Ac3
  • Ac3
Find the extreme values of the function subject to the given constraint. f(x,y,z)=x^(3)+y^(3)+z^(3) x^(2)+y^(2)+z^(2)=4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Ac3
  • Ac3
@IrishBoy123
Ac3
  • Ac3
i'm on the last part.
Ac3
  • Ac3
I go the point.....\[(\pm2/\sqrt{3},\pm2/\sqrt{3},\pm2/\sqrt{3})\]

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Ac3
  • Ac3
I also got x=y, x=z, which means y=z. So my question is since they are all technically equal should i just plug in the positive and negative version of that point into the original equation and that'll give me the max and min values or should I interchange the signs?
Ac3
  • Ac3
for instance I can plug in all positive 2/sqrt3 and all negative of that and i'm done? or am i supposed to go one by one (+,+,-) and then (+,-,+) then (-,+,+) etc
IrishBoy123
  • IrishBoy123
because \(\lambda = x = y = z\) logically i would go with the same value for each
Ac3
  • Ac3
I thought the same alright thanks man.
Ac3
  • Ac3
so i got for my max 8/root(3) and for my min -8/root(3)
Ac3
  • Ac3
is that what you got?
IrishBoy123
  • IrishBoy123
so you will get 2 extrema, at the - and then the +, i reckon. which makes practical sense, as you are mixing cubed terms......
IrishBoy123
  • IrishBoy123
i get \(f =x^3+y^3+z^3\) \(\nabla f = 3\) \(g = x^2+y^2+z^2- 4 = 0\) \(\nabla g = 2\) \(\nabla f = \lambda \nabla g \implies \lambda = x = y = z\) \(3x^2 = 4 \implies x = \pm \frac{2}{\sqrt{3}} = y = z\) \(f_{max} = 3( \frac{2}{\sqrt{3}})^3 = \dfrac{8}{ \sqrt{3}}\) ditto for the min, \(-\dfrac{8}{ \sqrt{3}}\)
Ac3
  • Ac3
nice! thanks irish boy

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