how to find minimum value of f(x, y)= 3x^4+3y^4-xy? should i make first derivative and make the equation equals 0?

Question

anonymous · Answer

yes

anonymous · Answer

okay so take the derivative once, and solve for when f' is 0, then take the derivative again and plug in the points you found for when f' was 0, if they are positive then you have a minimum point

zzr0ck3r · Answer

you need to take partials

zzr0ck3r · Answer

then you have a system of equations set to 0

zzr0ck3r · Answer

3x^4+3y^4-xy
so
partial with respect to x 12x^3-y=0
partial with respect to y 12y^2-x=0

zzr0ck3r · Answer

that should be y^3

anonymous · Answer

after partial derivative what should i do?

zzr0ck3r · Answer

x=cuberoot(y/12)
12y^3-cuberoot(y/12)=0
12y^3=cuberoot(y/12)
what is the only solution to this?

zzr0ck3r · Answer

I think I am doing this right, sorry its been a bit.

zzr0ck3r · Answer

do you know the answer?

zarkon · Answer

There will be 3 solutions to the system of equations...thought not all will correspond to a min

zarkon · Answer

*though

zzr0ck3r · Answer

what am I doing wrong? I thought we set the partials to 0, then solve?

zarkon · Answer

that is correct

anonymous · Answer

i dont know the answer..

zarkon · Answer

just as a reminder...the equation $x=x^3$ has 3 solutions ... -1,0,1

zarkon · Answer

for the same reasons you will get 3 solutions to your system

anonymous · Answer

still confused..