# Diminishing MRTS range

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??

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)

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 $$k > 0$$

• 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 – Sumukh Sai Nov 24 '18 at 12:59
• @SumukhSai Sorry I misread the question. I updated my answer, hopefully it makes more sense now – caverac Nov 24 '18 at 15:20