Wikipedia:Reference desk/Archives/Mathematics/2011 July 14

= July 14 =

Constrained minimization
Suppose a plant produces $$q$$ engines each week using $$K$$ assembly machines and $$L$$ workers.

The number of engines produced each week is

$$q=3K^{1/2}L^{1/2}$$

Each assembly machine rents for $$r$$ dollars per week and each worker costs $$w$$ dollars per week.

Part one of the question is, how much does it cost to produce $$q$$ engines, if the plant is cost-minimizing?

I am assuming that this is a constrained maximization problem where we need to maximize $$-rK-wL$$ subject to $$q=3K^{1/2}L^{1/2}$$

I'm trying to solve this with the Lagrangian $$G=-rK-wL-\lambda(q-3K^{1/2}L^{1/2})$$

that is, with

$$ G_K = -r - \lambda(\frac{3}{2} L^{1/2} K^{-1/2})=0$$

$$G_L = -w - \lambda(\frac{3}{2} L^{-1/2} K^{1/2})=0$$

$$ G_\lambda = q-3K^{1/2}L^{1/2}=0 $$

I think I worked out from the first two conditions that $$\lambda=-w \left(\frac{2}{3} K^{-1/2}L^{1/2} \right)=-r\left(\frac{2}{3} K^{1/2}L^{-1/2} \right) $$ and so $$L=\frac{r}{w}K$$

Then from the last condition $$q=3K^{1/2}L^{1/2}=3K \left(\frac{r}{w} \right)^{1/2} $$

I get $$K=\frac{q}{3} \left(\frac{w}{r} \right)^{1/2}$$ and $$L=\frac{q}{3} \left(\frac{r}{w} \right)^{1/2}$$

So the overall cost function is

$$cost = rK + wL = \frac{2q}{3} w^{1/2}r^{1/2} $$

Have I done this correctly, or is there another way to do this?

I am also asked what are the average and marginal costs for producing $$q$$ engines.

For the average cost I presume you just divide the total cost by $$q$$, while for the marginal cost you take the derivative of the total cost with respect to $$q$$

In this case the average and marginal costs are the same. Does this indicate constant returns to scale?

118.208.40.147 (talk) 02:28, 14 July 2011 (UTC)
 * The easiest way to solve this is to simply substitute $$L=\frac{q^2}{9K}$$ in $$rK+wL$$ and find the unconstrained extremum of this single-variable function. Using this I got the same result as you, so your derivation is probably correct. And yes, this result indicates that the cost is linear. -- Meni Rosenfeld (talk) 09:00, 14 July 2011 (UTC)

Matrix square root
I have read the Square root of a matrix article.

I have a real square positive-definite but possibly non-symmetric matrix M, and I seek a (preferably real) matrix S such that transpose(S)*S=M (the * indicates matrix product). I don't have to be able to construct S, it suffices to know that it exists (does it?).

Cholesky decomposition will not do, as the matrix M is not guaranteed to be symmetric.

A square root based on diagonalization won't help either, as that would give S*S=M and not transpose(S)*S=M

The Square root of a matrix article writes further

"In linear algebra and operator theory, given a bounded positive semidefinite operator (a non-negative operator) T on a complex Hilbert space, B is a square root of T if T = B* B, where B* denotes the Hermitian adjoint of B. According to the spectral theorem, the continuous functional calculus can be applied to obtain an operator T½ such that T½ is itself positive and (T½)2 = T. The operator T½ is the unique non-negative square root of T.  "

This seems very close to what I'm looking for, but, alas, "citation needed". I know next to nothing about continuous functional calculus and that article isn't very helpful.

213.49.89.115 (talk) 17:47, 14 July 2011 (UTC)


 * If M is non-symmetric then there is no such matrix S. Let M be a non-symmetric matrix, and assume that there exists S such that M = STS. Notice that STS is symmetric because (STS)T = STSTT = STS. If M = STS then M must also be symmetric; which is a contradiction. — Fly by Night  ( talk )  01:47, 15 July 2011 (UTC)


 * Thanks! (and my apologies for the stupid-in-hindsight question :) ) 213.49.89.115 (talk) 17:10, 15 July 2011 (UTC)


 * It's not a stupid question at all. It's just that you needed to add the assumption that M is symmetric. In fact, you need det(M) to be non-negative (which it is because M is positive definite, and so all of its eigenvalues are positive). Moreover, you need all of the entries on the leading diagonal to be non-negative too. These are all necessary conditions, but I'm not sure if they are sufficient. The bottom line is that M being positive definite is not enough for you to be able to solve M = STS. Even if M is positive definite, symmetric, and has non-negative entries along the leading diagonal; I'm not sure that that's enough. I'd be interested to see what progress you make. — Fly by Night  ( talk )  22:31, 15 July 2011 (UTC)