Asadaly J. answered • 10/30/20

College and High School Tutor Specialized in Algebra to Calculus

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

Abigail E.

For (b), how would we put the answer in interval notation?10/30/20