2008/012) x+y+z=6 , x^2+y^2+z^2=306 ,find maximum and minumum of XYZ

x+y+z = 6
x^2+y^2+ z^2 = 306
find minumum of maximum of xyz
the three unknowns need not be positive.

x+y + z = 6 ..1

x^2+y^2+z^2=306.. 2

square (1)
x^2+y^2 + z^2 + 2xy + 2yz + 2xz = 36

so xy+xz+yz = (36-306)/2 = -135

now let us take

p(t) = (t-x)(t-y)(t-z)
= t^3 - t^2(x+y+z) +t(xy+yz+xz) - xyz

= t^3 - 6t^2 + 135t - k
k = t^3 - 6t^2 - 135t
k is maximum or minimum when t^3 - 6t^2 - 135t
is

now you can differentiate wrt t and equate to zero

3 t^2 - 12 t - 135 = 0

or t^2 - 4t - 45 = 0

(t-9) ( t + 5 ) = 0

t =9 gives t^3 - 6t^2 - 135t = 729- 486- 1215 = -972 minimum
t =-5 gives -125-150+ 675 = 400 maximum