Question

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

Answer 1

@IrishBoy123

Answer 2

i'm on the last part.

Answer 3

I go the point.....\[(\pm2/\sqrt{3},\pm2/\sqrt{3},\pm2/\sqrt{3})\]

Answer 4

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?

Answer 5

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

Answer 6

because \(\lambda = x = y = z\) logically i would go with the same value for each

Answer 7

I thought the same alright thanks man.

Answer 8

so i got for my max 8/root(3) and for my min -8/root(3)

Answer 9

is that what you got?

Answer 10

so you will get 2 extrema, at the - and then the +, i reckon.

which makes practical sense, as you are mixing cubed terms......

Answer 11

i get
\(f =x^3+y^3+z^3\)
\(\nabla f = 3<x^2, y^2, z^2>\)

\(g = x^2+y^2+z^2- 4 = 0\)
\(\nabla g = 2<x, y, z>\)

\(\nabla f = \lambda \nabla g \implies \lambda = x = y = z\)

\(3x^2 = 4 \implies x = \pm \frac{2}{\sqrt{3}} = y = z\)

\(f_{max} = 3( \frac{2}{\sqrt{3}})^3 = \dfrac{8}{ \sqrt{3}}\)

ditto for the min, \(-\dfrac{8}{ \sqrt{3}}\)

Answer 12

nice! thanks irish boy