Calculus
Hi! So this question is on my midterm review and I'm not quite sure how to solve it.
Find the points on the surface xy3
+z2=4 that are closest to the origin.
Here's what I have so far:
Origin means (0,0,0)
So: D=√x2+y2+z2
D2= x2+y2+z2
and
z2=4-xy3
f(x,y)= x2+y2+4-xy3
fx= 2x-y3
fy= 2y-3xy2
fxx= 2
fyy= 2-6xy
fxy= -3y2
Working with these two first, I set fx and fy to 0
fx = 2x-y3 = 0
fy= 2y-32 = 0
For fx
2x=y3
so x = y3/2
Substitute y3/2
for x in fy
so y = (4/3)1/4
And I wasn't sure how to continue from there. I know it has something to do with D(0,0) being either > or < 0, and saddle points.
The answers given were (unless I've copied them down wrong):
3 critical points
(0,0), (2/3)(4/3^(-1/4)), (4/3)^(1/4)), and (-2/3)(4/3^(-1/4)), (4/3)^(1/4))
I've seen people online suggesting lagrange multipliers, and the only problem I have with that, is that my professor hasn't covered those.
Thanks for any and all help!