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Neko [114]
3 years ago
11

Minimum Average Cost

Mathematics
1 answer:
Dahasolnce [82]3 years ago
6 0

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

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