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March 31

micro economic development

1.A chemical manufacturer processes two chemicals C1 and C2 in varying propositions to produce three products A,B and C.He wishes to produce at least 150 units of A,200 units of B and 60 units of C.Each tonne of C1 yields 3 of A,5 of B and 3 of C.If C1 costs $40 per tonne and C2 costs $50 per tonne

(a)Formulate the problem as a minimisation LP model (5mks)

(b)Formulate the inverse or dual of the primal formulation in (a)above (1mk)

(c)Express the dual in (b)above as standard LP model (5mks)

(d)set up the initial simplex tableau for the solution to this problem(do not solve) (4mks)

(e)using the graphical method,identify the feasibility region for the solution in the LP model.(5mks) —Preceding unsigned comment added by 41.220.120.202 (talk) 09:01, 31 March 2008 (UTC)[reply]

First off, the reference desk will not do your homework, which this very much appears to be. Second, this is economics, not mathematics... I suggest you take it to the humanities desk or someting, as it's fairly likely no one here will understand the terms. -mattbuck (Talk) 09:11, 31 March 2008 (UTC)[reply]

Well, this looks like a standard linear programming exercise to me, in which case this would be the right desk to post it, except for the fact that it is obviously a homework assignment. If the questioner can show evidence that they have spent some time working on the problem, or at least thinking about what the various terms mean, then we might be able to provide some help. Our articles on linear programming, dual problem, simplex algorithm and feasible region may be good places to start. Gandalf61 (talk) 09:35, 31 March 2008 (UTC)[reply]

Yeah, this is maths, but we need to know where the OP is stuck in order to help. We're certainly not going to provide a complete solution. --Tango (talk) 14:24, 31 March 2008 (UTC)[reply]

$\cos {\frac {\pi }{2^{n}}}$ cos(π/2ⁿ)

I'm looking for the general (non-iterative) non-trigonometric expression for the exact trigonometric constants of the form: ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ , when n is natural (and is not given in advance). Do you know of any such general (non-iterative) non-trigonometric expression? (note that any exponential-expression-over-the-imaginaries is also excluded since it's trivially equivalent to a real-trigonometric expression).

Let me explain: if we choose n=1 then the term ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ becomes "0", which is a simple (non-trigonometric) constant. If we choose n=2 then the term ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ becomes ${\begin{aligned}{\frac {1}{\sqrt {2}}}\end{aligned}}$ , which is again a non-trigonometric expression. etc. etc. Generally, for every natural n, the term ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ becomes a non-trigonometric expression. However, when n is not given in advance, then the very expression ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ per se - is a trigonometric expression. I'm looking for the general (non-iterative) non-trigonometric expression equivalent to ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ , when n is not given in advance. If not for the cosine - then for the sine or the tangent or the cotangent.

Eliko (talk) 10:38, 31 March 2008 (UTC)[reply]

Not sure that this fits your definition of "non-iterative", but best I can offer is

\cos \left({\frac {\pi }{2^{n}}}\right)={\frac {\sqrt {2+{\sqrt {2+\cdots {\sqrt {2}}}}}}{2}},

where the expression on the right hand side has n nested square roots. You can derive this by iteratively applying

2\cos \left({\frac {x}{2}}\right)={\sqrt {2+2\cos x}}\,.

Gandalf61 (talk) 11:11, 31 March 2008 (UTC)[reply]

This is an iterative expression. Eliko (talk) 11:20, 31 March 2008 (UTC)[reply]

It's derived iteratively, but I wouldn't say it was iterative itself. Perhaps it would help if you explained why you need it to be non-iterative, then we could better understand what you mean by the phrase. --Tango (talk) 14:27, 31 March 2008 (UTC)[reply]

Ok. I'll explain. Any solution for my original question could have been helpful in answering other questions, like: finding a non-trigonometric proof for the following algebraic, non-trigonometric claim: Every real interval includes a point x having a natural number n such that

{\begin{aligned}(x+i)^{n}\end{aligned}}

is a real number. Note that this is an algebraic, non-trigonometric claim, so one may naturally expect that it may be proven by a non-trigonometric proof (note also that any proof based on an exponential-expression-over-the-imaginaries should also be excluded since such an expression is trivially equivalent to a real-trigonometric expression). Eliko (talk) 15:17, 1 April 2008 (UTC)[reply]

