Question

what is the Maximum and minimum of the f(x,y,z)=x^2 -2yz in {(x,y,z):z>=0,x^2+y^2+(z-1)^2<=3

Answer 1

\[|x|_{\max}=|y|_{\max}=|z-1|_{\max}=\sqrt{3}\]
zmax=sqrt(3)-1 and z min=0
\[x_{\max}=\sqrt3 ,x_{\min}=-\sqrt3;\]\[y_{\max}=\sqrt3 ,y_{\min}=-\sqrt3\]
so f(x,y,z)_{max}=(xmax)^2-2*ymin*zmax
=3+2*sqrt(3)*(sqrt(3)-1)
=9-2sqrt(3)

Answer 2

f_{min}=0^2-2*ymax*zmax
=-2*sqrt(3)*(sqrt(3)-1)
=-6+sqrt(3)

Answer 3

INTERNAL EXTREMA:
Partial Derivatives:
fx = 2x
fy = -2z
fz = -2y

Set each to zero and you get -> P(0,0,0)
f(0,0,0) = 0

BOUNDARY EXTREMA:
Because the region is a circle (center (0,0,1); radius sqrt(3)) your max will be along the boundary. Using common sense you can tell this will happen at:
\[x = \sqrt3\]
and f(sqrt(3),0,0) = 3 which is a max.
f(0,sqrt(3),0) and f(0,0,sqrt(3)) are both -2sqrt(3) corresponding to a minumum.
therefore the point found earlier, (0,0,0) is a saddle point.