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

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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 T^½ such that T^½ is itself positive and (T^½)² = T. The operator T^½ 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 = S^TS. Notice that S^TS is symmetric because (S^TS)^T = S^TS^TT = S^TS. If M = S^TS 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 = S^TS. 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)