<p>MRST can diminish in two ways: with increasing L or increasing K. For each we need the sign of the partial derivative with respect to the appropriate variable.</p>
<p>For MRST with increasing L we inspect its partial derivative with respect to L:</p>
<p><span class="math-container">$$\frac{\partial q}{\partial L} = \frac{\lambda (\lambda-1)L^{\lambda-2}}{\gamma 
 K^{\gamma-1}}$$</span></p>
<p>We can see that this derivative is negative if, and only if, <span class="math-container">$0 &lt; \lambda &lt; 1$</span>.</p>
<p>Likewise, the partial derivative of MRST with respect to K is negative if, and only if, <span class="math-container">$0 &lt; \gamma &lt; 1$</span>.</p>
<p>In addition, the second derivative of MRST with respect to L is positive for <span class="math-container">$0 &lt; \lambda &lt; 1$</span>.</p>
<p><span class="math-container">$$\frac{\partial^2 q}{\partial L^2} = \frac{\lambda (\lambda-1)(\lambda-2)L^{\lambda-3}}{\gamma 
 K^{\gamma-1}}$$</span></p>
<p>Likewise for <span class="math-container">$\frac{\partial^2 q}{\partial K^2}$</span> for <span class="math-container">$0 &lt; \gamma &lt; 1$</span>.</p>
<p>Together with the negative first derivatives, this implies that MRST is strictly convex.</p>

CC BY-SA 4.0
2024-10-22T15:54:56.393Z
Couple typos: It's MRTS (not MRST); and it's $\frac{\partial \color{red}K}{\partial L}$ as it measures the slope of the isoquant which lives in the $L$-$K$ plane (similarly for the second-order derivative).
Herr K.
CC BY-SA 4.0
2024-10-21T17:56:47.013Z
2024-10-21T17:56:47.013Z
Jerome Smith-Uldall
<p>For diminishing MRST, <span class="math-container">$\lambda &lt; \gamma &lt; 1$</span>. You can say that the function satisfies strong convexity and thus diminishing MRS holds.</p>

CC BY-SA 4.0
2024-09-21T17:06:21.263Z
2024-09-21T17:06:21.263Z
Mathemafrick
<p>For production function <span class="math-container">$q(L,K) = L^\lambda + K^\gamma$</span> The MRST is defined as
<span class="math-container">$\frac{\lambda L^{\lambda-1}}{\gamma K^{\gamma-1}}$</span>. Is it correct and sufficient to say that in order for MRTS to be diminishing: <span class="math-container">$\lambda$</span> and <span class="math-container">$\gamma$</span> must be between 0 and 1?
If not, what other conditions will this function exhibit diminishing MRTS? Does this function satisfy strong convexity if the parameters are between 0 and 1?</p>

<optimization><production-function>
How to determine if a production function in a functional form has diminishing marginal rate of technical substitution?
PoopyPoops
# How to determine if a production function in a functional form has diminishing marginal rate of technical substitution?

(Asked by: username_0 on 2024-09-21)

For production function $q(L,K) = L^\lambda + K^\gamma$ The MRST is defined as
$\frac{\lambda L^{\lambda-1}}{\gamma K^{\gamma-1}}$. Is it correct and sufficient to say that in order for MRTS to be diminishing: $\lambda$ and $\gamma$ must be between 0 and 1?
If not, what other conditions will this function exhibit diminishing MRTS? Does this function satisfy strong convexity if the parameters are between 0 and 1?

**Tags:** optimization, production-function

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## 2 answers

### Answer by username_1 on 2024-10-21. Score: 2

MRST can diminish in two ways: with increasing L or increasing K. For each we need the sign of the partial derivative with respect to the appropriate variable.

For MRST with increasing L we inspect its partial derivative with respect to L:

$$\frac{\partial q}{\partial L} = \frac{\lambda (\lambda-1)L^{\lambda-2}}{\gamma
K^{\gamma-1}}$$

We can see that this derivative is negative if, and only if, $0 < \lambda < 1$.

Likewise, the partial derivative of MRST with respect to K is negative if, and only if, $0 < \gamma < 1$.

In addition, the second derivative of MRST with respect to L is positive for $0 < \lambda < 1$.

$$\frac{\partial^2 q}{\partial L^2} = \frac{\lambda (\lambda-1)(\lambda-2)L^{\lambda-3}}{\gamma
K^{\gamma-1}}$$

Likewise for $\frac{\partial^2 q}{\partial K^2}$ for $0 < \gamma < 1$.

Together with the negative first derivatives, this implies that MRST is strictly convex.

> **Comments:**
> **username_2**: Couple typos: It's MRTS (not MRST); and it's $\frac{\partial \color{red}K}{\partial L}$ as it measures the slope of the isoquant which lives in the $L$-$K$ plane (similarly for the second-order derivative).  

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### Answer by username_3 on 2024-09-21. Score: 0

For diminishing MRST, $\lambda < \gamma < 1$. You can say that the function satisfies strong convexity and thus diminishing MRS holds.


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