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Math Help - Partial derivatives- production function

  1. #1
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    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
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    Quote Originally Posted by kelaroo View Post
    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
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    Quote Originally Posted by kelaroo View Post
    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
    \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,
    30 x^{-1/2}y^{2/3} = 40 x^{1/2} y^{-1/3}
    Thus,
    30 y = 40 x

    Thus, you got it right.
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