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October 31st, 2017, 08:43 AM   #1
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Production function.

A firm's production function is given by
Q=50KL
Unit capital and labour costs are 2 and 3 respectively. Find the values of K and L which minimise total input costs if the production quota is 1200.
Yeah, I hope somebody can help me solve this. Hardstuck. :/

Last edited by skipjack; December 1st, 2017 at 12:09 PM.
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October 31st, 2017, 06:04 PM   #2
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This problem represents a minimization problem subject to constraints.

First define the objective function that is to be minimized, namely

$C = 2K + 3L.$

Conceptually, there are three constraints, namely

$Q = 1200,\ 0 \le K,\ \text { and } 0 \le L,\ \text { given } Q = 50KL.$

However, in this case, it is obvious that K > 0 < L so only the first constraint is relevant.

Do you know the method of Lagrangian multipliers?

$M = C + \lambda (Q - 1200) = 2K + 3L + \lambda (50KL - 1200).$

$\dfrac{ \delta M}{ \delta K} = 2 + 50 \lambda L.$

$\dfrac{ \delta M}{ \delta L}= 3 + 50 \lambda K.$

$\dfrac{ \delta M}{ \delta \lambda }= 50KL - 1200.$

$\dfrac{ \delta M}{ \delta \lambda } = 0 \implies 50KL = 1200 \implies 0 < K = \dfrac{24}{L}.$

$\dfrac{ \delta M}{ \delta K } = 0 \implies 2 = -\ 50 \lambda L \implies \lambda = -\ \dfrac{1}{25L}.$

$\dfrac{ \delta M}{ \delta L} = 0 \implies 3 = -\ 50 \lambda K = (-\ 50) * \left ( -\ \dfrac{1}{25 L} \right ) * \dfrac{24}{L} = \dfrac{48}{L^2} \implies$

$3L^2 = 48 \implies L^2 = 16 \implies L = 4 \implies K = \dfrac{24}{4}= 6.$

Let's see if that makes sense.

$50 * 4 * 6 = 200 * 6 = 1200.$

We get the desired quantity.

$C = 2K + 3L = 2 * 6 + 3 * 4 = 12 + 12 = 24.$

Now vary L by x and K by y, keeping Q constant.

$1200 = 50(4 + x)(6 + y) \implies 24 = 24 + 4y + 6x + xy \implies y = -\ \dfrac{6x}{x + 4}, \text { where } -\ 4 < x.$

So the difference in cost by varying is

$e= \left \{3(4 + x) + 2 * \left ( 6 - \dfrac{6x}{x + 4} \right ) \right \} - 24 = 3x + \dfrac{-\ 12x}{x + 4} = \dfrac{3x^2 + 12x - 12x}{x + 4} = \dfrac{3x^2}{x + 4}.$

$-\ 4 < x \text { and } x \ne 0 \implies e > 0 \text {, but } -\ 4 < x = 0 \implies e = 0.$

So cost is indeed minimized at $K = 6$ and $L = 4$.

A great deal of micro-economics depends on Lagrangian multipliers. Learn them.
Thanks from greg1313

Last edited by skipjack; December 1st, 2017 at 11:44 AM.
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November 26th, 2017, 06:19 PM   #3
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The basic constrained maximization problem of the maximum area that be enclosed by a fixed rectangular perimeter has the length and width as equal. In this case 2K = 3L = 12. Can it be said generically that the two terms should be equal without doing all the work?
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November 26th, 2017, 06:42 PM   #4
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What do you mean enclosed by a rectangular perimeter?

Also, since the number of variables is small, we can solve graphically with level sets.

https://www.desmos.com/calculator/cbvhzbfvud
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November 28th, 2017, 05:28 AM   #5
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Quote:
Originally Posted by Joppy View Post
What do you mean enclosed by a rectangular perimeter?
I mean that given a perimeter of 8, having all the sides be equal (2 in this case) produces an area of 4, which is a greater area than a 1.5x2.5 rectangle or any other rectangle.
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November 28th, 2017, 12:48 PM   #6
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Quote:
Originally Posted by EvanJ View Post
I mean that given a perimeter of 8, having all the sides be equal (2 in this case) produces an area of 4, which is a greater area than a 1.5x2.5 rectangle or any other rectangle.
This has NOTHING to do with production functions or with Lagrangian multipliers. Why did you tack it on to this post?

