October 31st, 2017, 07:43 AM  #1 
Newbie Joined: Sep 2017 From: Latvia/Denmark Posts: 21 Thanks: 0  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 11:09 AM. 
October 31st, 2017, 05:04 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 997 Thanks: 410 
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 microeconomics depends on Lagrangian multipliers. Learn them. Last edited by skipjack; December 1st, 2017 at 10:44 AM. 
November 26th, 2017, 05:19 PM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 595 Thanks: 81 
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?

November 26th, 2017, 05:42 PM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,579 Thanks: 541 Math Focus: Yet to find out. 
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 
November 28th, 2017, 04:28 AM  #5 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 595 Thanks: 81  
November 28th, 2017, 11:48 AM  #6  
Senior Member Joined: May 2016 From: USA Posts: 997 Thanks: 410  Quote:
$A = x * \dfrac{8  2x }{2} = 0.5(8x  2x^2) = 4x  x^2 \implies$ $\dfrac{dA}{dx} = 4  2x \text { and } \dfrac{d^2A}{dx^2} = \ 2 \implies$ $A \text { is maximum if } x = 2.$ Basic calculus. Last edited by JeffM1; November 28th, 2017 at 12:02 PM.  
November 30th, 2017, 02:58 PM  #7 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 595 Thanks: 81 
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 03:03 PM. 
November 30th, 2017, 04:15 PM  #8  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,579 Thanks: 541 Math Focus: Yet to find out.  Quote:
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 10:51 AM.  
December 1st, 2017, 05:52 AM  #9  
Senior Member Joined: May 2016 From: USA Posts: 997 Thanks: 410  Quote:
$A = xy \text {, subject to } 2x + 2y = 8.$ 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 06:10 AM.  
December 1st, 2017, 12:51 PM  #10  
Global Moderator Joined: Dec 2006 Posts: 18,847 Thanks: 1568  Quote:
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 nonnegative, I didn't need to consider using p = 12.  

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