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We have the rule 'If the marginal is less than the average, then the average declines'.

So if our production function for an input x and output y is concave, does that mean that MP must be diminishing given AP is diminishing?

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    $\begingroup$ If the production function $y=f(x)$ is concave, i.e. $\frac{d^2y}{dx^2}=f''(x) \lt 0$, I would have thought that in itself would be equivalent to saying marginal product $MP=\frac{dy}{dx}=f'(x)$ was decreasing $\endgroup$ – Henry Jan 5 '17 at 16:39
  • $\begingroup$ @Henry please post answers as answers. Read this on meta. $\endgroup$ – Giskard Jan 5 '17 at 16:59
  • $\begingroup$ @denesp: since I ignored the first half of the question and challenged the last four words of the second half, I would regard what I said as not being an answer to the question asked. You are always free to turn somebody else's comment into an answer yourself $\endgroup$ – Henry Jan 5 '17 at 17:08
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Suppose the total product is given by $f(x)$. Thus, the average product is $$AP(x)=\frac{f(x)}{x}.$$

Differentiating: $$AP'(x)=\frac{x f'(x)-f(x)}{x^2},$$

which has the same sign as $x f'(x)-f(x)$. This is definitely negative if $f''(x)<0$ (i.e., if $f$ is concave).

But it can also be negative, even if $f$ is locally convex. Take a look at this figure:

enter image description here

At the point marked with a blue disc: the slope of the red line is $f'(x)$. The slope of the black line is $f(x)/x$. We can see geometrically that $f(x)/x>f'(x)$, or (after rearranging) that $f'(x)x<f(x)$.

Thus, at the point indicated by the blue disc we know that $AP'(x)<0$. But at this point we can also see in the figure that $f''(x)>0,$ implying that the marginal product is increasing.

Since we have a counter-example, we conclude that a decreasing marginal product is not a necessary condition for a decreasing average product.

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Taking @Henry's comment because he is too lazy to type it in as answer:

As long as $f$ is twice differentiable (which is likely if $MP(x)$ exists) just the fact that $f(x)$ is concave implies $\dfrac{d^2f(x)}{dx^2} < 0$. This is also the first differentiate of $MP(x)$, thus it is diminishing.

So all you need is concavity. The 'rule' you refer to has different implications.

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  • $\begingroup$ I think I've got an answer- if the production function isn't concave then it can be horizontal at the end. So MP is 0 whilst AP is still diminishing. If the production function is concave then you are perfectly right. $\endgroup$ – Dtomas Jan 5 '17 at 17:42
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    $\begingroup$ @Dtomas If that is the answer, I do not know the question. $\endgroup$ – Giskard Jan 5 '17 at 17:46
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    $\begingroup$ @Dtomas Also it seems we have both mixed up concavity and strict concavity. $\endgroup$ – Giskard Jan 5 '17 at 17:47
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I am a little confused by the phrasing of the question, but I'll take it to mean this:

Does diminishing average product imply that marginal product is also decreasing?

It is easy to show, as the OP acknowledges, that (under differentiability assumptions) that the average product of an input is decreasing in that input if and only if the marginal product of the said input is below the average product.

It turns out that this is essentially all you can say about the implications of diminishing average product on marginal product, as Ubiquitous shows in a separate answer.

I want to offer an alternative, simpler (albeit somewhat extreme) counter-example.

Consider a production function $F$ whose marginal product for input $x$ is $0$:

$$ \text{MP}_x = \frac{\partial F}{\partial x} = 0 $$

It is easy to see that as long as $F > 0$, $\text{AP}_x$ is decreasing in $x$. Hence it is possible to have a production function that exhibits diminishing average product but a non-decreasing marginal product.

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