If you want the number in a familiar, that is algebraic, form, this is the best you're gonna get. Each constant is expressed as a simple, finite nesting of radicals. Pretty much by the nature of square roots, no smaller formula will do. Black Carrot (talk) 15:56, 31 March 2008 (UTC)[reply]

I'm looking for the closed formula if any exists. However, if you say now that "this is the best I'm gonna get" - then I have to believe you. Anyways, thank you for your reply. By the way, Any solution for my original question could have been helpful in answering other questions, like: finding a non-trigonometric proof for the following algebraic, non-trigonometric claim: Every real interval includes a point x having a natural number n such that

{\begin{aligned}(x+i)^{n}\end{aligned}}

is a real number. Note that this is an algebraic, non-trigonometric claim, so one may naturally expect that it may be proven by a non-trigonometric proof (note also that any proof based on an exponential-expression-over-the-imaginaries should also be excluded since such an expression is trivially equivalent to a real-trigonometric expression). Eliko (talk) 15:17, 1 April 2008 (UTC)[reply]

The value of arg(x + i) as a function of real x is strictly monotonically decreasing. Let a and b be real numbers with a < b, and take integer n such that

n (arg(a+i) − arg(b+i) ≥ 2π.

This implies that there is a value x in the interval [a,b] such that

n arg(x+i) ≡ 0 (mod 2π).

Then Im (x+i)ⁿ = sin n arg(x+i) = 0. --Lambiam 00:05, 2 April 2008 (UTC)[reply]

A sine snuck in at the end there - does that count as non-trigonometric? --Tango (talk) 00:18, 2 April 2008 (UTC)[reply]

Tango is correct: this is a trigonometric proof since it uses trigonometric expressions, such as Sin (and Arg). Eliko (talk) 01:12, 2 April 2008 (UTC)[reply]

I don't agree that arg is a trigonometric function. The modulus and argument of a complex number z ≠ 0 are simply real numbers M and A such z = Me^iA, where M ≥ 0 and A is only determined modulo 2π. If you want to avoid sin, you just need a few more steps: n arg(x+i) = 2kπ for some integral k. Putting M = mod(x+i) and A = arg(x+i), we have:

(x+i)ⁿ = (Me^iA)ⁿ = Mⁿe^inA = Mⁿe^i2kπ = Mⁿ,

which is real. --Lambiam 10:03, 6 April 2008 (UTC)[reply]

Tango was correct. Look:

Arg is defined - either:

by a trigonometric function, e.g. by stating that every real x satisfies: Cos(Arg(x)) + iSin(Arg(x)) = x; or:
by an exponential-function-over-the-imaginaries, e.g. by stating that every real x satisfies: exp(iArg(x))=x.

However, since the beginning I've made it clear that I'm looking for a proof which does not assume the very existence of any trigonometric function, and since the beginning - I've also made it clear: "note that any exponential-expression-over-the-imaginaries - is also excluded, since it's trivially equivalent to a real-trigonometric expression".

When I wrote that "Arg" is a trigonometric function - I just meant that Arg assumes the existence of trigonometic functions, e.g. the existence of an exponential-function-over-the-imaginaries (which is trivially equivalent to a trigonometric function).

Let me explain why I insist on a non-trigonometric proof (hence - on a proof not invovling the exponential-function-over-the-imaginaries):

There are many theorems which can be proven by both trigonometric means and non-trigonometric means, e.g: the theorem stating that:

{\begin{aligned}(-1)^{n}\end{aligned}}

is either 1 or -1 for any natural n. Now look at the following theorem: "Every real interval includes a point x having a natural number n such that

{\begin{aligned}(x+i)^{n}\end{aligned}}

is a real number"; Note that the expression

{\begin{aligned}(x+i)^{n}\end{aligned}}

looks "naive", simply algebraic, non-trigonometric, and involving no imaginary exponent, so one may naturally expect that also this theorem may be proven by a non-trigonometric proof, just as the first theorem, having the naive simply algebraic expression:

{\begin{aligned}(-1)^{n}\end{aligned}}

. The abstract question is now: whether any such non-trigonometric proof does exist.

