<p>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>

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

<p>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).</p>

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

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

<p>$$(p-c_H)n_H\geq(p-c_L)n_L$$</p>

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

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

<hr>

<p>Let's look at an example:</p>

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

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

<p>Now we plug these numbers into the formula</p>

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

<p>$$n_H\geq\frac{(3-1.5) 1000}{(3-1.6)}$$</p>

<p>$$n_H\geq\frac{1500}{(1.4)}$$</p>

<p>$$n_H\geq1071.42.$$</p>

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

CC BY-SA 3.0
2016-08-09T13:30:33.627Z
2016-08-09T13:30:33.627Z
Ubiquitous
<p>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. </p>

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

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

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

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

<p>$$M\left(3- (C_w + 1)\right) &gt; N\left(3- (C_w + 0.5)\right)$$
$$M\left(2- C_w \right) &gt; N\left(2.5- C_w\right)$$
$$M &gt; N \frac{2.5- C_w}{2- C_w } $$</p>

<p>Example: $N = 10 000$, $C_w = 1.2$ </p>

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

<p><strong>EDIT:</strong></p>

<p>MR. A asks in comments for "the amount I increase/decrease"</p>

<p>The difference is just $M-N$, so just take the original formula and subtract $N$ from both sides:</p>

<p>$$M - N &gt; N \frac{2.5- C_w}{2- C_w } - N$$</p>

<p>Simplifies to: </p>

<p>$$M - N &gt; N \frac{0.5}{2- C_w }$$</p>

CC BY-SA 3.0
2016-08-08T21:17:26.980Z
But I want to know the relationship between the amount I increase/decrease the amount of lube in a widget and the amount of widgets I need to sell at a given price to break even after accounting for that increase/decrease in costs associated with the lube increase/decrease
Mr. A
CC BY-SA 3.0
2016-08-08T21:28:59.353Z
I see. Will add exactly that.
snoram
CC BY-SA 3.0
2016-08-09T08:22:48.793Z
I guess what I had thought was that if you raised the lube content, that would increase the cost of the lube content per widget by X.  Say from .5 to .6.  If I sold 1000 widgets that year, that would mean increasing my costs by (.6-.5)*1000 or $100.  If I am getting 3.00 per widget, that means I would have to sell 33.3 more widgets to cover that cost.  But here is where I am getting caught up.  I think I have to account for the increased cost for the additional widgets, meaning I'd have to add .1*(33.3) = +3.33 and then divide by 3 again to get 1.11 and add that to the previous 33.3 => 34.1
Mr. A
CC BY-SA 3.0
2016-08-08T20:30:42.073Z
2016-08-08T21:33:04.307Z
2016-08-08T21:33:04.307Z
snoram
<p>I need to know the name and formula for this problem.</p>

<p>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.</p>

<p>Price of widgets is constant.</p>

<p>What is the name of the formula I am seeking?  What is the formula I am seeking?  What is the answer?</p>

<microeconomics>
What is the formula for how many more I need to sell in order to justify raising cost?
Mr. A
# What is the formula for how many more I need to sell in order to justify raising cost?

(Asked by: username_0 on 2016-08-08)

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?

**Tags:** microeconomics


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## 2 answers

### Answer by username_1 on 2016-08-09. Score: 1 | Accepted answer

This is an algebra problem. My answer is quite similar to username_2s, 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.


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### Answer by username_2 on 2016-08-08. Score: 0

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:**

MR. A asks in comments for "the amount I increase/decrease"

The difference is just $M-N$, so just take the original formula and subtract $N$ from both sides:

$$M - N > N \frac{2.5- C\_w}{2- C\_w } - N$$

Simplifies to:

$$M - N > N \frac{0.5}{2- C\_w }$$

> **Comments:**
> **username_0**: But I want to know the relationship between the amount I increase/decrease the amount of lube in a widget and the amount of widgets I need to sell at a given price to break even after accounting for that increase/decrease in costs associated with the lube increase/decrease  
> **username_2**: I see. Will add exactly that.  
> **username_0**: I guess what I had thought was that if you raised the lube content, that would increase the cost of the lube content per widget by X.  Say from .5 to .6.  If I sold 1000 widgets that year, that would mean increasing my costs by (.6-.5)*1000 or $100.  If I am getting 3.00 per widget, that means I would have to sell 33.3 more widgets to cover that cost.  But here is where I am getting caught up.  I think I have to account for the increased cost for the additional widgets, meaning I'd have to add .1*(33.3) = +3.33 and then divide by 3 again to get 1.11 and add that to the previous 33.3 => 34.1  

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