Answer
a. $\quad C(3)=C(6)=\displaystyle \frac{25}{3}$
b. $\quad\approx 410$ components
Work Step by Step
(a)
$C(3)=10(\displaystyle \frac{1}{3}+\frac{3}{3+3})=\frac{25}{3}$
$C(5)=10(\displaystyle \frac{1}{6}+\frac{6}{6+3})=\frac{25}{3}$
$C(3)=C(6)=\displaystyle \frac{25}{3}$
(b)
$... $there must be a value x=c in the interval such that $C^{\prime}(x)=0$
$C(x)=10(x^{-1}+x(x+3)^{-1})$
$C^{\prime}(x)=10[-x^{-2}+(1\cdot(x+3)^{-1}+x\cdot(-1)(x+3)^{-2}\cdot 1)$
$=10(-\displaystyle \frac{1}{x^{2}}+\frac{1}{x+3}-\frac{x}{(x+3)^{2}})$
$=10(-\displaystyle \frac{1}{x^{2}}+\frac{x+3-x}{(x+3)^{2}})$
$=10(-\displaystyle \frac{1}{x^{2}}+\frac{3}{(x+3)^{2}})$
$10(-\displaystyle \frac{1}{x^{2}}+\frac{3}{(x+3)^{2}})=0$
$\displaystyle \frac{3}{x^{2}+6x+9}=\frac{1}{x^{2}}$
$3x^{2}=x^{2}+6x+9$
$2x^{2}-6x-9=0$
$x=\displaystyle \frac{6\pm\sqrt{108}}{4}$
$=\displaystyle \frac{6\pm 6\sqrt{3}}{4}$
$=\displaystyle \frac{3\pm 3\sqrt{3}}{2}$
We take the value in the interval $(3, 6)$,
$c=\displaystyle \frac{3+3\sqrt{3}}{2}\approx 4.098$
The order size is in hundreds, so this is amounts to
$\approx 410$ components
