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#1 Re: Help Me ! » Linear Programing - word problem confusing me ! » Yesterday 06:10:00

hi Martha

Good effort.  smile

The sales need x, y and z in them.

And I think you'll find it easier to write the costs and sales as decimals  eg    $0.20x .  Then the 'sums' are easier.

The final expression will have 0.1x etc.

But don't worry.  There is a simple reason for this: I suggested working in Kg.  I actually started with 100s of Kg but it was getting too complicated when 1000s came up as well so I simplified.  But it wouldn't be wrong to work in 100s or even in 1000s.  The question doesn't say.  With a two variable problem you can plot the constraints as lines and shade areas.  The 'profit' expression can be made into an equation by choosing an arbitrary profit value, plotting that line, and then tracking up the graph with parallel lines until we find the last allowable point as the parallel lines run out of the solution space.  So you can take your profit line, multiply every term by 10 or 100 or 1000 and it will still be a profit line and you can still find the maximum.

With 3 variables you cannot draw a graph but the same principle applies.  Just scale up all the terms until you have the book values.  If you actually went all the way to a solution, one person might have x = 2500 Kg say and another might have x = 2.5 hundreds of Kg.  Same answer.

If you want to complete the solution of  this problem you could go to a site I found recently for someone else:


#2 Re: Help Me ! » Is Monty Hall problem still recognized to be corrected? » Yesterday 04:21:13

"1/2" chance to get the CAR is closer to probability in practice instead of "2/3"

No!  These two events (car or goat) are NOT equally likely.  Have you looked at all the analysis ?


#3 Re: Help Me ! » Linear Programing - word problem confusing me ! » Yesterday 04:00:39

hi Miss_band

Welcome to the forum.

What you have is ok so far.  It is a bit confusing that they keep jumping from 100s to 1000s to 10000s.  So I would stick to Kg and adjust everything as you make the equations.

Consider what are your variables.  The question wants us to decide how much of each feed to make, so call the amounts (in Kg) x, y and z.

Now you have to make the constraints.

If you make x Kg  of feed 1, then you'll need A =  0.40x Kg,     B = 0.40x Kg,        C =  0.20x Kg

Do similar for feed 2 and feed 3.

So the total of material A you'll need will be 0.40x + 0.20y  + 0z.  This must be less than 50000

Get two more constraints like this.

Now what about your profit.  The cost of x will be  $ x/1000 times 200 = $ x/5 = $ 0.20x

Get two more expressions like this.

Then work out your sales (assuming you sell it all)

For x you'll sell at $ 30x/100 = $ 0.30x.  Do the same for y and z.

Then get the profit expression which will be sales minus costs.  That's what you want to maximise.


#4 Re: Help Me ! » Is Monty Hall problem still recognized to be corrected? » Yesterday 03:40:28

I believe that it is assumed you would not choose the door that had been revealed as having a goat.  So interpretation (2) is correct.


whoops.  hi bobbym.  You got there first.  smile

#5 Re: Help Me ! » Formal Limits » Yesterday 03:34:27

First you choose an e so that your x point is close to zero.  Then you generate f(0+e) .  If it's close to zero, then choose a smaller e.  The limit rule requires that you can get arbitrarily close.  Something like this:

   x           e             f(x+e)

   0           -3              0                 (oh dear I seem to have reached zero, but now to reduce e ..)
   0           -2              1
   0           -1              2
   0           -0.5           2.5
   0           -0.25         2.75
   0           -0.1           2.9
   0           -0.01         2.99


As e gets smaller and I get closer to x = 0, f(x) does not get closer to zero. (It gets closer to 3 of course.)

So you might be able to choose an e that makes f(x) = 0, but that's not enough.  You have to show that as the x gets closer to zero, the f(x) gets closer to whatever.


#6 Re: Help Me ! » Formal Limits » Yesterday 02:13:11

If e is small then x +/-e is a neighbourhood of x.

And f(x +/-e) is the corresponding neighbourhood of f(x).

Let's take +e.  -e is similar.

Then f(0 + e) = 0 + e + 3

No matter how small e gets,  e + 3 is not close to 0.


