What is the formula for how many more I need to sell in order to justify raising cost?
This is an algebra problem. My answer is quite similar to snorams, but without the numbers plugged in (which hopefully makes it a bit easier to see what is going on). Let


$p$ be the price of a widget
$c_L$ be the old cost for producing one widget (i.e. with a low amount of lube)
$c_H$ be the new cost for producing one widget (i.e. including the high amount of lube)
$n_L$ be the number of units sold when you use a low amount of lube
$n_H$ the number of units solve when you use a high amount of lube.


With a low amount of lube, your profit is $(p-c_L)n_L$ (in words: you earn a margin of $p-c_L$ on each unit, and sell $n_L$ units).

With a high amount of lube, your profit is $(p-c_H)n_H$.

You would like to know when is it true that the higher lube content is more profitable, i.e.

$$(p-c_H)n_H\geq(p-c_L)n_L$$

Now, to find the number of units, $n_H$, at which the change results in break even, we can simply solve the inequality:

$$n_H\geq\frac{(p-c_L) n_L}{(p-c_H)}$$



Let's look at an example:


price of a widget is $p=3$.
unit costs of everything but the lube are $1$.
cost of low amount of lube is $0.5$
cost for high amount of lube is $0.6$
low-lube sales are $1000$.


We calculate the unit cost as $c_L=1+0.5=1.5$ and $c_H=1+0.6=1.6$.

Now we plug these numbers into the formula

$$n_H\geq\frac{(p-c_L) n_L}{(p-c_H)}$$

$$n_H\geq\frac{(3-1.5) 1000}{(3-1.6)}$$

$$n_H\geq\frac{1500}{(1.4)}$$

$$n_H\geq1071.42.$$

So if you sold at least 1072 units (i.e. an extra 72 units) then the change would pay for itself.