## Calculus 10th Edition

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.
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.