It was trial and error method.
Since we want the max value, I did not chose 1 for x,y or z. Similarly no 0.
So we are left with the numbers, 2,3,4,5,6,7,8
If we take 3 numbers that add up to 12 (say 4,6,2), we would assign the highest value to y(since the term has y^3), the second highest value to x (since it has a square) and the least value to z.
So I tried the following,
x y z x+y+z (x^2)(y^3)(z)
2 8 2 12 4096
4 6 2 12 6912
3 7 2 12 6174
5 5 2 12 6250
Hope this helps! But I am sure that should be some other easy way of doing it.
winnerhere wrote:nithi_mystics wrote:Is the answer B?
yes..could you please explain how u got this answer?
Thanks
Nithi
