<p>I obtained the definition of the principle of variable proportion from this <a href="http://trcollege.edu.in/study-material/24-economics/41-law-of-variable-proportions.html" rel="nofollow noreferrer">site</a></p>
<p>It states:</p>
<blockquote>
<p>... means that upto the use of a certain amount of variable factor, marginal product of the factor may increase and after a certain stage it starts diminishing.</p>
</blockquote>
<p>The following is a technical derivation. See the picture at the end for an illustration of the various results.</p>
<p>Denote by <span class="math-container">$X$</span> the variable input and by  <span class="math-container">$Q = {\cal Q}(X)$</span> the output produced by the amount <span class="math-container">$X$</span>. The marginal product can be expressed using the first derivative <span class="math-container">$MP(X) = {\cal Q}'(X)$</span>. The increase or decrease of the marginal product then says something about the second order derivative <span class="math-container">$MP'(X) = {\cal Q}''(X)$</span>.</p>
<p>The principle of variable proportions then states that <span class="math-container">$MP'(X)$</span> is first positive and then negative.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>period</th>
<th>change</th>
<th>sign of <span class="math-container">$MP(X) = {\cal Q}'(X)$</span></th>
<th>sign of <span class="math-container">$MP'(X) = {\cal Q}''(X)$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>marginal product increases</td>
<td><span class="math-container">$&gt;0$</span></td>
<td><span class="math-container">$&gt; 0$</span></td>
</tr>
<tr>
<td>2</td>
<td>marginal product decreases</td>
<td><span class="math-container">$&gt; 0$</span></td>
<td><span class="math-container">$&lt; 0$</span></td>
</tr>
</tbody>
</table>
</div>
<p>Given that <span class="math-container">${\cal Q}'(X) &gt; 0$</span> we can invert <span class="math-container">${\cal Q}(X)$</span> to obtain  <span class="math-container">$X = {\cal X}(Q)$</span> which gives the amount of variable input that produces <span class="math-container">$Q$</span> Then as the following identity holds for all <span class="math-container">$X$</span>:
<span class="math-container">$$
{\cal X}(Q(X)) = X,
$$</span>
We can take the derivative on both sides and obtain:
<span class="math-container">$$
{\cal X}'({\cal Q}(X))\,\, {\cal Q}'(X) = 1 \iff {\cal X}'(Q) = \frac{1}{{\cal Q}'(X)}.
$$</span>
and taking derivatives once more gives:
<span class="math-container">$$
{\cal X}''({\cal Q}(X)) \,\,({\cal Q}'(X))^2 + {\cal X}'({\cal Q}(X))\,\,{\cal Q}''(X) = 0 \iff {\cal X}''(Q) = -\frac{{\cal X}'({\cal Q}(X))\,\, {\cal Q}''(X)}{({\cal Q}'(X))^2}.
$$</span>
This allows us to complete the table:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>period</th>
<th>change in <span class="math-container">$Q$</span></th>
<th>sign of <span class="math-container">${\cal Q}'(X)$</span></th>
<th>sign of <span class="math-container">${\cal Q}''(X)$</span></th>
<th>sign of <span class="math-container">${\cal X}'(Q)$</span></th>
<th>sign of <span class="math-container">${\cal X}''(Q)$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>marginal product increases</td>
<td><span class="math-container">$&gt;0$</span></td>
<td><span class="math-container">$&gt; 0$</span></td>
<td><span class="math-container">$&gt; 0$</span></td>
<td><span class="math-container">$&lt;0$</span></td>
</tr>
<tr>
<td>2</td>
<td>marginal product decreases</td>
<td><span class="math-container">$&gt; 0$</span></td>
<td><span class="math-container">$&lt; 0$</span></td>
<td><span class="math-container">$&gt; 0$</span></td>
<td><span class="math-container">$&gt;0$</span></td>
</tr>