Eliko (talk) 13:14, 6 April 2008 (UTC)[reply]

FWIW, here's a JSTOR paper On Defining the Sine and Cosine that defines sine and cosine using this dyadic subdivision.

Nbarth (email) (talk) 00:32, 2 April 2008 (UTC)[reply]

No definition of functions is a proof for the very existence of functions satisfying the definition.

Unfortunately, I couldn't enter the JSTOR paper. Does it suggest an iterative definition or a closed definition? Eliko (talk) 01:12, 2 April 2008 (UTC)[reply]

More precisely, it's a construction, of sine and cosine; it does prove that the construction works and thus that the functions exist.

It is an iterative definition, so it doesn't satisfy your desiderata.

As mentioned by other previously, it's unlikely that a closed form solution exists; I would consider your question about "a real interval containing a real solution" to be fundamentally analytic, (it's rather like the Intermediate Value Theorem), not algebraic, so I don't find it surprising that one must use analytic methods.

Nbarth (email) (talk) 07:22, 2 April 2008 (UTC)[reply]

More precisely, the statement:

Every real interval includes a point x having a natural number n such that

(x+i)^{n}

is a real number.

is not purely algebraic: the use of interval means you're using the order structure on the reals, hence their topology. Thus the statement is fundamentally a topological / analytic statement, and one should expect an analytic proof. Indeed, as it's a statement about iterating a function of complex variables, it's properly a question in (discrete) complex dynamics. I'm no expert in this field, but if you wish to pursue this specific question further, that's where I would recommend turning.

Nbarth (email) (talk) 09:51, 2 April 2008 (UTC)[reply]

Yeah, it probably has to be based on an analytic proof, due to the interval. However, must its analytic proof be based on trigonometric constructions (or any equivalent analytic construction being sufficient for the very existence of any trigonometric function)? Note that the naive (simply algebraic) expression

(x+i)^{n}

doesn't make us "recall" trigonometric properties, does it? this is not the case in the alternative definitions for the trigonometric functions you've found in JSTOR, is it? Anyways, thank you for your proposal of referring to: complex dynamics.

Eliko (talk) 13:44, 2 April 2008 (UTC)[reply]

(x+i)^{n}

is a complex exponential, so it is closely related to the behavior of the exponential function; De Moivre's formula is the most natural way of seeing why trig functions naturally come in. More generally, considering complex numbers in polar coordinates is very natural, as seen in the multiplication of complex numbers and Euler's formula.

Thus it's not surprising that when looking at multiplying complex numbers, you get trig functions.

Nbarth (email) (talk) 23:27, 16 April 2008 (UTC)[reply]

Also, for the question about

(x+i)^{n}

, you could express it in terms of binomial coefficients (the imaginary part is a sum of the terms with odd powers of

i

), so it's possible that there's a combinatorial proof.

Nbarth (email) (talk) 23:27, 16 April 2008 (UTC)[reply]

Further, the geometric POV gives a stronger result: for any given interval

[a,b]

, for all n sufficiently large (

n\geq N

: not just a single n),

S=\left\{(x+i)^{n}\mid x\in [a,b]\right\}

contains a real number — the general bound is

N={\frac {1}{\arg(a+i)-\arg(b+i)}}

: since exponentiation multiplies argument, after this point S wraps all the way around the circle (contains points of all arguments).

Nbarth (email) (talk) 23:27, 16 April 2008 (UTC)[reply]

You used the "arg" function, as User talk:Lambiam did (on this page), but (as I explained to him) the "arg" function mustn't be used here. Why? See above (my explanation to Lambiam), and see also my next comment.
As I've indicated above, I'm not looking for the trigonometric proof, but rather for an algebraic (non-analytic non-trigonometric) proof. e.g. for a combinatorial proof. I still don't know whether any such combinatorial proof exists (after having expressed $(x+i)^{n}$ in terms of binomial coefficients).
I could certainly find a non-trigonometric proof if I could find a series $(a_{n})$ satisfying the following three properties:

The series $(a_{n})$ converges to zero (e.g. if it were the series $2^{-n}$ , but not necessarily).
The series is expressible algebraically (e.g. if it were the series $2^{-n}$ , but not necessarily).
The series $\cos(a_{n})$ is expressible algebraically (e.g. if $(a_{n})$ were the series $\arccos(1/n)$ , but not necessarily).

