Question

Minimum Average Cost

Answer 1

Answer:

a)$\overline{C(x)} = 500x^{-1} + 300 - 300\dfrac{\ln({x})}{x}$

b)$\overline{C(e^{\frac{8}{3}})} = 279.1549$

Step-by-step explanation:

Given the cost function as C(x):

$C(x) = 500 + 300x - 300\ln{x} \quad\quad,x\geq1$

a) Find the average cost function, $(\overline{C(x)})$

if C is the cost of selling x units, The Average can be denoted by:

$\overline{C(x)} = \dfrac{\text{total cost of selling x units}}{\text{x units}}$

$\overline{C(x)} = \dfrac{C(x)}{x}$

$\overline{C(x)} = \dfrac{500 + 300x - 300\ln{(x)}}{x}$

$\overline{C(x)} = 500x^{-1} + 300 - 300\dfrac{\ln({x})}{x}$

this is the average cost function

b) The minimum average cost:

To find the minimum average cost, we'll have to differentiate the average cost function $(\overline{C(x)})$. and equate it to zero. (like finding the stationary point of any function)

$\dfrac{d}{dx}(\overline{C(x)}) = \dfrac{d}{dx}\left(500x^{-1} + 300 - 300\dfrac{\ln({x})}{x}\right)$

$\overline{C'(x)} = -500x^{-2} + 0 - 300\dfrac{x\frac{1}{x} - \ln{(x)}}{x^2}}$

now just simplify:

$\overline{C'(x)} = -\dfrac{500}{x^2}-300\dfrac{1 - \ln{(x)}}{x^2}}$

$\overline{C'(x)} = -\dfrac{1}{x^2}(500+300(1 - \ln{(x)}))$

we've found the derivative of C(x), now to find the minimum we'll equate this derivative to zero. $\overline{C'(x)} = 0$

$0 = -\dfrac{1}{x^2}(500+300(1 - \ln{(x)}))$

and now solve for x

$0 = 500+300(1 - \ln{(x)})$

$-500 = 300(1 - \ln{(x)})$

$1 + \dfrac{5}{3}=\ln{(x)}$

$\ln{(x)}=\dfrac{8}{3}$

$x=e^{\frac{8}{3}}\approx 14.392$

at this value of x the average cost is minimum.

$\overline{C(x)} = 500x^{-1} + 300 - 300\dfrac{\ln({x})}{x}$

$\overline{C(e^{\frac{8}{3}})} = 500{e^{-\frac{8}{3}}} + 300 - 300\dfrac{\ln({e^{\frac{8}{3}}})}{e^{\frac{8}{3}}}$

$\overline{C(e^{\frac{8}{3}})} = 500{e^{-\frac{8}{3}}} + 300 - 300\dfrac{8}{3e^{\frac{8}{3}}}}$

$\overline{C(e^{\frac{8}{3}})} = 34.7417 + 300 -55.5868$

$\overline{C(e^{\frac{8}{3}})} = 279.1549$

This is the minimum average cost!