Answer
$C^{\prime}(x) =10x$
$R^{\prime}(x)=3x^{2}+7$
$P^{\prime}(x)=3x^{2}-10x+7$
The profit peaks at production levels of
x= $1$ and x$\approx 2.3333$
Work Step by Step
Profit function: $P(x)=R(x)-C(x).$
Marginal cost function: $C^{\prime}(x)$
Marginal revenue and profit functions: $R^{\prime}(x)$ and $P^{\prime}(x)$
What it means when the marginal profit is zero is described on p. 802:
Solving $P^{\prime}(x)=0$ for x gives the exact value of x, if such a value exists, for which the profit peaks (neither increases nor decreases) with respect to production level.
-----------------
Marginal cost function: $C^{\prime}(x) =5(x)=10x$
Marginal revenue function:
$R^{\prime}(x)=3x^{2}+7$
Marginal profit function$: $
$P^{\prime}(x)=3x^{2}+7-10x$
$P^{\prime}(x)=0$
$3x^{2}-10x+7=0$
... quadratic formula ...
$x=\displaystyle \frac{10\pm\sqrt{(-10)^{2}-4(3)(7)}}{2(3)}=\frac{10\pm\sqrt{100-84}}{6}$
$x=\displaystyle \frac{10\pm 4}{6}$
$x=1$ or $x=\displaystyle \frac{14}{6}=\frac{7}{3}\approx 2.3333$
The profit peaks at production level of
x= $1$ and x$\approx 2.3333$
$C^{\prime}(x) =10x$
$R^{\prime}(x)=3x^{2}+7$
$P^{\prime}(x)=3x^{2}-10x+7$
The profit peaks at production level of
x= $1$ and x$\approx 2.3333$
