Marginal rate of technical substitution for a production function

[claratanone]

Find the marginal rate of technical substitution for the following production function:

\(\displaystyle Q=10(0.2L^{-0.5} +0.8K^{-0.5})^{-2}\)

Here is my attempt so far:

\(\displaystyle \frac{\delta Q}{\delta L}=[10(-2)][0.2K^{-0.5}+0.8L^{-0.5})^{(-2-1)}*[0.8*(-0.5)]L^{(\frac{-1}{2}-1)}=[(-20)*(-0.4)](0.2K^{-0.5} +0.8L^{-0.5})^{-3} *L^{\frac{-3}{2}}=8(0.2K^{-0.5}+0.8L^{-0.5})^{-3} *L^{\frac{-3}{2}}\)

\(\displaystyle \frac{\delta Q}{\delta K}=[10(-2)](0.2K^{-0.5} + 0.8L^{-0.5})^{(-2-1)}*[0.2*(-0.5)]K^{(\frac{-1}{2}-1)}\)
\(\displaystyle =[(-20)*(-0.1)](0.2K^{-0.5} +0.8L^{-0.5})^{-3}* K^{\frac{-3}{2}}=2(0.2K^{-0.5}+0.8^{-0.5})^{-3}K^{\frac{-3}{2}}\)

Is this all correct so far?
What do I do next?
Would anyone be able to guide me through the rest?

 

[jvicens]

Yes, your calculations so far are correct. To find the marginal rate of technical substitution (MRTS), we need to take the ratio of the partial derivatives you have calculated:

MRTS = \frac{\frac{\delta Q}{\delta L}}{\frac{\delta Q}{\delta K}} = \frac{8(0.2K^{-0.5}+0.8L^{-0.5})^{-3} *L^{\frac{-3}{2}}}{2(0.2K^{-0.5}+0.8L^{-0.5})^{-3}* K^{\frac{-3}{2}}}

Simplifying this expression, we get:

MRTS = \frac{4L^{-3}}{K^{-3}} = 4\left(\frac{L}{K}\right)^{-3}

Therefore, the marginal rate of technical substitution is equal to 4 times the ratio of labor to capital raised to the power of -3. This means that as we increase the amount of labor used in production, the amount of capital needed decreases by 4 times the ratio of labor to capital. Similarly, as we increase the amount of capital used, the amount of labor needed decreases by 4 times the ratio of capital to labor.

I hope this helps guide you through the rest of the problem. Let me know if you have any other questions. Good luck!