# Brain breaking: Diamond price.

A diamon's price is ratio to square of its mass(not sure if it true in reality). If the diamon was cut to 3 smaller diamonds  the ratio remain the same. Is the owner got more money or less , if he sold the diamond as a big one or cut it to three smaller ones?If it it true, that he will lost money if cut the diamond by three smaller one, in which case, the money lost it most?
A little suggestion: In my opinion, it is a lot easier for one to read and understand other's solutions if all of us use same variables refer to the same problem. So if  you thing my suggestion is not violate your "free of choice" please use the following:

Let M be the mass of the big diamond, and a, b and c be the masses of the 3 smallers one after cut.
Let P be the price of M and x,y,z be prices of a, b, c respectively.
Then what was given is:
P/M^2 = x/a^2 = y/b^2 = z/c^2
find if:         P > x + y + z   *
and if *  is true, find  Max(P - (x + y + z) )
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Commented:
A simple example shows that splitting is a bad choice:
If M=9 and a=3, b=3, c=3 and if we assume the constant P/M^2 to be 1, then P=81 and x=y=z=9 and 81 > 27.

Changing to a=4 and b=2, results in x=16 and y=4. So the increase of a leads to a bigger increase of x (+7) than the decrease of y (-5) caused by the decrease of b.

Hence the price difference is maximized when a=b=c. Because of the squaring, every little increase of one of the weights leads to a bigger increase in price than the decrease in price of the diamond that becomes smaller.

I tried to prove it with the formulas, but I'm stuck at proving that the Max(ab+ac+bc) with M=a+b+c is at a=b=c...
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Commented:
If I understand correctly the question is:

is( m squared) more than 3 times ((m/3) squared)?

By my old algebra I get:

m squared minus three (m/3) squared = 2/3 m squared

That means m squared will always exceed the other by a factor of 2/3 m squared.

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Commented:
Lagrange multipliers gives me split it into 3 for the max loss but I think there is probably a prettier way of doing this.

The problem reduces to finding (if E is the price difference)

Max E =  M² -(a² + b² + c²)

subject to a + b + c = M.  Equivalently as a Lagrange multiplier this is

Max E =  M² -(a² + b² + c²) + L(M-a-b-c)

the Max/Mins are given by dE/da = dE/db = dE/db = dE/dL =0

ie -2a-L=0   -2b-L=0  -2c-L=0  M-a-b-c=0

these have solution a=b=c=M/3  L=-2M/3 (clearly this function only has  1 turning point and is a Max)

As I said there is probably a prettier way and I will look later.

GfW
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Commented:
Sorry E is the price difference  divided by the constant of propotionality to be precise
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Commented:
look at squares:
if        a^2  =  b
then   (a+1)^2  =  b +(2a+1)
and    (a-1)^2   =  b  -(2a-1)

thus increasing the size of a diamond will allways give more 'gain' than an equal reduction of the size of an same-sized or smaller diamond will give 'loss'.

re "not sure if it true in reality"  >  not true, if it were the profession of 'diamondcutter' would not exist.
partly true in the early stages of gathering where the whethered matte surface of the raw diamonds make it hard to inspect then for color and other 'wonderfull flaws' :-))

regards JakobA
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Commented:
This is a bit prettier. To maximise E

E  =  M² - (a² + b² + c²)

= -2M²/3 -( a² + b² + c²) + 2M² - M²/3

= 2M²/3 - (a² + b² + c²) + 2M(a + b + c) - 3M²/9

= 2M²/3 - (a-M/3)²- (b-M/3)² - (c-M/3)²

Max clearly occuring at a = b= c= m/3
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Commented:
something wrong in the steps there should end up as

E  =  M² - (a² + b² + c²)

= M²/3 - (a-M/3)² - (b-M/3)² - (c-M/3)²

Max  occuring at     a =  b =  c =  M/3
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Commented:
This is it (phew!)

E = (a+b+c)² - (a² + b² + c²)

= M² - (a² + b² + c²)

= 2M²/3 -(a² + b² + c²) + M²/3

= 2M²/3 - (a² + b² + c²) + 2M²/3 - M²/3

= 2M²/3 - (a² + b² + c²) + 2M/3(a + b + c) - 3M²/9

= 2M²/3 - (a-M/3)²- (b-M/3)² - (c-M/3)²

with Max of  2M²/3  occuring at  a = b = c = M/3

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Author Commented:
Hi all
thank for participating
I dont get why you have E = M^2 - (a^2+b^2+c^2)While  M and a, b, c are masses(unit karat in diamond business, may be called  ounce in science) and  E, unit should be dollars (\$) Please read the post again. And please show all the steps
I dont know why you all ignore the variable P (which is the value of the diamond before cut).
By the way how do you write the square sign like that( not a^2).  Did you write from a word document then paste into this?Or can you write directly
To JacobA. And it should be the variable use to compute the money lost too.
I think it true, JacobA. I just remember the story "It good to be Michel Duglas and Catherin Zethazon (something like that : ))) he bought for her 10 karats diamond for they weeding for \$200,000 , I knew that a 2karats_diamond now cost about \$5,000. So the bigger the rock, the more valueble it get???

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Commented:
Sorry my I should have been more explicit in my analysis I was working in non-dimensional values. ( I get the a² etc by cutting and pasting from a notepad file with the ASCII  set in it which I keep handy, I have not found a key baord method for a text area. I posted that set earlier today for all to use)

Since price is  k*mass² then price of large diamond is P =  kM² = k(a+b+c)² and the prices of the smaller are  ka², kb², and kc² then the problem is to find

Max E = kM² - (ka² + kb² +  kc²)

which is equivalent to finding (since k is +ve)

Max ( M² - (a² + b² +  c²) )

which I have show above   (using M=a + b +c )

Max ( M² - (a² + b² +  c²) ) = Max ( 2M²/3 - (a-M/3)² - (b-M/3)² - (c-M/3)² )

and clearly occurs at a = b = c = M/3

price of large P = kM²

price othe 3 smaller diamonds at weight  M/3 each is

3*k(M/3)² = kM²/3 = P/3

hence the difference is   P - P/3 = 2P/3

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Commented:
If you do not like me dropping the k then we can do like this as well ( using M=a + b +c )

Max E = kM² - (ka² + kb² +  kc²)

= k( M² - (a² + b² +  c²) )

= k( 2M²/3 - (a-M/3)² - (b-M/3)² - (c-M/3)² )

Max occuring at   a = b = c = M/3

with max being    2kM²/3  = 2P/3

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Author Commented:
You got it.
However, in my work, the different between P and (x+Y=y+z)
is: P - (x+y+z)= 2P(ab+bc+ca)/M^2.
I did not see any wrong in your work.  So you got the points
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