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Comments:
<0> nevermind
<1> Regina, so the constraints are 80x+90y<=800.000
<0> the first term is 1/2 in that second sum
<2> wait wait
<1> Regina, what ?
<2> Why are you able to have it be 80x+90y when it's a percentage?
<2> can you Can I just multipy them by 100
<1> Regina, i say we forget the "%" percentage because a 80% of 100 pounds means exactly 80
<2> because of their weight?
<1> Regina, yes ofcourse
<1> Regina, so we want to maximize our function f
<1> Regina, so how do we do that ?
<1> Regina, well we use partial derivatives
<2> wait wait that's so awesome
<2> lol
<2> one second
<1> Regina, yes tell me
<2> well after I have my equations I solve for x and y by finding where they cross x and y axis
<1> Regina, no , you use partial derivatives
<2> that's simple enough, and then you have your feasible region and your corners where you can solve by inserting the corner points into your money equations.
<2> What are those?
<1> oh and i mistyped the function f(x,y) = x*90 + y*100
<2> oh yea, that's where I plug in my points after I graph my feasible region.
<1> Regina, i dont know linear programming
<1> Regina, can you teach me ?
<2> to find my maximum revenue.
<2> REALLY? lol
<1> Regina, really
<2> well I hope my teacher accepts just multiplying it by 100.
<1> it will accept
<2> Well after you have your equations for the premium and standard. you graph those equations
<1> Regina, teach me how you do it
<1> Regina, how do you graph it ? in 3D ?
<2> by solving for x and Y.
<2> 90(0)+80Y<=800000
<2> is where your first line crosses the y axis.
<2> and then do that for x and Y for both your equations
<2> Do you know how to graph inequalities?
<1> Regina, i guess,i just color the region of space wich is on the left of the inequality,or i just choose a point not on the line...and see on wich side it is and color that side
<1> Regina, semiplane coloring i guess
<2> Yea
<1> Regina, are you a girl ?
<2> so that's your region where you can have a profit
<2> and the points where the lines cross is your corners (where your maximum profit will be)
<2> X and Y are constraints because the values cant be negative.
<2> dang, it's tough to explain things without writing them on paper. Bravo to you guys for helping people.
<1> Regina, are you a girl ?
<2> ..I am now ;)
<3> she didnt always used to be
<3> it's called an operation not covered by insurance
<4> She used to be Rex
<2> lol
<1> so we have girls here in #math , this gives me hope that i just may have a chance of getting a proper girlfriend
<1> not from #math ..
<1> ofcourse people present in the channel are excluded..
<1> i was just talking in general
<2> awww lol
<2> I partially go to the math lab for the guys
<1> aren't they geeks ?
<1> cause i'm not
<2> lol, no that's the thing. Most are NOT socially retarded, they are very well ajusted
<1> thats nice to hear :)
<1> most girls from our college are soc. tarded ... and the few that are nice looking ... are ofcourse already taken by some other guy from another college heh ...
<1> anyway ... i have a friend who's tryin to pick up girls in clubs and so on ...
<1> i dont try to catch attention ... but heh ...girls come to me :)
<2> Yea, I think most of the good ones are taken.
<5> so every function defined on a closed interval is bounded correct?
<5> defined at every point in that interval
<6> Any chance we can pretend this isn't #dating?
<7> |Steve| : asl?
<5> you can, by answering my question
<2> ROFL
<7> im 18/f/cali
<5> ...
<5> ok im trying to do some maths here..
<6> JabberWalkie: I can't think of an immediate counterexample.
<7> according to my personal investigation on aol and yahoo chat rooms, approximately 95% of females on planet earth are 18 and live in california
<7> ok im done
<5> well if it is infinite......then it wouln't be definined.....would that count as a proof?
<4> Every countinuous function on a closed and bounded interval is bounded.
<5> provided we are mapping to the reals
<5> not the extened reals
<4> R, for example, is a closed set
<4> But continuous functions on R need not be bounded
<5> Kasadkad: well it wouln't be defined it if was unbounded..
<4> Sure it would
<4> f(x) is defined on all of R
<6> Ah, good point.
<4> er
<5> example?
<4> f(x) = x
<6> f(x)=x.
<5> on what interval?
<4> On R
<1> whats_in_a_name, so do we conclude those are girls ?
<4> (-inf, inf), if you like; it's a closed set
<1> whats_in_a_name, sorry bots...
<5> looks open to me...
<6> It isn't.
<4> Well
<6> Well, it is.
<4> It is :P
<6> But it's both.
<4> It's just also closed
<5> ok on [a,b] then
<5> or strictly closed..
<7> spx2: whatever they were, they got the job done
<8> hmm
<6> The complement of R is the empty set which is open, so R is closed.
<4> Also, it doesn't even work for bounded closed sets if the function isn't continuous: Take the interval [-1,1] and f(x) = {1/x for x != 0; 0 for x = 0}
<8> I was constructing an example, but that one's way better
<6> In fact, given any topological space, the empty set and the whole space are always both open and closed.
<1> |Steve|, R is closed respective to the topology on R , if we have topology on C it weren't closed any more
<8> I was thinking something like the reciprocal of the popcorn function
<5> Kasadkad: but the function isn't defined at x=0
<4> JabberWalkie: I defined it to be 0 at x = 0
<6> spx2: Uh, yes it is.
<8> Which would be neat because it'd be unbounded on all open sets (and all closures of those open sets)
<5> ahh
<6> It contains all of its limit points.
<5> ok
<9> if f(x) is continuous on [1,3] and 2<=f(x)<=4, what is the greatest value of the integration of f(x) from 1 to 3... I think it's 8, just want to check
<8> correct, doofy2
<4> It is 8
<9> k good
<5> ahhh ok, its integral existing garentees that it is bounded
<6> % E^Pi-Pi//N
<4> mbot's not falling for it
<6> Indeed not.
<6> I'm afraid I'm not catching the joke of the latest xkcd.
<10> |Steve|: It's not _really_ 20.
<10> |Steve|: He was pulling their leg.
<11> how do you create functions?
<7> you declare and then define them
<10> dn4: As in... define them by sets?
<6> Yes, I caught that. I figured there had to be something more to it.
<12> can someone help me with the following question: g is a line and E is a plane in R^3. Find a parametric form for g and E. g: {a-5b=-4, 3b-c=2} and E: -a+4b-2c=2
<10> |Steve|: Nope.
<10> |Steve|: Just not so funny.
<6> Bah. I usually have such high hopes for xkcd.
<13> my friend me sent me that link and i told him it wasn't funny
<13> i think he was disappointed at my reaction
<6> dn4: What are you trying to do, exactly?
<6> I really liked the e^{-i pi} one.
<10> dn4: Something like: g = {(x,y) | y=f(x) and x in A} ?
<1> |Steve|, xkcd is lame anywya
<6> spx2: I disagree.
<12> anyone know how to do this? g is a line and E is a plane in R^3. Find a parametric form for g and E. g: {a-5b=-4, 3b-c=2} and E: -a+4b-2c=2
<4> A function f : S -> T is a subset of the cartesian product S x T with the property that if (x,y) and (x,z) are in f, then y = z
<1> |Steve|, do you know of any good comic books,id like to look on some,but not typical ones..
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