Answer:
x=12 maximizes the profit function.
Step-by-step explanation:
We are given that the profit is
for ![x\geq 5](https://tex.z-dn.net/?f=x%5Cgeq%205)
To find x that maximizes P, we will find the derivative of P(x) and find x such that P'(x) =0. Recall that the derivative of a function of the form
is
, and that the derivative of a constant is zero. Then, by using the properties of derivatives, we get (the details of the calculation is omitted).
.
We want to solve
. By dividing the equation by -3, we get
![x^2-10x-24=0=(x-12)(x+2)](https://tex.z-dn.net/?f=x%5E2-10x-24%3D0%3D%28x-12%29%28x%2B2%29)
So we have that x=12 and x=-2 are solutions. In this case, we are only considering x greater than o equals 0. So, we take x=12.
We will check that x=12 is a maximum of P.
To do so, we will use the second derivative criteria, which is as follows. Given a function f whose first and second derivative exist, a point x is a maximum if
. In our case,
Note that
. So x=12 is a maximum of P.