anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
yes
anonymous
  • anonymous
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
  • zzr0ck3r
you need to take partials

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zzr0ck3r
  • zzr0ck3r
then you have a system of equations set to 0
zzr0ck3r
  • zzr0ck3r
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
  • zzr0ck3r
that should be y^3
anonymous
  • anonymous
after partial derivative what should i do?
zzr0ck3r
  • zzr0ck3r
x=cuberoot(y/12) 12y^3-cuberoot(y/12)=0 12y^3=cuberoot(y/12) what is the only solution to this?
zzr0ck3r
  • zzr0ck3r
I think I am doing this right, sorry its been a bit.
zzr0ck3r
  • zzr0ck3r
do you know the answer?
Zarkon
  • Zarkon
There will be 3 solutions to the system of equations...thought not all will correspond to a min
Zarkon
  • Zarkon
*though
zzr0ck3r
  • zzr0ck3r
what am I doing wrong? I thought we set the partials to 0, then solve?
Zarkon
  • Zarkon
that is correct
anonymous
  • anonymous
i dont know the answer..
Zarkon
  • Zarkon
just as a reminder...the equation \(x=x^3\) has 3 solutions ... -1,0,1
Zarkon
  • Zarkon
for the same reasons you will get 3 solutions to your system
anonymous
  • anonymous
still confused..

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