pre-calculus, ranges and revenues.

Question

Suppose that the manufacturer of a gas clothes dryer has found that when the unit price is p​ dollars, the revenue R​ (in dollars) is R(p)= −6p^2 + 18,000p.

​(a) At what prices p is revenue​ zero?

​(b) For what range of prices will revenue exceed $1,200,000​?

Please help!

Answer 1

Hey Abigail!

A) Here we need to find when the revenue is 0, so we can set R(p) to 0 since that is our revenue, and then solve:

R(p)= −6p^2 + 18,000p

0= −6p^2 + 18,000p

6p^2 - 18,000p = 0

6p(p - 3000) = 0

6p=0 and p-3000=0

p=0 and p=3000

Therefore, our revenue is 0 when the price is at 0 or 3000.

B) Here we need to find when the revenue is greater than 1,200,000, so let's plug in that number for our R(p) and solve:

R(p)= −6p^2 + 18,000p

12,000,000 = −6p^2 + 18,000p

6p^2 - 18,000p + 12,000,000 = 0

6(p^2 - 3,000p + 2,000,000) = 0

6(p - 1,000)(p - 2,000) = 0

p-1,000=0 and p-2,000=0

p=1000 and p=2000

We know that the Revenue is at 1,200,000 at p=1000 and p=2000, so let's check the value for Revenue when p is at say 1,500.

R(p)= −6p^2 + 18,000p

R(p)= −6(1500)^2 + 18,000(1500)

R(p) = 13,500,000

We know its greater than our target in the middle of these two points, so the range is between 1000 and 2000