Good effort.

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 solve this problem you could go to a site I found recently for someone else:

http://www.zweigmedia.com/RealWorld/simplex.html

Bob

]]>"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 ?

Bob

]]>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 ?

Bob

]]>volume of P / surface area of P

In other words, what is the smallest real number t such that

volume of P/ surface area of P is less than or equal to t

must be true for all polyhedra P that can be inscribed in a sphere of radius 36?

Solar.

]]>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. )

Bob

]]>]]>

Solar

(btw this is aops geo stuff)

]]>Also, thanks bobbym for straightening my LATEX.

]]>I said that you need all primes X/2 to sieve X. Isnt this obivious?

Don't you only need the primes that are less than sqrtX to sieve X?

@martnSpruce I think your idea is interesting but I find it slightly hard to distinguish it from the Sieve of Eratosthenes.

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

gives

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.

Bob

]]>