Answer
See explanations.
Work Step by Step
Step 1. Given the revenue function as $r(x)=6x$ and the cost function as $c(x)=x^3-6x^2+15x$, the profit function can be written as $p(x)=r(x)-c(x)=-x^3+6x^2-9x$
Step 2. Trying to make the maximum profit, we will need to maximize the profit function. Take its derivative to get $p'(x)=-3x^2+12x-9$ and let $p'=0$ to get $x^2-4x+3=0$, which gives $x=1,3$. We have $p(1)=-(1)^3+6(1)^2-9(1)=-2$ (negative means a loss) and $p(3)=-(3)^3+6(3)^2-9(3)=0$ (0 means break even).
Step 3. $p''=-6x+12$ and $p''(3)=-18+12=-6\lt0$ indicating a concave down region with a maximum. Thus the maximum of profit happens at $x=3$ with $p(3)=0$, meaning the best we can do is break even (have revenue equal cost).
