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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.
  1. 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?

  2. 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"
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  • $\begingroup$ 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 $\Delta r$? $\endgroup$ Commented May 9, 2023 at 14:41
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    $\begingroup$ Apologies yes that's a silly typo. I'll change it now! $\endgroup$
    – CormJack
    Commented May 9, 2023 at 18:01
  • $\begingroup$ Ok, I answered as I thought it was a typo! $\endgroup$ Commented May 9, 2023 at 18:23

1 Answer 1

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  1. 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

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 $C(r)$).

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All this answers also question $3$:

  1. 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.

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