Find prize per unit that will maximize profit at a given $x$-value Struggling while reviewing my old math books. The problem has a prize-function and wants to know how the prize-per-unit should be chosen to maximize the profit at $\mathbf{x=160}$.
First I look analyze the given equation. The given prize function is:
$$
\operatorname{Prize}(x) = 0.003x^3-0.54x^2+96.6x+8100
$$
I calculated the prize-per-unit by dividing by $x$:
$$
\operatorname{PrizePerUnit}(x) = 0.003x^2-0.54x+96.6 + 8100/x
$$
First derivative is:
$$
\operatorname{PrizePerUnit}'(x) = 0.006x-0.54 + 8100/x^2
$$
Only real solution is $x=150$ which gives a prize-per-unit of $137.10$.
The usual way to calculate the profit is:
$$
\operatorname{Profit} = \operatorname{Returns} - \operatorname{Prize}
$$
If I use the prize-per-unit from the given function:
$$
\operatorname{Profit} = \operatorname{Returns} - \operatorname{Prize}
$$
$$
\operatorname{Profit} = x - (0.006x-0.54 + 8100/x^2)
$$
I get:
$$
\operatorname{Profit}(x)=-96.6-0.003 x^2+1.54 x-8100/x
$$
Which has a maximum at $\mathbf{274.573}$.
And this is where I'm stuck. The solution according to the book is $\mathbf{154.20}$ prize-per-unit. Anybody know how to reach that number?
Integration is apparently not part of the solution, because that is introduced much later in the book. Can this be done with some kind of equivalence-equation?

You price function gives the total costs to make a number of $x$ products. Call this $C(x)$. Your income is given by the number of sold items times the sale price. If we denote the sale price by $\alpha$, then the income is given by $\alpha x$.
Your profit $P$ can be calculated by subtracting the costs from the income:
$$P(x) = \alpha x - C(x).$$
If $P$ has a maximum then it's derivative must be zero there, i.e.:
$$P^\prime(x) = \alpha - 0.009x^2 + 1.08x - 96.6 = 0.$$
Now you can simply move all the $x$'s to the right-hand side and substitute $x=160$. Then you will find that $\alpha = 154.20$.
Formally you should check that this indeed gives you a maximum and not a minimum, since we only used that the derivative is zero, which can imply both.
