Question

CAL 3 Problem
Using Lagrange multipliers.

Answer 1

Find the critical points of
\[f(x,y,z)=x+2y-z\]
on the surface \[z=1-x^2-y^2\]
Check by substitution of the constraint into the function

Answer 2

Have you considered substitution?

\(f(x,y) = x + 2y - (1 - x^{2} - y^{2})\)

Answer 3

Sorry, I didn't put the complete question. I made the correction

Answer 4

I have to solve using Lagrange multipliers

Answer 5

Fair enough. With substitution, we went down one variable, With Lagrange, we go up one.

\(f(x,y,z,\lambda) = x + 2y - z + \lambda(x^{2} + y^{2} + z - 1)\)

Okay, now make all four of those partial derivates zero at the same time!

Answer 6

how did you find \[f(x,y,z, \lambda)\] ?

Answer 7

Your constraint is \(g(x,y,z) = c\). You must see this.

\(f(x,y,z,\lambda) = f(x,y,z) + \lambda (g(x,y,z) - c)\)

It's a relatively simple construction once you see it.

Answer 8

Thank you for explaining, thats exactly how i thought you found it. so the four partials would be:
\[f _{x}=1+2\lambda x\]
\[f _{y}=2+2\lambda y\]
\[f _{z}=1+\lambda\]
\[f _{\lambda}= x^2+y^2+z-1\]

Answer 9

four equations and four unknowns. do i then solve for lambda by fsubx - fsuby

Answer 10

You must first get \(f_{z}\) correct. ;-)

Answer 11

ohh -1

Answer 12

It's kind of traditional to solve the first three for x, y, and z in terms of \(\lambda\) and substitute into the fourth.

Answer 13

this is what i came up with
\[x=-\frac{ 1 }{ 2\lambda }\]
\[y=-\frac{ 1 }{ \lambda }\]
but no "z" in the third system?

Answer 14

\[\lambda=1\]

Answer 15

so f sub x - f sub y?

Answer 16

Forget that forget that. Setups vary. I was thinking upside down. It's find the way it is.

Answer 17

so how do i find the critical points? or were they found above?

Answer 18

What do you think? With z dropping out, we do not get a unique solution. The 3rd system leads to \(\lambda = -1\). This gives values for x and y.

Where does that leave us?

Answer 19

lol how do i make lambda w/o clicking on "Equation" below?

Answer 20

this is as far as i've gotten

Answer 21

Square x and y and then we can think about what it means.

Answer 22

am i squaring \[x=-\frac{ 1 }{ 2 }\] and

Answer 23

\[y=-1\]

Answer 24

That is what your equation #4 suggests.

Answer 25

OOOOOOHHHHH

Answer 26

\[z=-\frac{ 1 }{ 4 }\]

Answer 27

You have it. We were just a little freaked out when equation #3 lost its z . Don't let that scare you next time. Just go with it! It solves how it solves. I think we're done.

Answer 28

THANK YOU, WISH I COULD GIVE MORE THAN 1 MEDAL

Answer 29

This is a tough section. The setup is so simple, but most examples are TOO simple. Real problems rarely cooperate quite so nicely as the examples.

Answer 30

yes it was very simple now that you explained it. i should've noticed the square's in equation 4. thank you again