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

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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. Template:Citation needed 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. Template:Citation needed"

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 Template:Nowrap. Notice that S^TS is symmetric because Template:Nowrap If Template:Nowrap 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 Template:Nowrap 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)