</tbody>
</table>
</div>
<p>Now, let's have a look at the variable cost. If <span class="math-container">$w$</span> is the unit price of an input <span class="math-container">$X$</span>,  then total variable costs is:
<span class="math-container">$$
VC(Q) = w {\cal X}(Q).
$$</span>
Marginal costs is the derivative of the variable costs:
<span class="math-container">$$
MC(Q) = VC'(Q) = w {\cal X}'(Q)
$$</span>
<strong>Th</strong> Marginal costs are first decreasing and then increasing.</p>
<blockquote>
<p><em>proof</em>: follows from the previous table as <span class="math-container">${\cal X}''(Q)$</span> is first negative and then positive.</p>
</blockquote>
<p>Average variable cost is variable costs divided by output:
<span class="math-container">$$
AVC(Q) = \frac{w {\cal X}(Q)}{Q}.
$$</span></p>
<p>Now let's have a look at how the  <span class="math-container">$AVC(Q)$</span> varies with <span class="math-container">$Q$</span> (which can be expressed using the derivative <span class="math-container">$AVC'(Q)$</span>). We have:
<span class="math-container">$$
AVC'(Q) = \frac{(w {\cal X}'(Q))Q  - (w{\cal X}(Q))}{Q
^2} = \frac{MC(X) - AVC(Q)}{Q},\\
$$</span>
<strong>Th</strong> <span class="math-container">$AVC(Q)$</span> is decreasing if  <span class="math-container">$MC(Q) &lt; AVC(Q)$</span> and increasing if <span class="math-container">$MC(Q) &gt; AVC(Q)$</span>. If <span class="math-container">$AVC(Q)$</span> is at a minimum (or maximum) then <span class="math-container">$MC(Q) = AVC(Q)$</span>.</p>
<blockquote>
<p><em>proof:</em> follows from the equation above as the sign of <span class="math-container">$AVC'$</span> depends on the sign of <span class="math-container">$MC - AVC$</span>.</p>
</blockquote>
<p>The following shows that for <span class="math-container">$Q = 0$</span>, average and marginal costs are identical.</p>
<p><strong>Th</strong> If <span class="math-container">$MC(0)$</span> exists, then <span class="math-container">$MC(0) = AVC(0)$</span>.</p>
<blockquote>
<p><em>proof</em>: We have that <span class="math-container">$AVC(Q) = w {\cal X}(Q)/Q$</span>. Using l'Hospital, when taking the limit of <span class="math-container">$Q \to 0$</span>, gives  <span class="math-container">$AVC(0) = w {\cal X}'(0) = MC(0)$</span>.</p>
</blockquote>
<p>The following shows that initiall, for low values of <span class="math-container">$Q$</span>, the <span class="math-container">$AVC$</span> will be decreasing.</p>
<p><strong>Th</strong> for <span class="math-container">$Q$</span> small enough, <span class="math-container">$MC &lt; AVC$</span>, and therefore <span class="math-container">$AVC$</span> is decreasing.</p>
<blockquote>
<p><em>proof</em>, Assume not, then <span class="math-container">$MC &gt; AVC$</span> in some neighbourhood <span class="math-container">$]0, Q[$</span> which means that <span class="math-container">$AVC$</span> is increasing. Also, we know that <span class="math-container">$MC$</span> is initially decreasing. As <span class="math-container">$MC(0) = AVC(0)$</span>, this means that <span class="math-container">$AVC &gt; MC$</span> in <span class="math-container">$]0,Q[$</span>, a contradiction.</p>
</blockquote>
<p>The following shows that the <span class="math-container">$MC$</span>-curves cuts the <span class="math-container">$AVC$</span>-curve from below.</p>
<p><strong>Th</strong> let <span class="math-container">$Q^\ast$</span> be the first point where <span class="math-container">$MC(Q^\ast) = AVC(Q^\ast)$</span>, then <span class="math-container">$MC'(Q^\ast) &gt; 0$</span> , so the <span class="math-container">$MC$</span>-curve cuts the <span class="math-container">$AVC$</span>-curve from below.</p>
