Respuesta :
Answer:
a)[tex]\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}[/tex]
b)[tex]\bar{C}(x)=5.81[/tex]
Step-by-step explanation:
Given that
C = 100 + 25 x - 120 ln x ,x ≥ 1.
The average cost function given as
[tex]\bar{C}(x)=\dfrac{C(x)}{x}[/tex]
[tex]\bar{C}(x)=\dfrac{100 + 25 x - 120 \ln x}{x}[/tex]
[tex]\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx{x}[/tex]
Therefore
[tex]\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}[/tex]
To find average minimum cost
[tex]\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x} [/tex]
[tex]\dfrac{d\bar{C}(x)}{dx} = -\dfrac{100}{x^2} +0-120\times \dfrac{1-lnx}{x^2}[/tex]
[tex]0 = -\dfrac{100}{x^2} +0- 120\times \dfrac{1-lnx}{x^2}[/tex]
100 + 120 (1-lnx) = 0
[tex]lnx=\dfrac{220}{120}[/tex]
ln x =1.833
[tex]x=e^{1.833}[/tex]
x=6.25
[tex]\bar{C}(x)=\dfrac{100}{6.25}+25-120\dfrac{ln6.25}{6.25}[/tex]
[tex]\bar{C}(x)=5.81[/tex]
