Answer
a) $\frac{dC}{dx} = -38.1250$
b) $\frac{dC}{dx} = -10.3704$
c) $\frac{dC}{dx} = -3.8000$
These rates imply that the ordering and transportation cost decreases as the order size increases.
Work Step by Step
Rewrite the function $C$ by using the common denominator:
$C = 100(\frac{200}{x^{2}} + \frac{x}{x + 30})$
$= 100(\frac{200(x + 30) + x(x^{2})}{x^{2}(x + 30)})$
$= 100(\frac{200x + 6000 + x^{3}}{x^{3} + 30x^{2}})$
$= \frac{20000x + 600000 + 100x^{3}}{x^{3} + 30x^{2}}$
$= \frac{100x^{3} + 20000x + 600000}{x^{3} + 30x^{2}}$
Find the derivative of the function $C$ using the Quotient Rule:
$\frac{dC}{dx} = \frac{(300x^{2} + 20000)(x^{3} + 30x^{2}) - (100x^{3} + 20000x + 600000)(3x^{2} + 60x)}{(x^{3} + 30x^{2})^{2}}$
For part a), plug in $x = 10$ into $\frac{dC}{dx}$:
$\frac{dC}{dx} |_{x = 10} = \frac{(300(10)^{2} + 20000)((10)^{3} + 30(10)^{2}) - (100(10)^{3} + 20000(10) + 600000)(3(10)^{2} + 600))}{((10)^{3} + 30(10)^{2})^{2}}$
$= -38.1250$
For part b), plug in $x = 15$ into $\frac{dC}{dx}$:
$\frac{dC}{dx} |_{x = 15} = \frac{(300(15)^{2} + 20000)((15)^{3} + 30(15)^{2}) - (100(15)^{3} + 20000(15) + 600000)(3(15)^{2} + 900))}{((15)^{3} + 30(15)^{2})^{2}}$
$= -10.3704$
For part c), plug in $x = 20$ into $\frac{dC}{dx}$:
$\frac{dC}{dx} |_{x = 20} = \frac{(300(20)^{2} + 20000)((20)^{3} + 30(20)^{2}) - (100(20)^{3} + 20000(20) + 600000)(3(20)^{2} + 1200))}{((20)^{3} + 30(20)^{2})^{2}}$
$= -3.8000$
Since $C$ decreases as $x$ increases (because $\frac{dC}{dx} \lt 0$), these rates imply that the ordering and transportation cost decreases as the order size increases.
