We’re given a Cobb-Douglas Production function P of L comma K. We want to find in the

marginal productivity of labor and the marginal productivity

of capital functions which would be the partial

derivative with respect to L and the partial derivative

with respect to K. Before we do this though, let’s talk more about the Cobb-Douglas

Production function. In economics, the Cobb-Douglas

Production function is a form of the production

function often used to represent the relationship between physical capital and labor and the amount of output that can be produced by those inputs. So this is the form of the

Cobb-Douglas Production function where P is the total production, L is the labor input,

K is the capital input A is the total factor productivity. And notice alpha and beta, the exponents, are the output and elasticity of labor and the output and elasticity of capital. So going back to our example, we’ll now find the marginal

productivity of labor function by determining the partial

derivative with respect to L. So to do this, we’ll

differentiate with respect to L treating K as a constant. So we’d have 12 times, if we differentiate with respect to L, we

would multiply by 0.7, and then we’d have L raised

to the power of 0.7 minus one, that’s negative 0.3 and we’re

treating K as a constant. Now let’s go ahead and simplify this and write this only

using positive exponents. So 12 times 0.7 is 8.4. So we’d have 8.4. K has a positive exponent, so that would be in the numerator. So K to the 0.3. And because L has a negative exponent, we’ll move it down to the denominator which would change the

sign of the exponent. So we’d have L to the positive

0.3 in the denominator. So this is the marginal

productivity of labor function. And then we’ll find the partial

derivative with respect to K which would be the marginal productivity of capital function. So we differentiate with respect to K, now treating L as a constant. So we’d have 12 and we

multiply by the exponent on K which would be 0.3. L will stay this same, L to the 0.7. And then for the exponent

on K, we would subtract one, 0.3 minus one is negative 0.7. Let’s go ahead and simplify. 12 times 0.3 is 3.6. L has a positive exponent. The denominator would be K

raised to the power of 0.7. Now that we have our two functions, we’re asked to evaluate them when L equals 400 and K equals 1,000. After we do this, we’ll

explain what this means. So the partial with respect to L, notice how the coordinates are L comma K. So it’d be 400 comma 1,000. Again, where L equals

400 and K equals 1,000. So for the partial with respect to L we would have 8.4 times K or 1,000 raised to the power of

0.3 divided by L or 400 to the 0.3. We’ll evaluate this in just a moment. Let’s go ahead and set up

the partial with respect to K at the same point. So we’ll have 3.6 times L or 400 raised to the power of 0.7 divided by K or 1,000 raised to the power of 0.7. Now we’ll evaluate

these on the calculator. So first we have in the numerator 8.4 and then times 1,000. This is raised to the power of .3, right arrow, closed parenthesis,

that’s the numerator, divided by 400 raised to the power of .3, Enter. If we round to three decimal places, this would be approximately 11.058. This is the change in production with respect to labor at this point. And next we have, and now for the partial with respect to K, we

have in the numerator 3.6 times 400 raised to the power of .7, right arrow, closed parenthesis,

there’s the numerator, divided by 1,000 raised to the power of .7, Enter. To three decimal places we’d

have approximately, 1.896. Now let’s emphasize what

these two values represent. The value from the

partial with respect to L which we found here is additional output that results from employing

an additional unit of labor when L equals 400 and K equals 1,000. So we can say at this point

production is increasing at approximately 11 units per worker. And then for the partial

with respect to K. This represents the additional output resulting from the use of an

additional unit of capital, again when L equals

400 and K equals 1,000. So the units on this value would be units of production per let’s say dollar. I hope you found this helpful.

## Juni Tandra says:

Why isn't product rule being used here? Don't we treat K^0,3 as a function itself?

## mm22 sapphire says:

amazing

## Wisani Kenny says:

It was very helpful indeed. Thanks a lot.

## Aaron Barragan says:

thanks, i used this for my final project, stay smart!

## ZmayorTag says:

5Head of course