A factory's output is Q=600K^1/2L^1/3 units, (read 600 x K to the 1/2 power x L to the 1/3 power). K denotes capital investment and L denotes the size of the labor force. Estimate the percentage increase in output that will result in a 2% increase in the size of the labor force if capital investment is not changed.
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
It looks like you have made a mistake in your calculations. Let's go through the steps again.
First, we need to find the derivative of the output function with respect to the labor force, L.
Q = 600K^1/2L^1/3
Taking the derivative with respect to L, we get:
dQ/dL = 200K^1/2 (1/3)L^-2/3
Next, we need to calculate the percentage change in output that will result from a 2% increase in the size of the labor force, assuming capital investment remains unchanged.
Percentage change = (dQ/dL) / Q * 100
Substituting the values we have:
Percentage change = (200K^1/2 (1/3)L^-2/3) / (600K^1/2L^1/3) * 100
Now, simplify the expression:
Percentage change = (1/3K^1/2L^-2/3) * 100
Given that the increase in the size of the labor force, dL/L, is 0.02 (2%), we substitute this value into the expression:
Percentage change = (1/3 * (0.02)) * 100
Percentage change = 1.33%
Therefore, a 2% increase in the size of the labor force, without changing the capital investment, will result in approximately a 1.33% increase in output.