#7 Re: Help Me ! » Formal Limits » Yesterday 01:42:47

Oh whaat.gif

You have to show that in a small neighbourhood of x = 0, f(x) is not in a small neighbourhood of zero.  As f(0) = 3, that shouldn't be hard.


#8 Re: Help Me ! » Is Monty Hall problem still recognized to be corrected? » Yesterday 01:26:42

Monty Hall?  That means Christmas must be coming.  For like Santa, Monty surfaces once a year.  smile

hi oem7110

What didn't you like about the mathforum explanations?

Here's my suggestion:  Take a look at some past threads: from post 12 on.


#9 Re: Help Me ! » Statistics PLEASE! » 2014-10-29 20:42:30

hi jeff 23

Thanks for the information.  I'll start by looking at a simple case.

Suppose there are 10 cards and you have bought 3.  One of the ten is special.  What's the chances of you having it ?  You have three chances out of ten of getting it so the probability of owning the card = 3/10 = 0.3 = 30% 

Now I've done some thinking on this, I don't think it matters who owns the others or even if they are unsold.  Nor does it matter what the world population is; nor how many fans there are.

Note:  You need to know how many cards the company made; not how many the card authenticators have analysed.

So what will you do with it ?  Get it encased and display it ?  Or save it as part of your pension fund ? 


#10 Re: Help Me ! » Formal Limits » 2014-10-29 20:22:35

Thanks Stefy.  smile

I thought I was just being stupid.


#11 Re: Help Me ! » Statistics PLEASE! » 2014-10-29 00:51:50

hi Jeff 23

Welcome to the forum. 

You have described a whole new world to me.  I just do the math.

Some questions first so that I can get a clear view of the situation.

(1)  Who makes and numbers the cards ?

(2)  How are they distributed to fans ?

(3)  Do you have any control over what player you get or is it random ?

(4) Do you have any control over what number is on the card or it that random ?

(5) Who says there are 23,461,850 cards ? How do they know ?

First thoughts:

Other numbered cards may also have these digits, but in a different order:  eg. 45848423; 23238445; 23458484 etc etc.

Also fans may have numbers that are considered special for other reasons, such as the number of goals he has scored, number of matches etc.

If N cards are sold, each with a unique and sequential number, then the probability of getting one (at random) and it having a particular number, is 1/N

But, even now, there are complications.  Suppose you were wishing to get a special number but no card exists with this number.  Then the probability of getting it is zero.

I suspect there are too many unknowns for an accurate calculation.

I don't think the population of the world has any bearing.  It may come as a shock to you, but many people do not collect the cards.  You would need to restrict your calculation to just the card collectors, and you'd need to know how many such cards there are.  After all, if there were only two collectors, and you had all the available cards between you, then it would be fairly likely that you would be the one with a particular card.


#12 Re: Help Me ! » Tautologies » 2014-10-28 23:14:12

Reduction to DNF.  In this case I am trying to put all terms in the form (A^B^C) V etc

Left hand side :

≡ ( A→(B→C)) ≡  ¬A V (B→C) ≡  ¬A V (¬BVC) ≡ (¬A^TRUE) V (A^(¬BVC))

≡   ¬A^[(B^C) V (B^¬C) V (¬B^C) V (¬B^¬C)]     V     (A ^ [(¬ B^C) V (¬ B^¬ C) V (B ^ C) ]

≡  (¬A^B^C) V (¬A^B^¬C) V (¬A^¬B^C) V (¬A^¬B^¬C) V (A^¬ B^C)  V  (A^¬ B^¬ C)  V  (A^B ^ C)

Right hand side :

≡ ((A→B)→C) ≡ ¬(A→B) V C  ≡ ¬(¬AVB) V C  ≡ (A^¬B) V C     

≡  (A^¬ B^C)  V (A^¬ B^¬C)   V  (A^B V C)  V (¬A^B ^ C) V  (A^¬B ^ C)  V (¬A^¬B ^ C)

Remove duplicate and re-order

≡) (¬A^B ^ C) V                      (¬A^¬B ^ C) V                          (A^¬B ^ C)  V  (A^¬ B^¬C) V   (A^B V C)

Compare with LHS
≡   (¬A^B^C) V (¬A^B^¬C) V (¬A^¬B^C) V (¬A^¬B^¬C) V (A^¬ B^C)  V    (A^¬ B^¬ C)  V  (A^B ^ C)

Clearly, these are not the same.

