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So I was solving a question which says find the range in which MRTS is diminishing $f(k,l) = 600k^2l^2-k^3l^3$ is the production function I got the MRTS = $ -(1200kl^2-3k^2l^3) \over 1200k^2l-3k^3l^2$

Now, i'm getting $\frac{\partial f'(k,l)}{\partial k} = \frac{l}{k^2}$ then range is $l <0 $ which doesn't seem to make sense. Is my answer correct or is there some mistake??

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Fix the production function to a constant value $f_0$ so that

$$ f_0 = 600k^2l^2 - k^3 l^3 \tag{1} $$

This is a figure for different values of $f_0$ (isoquants)

enter image description here

Note that with increasing $k$ the rate of change of $l$ becomes smaller for a given $f$ and a fixed change in $k$. We can show this more formally by taking the derivative w.r.t $l$ at both sides

$$ 0 = 600\left(2kl^2 + 2k^2 l \frac{\partial l}{\partial k}\right) - \left(3k^2l^3 + 3k^3l^2\frac{\partial l}{\partial k}\right) \tag{2} $$

You can further simplify this to

$$ 0 = -3k l (-400 + k l)\left(l + k \frac{\partial l}{\partial k}\right) \tag{3} $$

The solutions are $k = 0$, $l = 0$, $kl = 400$ or

$$ \frac{\partial l}{\partial k} = -\frac{l}{k} \tag{4} $$

Which confirms that that increasing $k$ decreases the MRTS, in other words, the range in which the marginal diminishes is $k > 0$

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  • $\begingroup$ I know I got that, now to find the range of diminishing MRTS , if you differentiate mrts you get $\frac{l}{k^2}$ same thing what I got. But, what is the range of values such that mrts is diminishing? I'm getting $l<0$ which doesn't make sense $\endgroup$ – Sumukh Sai Nov 24 '18 at 12:59
  • $\begingroup$ @SumukhSai Sorry I misread the question. I updated my answer, hopefully it makes more sense now $\endgroup$ – caverac Nov 24 '18 at 15:20

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