<blockquote>
<ol start="2">
<li>If this is correct, is it unsurprising that it is wildly different
from the estimate given that the change in capital was a 50% increase,
and calculus is designed to approximate small changes? Therefore was
calculus ever really an appropriate solution to this question?</li>
</ol>
</blockquote>
<p>You are right when you say that <strong>calculus with the derivative</strong></p>
<p><span class="math-container">$$\frac{\partial C}{\partial r}Δ r = k^c Δ r \qquad (1)$$</span></p>
<p><strong>overestimates the  solution,</strong> and it is not surprising.</p>
<p>The point is what you said, that the use of the derivative as in <span class="math-container">$(1)$</span> is a <strong>linear approximation</strong> of the function <span class="math-container">$C(r)$</span>, and it is <strong>local,</strong> at point <span class="math-container">$r=10$</span>. So, the  farther you go from <span class="math-container">$r=10$</span> , the more imprecise will be the approximation.</p>
<p>Approximation with polynomials, and therefore also linear approximations,  are local (remember the Taylor’s formula).</p>
<p>In our case above, <strong>with formula <span class="math-container">$(1)$</span>  we are approximating the function of cost with respect to <span class="math-container">$r$</span> with the tangent line in <span class="math-container">$r=10$</span></strong>. Actually,  the derivative</p>
<p><span class="math-container">$$\frac{\partial C}{\partial r}=k^c$$</span></p>
<p>is the derivative of the function <span class="math-container">$C(r)$</span> with respect to <span class="math-container">$r$</span>, taking all the other variables fixed.</p>
<p><strong>The function <span class="math-container">$C(r)$</span> is increasing with respect to  <span class="math-container">$r$</span></strong> (the first derivative is positive), but <strong>the derivative is decreasing with respect to <span class="math-container">$r$</span>.</strong></p>
<p>The derivative <span class="math-container">$\frac{\partial C}{\partial r}$</span>, of course, geometrically, is the slope of the tangent line to <span class="math-container">$C(r)$</span>.</p>
<p>Therefore, when we use formula <span class="math-container">$(1)$</span> to get</p>
<p><span class="math-container">$\frac{\partial C}{\partial r}Δ r = k^c Δ r$</span>  = 0.41 • 5 = 2.05,</p>
<p>we  are approximating the function <span class="math-container">$C(r)$</span> with its tangent, and we are actually calculating the increase of the tangent.</p>
<p>It can be easily seen in a picture as below:
<a href="https://i.sstatic.net/KXD6r.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/KXD6r.png" alt="enter image description here" /></a></p>
<p>You can see that when you use the derivative you are <strong>calculating the increase of the tangent, represented by the segment <span class="math-container">$CD$</span>, whereas the increase in <span class="math-container">$C(r)$</span>, <span class="math-container">$\Delta C$</span>,  is the segment <span class="math-container">$BD$</span>.</strong></p>
<p>And, of course, the greater is the increase <span class="math-container">$\Delta r$</span>, the worse is the approximation of the increase of <span class="math-container">$C(r)$</span> using the derivative.</p>
<p>(In the picture you can also recognize the <em>differential</em> of the function of one variable <span class="math-container">$C(r)$</span>).</p>
<p><span class="math-container">$$\;$$</span></p>
<p>All this answers also <strong>question <span class="math-container">$3$</span>:</strong></p>
<blockquote>
<ol start="3">
<li>Is the following a correct statement:</li>
</ol>
<ul>
<li><em>&quot;Given that the cost function is increasing in input prices. Using calculus will always <strong>over- estimate the increase in cost</strong></em></li>
</ul>
</blockquote>
<p>Yes, because the function of the cost <span class="math-container">$C(r)$</span> has a <strong>decreasing derivative.</strong></p>
<p>Of course, the converse will be true, that is, we had an underestimate if we had a function with an increasing derivative.</p>