If you can find such a series, which satisfies all of the preceding three properties, I'll be very grateful.

Eliko (talk) 08:23, 17 April 2008 (UTC)[reply]

A metrisation theorem

I want to prove the following version of the Urysohn metrization theorem: A normal second-countable Hausdorff space is metrisable. I'd be grateful for a cross-check of this proof.

Let X be such a space. Let $\{U_{n}\}$ be a base for the topology. Let $(U_{n_{i}},U_{m_{i}})$ be an enumeration of all pairs of elements in this base such that ${\overline {U}}_{n_{i}}\subset U_{m_{i}}.$ For each i let $f_{i}$ be a continuous function satisfying $0\leq f_{i}\leq 1$ and such that $f_{i}$ is 0 on ${\overline {U}}_{n_{i}}$ and 1 on the complement of $U_{m_{i}}.$ Let

d(x,y)=\sum _{i=1}^{\infty }{\frac {1}{2^{i}}}|f_{i}(x)-f_{i}(y)|.

We observe that given $x\in X$ and some open set $U_{m}$ in the base containing x, there exists another set $U_{n}$ in the base such that $x\in U_{n}\subset {\overline {U}}_{n}\subset U_{m}.$

Then the following assertions hold:

1. d is a metric on X. We let $X_{d}$ denote X viewed with the metric topology.

2. All open subsets of X are open in $X_{d}.$ Specifically, given an open set U in the base and a point x in U, there exists $\epsilon >0$ such that the ball $B_{\epsilon }(x)\subset U.$

3. All open subsets of $X_{d}$ are open in X. Specifically, given B open in $X_{d}$ and x in B, there exists an open subset U of X such that $x\in U\subset B.$

For (1), it is clear that d is nonnegative symmetric, and the triangle inequality is also immediate. Now if $x\neq y$ we can find an open set $U_{m}$ containing x and not containing y. Applying the observation above we can then find $U_{n}$ such that $x\in U_{n}\subset {\overline {U}}_{n}\subset U_{m}.$ Thus in the sum for d there will be some i for which $|f_{i}(x)-f_{i}(y)|=1$ , so that $d(x,y)>0.$ Thus d is a metric on X.

For (2), find $U_{n}$ with $x\in U_{n}\subset {\overline {U}}_{n}\subset U$ and consider the corresponding i in the sum for d. If $y\not \in U$ then $|f_{i}(x)-f_{i}(y)|=1,$ and hence $d(x,y)\geq 1/2^{i}.$

For (3), let $\epsilon >0.$ For each i there exists an open set $V_{i}$ containing x such that $f_{i}(V_{i})\subset B_{\epsilon }(f_{i}(x)).$ Hence for $y\in V_{i}$ we have $|f_{i}(x)-f_{i}(y)|<\epsilon .$ The intersection $V_{1}\cap \cdots \cap V_{n}$ is an open set containing x. For y in this intersection, we have

d(x,y)\leq \sum _{i=1}^{n}{\frac {\epsilon }{2^{i}}}+\sum _{i=n+1}^{\infty }{\frac {1}{2^{i}}}|f_{i}(x)-f_{i}(y)|.

It follows that we can make $d(x,y)$ as small as we wish by taking n sufficiently large. That is, given $\delta >0$ we can find an open subset V of X containing x with $V\subset B_{\delta }(x).$ — merge 13:31, 31 March 2008 (UTC)[reply]

I can't see any problems. Good work! Algebraist 14:26, 31 March 2008 (UTC)[reply]

Thank you very much! — merge 14:31, 31 March 2008 (UTC)[reply]

I find (2) a little unclear; I find your notation is a little confusing. You've ordered the base as

U_{n}

and then ordered pairs of bases. So to each base element

U_{n}

there exist many (probably countably infinite) pairs

(U_{n_{i}},U_{m_{i}})

. Hence the phrase "consider the corresponding

i

" is ambiguous; there are many such

i

's. Furthermore, if you're proving that "there exists

\epsilon

so that

B(x,\epsilon )\subset U

," there should be some explicit connection between

\epsilon

and your

{\frac {1}{2^{i}}}

(and note that the

{\frac {1}{2^{i}}}

isn't well-defined until you tell us how you choose

i

). The ideas of the proof all seem ok; it's just that I find the notation a little confusing. SmaleDuffin (talk) 17:34, 31 March 2008 (UTC)[reply]

