1. ## functions

Hello,
I've been studying for my test and got stuck on this question, can anybody explain it to me?

The demand function for a new product is P(x) = -5x + 22 where x is the number of items sold in thousands and p is the rice in dollars. The cost function is
C(x) = 3x + 15.
a) state the corresponding revenue function (I'm pretty sure it's -5x^2 + 22x)
b) Find the corresponding profit function(I'm having problem with that)
c) Complete the square to find the value that will maximize the profits.
d) Find the break even quantities.

Thank you,

2. The demand function for a new product is P(x) = -5x + 22 where x is the number of items sold in thousands and p is the price in dollars.
Umm. Shouldn't that be called the price function?

a) state the corresponding revenue function (I'm pretty sure it's -5x^2 + 22x)
Yes, it is.
Because revenue is (demand)*(price).

b) Find the corresponding profit function(I'm having problem with that)
Profit = Revenue -Cost
So,
Profit(x) = (-5x^2 +22x) -(3x +15)
Profit(x) = -5x^2 +19x -15 ---------------------answer.

c) Complete the square to find the value that will maximize the profits.
-5x^2 +19x -15
= -5[x^2 +(19/5)x +3]
= -5[x^2 +(19/5)x +(19/10)^2 -(19/10)^2 +3]
= -5[(x +(19/10))^2 -(19/10)^2 +3]

So, the x at maximum profit is -19/10
What?
Negative demand? No product?

There is something wrong with your Question as posted.

Edit:
Oopps, my mistake!
It should have been:
-5x^2 +19x -15
= -5[x^2 -(19/5)x +3]
= -5[x^2 -(19/5)x +(19/10)^2 -(19/10)^2 +3]
= -5[(x -(19/10))^2 -(19/10)^2 +3]
So the x at maximum profit is 19/10 thousand = 1900 products

Therefore, to maximize profit, 1900 of the products must me made. ---------------answer.

d) Find the break even quantities.
At break-even, the profit is zero because the revenue equals the cost only.

So,
-5x^2 +22x = 3x +15
-5x^2 +22x -3x -15 = 0
-5x^2 +19x -15 = 0
Divide both sides by -5,
x^2 -3.8x +3 = 0