Here are the truth tables for the same.  (It took me a fraction of the time to do it this way.  smile )



#13 Re: Help Me ! » stuff in a box » 2014-10-28 20:47:41

The height has to be 'perpendicular to the base'.  For this polyhedron, I'm unclear how you are calculating this.  Look at post 3.  My method avoids this difficulty by getting the volume of four 'easy' pyramids and subtracting from the volume of the cuboid.  If I've got that right then the answer is an integer.


#14 Re: Help Me ! » some 3d stuff :D » 2014-10-27 20:46:43

SolarDevil wrote:

When I enter it in it says that x is incorrect

Where did you get this problem from?  I have not tried 'entering it' anywhere, but I will if you post the link.


#15 Re: Help Me ! » Tautologies » 2014-10-27 20:36:15

hi SolarDevil,

Do you want me to reduce both expressions to disjunctive normal form?


#16 Re: Help Me ! » Calc Problem for Class » 2014-10-26 20:51:35

hi Solardevil,

I couldn't do this so I tried it on Wolfram.  Here is the answer I got:


Please check whether I have the right question.


#17 Re: Help Me ! » Tautologies » 2014-10-26 20:38:57

Don't want to get into a deep religious discussion on this, but, strictly, it will repeat until the event occurs, irrespective of what either of us says.


#18 Re: Help Me ! » stuff in a box » 2014-10-26 20:35:52

hi SolarDevil

That's not what I'm getting.  Please post all your working, so I can compare with mine.


#20 Re: Help Me ! » Solve Equation » 2014-10-26 02:22:31

hi PhuongMath

[math] \Leftrightarrow (x+2)^3=-6x^3\Leftrightarrow x+2=x\sqrt[3]{-6}\Leftrightarrow x=\frac{2}{\sqrt[3]{-6}-1}[/math]


Yes, that is consistent with my graphing attempt.  x is approximately -0.70994

Cube root (6) = 1.817121  = r

The three cube roots of -6 are therefore -1.817121, and complex solutions with the same modulus, r and arguments, + or - pi/3

This places them 1/3 of a turn apart on a circle radius r.

So you can construct the other two values of x from this.

If this is not what you were expecting, then you'll need to go back to the start of the problem so I can see how you got to that equation.


#21 Re: Help Me ! » Solve Equation » 2014-10-25 23:32:01

hi PhuongMath

Welcome to the forum.  smile

I couldn't spot an easy factorisation, so I went to the function grapher: … x)&func2=2

Both cubics have only one intersection with the x axis and zooming in shows it's not an integer.

So what methods do you know for getting a value for this ? eg. Newton's (approximation) method

Once you have one factor you could go on to find two complex values but it will not be easy arithmetic.


#22 Re: Help Me ! » Tautologies » 2014-10-25 22:54:03

Why is that? Does your computer get bored after a bit?  How about this:

                          Me: Here's my answer.
                  Lecturer: Don't do it like that.

UNTIL Hell freezes over


#23 Re: Help Me ! » Tautologies » 2014-10-25 22:47:28


WHILE 1 = 0
                          Me: Here's my answer.
                  Lecturer: Don't do it like that.


#24 Re: Help Me ! » algebra » 2014-10-25 22:43:21

Arhh.  Got it.  You mean wz not equal to 1.

I wrote w = a + bi and z = c + di

Then form the fraction and (without yet trying to simplify) multiply top and bottom by the complement of the bottom.

The new denominator will be real and the imaginary part of the numerator is

(b+d)(1 + ac - bd) - (a+c)(bc+ad)

You know that a^2+b^2 = c^2+d^2=1

So expand this, cancel and simplify until you're left with zero.

So the fraction is (a real)/(a real).


#25 Re: Help Me ! » algebra » 2014-10-25 22:32:06


What does this mean:  wz\ne -1   ?


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