<p>Consider the following production function:</p>
<p><span class="math-container">$f(x)=\begin{cases} x^2 &amp; x &lt; 1 \\ x^{\frac{3}{2}} &amp; x \geq 1\end{cases}$</span></p>
<p>In this case, the average product is
<span class="math-container">$\text{AP}(x)=\dfrac{f(x)}{x}=\begin{cases} x &amp; x &lt; 1 \\ x^{\frac{1}{2}} &amp; x \geq 1\end{cases}$</span></p>
<p>which is increasing. However, the marginal product is</p>
<p><span class="math-container">$\text{MP}(x)=\dfrac{df(x)}{dx}=\begin{cases} 2x &amp; x &lt; 1 \\ \frac{3}{2}x^{\frac{1}{2}} &amp; x &gt; 1\end{cases}$</span></p>
<p>which is clearly not increasing.</p>

CC BY-SA 4.0
2024-07-17T16:56:51.493Z
Why is MP(x) not increasing?
Jesper Hybel
CC BY-SA 4.0
2024-07-17T16:58:56.500Z
Because $\lim_{x\rightarrow 1^{-}}\text{MP}(x)=2> \frac{3}{2} = \lim_{x\rightarrow 1^{+}}\text{MP}(x)$. So, for values of $x$ close to $1$, but less than $1$, MP is close to 2, but for values of $x$ close to $1$, but greater than $1$, MP is close to 3/2.
Amit
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2024-07-17T17:14:54.073Z
Thanks for this counterexample. This demonstrates that the original statement doesn't hold for non-continuous MP. May I ask is there a known counterexample with a continuous MP?
Gaoge Zhang
CC BY-SA 4.0
2024-07-17T17:49:39.067Z
Here is an example with continuous MP: $\text{MP}(x)=\dfrac{df(x)}{dx}=\begin{cases} 2x & x < 5 \\ 11 - \frac{x}{5} &  5 \leq x \leq 6 \\  3.8 + x &  x \geq 6\end{cases}$;  $f(x)=\begin{cases} x^2 & x < 5 \\ -27.5+11x - \frac{x^2}{10} &  5 \leq x \leq 6 \\  -5.9+3.8x + \frac{x^2}{2} &  x \geq 6\end{cases}$
Amit
CC BY-SA 4.0
2024-07-17T17:59:31.803Z
Excellent! In this counter example, $AP(x)=f(x)/x=\begin{cases}x & x<5 \\ 11 .-\frac{27.5}{x}-0.1 x & 5 \leq x \leq 6 \\ 3.8-\frac{5.9}{x}+0.5 x & \text { True }\end{cases}$, which increases; $MP(x)$ decreases when $5 \leqslant x \leqslant 6$; and $MP=AP$ when $x=0$. $MP$ is also continuous. Thanks for your answer. You helped me a lot.
Gaoge Zhang
CC BY-SA 4.0
2024-07-17T18:01:07.733Z
You may also want to check [this similar question](https://economics.stackexchange.com/questions/14913/does-the-marginal-product-need-to-be-diminishing-to-have-a-diminishing-average-p?rq=1)
Gaoge Zhang
CC BY-SA 4.0
2024-07-17T16:51:20.517Z
2024-07-17T16:51:20.517Z
Amit
<p>I'm studying microeconomics and came across a statement that I'm not sure is correct:</p>
<p>&quot;If average product (<span class="math-container">$AP = F(X)/X$</span>) always increases from <span class="math-container">$X=0$</span>, then marginal product (<span class="math-container">$MP = F'(X)$</span>) also always increases from <span class="math-container">$X=0$</span>.&quot;</p>
<p>Where:</p>
<ul>
<li><span class="math-container">$F(X)$</span> is the production function</li>
<li><span class="math-container">$X$</span> is the input</li>
<li><span class="math-container">$AP$</span> is the average product</li>
<li><span class="math-container">$MP$</span> is the marginal product</li>
</ul>
<p>I understand that when AP is increasing, MP must be greater than AP (<span class="math-container">$MP &gt; AP$</span>). However, I'm not certain if this necessarily means that MP itself must always be increasing.</p>
<p>Is this statement correct? If so, how can it be proven? If not, is there a counterexample?</p>
<p>I'd appreciate any insights, proofs, or counterexamples that can help clarify this relationship between AP and MP.</p>