<blockquote>
<p><em>proof:</em> From the previous result, we know that slightly before <span class="math-container">$Q^\ast$</span>, <span class="math-container">$MC &lt; AVC$</span>  and at <span class="math-container">$Q^\ast$</span>, <span class="math-container">$MC = AVQ$</span> so <span class="math-container">$(MC - AVQ)' &gt; 0$</span> at <span class="math-container">$Q^\ast$</span>.  Working out this derivative, shows that <span class="math-container">$MC' - \frac{MC - AVC}{Q} &gt; 0$</span>. As <span class="math-container">$MC = AVC$</span> at <span class="math-container">$Q^\ast$</span>, we have <span class="math-container">$MC'(Q^\ast) &gt; 0$</span>.</p>
</blockquote>
<p>The following shows that the <span class="math-container">$MC$</span>-curve cuts the <span class="math-container">$AVC$</span>-curve at most once.</p>
<p><strong>Th</strong> let <span class="math-container">$Q^\ast$</span> be the first point where <span class="math-container">$MC(Q^\ast) = AVC(Q^\ast)$</span> and let <span class="math-container">$Q &gt; Q^\ast$</span>  then <span class="math-container">$MC(Q) &gt; AVC(Q)$</span> and therefore <span class="math-container">$AVC$</span> is increasing. As such, there is at most one point where the <span class="math-container">$MC$</span> is equal to the <span class="math-container">$AVC$</span>.</p>
<blockquote>
<p><em>proof</em>: As at <span class="math-container">$Q^\ast$</span> , <span class="math-container">$MC$</span> is in its increasing part, we have (from the table above) that <span class="math-container">$MC(Q) &gt; 0$</span> for all <span class="math-container">$Q \ge Q^\ast$</span>,  Also, as  <span class="math-container">$AVC(Q^\ast) = MC(Q^\ast)$</span> and <span class="math-container">$MC'(Q^\ast) &gt; 0$</span> and <span class="math-container">$AVC'(Q^\ast) = 0$</span>, we have for <span class="math-container">$Q &gt; Q^\ast$</span> close enough to <span class="math-container">$Q^\ast$</span> that <span class="math-container">$MC(Q) &gt; AVC(Q)$</span>. So <span class="math-container">$AVC(Q)$</span> is increasing for all <span class="math-container">$Q&gt; Q^\ast$</span> close enought to <span class="math-container">$Q$</span>.</p>
<p>Next, for such <span class="math-container">$Q$</span>,  <span class="math-container">$(MC(Q) - AVC(Q))' = MC'(Q) + (MC(Q) - AVC(Q))/Q &gt; 0$</span>, so the distance between <span class="math-container">$MC(Q)$</span> and <span class="math-container">$AVC(Q)$</span> is increasing. this means that the distance between <span class="math-container">$MC(Q)$</span> and <span class="math-container">$AVC(Q)$</span> will grow bigger and bigger. As such, the two curves will never again be equal so <span class="math-container">$MC(Q) &gt; AVC(Q)$</span> for all <span class="math-container">$Q &gt; Q^\ast$</span>.</p>
</blockquote>
<p>The final situation can be seen in the picture below. The <span class="math-container">$MC$</span> curve is first decreasing and then increasing. At <span class="math-container">$Q = 0$</span> the two are the same. First the <span class="math-container">$MC$</span> curve obtains its minimum. Then in its increasing part, it cuts the <span class="math-container">$AVC$</span> curve at the point where the <span class="math-container">$AVC$</span> curve is at its minimum. After this point the distance between the <span class="math-container">$MC$</span> and <span class="math-container">$AVC$</span> curve grows bigger and bigger, so the two curves never intersect again, and both courves keep increasing.</p>
<p><a href="https://i.sstatic.net/yI5h2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yI5h2.png" alt="Cost curves" /></a></p>

