Math Challenge

[xtreme2k]

hehe
i get it now

my mathz arent that bad really 😉
doing civil engineering at uni

 

[Repa]

The problem has not a unique solution if the numbers are real. In fact it has infinite solutions.

For example
1,254652
0,964
2,125258
2,766
is an approximate solution as well.

If one uses the constraint of integer cents then it is a much more interesting problem at which one could exploit the fact that 711000000 has a large prime factor and it makes the solution unique.

 

[Duckers]

I know this is an old thread.

I would like to ask if anyone understood how Dr Math obtained this last equation:


493039*a^4 - 13312053*a^3 + 119808477*a^2 - 359425431*a + 243000000 < 0

 

[Duckers]

yes, I e-mailed Dr. Math and he didn't reply 🙁

 

[thEnEuRoMancER]

As far as the units are concerned, ABCD doesn't equal 7.11 dollars but 7.11 dollars to the power of 4 🙂

 

[Duckers]

It's nice to see you 🙂

 

[thEnEuRoMancER]

OK Duckers, let me save your ass again 🙂

...The fact that the arithmetic mean of B, C, and D is greater than their
geometric mean gives

(B+C+D)/3 = (711-79*a)/3 < (B*C*D)^(1/3) = [711000000/(79*a)]^(1/3)...

this equals to:
(711-79*a)/3 < [711000000/(79*a)]^(1/3)

now just put the left and the right side to the power of three:

((711-79*a)/3)^3 < ([711000000/(79*a)]^(1/3))^3 = 711000000/(79*a)

this is same as

((711-79*a)/3)^3 - 711000000/(79*a) < 0

arrange it and you'll get

493039*a^4 - 13312053*a^3 + 119808477*a^2 - 359425431*a + 243000000 < 0

 

[thEnEuRoMancER]

Another thing, if the number of unknows is bigger then the number of linear equations, there is an infinite number of results. Only the magnitude of unknows is limited by 0$<x<7.11$ . So you could make two unknows up and just calculate the remaining two unknows from two equations.

But it is unacceptable for a price to be sth like 3.4565567474$ so we have another restraint: price*100 = price in cents must be an integer. That's why Doctor Rob had to use some &quot;sophisticated&quot; math. The rounded solutions are no good here.

 

[Duckers]

lol, I get it now 🙂

I missed this part:

((711-79*a)/3)^3 - 711000000/(79*a) < 0