The 'corresponding i' is the i corresponding to the pair

U_{n}\subset U

. Algebraist 17:48, 31 March 2008 (UTC)[reply]

Ah, ok. Thank you. So: Let

U

be a base set; then there exists numbers

n_{i}

and

m_{i}

so that

U=U_{m_{i}}

and

x\in U_{n_{i}}\subset {\overline {U_{n_{i}}}}\subset U_{m_{i}}

. Are there other indices that work as well? Do there exist numbers

n_{j}

and

m_{j}

so that

U=U_{m_{j}}

and

x\in U_{n_{j}}\subset {\overline {U_{n_{j}}}}\subset U_{m_{j}}

with

i\neq j

? If this is possible, consider the following construction. Let, for each

k

,

U_{m_{k}}=U

and

x\in U_{n_{k}}\subset {\overline {U_{n_{k}}}}\subset U_{m_{k}}

, and let

y_{k}

be a point not in

U

. Then

d(x,y_{k})>{\frac {1}{2^{k}}}

, and so

\lim _{k\to 0}d(x,y_{k})=0

. This certainly doesn't prove that

x

is a limit point of points outside of

U

, but the argument provided doesn't seem to rule this out. There should be some argument (probably involving

\epsilon

) that shows that the infimum of

d(x,y_{k})

is strictly greater than zero. Of course, it's been a long while since I've thought about these sorts of things, so I could very well be missing something obvious. SmaleDuffin (talk) 19:08, 31 March 2008 (UTC)[reply]

Huh?

\lim _{k\to 0}d(x,y_{k})=0

doesn't appear to make sense (k is ranging over the natural numbers, yes?) and if you meant

\lim _{k\to \infty }d(x,y_{k})=0

then it doesn't follow from

d(x,y_{k})>{\frac {1}{2^{k}}}

and isn't true, either. All you need to do (given x and U) is pick an i such that x is in

U_{n_{i}}

and

U_{m_{i}}=U

, and then by construction, if z is in

{\overline {U}}_{n_{i}}

and y is outside U, then f_i(z)=0 and f_i(y)=1, so d(z,y) is at least 1/2ⁱ. I really don't see what the problem is here. Algebraist 19:33, 31 March 2008 (UTC)[reply]

Never Mind; I figured out where I was confused. Thanks. SmaleDuffin (talk) 20:35, 31 March 2008 (UTC)[reply]

microeconomic development

A chemical manufacturer processes two chemicals C1 and C2 in varying propositions to produce three products A,B and C.He wishes to produce at least 150 units of A,200 units of B and 60 units of C.Each tonne of C1 yields 3 of A,5 of B and 3 of C.If C1 costs $40 per tonne and C2 costs $50 per tonne

(a)Formulate the problem as a minimisation LP model

(b)Formulate the inverse or dual of the primal formulation in (a)above

(c)Express the dual in (b)above as standard LP model

(d)set up the initial simplex tableau for the solution to this problem(do not solve)

(e)using the graphical method,identify the feasibility region for the solution in the LP model. —Preceding unsigned comment added by 41.220.120.202 (talk) 09:01, 31 March 2008 (UTm

C1 C2

 A  B   C

A-150
B-200
C-60
SUBJECT TO
C1=3A+5B+3C
C2=5A+5B+C

C1+C2>OR =150A+200B+60C

C1=3A+5B+3C
C2=5A+5B+C

I think that you accidentally omitted one clause from the original assignment, namely:

Each tonne of C2 yields 5 units of A, 5 of B, and 1 of C.

Your model doesn't look right. I assume that by "A" you mean the number of units of A, and by "C1" the number of tonnes of C1. It is a bit confusing to let A, B etcetera stand both for commodities and for quantities, but let's stick with that for now. You express C1 and C2 in terms of A, B and C, but it should be the other way around:

A = 3C1 + 5C2

etcetera. Or are these equations already meant to be the dual problem?

I hope this will help you to go further. --Lambiam 23:47, 31 March 2008 (UTC)[reply]

Following Lambian's remark on a missing premise, the primal could be stated as