zmudz
  • zmudz
Find the maximum \(p\) such that \(2x^4y^2 + 9y^4z^2 + 12z^4x^2 - px^2y^2z^2\) is always nonnegative for all \(x\), \(y\), and \(z\) real.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
hmm, since all of our exponents are even, the results will always be postive for the first 3 terms ...
amistre64
  • amistre64
i wonder if a gradient would help out any ...
amistre64
  • amistre64
\[F(x,y,z)=2x^4y^2 + 9y^4z^2 + 12z^4x^2 - px^2y^2z^2\] \[F_x=8x^3y^2 + 24z^4x - 2pxy^2z^2\] \[F_y=4x^4y + 36y^3z^2 - 2px^2yz^2\] \[F_z=18y^4z + 48z^3x^2 - 2px^2y^2z\] wonder where i would go from there if this notion has any worth

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
might want to work this for p instead of F 2 x^4 y^2 + 9 y^4 z^2 + 12 z^4 x^2 - p x^2 y^2 z^2 >= 0 2 x^4 y^2 + 9 y^4 z^2 + 12 z^4 x^2 >= p x^2 y^2 z^2 2 x^4 y^2 + 9 y^4 z^2 + 12 z^4 x^2 ----------------------------------- >= p x^2 y^2 z^2 2 x^2 + 9 y^2 + 12 z^2 ---- ---- --- >= p z^2 x^2 y^2 then my idea of min/max might be useful, but then again it might not ...
thomas5267
  • thomas5267
It seems that p=0. Using Mathematica, the equation can be smaller than 0 even if p=0.0000000000000000001.
thomas5267
  • thomas5267
Alternatively, you can show that for all \(p>0\) there always exist \(x,y,z\) such that: \[ \frac{2 x^2}{p z^2} > 1\land \frac{9 y^2}{p x^2} > 1\land \frac{12 z^2}{p y^2}>1 \]
thomas5267
  • thomas5267
I did something wrong.
zmudz
  • zmudz
@thomas5267 yea, p=0 isn't working :(
thomas5267
  • thomas5267
Showing that for all \(p>0\) there exists \(x,y,z\) such that: \[ \frac{p z^2}{2 x^2} > 1\land \frac{p x^2}{9 y^2} > 1\land \frac{p y^2}{12 z^2}>1 \] will show that p=0 is the maximum p.
thomas5267
  • thomas5267
The idea is to show that \(x^2y^2z^2\) dominates all other term for some \(x,y,z\).
thomas5267
  • thomas5267
I put the wrong function in Mathematica lol.
thomas5267
  • thomas5267
It seems like p=18.
zmudz
  • zmudz
aha! Yay, so x^2 =3, y^3 = 2, x^2 = 1. that makes sense, and you can find it using am-gm. thanks for your help!
thomas5267
  • thomas5267
How to find it using AM-GM?
zmudz
  • zmudz
when you apply the inequality, you should get something like \(\frac{2x^4y^2 + 9y^4z^2 + 12z^4x^2}{3} \geq 6x^2y^2z^2\) this simplifies to \(2x^4y^2 + 9y^4z^2 + 12z^4x^2 - 18x^2y^2z^2 \geq 0\) then that only works if \(2x^4y^2 = 9y^4z^2 = 12z^4x^2\)
zmudz
  • zmudz
and that only works if \(x^2 = 3, y^2 = 2, z^2 = 1\) which gives p=18 once you plug it back in
thomas5267
  • thomas5267
Oh I see!

Looking for something else?

Not the answer you are looking for? Search for more explanations.