# Math Help - A quick question about Increasing and decreasing Functions.

1. ## A quick question about Increasing and decreasing Functions.

Ok so it says here

A manufacturing company finds the profit for a production level of x motors per hour is

p = 400 x sqr(12x - x^2) - 100 x e (0,12]
For what range of production is the profit
Increasing
Decreasing

Now the answer is (0, 6) increasing and (6, 12) decreasing.

When I found the derivative, I got

(-200x +2400) (12x-x^2)^-1/2

Now the critical number for the first bracket is 12, which I found, makes sence because the function decreases at 12. But I cant find where the 6 comes from.

Can anyone help?

2. Originally Posted by Sky12
Ok so it says here

A manufacturing company finds the profit for a production level of x motors per hour is

p = 400 x sqr(12x - x^2) - 100 x e (0,12]
For what range of production is the profit
Increasing
Decreasing
Thus you have,
$P=400x\sqrt{12x-x^2}-100$
And I assume you mean, $x\in (0,12]$

So find the derivative und make it zero.
$P'=(400x)'(\sqrt{12x-x^2})+(400x)(\sqrt{12x-x^2})'$
Thus,
$P'=400\sqrt{12x-x^2}+\frac{400x(12-2x)}{2\sqrt{12x-x^2}}=0$
Multiply through by the radical,
$400(12x-x^2)+200x(12-2x)=0$
Divide by 200,
$2(12x-x^2)+x(12-2x)=0$
$24x-2x^2+12x-2x^2=0$
$-4x^2+36x=0$
$-4x(x-9)=0$
Thus,
$x=0,9$
But,
$x\in (0,12]$.
Thus,
$x=9$
Is the only solution.

Now take any point to the left of 9 (and more than zero) and see what the sign is. Positive increasing and negative decreasing. Same for any point to the right of 9 (and less than 12)

3. Ah sorry for the confusion but i ment the x as (times) so it would be more like

400sqr(12x - x^2) - 100