# Thread: Partial derivatives- production function

1. ## Partial derivatives- production function

The question is: f(x,y) is a production function and f(x,y) = 60x^1/2y^2/3

I have to find the pairs where marginal productivity of capital and labor are equal. I know these are the partial derivatives, so I am setting them equal to each other.

For the partials, I got:

30(y^2/3)/(x^1/2) and 40(x^1/2)/(y^1/3)

If I set them equal I can get y= 4/3 x but is this the answer??

Thanks

2. Originally Posted by kelaroo
The question is: f(x,y) is a production function and f(x,y) = 60x^1/2y^2/3

I have to find the pairs where marginal productivity of capital and labor are equal. I know these are the partial derivatives, so I am setting them equal to each other.

For the partials, I got:

30(y^2/3)/(x^1/2) and 40(x^1/2)/(y^1/3)

If I set them equal I can get y= 4/3 x but is this the answer??

Thanks
please use parentheses in your original function. i do not know what you are saying. it can be interpreted many ways

3. Originally Posted by kelaroo
The question is: f(x,y) is a production function and f(x,y) = 60x^1/2y^2/3

I have to find the pairs where marginal productivity of capital and labor are equal. I know these are the partial derivatives, so I am setting them equal to each other.

For the partials, I got:

30(y^2/3)/(x^1/2) and 40(x^1/2)/(y^1/3)

If I set them equal I can get y= 4/3 x but is this the answer??

Thanks
$\displaystyle \frac{\partial f}{\partial x} = 30 x^{-1/2}y^{2/3} \text{ and }\frac{\partial f}{\partial y} = 40x^{1/2} y^{-1/3}$

Make them equal,
$\displaystyle 30 x^{-1/2}y^{2/3} = 40 x^{1/2} y^{-1/3}$
Thus,
$\displaystyle 30 y = 40 x$

Thus, you got it right.

4. Thanks