Question

Minimum Average Cost

Answer 1

Answer:

a)$\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}$

b)$\bar{C}(x)=5.81$

Step-by-step explanation:

Given that

C = 100 + 25 x - 120 ln x   ,x ≥ 1.

The average cost function given as

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

$\bar{C}(x)=\dfrac{100 + 25 x - 120 \ln x}{x}$

$\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx{x}$

Therefore

$\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}$

To find average minimum cost

$\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}$

$\dfrac{d\bar{C}(x)}{dx} = -\dfrac{100}{x^2} +0-120\times \dfrac{1-lnx}{x^2}$

$0 = -\dfrac{100}{x^2} +0- 120\times \dfrac{1-lnx}{x^2}$

100 + 120 (1-lnx) = 0

$lnx=\dfrac{220}{120}$

ln x =1.833

$x=e^{1.833}$

x=6.25

$\bar{C}(x)=\dfrac{100}{6.25}+25-120\dfrac{ln6.25}{6.25}$

$\bar{C}(x)=5.81$