• Feb 8th 2010, 03:19 PM
jasimmons86
ok im having the hardest time getting an answer for this problem

A marketing firm estimates that n monthsafter the introduction of a clients new product, f(n) thousand household will use it, where

f(n) = 10/9n(12 - n), 0<=n<=12

<=, less than equal, or greater than equal

10/9 is a fraction.

b. How long will it take for household use to reach a maximum?
c. How many households will use the product when it reaches its maximum?
d. The client must have 3 thousand households using the product in order to break even. How many months will it take before 3 thousand households use the product? (Give the decimal value of n.)

i got answers for b and c, but i can't get d... ive tried the quadratic formula and im stumped
a.) was graph the equation which i got already

Thanks for anyhelp
• Feb 8th 2010, 05:38 PM
apple123
Well, look at the equation you're given:

f(n) = (10/9)(12-n)

Think about what this function does. After 1 month, we can find how many households use the product:

f(n) = (10/9)(12 - 1) = 12.2 thousand households

Now, if we KNOW that 3 thousand households use the product, we simply work backwards to find how many months that's at. What part of the function does 3 thousand correspond to?
• Feb 8th 2010, 06:19 PM
jasimmons86
i think you miss quoted part of the equation, but thats fine.
you forgot the n after 10/9
heres what i come up with as far as working with 3000

f(n) = 10/9n(12 - n)
f(n) = -10/9n^2 + 13,1/3n

3000=-10/9n^2 + 13,1/3n
0=-10/9n^2 + 13,1/3n - 3000

13,1/3 = is a whole number with a fraction
• Feb 8th 2010, 07:12 PM
apple123
You're right, I did forget that n! Thanks for pointing that out.

What you've done here is mostly right, except that f(n) is in THOUSANDS! So you don't need to use 3000 here, just use 3. Does that make sense?

Now, can you solve 0 = (-10/9)n^2 + 13.3n - 3?

(I wrote 13.3 as your 13,1/3 just to make it look a bit neater. You can use 13 1/3 on paper though if you like).