$A = x * \dfrac{8 - 2x }{2} = 0.5(8x - 2x^2) = 4x - x^2 \implies$

$\dfrac{dA}{dx} = 4 - 2x \text { and } \dfrac{d^2A}{dx^2} = -\ 2 \implies$

$A \text { is maximum if } x = 2.$

Basic calculus.
Thanks from Joppy

Last edited by JeffM1; November 28th, 2017 at 01:02 PM.
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November 30th, 2017, 03:58 PM   #7
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What I posted would make a problem in this form:

Maximize AB (the area of a rectangle) with 2A + 2B = to the perimeter given in a problem.

How is that different from what this topic asks? Let's say the constraints remain the same, but we wanted to minimize C = 4K + 3L instead of C = 2K + 3L. What would the answer be? I'm trying to learn.

Last edited by EvanJ; November 30th, 2017 at 04:03 PM.
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November 30th, 2017, 05:15 PM   #8
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Quote:
Originally Posted by EvanJ View Post
What I posted would make a problem in this form:

Maximize AB (the area of a rectangle) with 2A + 2B = to the perimeter given in a problem.

How is that different from what this topic asks?
It's almost the complete opposite. Your objective function is a product, and the constraints are a sum.

Quote:
Originally Posted by EvanJ View Post
Let's say the constraints remain the same, but we wanted to minimize C = 4K + 3L instead of C = 2K + 3L. What would the answer be? I'm trying to learn.
The constraints and the objective function don't function independently from each other. Did you try follow through with JeffM1's solution with the new numbers?

Last edited by skipjack; December 1st, 2017 at 11:51 AM.
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December 1st, 2017, 06:52 AM   #9
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Quote:
Originally Posted by EvanJ View Post
What I posted would make a problem in this form:

Maximize AB (the area of a rectangle) with 2A + 2B = to the perimeter given in a problem.

How is that different from what this topic asks? Let's say the constraints remain the same, but we wanted to minimize C = 4K + 3L instead of C = 2K + 3L. What would the answer be? I'm trying to learn.
Well, yes you could turn what is a simple calculus problem into a problem using a Lagrangian multiplier if you are masochistic enough. (It still would have NOTHING to do with production functions.)

$A = xy \text {, subject to } 2x + 2y = 8.$

So set up L as

$L= xy - \lambda \{8 - 2(x + y)\}.$

$\therefore \dfrac{ \delta L}{ \delta \lambda } = 8 - 2(x + y).$

$\dfrac{ \delta L}{\delta x} = y + 2 \lambda.$

$ \dfrac{ \delta L}{ \delta y} = x + 2 \lambda.$

$\therefore \dfrac{ \delta L}{ \delta \lambda } = 0 \implies 8 - 2(x + y) = 0 \implies x + y = 4 \implies y = 4 - x.$

$\therefore \dfrac{ \delta L }{ \delta x } = 0 \implies y + 2 \lambda = 0 \implies \lambda = \dfrac{-\ y}{2} = \dfrac{x - 4}{2}.$

$\therefore \dfrac{ \delta L }{ \delta y} = 0 \implies 0 = x + 2 \lambda = x + 2 * \dfrac{x - 4}{2} = x + x - 4 \implies 2x = 4 \implies$

$x = 2,\ y = 4 - 2 = 2.$

EDIT 1: Notice that I should be calculating second derivatives as well.

EDIT 2: I suppose that you CAN use Lagrangian multipliers in any problem requiring optimization of a differentiable objective function under constraint. Joppy seems to know a criterion that indicates when you MUST use LaGrangian multipliers.

Last edited by JeffM1; December 1st, 2017 at 07:10 AM.
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December 1st, 2017, 01:51 PM   #10
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Quote:
Originally Posted by EvanJ View Post
How is that different from what this topic asks? Let's say the constraints remain the same, but we wanted to minimize C = 4K + 3L instead of C = 2K + 3L.
Using the various lengthy methods given above, the new answers would be K = 3√2 and L = 4√2.
Notice that 4K = 3L = 12√2.

However, the above values are irrational, which is presumably unsatisfactory.

If K and L must be integers, K = 4 and L = 6, but I don't see an easy method (other than trial and error) of finding these.

For the original problem, you need to minimize 2K + 3L = 2K + p²/(2K),
where p can be an integer, namely 12, because (2K)(3L) = (6/50)1200 = 144 = 12².

As 2K + p²/(2K) = (2K - p)²/(2K) + 2p,
K = p/2 = 6 (which implies L = 4) minimizes the sum and these values happen to be integers. As K and L have to be non-negative, I didn't need to consider using p = -12.
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