# Is increasing Average Product(AP) always implying increasing Marginal Product(MP) in microeconomics?

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.

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.

• Why is MP(x) not increasing? Commented Jul 17 at 16:56
• 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
Commented Jul 17 at 16:58
• 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? Commented Jul 17 at 17:14
• 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
Commented Jul 17 at 17:49
• 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. Commented Jul 17 at 17:59