<microeconomics><production-function>
Is increasing Average Product(AP) always implying increasing Marginal Product(MP) in microeconomics?
Gaoge Zhang
# Is increasing Average Product(AP) always implying increasing Marginal Product(MP) in microeconomics?

(Asked by: username_0 on 2024-07-17)

I'm studying microeconomics and came across a statement that I'm not sure is correct:

"If average product ($AP = F(X)/X$) always increases from $X=0$, then marginal product ($MP = F'(X)$) also always increases from $X=0$."

Where:

* $F(X)$ is the production function
* $X$ is the input
* $AP$ is the average product
* $MP$ is the marginal product

I understand that when AP is increasing, MP must be greater than AP ($MP > AP$). However, I'm not certain if this necessarily means that MP itself must always be increasing.

Is this statement correct? If so, how can it be proven? If not, is there a counterexample?

I'd appreciate any insights, proofs, or counterexamples that can help clarify this relationship between AP and MP.

**Tags:** microeconomics, production-function


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## 1 answers

### Answer by username_1 on 2024-07-17. Score: 5 | Accepted answer

Consider the following production function:

$f(x)=\begin{cases} x^2 & x < 1 \\ x^{\frac{3}{2}} & x \geq 1\end{cases}$

In this case, the average product is
$\text{AP}(x)=\dfrac{f(x)}{x}=\begin{cases} x & x < 1 \\ x^{\frac{1}{2}} & x \geq 1\end{cases}$

which is increasing. However, the marginal product is

$\text{MP}(x)=\dfrac{df(x)}{dx}=\begin{cases} 2x & x < 1 \\ \frac{3}{2}x^{\frac{1}{2}} & x > 1\end{cases}$

which is clearly not increasing.

> **Comments:**
> **username_2**: Why is MP(x) not increasing?  
> **username_1**: Because $\lim_{x\rightarrow 1^{-}}\text{MP}(x)=2> \frac{3}{2} = \lim_{x\rightarrow 1^{+}}\text{MP}(x)$. So, for values of $x$ close to $1$, but less than $1$, MP is close to 2, but for values of $x$ close to $1$, but greater than $1$, MP is close to 3/2.  
> **username_0**: Thanks for this counterexample. This demonstrates that the original statement doesn't hold for non-continuous MP. May I ask is there a known counterexample with a continuous MP?  
> **username_1**: Here is an example with continuous MP: $\text{MP}(x)=\dfrac{df(x)}{dx}=\begin{cases} 2x & x < 5 \\ 11 - \frac{x}{5} &  5 \leq x \leq 6 \\  3.8 + x &  x \geq 6\end{cases}$;  $f(x)=\begin{cases} x^2 & x < 5 \\ -27.5+11x - \frac{x^2}{10} &  5 \leq x \leq 6 \\  -5.9+3.8x + \frac{x^2}{2} &  x \geq 6\end{cases}$  
> **username_0**: Excellent! In this counter example, $AP(x)=f(x)/x=\begin{cases}x & x<5 \\ 11 .-\frac{27.5}{x}-0.1 x & 5 \leq x \leq 6 \\ 3.8-\frac{5.9}{x}+0.5 x & \text { True }\end{cases}$, which increases; $MP(x)$ decreases when $5 \leqslant x \leqslant 6$; and $MP=AP$ when $x=0$. $MP$ is also continuous. Thanks for your answer. You helped me a lot.  
> **username_0**: You may also want to check [this similar question](https://economics.stackexchange.com/questions/14913/does-the-marginal-product-need-to-be-diminishing-to-have-a-diminishing-average-p?rq=1)  

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