<p>If you use derivative to calculate marginal cost, then marginal cost is how much cost increases given an infinitesimal increase in quantity. This implies that quantity of good is continuous. Indeed, if you have a cost function $C:\mathbb{R}\to\mathbb{R}$, then that implies that the quantity of good is continuous already.</p>

<p>If the quantity is discrete, then you should stick with just $C(q+1) - C(q)$.</p>

CC BY-SA 3.0
2016-02-03T07:58:34.930Z
Actually, I guess they approximate both being the derivative and the concept of the increase in cost based on an increase of the production by 1 unit. And I think I managed to combine both ideas together!
YoussefDir
CC BY-SA 3.0
2016-02-03T05:50:52.427Z
2016-02-03T05:50:52.427Z
Art
<p>I am not sure I understand your question but if I do, then I do not agree. </p>

<p>Marginal cost is defined as the derivative of the cost function. If the producible units are too large then marginal cost is less likely be a good approximation of non-marginal cost changes. The importance of this is that the derivative of an estimated cost function will probably not correspond to the real cost increase when non-marginally increasing the output.</p>

CC BY-SA 3.0
2016-01-31T23:56:59.863Z
2016-02-01T11:42:42.240Z
2016-02-01T11:42:42.240Z
Giskard
<p>The cost function depends on the number of produced items.</p>

<p>Let the value of the cost function C be in hundreds of thousands $ and the quantity is in 1,000,000 items, thus q = 1 indicates 1,000,000 items and q = 0.000001 indicates 1 item.</p>

<p>The marginal cost at a certain number of items (say 1560 items corresponding to q = 0.00156) is in one hand interpreted as the cost of an additional produced item (so it is the cost of the 1561th item), and from another hand it is the derivative of the cost function </p>

<p>$\frac{\partial \text{C(q)}}{\partial q}$ and then applying q to be 0.00156.</p>

<p>I see a good approximation for $\frac{\partial \text{C(q)}}{\partial q}$ to be $\frac{\partial \text{C(q)}}{1additionalitem}$ since 1 additional item is (0.001561 - 0.001560) which is an infinitly small quantity suitable to be considered as ${\partial q}$.</p>

<p>But what if in some other case, q actually represented just item units and not normalized item numbers e.g.: </p>

<p>q = 1 indicates 1 item, q = 1560 indicates 1560 produced items, and so on, so in this case $\frac{\partial \text{C(q)}}{\partial q}$ can't be $\frac{\partial \text{C(q)}}{1additionalitem}$ since 1 additional item is (1561 - 1560) which is 1 and not an infinitly small quantity.</p>

<p>would it be possible then to use the derivative of the cost function?</p>

<p>What I'm saying is that considering the marginal cost to be the derivative of the cost function is conditioned by having q a normalized value of a very big number of produced items. Do you agree?</p>

CC BY-SA 3.0
2016-02-01T20:32:41.340Z
The background if this question is whether the derivative, as we usually use it, is still valid to correctly express the increase in the cost for exactly one additional item or not.
YoussefDir
<cost>
The marginal cost function is meaningful only when the produced quantity is normalized
YoussefDir
# The marginal cost function is meaningful only when the produced quantity is normalized

(Asked by: username_0 on 2016-01-31)

The cost function depends on the number of produced items.

Let the value of the cost function C be in hundreds of thousands $ and the quantity is in 1,000,000 items, thus q = 1 indicates 1,000,000 items and q = 0.000001 indicates 1 item.

The marginal cost at a certain number of items (say 1560 items corresponding to q = 0.00156) is in one hand interpreted as the cost of an additional produced item (so it is the cost of the 1561th item), and from another hand it is the derivative of the cost function

$\frac{\partial \text{C(q)}}{\partial q}$ and then applying q to be 0.00156.

I see a good approximation for $\frac{\partial \text{C(q)}}{\partial q}$ to be $\frac{\partial \text{C(q)}}{1additionalitem}$ since 1 additional item is (0.001561 - 0.001560) which is an infinitly small quantity suitable to be considered as ${\partial q}$.

But what if in some other case, q actually represented just item units and not normalized item numbers e.g.:

q = 1 indicates 1 item, q = 1560 indicates 1560 produced items, and so on, so in this case $\frac{\partial \text{C(q)}}{\partial q}$ can't be $\frac{\partial \text{C(q)}}{1additionalitem}$ since 1 additional item is (1561 - 1560) which is 1 and not an infinitly small quantity.

would it be possible then to use the derivative of the cost function?

What I'm saying is that considering the marginal cost to be the derivative of the cost function is conditioned by having q a normalized value of a very big number of produced items. Do you agree?

**Tags:** cost

> **Comments:**
> **username_0**: The background if this question is whether the derivative, as we usually use it, is still valid to correctly express the increase in the cost for exactly one additional item or not.  

---

## 2 answers

### Answer by username_1 on 2016-02-03. Score: 1

If you use derivative to calculate marginal cost, then marginal cost is how much cost increases given an infinitesimal increase in quantity. This implies that quantity of good is continuous. Indeed, if you have a cost function $C:\mathbb{R}\to\mathbb{R}$, then that implies that the quantity of good is continuous already.

If the quantity is discrete, then you should stick with just $C(q+1) - C(q)$.

> **Comments:**
> **username_0**: Actually, I guess they approximate both being the derivative and the concept of the increase in cost based on an increase of the production by 1 unit. And I think I managed to combine both ideas together!  

---

### Answer by username_2 on 2016-01-31. Score: 1

I am not sure I understand your question but if I do, then I do not agree.

Marginal cost is defined as the derivative of the cost function. If the producible units are too large then marginal cost is less likely be a good approximation of non-marginal cost changes. The importance of this is that the derivative of an estimated cost function will probably not correspond to the real cost increase when non-marginally increasing the output.


---

