# What is the formula for how many more I need to sell in order to justify raising cost?

I need to know the name and formula for this problem.

I have widgets which have a certain lube content to them. We currently are selling the widgets for 3.00 each. The lube costs .50 cents for each widget, but we are considering a change in the lube content from the current state of 1 oz. We currently are making sales of 1000 widgets per year. I want to know how many more widgets we would have to sell in order to justify a given increase in the lube content of a widget. I am not asking how many more we would sell if we increased the lube content or what the revenue would actually be. I am merely asking what the number of additional widgets sold must be in order to justify a given lube content increase.

Price of widgets is constant.

What is the name of the formula I am seeking? What is the formula I am seeking? What is the answer?

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.

Assume that today you sell $N$ widgets for price of $3$. Your cost is the cost of the widget plus the lube: $C_w + 0.5$ assuming all other marginal costs are small enough to ignore.

Your profit is $N\left(3- (C_w + 0.5)\right)$.

After changing the lube you sell $M$ widgets for a price of $3$ but cost has increased to $C_w + 1$.

Your profit is: $M\left(3- (C_w + 1)\right)$.

Your question rephrased: How big has $M$ to be so profit is higher after the change. Doing the math:

$$M\left(3- (C_w + 1)\right) > N\left(3- (C_w + 0.5)\right)$$ $$M\left(2- C_w \right) > N\left(2.5- C_w\right)$$ $$M > N \frac{2.5- C_w}{2- C_w }$$

Example: $N = 10 000$, $C_w = 1.2$

Then M needs to be bigger than $10 000 \frac{2.5- 1.2}{2- 1.2} = 16 250$

EDIT:

The difference is just $M-N$, so just take the original formula and subtract $N$ from both sides:
$$M - N > N \frac{2.5- C_w}{2- C_w } - N$$
$$M - N > N \frac{0.5}{2- C_w }$$