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. :/ 
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. 
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? 
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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$A = x * \dfrac{8  2x }{2} = 0.5(8x  2x^2) = 4x  x^2 \implies$ $\dfrac{dA}{dx} = 4  2x \text { and } \dfrac{d^2A}{dx^2} = \ 2 \implies$ $A \text { is maximum if } x = 2.$ Basic calculus. 
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. 
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$A = xy \text {, subject to } 2x + 2y = 8.$ 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. 
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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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"Can it be said generically that the two terms should be equal without doing all the work?" 
If noninteger values are acceptable, my solution method shows quickly that both terms should equal the p that I defined. Unfortunately, p is typically irrational. 
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