (constant term).

The factored form of the equation will be:

R(p) = -(p - 10)(p - 20)

This represents the revenue equation with 200 as the revenue in factored form.