Word problem and polynomials

[Vempy]

I'm stuck, completely and utterly stuck.

Here's the question:
The total revenue, R, received from selling, p items of a product is R(p) = 42p -0.15p^2 where P < or = 500. How many items were sold if the revenue was less than 8796$?

Here's my work so far:

R= total revenue
p= number of items sold to obtain R

So, if R(p)= 42p - 0.15p^2 and p is under the limitations p < or = 500, how many items were sold if the revenue is less than 8796$.

Therefore, R < 8796$

Now, if you set it up as an inequality you get:

8796 > -0.15p^2 + 42p

And then you have to solve it.

0 > -0.15p^2 + 42p - 8796

Which doesn't factor because (a)(c) is 1319.4 and (b) is 42.

But the discriminate for the quadratic formula is negative, which means they are unreal roots.

b^2 - 4ac
=42^2 - (4)(-0.15)(-8796)
=-3513.6

Therefore, I'm clueless.

Unless, it doesn't need to be set up as an inequality?



I'd rather someone set me in the right direction if I'm wrong, but not solve it for me or anything like that. Thank you! ^^

 

[skeeter]

Here's the question:
The total revenue, R, received from selling, p items of a product is R(p) = 42p -0.15p^2 where P < or = 500. How many items were sold if the revenue was less than 8796$?

R(p) < 8796

42p - 0.15p² < 8796

0.15p² - 42p + 8796 > 0

since the discriminate < 0, that means that this quadratic is never equal to 0, and that it is always either greater than or less than 0 for all p, 0 < p < 500.

plug in any value of p in the domain ... you get a value for the quadratic > 0 ... so this says that any number of items sold (from 0 to 500) will yield a revenue less than $8796.

the maximum revenue occurs when R'(p) = 0 ...
42 - 0.30p = 0
p = 140 items sold maximizes revenue.
so, max revenue = R(140) = $2940 < $8796

my question ...
sure you copied the problem correctly?

 

[Vempy]

The question is copied correctly, but it doesn't say it has to be an inequality, that's just how I thought it made the most sense to set it up. Also, p is the only value squared in -0.15p^2, if that makes a difference.