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April 18, 2014

April 18, 2014

Posted by **becky** on Sunday, August 13, 2006 at 10:31pm.

Q = 600K^1/2L^1/3

Q' = 300K^-1/2L-2/3 (read 300 x K to the negative 1/2 power x L to the negative 2/3 power.)

Percentage Change equation

100[Q'(.02)/Q]

100[300K^-1/2L^2/3(0.02)/600K^1/2L^1/3]=

100[6K^-1/2L^-2/3]/600K^1/2L^1/3=

600K^-1/2L^-2/3/600K^1/2L^1/3

I'm not sure if this is correct so far and where do I go from here?

Your derivative should be with respect to labor force L.

Q = 600(K^1/2)*(L^1/3)

dQ/dL=600(K^1/2)(1/3)(L^-2/3)

200(K^1/2) (L^-2/3)

check that.

Percent change..

dQ=100[Q'(.02)/Q]

100[200*K^1/2)(L^-2/3)dL/600(K^1/2)*(L^1/3)

]

=1/3*1/L*dL]

check that.

And from there..

percent change=1/3*1/L*dL]

= 1/3 (dL/L) where you are given

dL/L=.02

comments? check it all, it is hard to type an error when in ASCII

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