CC BY-SA 4.0
2023-05-09T14:48:24.120Z
2023-05-09T17:39:21.043Z
2023-05-09T17:39:21.043Z
BakerStreet
<p>I have the following model, and i solve for my optimised conditional factor demands, and minimised cost functions <span class="math-container">$C$</span>. (<em>Note: I have turned a minimisation problem into a maximisation problem)</em>. Let's assume <span class="math-container">$k&gt;0, l&gt;0$</span></p>
<p><span class="math-container">$Max \; L(k, l, λ) = -rk - wl + \lambda [(k + 1)^\alpha l^{1 - \alpha} - y]$</span></p>
<ul>
<li><p><span class="math-container">$k^c = y(\frac{w α}{r (1-α)})^{1-α} - 1$</span></p>
</li>
<li><p><span class="math-container">$l^c = y(\frac{r(1- α)}{wα})^α$</span></p>
</li>
<li><p><span class="math-container">$C(k^c, l^c) = rk^c + wl^c$</span></p>
</li>
</ul>
<p><strong>A question then asks:</strong></p>
<blockquote>
<p><span class="math-container">$w = 20, r = 10,  α = 0.5, y = 1$</span>
If <span class="math-container">$r$</span> increased by 5. How does cost change?</p>
</blockquote>
<p>No hints of approach are given.  The model answer to the question immediately assumes calculus and specifically Shepherds Lemma.</p>
<p>The answer we get is:
<span class="math-container">$\frac{\partial C}{\partial r}Δ r = k^c Δ r$</span>  = 0.41 • 5 = 2.05</p>
<p><strong>My Questions:</strong></p>
<ol>
<li>If i wanted to calculate that change exactly would the following calculation be correct:</li>
</ol>
<ul>
<li><span class="math-container">$C_2 - C_1 = [k_2^c(15,20)15 + l_2^c(15,20)20] - [k_1^c(10,20)10 + l_2^c(10,20)20]$</span> = 1.365.</li>
</ul>
<ol start="2">
<li><p>If this is correct, is it unsurprising that it is wildly different from the estimate given that the change in capital was a 50% increase, and calculus is designed to approximate small changes? Therefore was calculus ever really an appropriate solution to this question, apart from the fact that it happens to be a Mathematical Economics Exam?</p>
</li>
<li><p>Is the following a correct statement:</p>
</li>
</ol>
<ul>
<li><em>&quot;Given that the cost function is increasing in input prices. Using calculus will always <strong>over estimate the increase in cost</strong>, because it uses the initial optimal factor demand level as 'reference' [<span class="math-container">$k_1^c(10,20)$</span>], and therefore <strong>under estimates the fall in demand for that factor</strong> when it's price rises. Therefore it over estimates the impact of this now more expensive input in the cost function&quot;</em></li>
</ul>

CC BY-SA 4.0
2023-05-09T14:41:42.053Z
Are you sure that here "The answer we get is:
$\frac{\partial C}{\partial r}Δ k = k^c Δ k$  = 0.41 • 5 = 2.05" is $\Delta k$ and not $\Delta r$?
BakerStreet
CC BY-SA 4.0
2023-05-09T18:01:49.330Z
Apologies yes that's a silly typo. I'll change it now!
CormJack
CC BY-SA 4.0
2023-05-09T18:23:21.273Z
Ok, I answered as I thought it was a typo!
BakerStreet
<mathematical-economics><optimization><cost-functions><lagrangian>
Inappropriate use of Calculus in estimating ΔCost?
CormJack
# Inappropriate use of Calculus in estimating ΔCost?

(Asked by: username_0 on 2023-05-08)

I have the following model, and i solve for my optimised conditional factor demands, and minimised cost functions $C$. (*Note: I have turned a minimisation problem into a maximisation problem)*. Let's assume $k>0, l>0$

$Max \; L(k, l, λ) = -rk - wl + \lambda [(k + 1)^\alpha l^{1 - \alpha} - y]$

* $k^c = y(\frac{w α}{r (1-α)})^{1-α} - 1$
* $l^c = y(\frac{r(1- α)}{wα})^α$
* $C(k^c, l^c) = rk^c + wl^c$

**A question then asks:**

> $w = 20, r = 10,  α = 0.5, y = 1$
> If $r$ increased by 5. How does cost change?

No hints of approach are given. The model answer to the question immediately assumes calculus and specifically Shepherds Lemma.