CC BY-SA 4.0
2021-05-01T13:57:06.530Z
2021-05-01T13:57:06.530Z
tdm
<p>How is AVC(Average Variable Cost) ‘U’ shaped because of the principle of variable Proportions?</p>
<p>How are average variable cost(AVC) and marginal product(MP) related?</p>

<microeconomics>
How is AVC(Average Variable Cost) ‘U’ shaped because of the principle of variable Proportions?
kob003
# How is AVC(Average Variable Cost) ‘U’ shaped because of the principle of variable Proportions?

(Asked by: username_0 on 2021-05-01)

How is AVC(Average Variable Cost) ‘U’ shaped because of the principle of variable Proportions?

How are average variable cost(AVC) and marginal product(MP) related?

**Tags:** microeconomics


---

## 1 answers

### Answer by username_1 on 2021-05-01. Score: 4

I obtained the definition of the principle of variable proportion from this [site](http://trcollege.edu.in/study-material/24-economics/41-law-of-variable-proportions.html)

It states:

> ... means that upto the use of a certain amount of variable factor, marginal product of the factor may increase and after a certain stage it starts diminishing.

The following is a technical derivation. See the picture at the end for an illustration of the various results.

Denote by $X$ the variable input and by $Q = {\cal Q}(X)$ the output produced by the amount $X$. The marginal product can be expressed using the first derivative $MP(X) = {\cal Q}'(X)$. The increase or decrease of the marginal product then says something about the second order derivative $MP'(X) = {\cal Q}''(X)$.

The principle of variable proportions then states that $MP'(X)$ is first positive and then negative.

| period | change | sign of $MP(X) = {\cal Q}'(X)$ | sign of $MP'(X) = {\cal Q}''(X)$ |
| --- | --- | --- | --- |
| 1 | marginal product increases | $>0$ | $> 0$ |
| 2 | marginal product decreases | $> 0$ | $< 0$ |

Given that ${\cal Q}'(X) > 0$ we can invert ${\cal Q}(X)$ to obtain $X = {\cal X}(Q)$ which gives the amount of variable input that produces $Q$ Then as the following identity holds for all $X$:
$$
{\cal X}(Q(X)) = X,
$$
We can take the derivative on both sides and obtain:
$$
{\cal X}'({\cal Q}(X))\,\, {\cal Q}'(X) = 1 \iff {\cal X}'(Q) = \frac{1}{{\cal Q}'(X)}.
$$
and taking derivatives once more gives:
$$
{\cal X}''({\cal Q}(X)) \,\,({\cal Q}'(X))^2 + {\cal X}'({\cal Q}(X))\,\,{\cal Q}''(X) = 0 \iff {\cal X}''(Q) = -\frac{{\cal X}'({\cal Q}(X))\,\, {\cal Q}''(X)}{({\cal Q}'(X))^2}.
$$
This allows us to complete the table:

| period | change in $Q$ | sign of ${\cal Q}'(X)$ | sign of ${\cal Q}''(X)$ | sign of ${\cal X}'(Q)$ | sign of ${\cal X}''(Q)$ |
| --- | --- | --- | --- | --- | --- |
| 1 | marginal product increases | $>0$ | $> 0$ | $> 0$ | $<0$ |
| 2 | marginal product decreases | $> 0$ | $< 0$ | $> 0$ | $>0$ |

Now, let's have a look at the variable cost. If $w$ is the unit price of an input $X$, then total variable costs is:
$$
VC(Q) = w {\cal X}(Q).
$$
Marginal costs is the derivative of the variable costs:
$$
MC(Q) = VC'(Q) = w {\cal X}'(Q)
$$
**Th** Marginal costs are first decreasing and then increasing.

> *proof*: follows from the previous table as ${\cal X}''(Q)$ is first negative and then positive.

