Respuesta :
Answer:
x=12 maximizes the profit function.
Step-by-step explanation:
We are given that the profit is [tex]P(x) = -x^3+15x^2+72x-200[/tex] for [tex]x\geq 5[/tex]
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 [tex]x^k[/tex] is [tex]kx^{k-1}[/tex], 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).
[tex] P'(x) =-3x^2+30x+72[/tex].
We want to solve [tex]P'(x) = 0[/tex]. By dividing the equation by -3, we get
[tex]x^2-10x-24=0=(x-12)(x+2)[/tex]
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 [tex]f''(x)<0[/tex]. In our case,
[tex]P''(x) = -6x+30[/tex]
Note that [tex]P''(12) = -42<0[/tex]. So x=12 is a maximum of P.
