constrained optimization 3 variables

burt

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Aug 1, 2019
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The following image includes the question and my work. The reason why I am posting it here is because I am starting to feel comfortable with 2 variables, but the third one is throwing me off. I am pretty certain about my equations, I just am not sure that my method for solving for my points was valid.
Any input is appreciated!
Thank you for all your help as I am trying to learn this complex subject!
1586718194778.png
 
I'm showing a minimum of -12 at

[MATH]\left\{-\sqrt{2},-1,\sqrt{2}\right\},\left\{ \sqrt{2},-1, -\sqrt{2}\right\}[/MATH]
I'm not sure how you've been taught but I've always done Lagrange multipliers as solving

[MATH]\nabla (f+\lambda g) = 0[/MATH]
i.e. you have to combine the function to be maximized and the constraint when taking the gradient and solving.

You seem to do this separately for the two of them.
 
I'm showing a minimum of -12 at

[MATH]\left\{-\sqrt{2},-1,\sqrt{2}\right\},\left\{ \sqrt{2},-1, -\sqrt{2}\right\}[/MATH]
I didn't use these because I found that \(z=x\).
Was that assumption incorrect on my part?
 
I didn't use these because I found that \(z=x\).
Was that assumption incorrect on my part?

flip the sign on \(\displaystyle \lambda\) and \(\displaystyle z=-x\) becomes a valid solution as well.
 
So you add the two functions first and then take the gradient?

you want to maximize/minimize \(\displaystyle f(x,y,z)\) subject to \(\displaystyle g(x,y,z) = 0\)

Solve \(\displaystyle \nabla (f + \lambda g) = 0\)

and choose the critical points that produce the maximum/minimum

If there is no solution then you need to explore the boundaries as you were doing.
 
you want to maximize/minimize \(\displaystyle f(x,y,z)\) subject to \(\displaystyle g(x,y,z) = 0\)

Solve \(\displaystyle \nabla (f + \lambda g) = 0\)

and choose the critical points that produce the maximum/minimum

If there is no solution then you need to explore the boundaries as you were doing.
I think I understand. Basically, what I was doing was fine - but it's unnecessary work if it can be found in your simpler way.
 
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