Question

Find the production level that minimizes the average cost. c(x) = 10e^0.02x

Answer 1

@amistre64 can u help?

Answer 2

the word average implies that we divide by c(x) by x

Answer 3

okay

Answer 4

so we would want the deriavtive of C(x)/x:\[\frac{xC'-x'C}{x^2}=0\]

Answer 5

okay so 10e^002x -10.202 / x^2 ?

Answer 6

something like that yes, but the x^2 is useless to us since \(\frac 0n=0\) so lets focus on the top
\[xC'-C=0\]
\[ .2x e^{0.02x}-10e^{0.02x}=0\]
since e^whatever is never 0, lets factor it out and solve\[.2x-10=0\]

Answer 7

its either that, or when the derivaitve is undefined ... namely at x=0, but that is not a good value to use in any case

Answer 8

ok

Answer 9

x = 50 i got after i solved

Answer 10

yes

Answer 11

okay