Average variable cost is variable costs divided by output:
$$
AVC(Q) = \frac{w {\cal X}(Q)}{Q}.
$$

Now let's have a look at how the $AVC(Q)$ varies with $Q$ (which can be expressed using the derivative $AVC'(Q)$). We have:
$$
AVC'(Q) = \frac{(w {\cal X}'(Q))Q - (w{\cal X}(Q))}{Q
^2} = \frac{MC(X) - AVC(Q)}{Q},\\
$$
**Th** $AVC(Q)$ is decreasing if $MC(Q) < AVC(Q)$ and increasing if $MC(Q) > AVC(Q)$. If $AVC(Q)$ is at a minimum (or maximum) then $MC(Q) = AVC(Q)$.

> *proof:* follows from the equation above as the sign of $AVC'$ depends on the sign of $MC - AVC$.

The following shows that for $Q = 0$, average and marginal costs are identical.

**Th** If $MC(0)$ exists, then $MC(0) = AVC(0)$.

> *proof*: We have that $AVC(Q) = w {\cal X}(Q)/Q$. Using l'Hospital, when taking the limit of $Q \to 0$, gives $AVC(0) = w {\cal X}'(0) = MC(0)$.

The following shows that initiall, for low values of $Q$, the $AVC$ will be decreasing.

**Th** for $Q$ small enough, $MC < AVC$, and therefore $AVC$ is decreasing.

> *proof*, Assume not, then $MC > AVC$ in some neighbourhood $]0, Q[$ which means that $AVC$ is increasing. Also, we know that $MC$ is initially decreasing. As $MC(0) = AVC(0)$, this means that $AVC > MC$ in $]0,Q[$, a contradiction.

The following shows that the $MC$-curves cuts the $AVC$-curve from below.

**Th** let $Q^\ast$ be the first point where $MC(Q^\ast) = AVC(Q^\ast)$, then $MC'(Q^\ast) > 0$ , so the $MC$-curve cuts the $AVC$-curve from below.

> *proof:* From the previous result, we know that slightly before $Q^\ast$, $MC < AVC$ and at $Q^\ast$, $MC = AVQ$ so $(MC - AVQ)' > 0$ at $Q^\ast$. Working out this derivative, shows that $MC' - \frac{MC - AVC}{Q} > 0$. As $MC = AVC$ at $Q^\ast$, we have $MC'(Q^\ast) > 0$.

The following shows that the $MC$-curve cuts the $AVC$-curve at most once.

**Th** let $Q^\ast$ be the first point where $MC(Q^\ast) = AVC(Q^\ast)$ and let $Q > Q^\ast$ then $MC(Q) > AVC(Q)$ and therefore $AVC$ is increasing. As such, there is at most one point where the $MC$ is equal to the $AVC$.

> *proof*: As at $Q^\ast$ , $MC$ is in its increasing part, we have (from the table above) that $MC(Q) > 0$ for all $Q \ge Q^\ast$, Also, as $AVC(Q^\ast) = MC(Q^\ast)$ and $MC'(Q^\ast) > 0$ and $AVC'(Q^\ast) = 0$, we have for $Q > Q^\ast$ close enough to $Q^\ast$ that $MC(Q) > AVC(Q)$. So $AVC(Q)$ is increasing for all $Q> Q^\ast$ close enought to $Q$.
>
> Next, for such $Q$, $(MC(Q) - AVC(Q))' = MC'(Q) + (MC(Q) - AVC(Q))/Q > 0$, so the distance between $MC(Q)$ and $AVC(Q)$ is increasing. this means that the distance between $MC(Q)$ and $AVC(Q)$ will grow bigger and bigger. As such, the two curves will never again be equal so $MC(Q) > AVC(Q)$ for all $Q > Q^\ast$.

The final situation can be seen in the picture below. The $MC$ curve is first decreasing and then increasing. At $Q = 0$ the two are the same. First the $MC$ curve obtains its minimum. Then in its increasing part, it cuts the $AVC$ curve at the point where the $AVC$ curve is at its minimum. After this point the distance between the $MC$ and $AVC$ curve grows bigger and bigger, so the two curves never intersect again, and both courves keep increasing.

[![Cost curves](https://i.sstatic.net/yI5h2.png)](https://i.sstatic.net/yI5h2.png)


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