How is AVC(Average Variable Cost) ‘U’ shaped because of the principle of variable Proportions? 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?
I obtained the definition of the principle of variable proportion from this site
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
$&gt;0$
$&gt; 0$


2
marginal product decreases
$&gt; 0$
$&lt; 0$




Given that ${\cal Q}'(X) &gt; 0$ 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
$&gt;0$
$&gt; 0$
$&gt; 0$
$&lt;0$


2
marginal product decreases
$&gt; 0$
$&lt; 0$
$&gt; 0$
$&gt;0$




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  $MC(Q) &lt; AVC(Q)$ and increasing if $MC(Q) &gt; AVC(Q)$. 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, $MC &lt; AVC$, and therefore $AVC$ is decreasing.

proof, Assume not, then $MC &gt; AVC$ 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 $AVC &gt; MC$ 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 $MC'(Q^\ast) &gt; 0$ , so the $MC$-curve cuts the $AVC$-curve from below.

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

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 $Q &gt; Q^\ast$  then $MC(Q) &gt; AVC(Q)$ 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 $MC(Q) &gt; 0$ for all $Q \ge Q^\ast$,  Also, as  $AVC(Q^\ast) = MC(Q^\ast)$ and $MC'(Q^\ast) &gt; 0$ and $AVC'(Q^\ast) = 0$, we have for $Q &gt; Q^\ast$ close enough to $Q^\ast$ that $MC(Q) &gt; AVC(Q)$. So $AVC(Q)$ is increasing for all $Q&gt; Q^\ast$ close enought to $Q$.
Next, for such $Q$,  $(MC(Q) - AVC(Q))' = MC'(Q) + (MC(Q) - AVC(Q))/Q &gt; 0$, 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 $MC(Q) &gt; AVC(Q)$ for all $Q &gt; Q^\ast$.

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.