The answer we get is:
$\frac{\partial C}{\partial r}Δ r = k^c Δ r$ = 0.41 • 5 = 2.05

**My Questions:**

1. If i wanted to calculate that change exactly would the following calculation be correct:

* $C\_2 - C\_1 = [k\_2^c(15,20)15 + l\_2^c(15,20)20] - [k\_1^c(10,20)10 + l\_2^c(10,20)20]$ = 1.365.

2. If this is correct, is it unsurprising that it is wildly different from the estimate given that the change in capital was a 50% increase, and calculus is designed to approximate small changes? Therefore was calculus ever really an appropriate solution to this question, apart from the fact that it happens to be a Mathematical Economics Exam?
3. Is the following a correct statement:

* *"Given that the cost function is increasing in input prices. Using calculus will always **over estimate the increase in cost**, because it uses the initial optimal factor demand level as 'reference' [$k\_1^c(10,20)$], and therefore **under estimates the fall in demand for that factor** when it's price rises. Therefore it over estimates the impact of this now more expensive input in the cost function"*

**Tags:** mathematical-economics, optimization, cost-functions, lagrangian
> **Comments:**
> **username_1**: Are you sure that here "The answer we get is: $\frac{\partial C}{\partial r}Δ k = k^c Δ k$  = 0.41 • 5 = 2.05" is $\Delta k$ and not $\Delta r$?  
> **username_0**: Apologies yes that's a silly typo. I'll change it now!  
> **username_1**: Ok, I answered as I thought it was a typo!  

---

## 1 answers

### Answer by username_1 on 2023-05-09. Score: 2 | Accepted answer

> 2. If this is correct, is it unsurprising that it is wildly different
>    from the estimate given that the change in capital was a 50% increase,
>    and calculus is designed to approximate small changes? Therefore was
>    calculus ever really an appropriate solution to this question?

You are right when you say that **calculus with the derivative**

$$\frac{\partial C}{\partial r}Δ r = k^c Δ r \qquad (1)$$

**overestimates the solution,** and it is not surprising.

The point is what you said, that the use of the derivative as in $(1)$ is a **linear approximation** of the function $C(r)$, and it is **local,** at point $r=10$. So, the farther you go from $r=10$ , the more imprecise will be the approximation.

Approximation with polynomials, and therefore also linear approximations, are local (remember the Taylor’s formula).

In our case above, **with formula $(1)$ we are approximating the function of cost with respect to $r$ with the tangent line in $r=10$**. Actually, the derivative

$$\frac{\partial C}{\partial r}=k^c$$

is the derivative of the function $C(r)$ with respect to $r$, taking all the other variables fixed.

**The function $C(r)$ is increasing with respect to $r$** (the first derivative is positive), but **the derivative is decreasing with respect to $r$.**

The derivative $\frac{\partial C}{\partial r}$, of course, geometrically, is the slope of the tangent line to $C(r)$.

Therefore, when we use formula $(1)$ to get

$\frac{\partial C}{\partial r}Δ r = k^c Δ r$ = 0.41 • 5 = 2.05,

we are approximating the function $C(r)$ with its tangent, and we are actually calculating the increase of the tangent.

It can be easily seen in a picture as below:
[![enter image description here](https://i.sstatic.net/KXD6r.png)](https://i.sstatic.net/KXD6r.png)

You can see that when you use the derivative you are **calculating the increase of the tangent, represented by the segment $CD$, whereas the increase in $C(r)$, $\Delta C$, is the segment $BD$.**

And, of course, the greater is the increase $\Delta r$, the worse is the approximation of the increase of $C(r)$ using the derivative.

(In the picture you can also recognize the *differential* of the function of one variable $C(r)$).

$$\;$$

All this answers also **question $3$:**

> 3. Is the following a correct statement:
>
> * *"Given that the cost function is increasing in input prices. Using calculus will always **over- estimate the increase in cost***

Yes, because the function of the cost $C(r)$ has a **decreasing derivative.**

Of course, the converse will be true, that is, we had an underestimate if we had a function with an increasing derivative.


---

