Question

find the extreme values using Lagrange multiplier subject to the given constraints

f(x,y,z) = x + 2y,

x + y + z = 1

y^2 + z^2 = 4

Answer 1

I used to be good at this at one point. Let me see if I remember anything.

Answer 2

okay, i know how to do them. i'm just having some trouble with this one

Answer 3

i'll cal lambda L
fx = L*gx
fy=L*gy

Answer 4

hold on i dont even remember what to do next lol, getting my notes.

Answer 5

Okay:
fx = L*gx
fy = L*gy
fz = L*gz
So:
1=L*1
.. hmm i'm lost why do u have two constraints. Zarkon can u help us out?

Answer 6

\[g(x,y,z)=x+y+z-1\]
\[h(x,y,z)=y^2+z^2-4\]

you want

\[\nabla f=\lambda\nabla g+\gamma\nabla h\]

Answer 7

do you see how to proceed?

Answer 8

is that lambda h?

Answer 9

I used gamma...but you could use lambda1 and lambda 2

Answer 10

so i'm going to have 4 unknowns? can you solve it so i can see how to do it?

Answer 11

I have a solution...you show me what you can do and I'll let you know if it is correct.

Answer 12

using \[\nabla f=\lambda_1\nabla g+\lambda_2\nabla h\]

can you find

\[f_x=\lambda_1g_x+\lambda_2 h_x\]

Answer 13

okay so i have (0,2,0) = \[\lambda(1,1,1) + \beta(0,2y,2z)\]

Answer 14

why (0,2,0)

Answer 15

(1,2,0)

Answer 16

ok

Answer 17

so what is \(\lambda\)

Answer 18

1, 2-2ybeta,-2zbeta

Answer 19

the first equation gives us \[1=1\cdot\lambda=0\cdot\beta\]

thus \[\lambda=1\]

Answer 20

yup got that

Answer 21

\[1=1\cdot\lambda+0\cdot\beta\]

Answer 22

ok...solve the other 2 equations for beta...then set them equal to each other

Answer 23

should i ignore lambda?

Answer 24

lambda=1

Answer 25

oh yeah!

Answer 26

you should get a relationship between y and z

Answer 27

b = 1/2y and b = -1/2z

Answer 28

so i set those equal

Answer 29

y = -z

Answer 30

yes...since they are both equal to beta

Answer 31

correct

Answer 32

nopw look at the equation \[y^2+z^2=4\]

Answer 33

z = +/- root 2

Answer 34

yes

Answer 35

so what is y then?

Answer 36

so y is the same, well -/+ root 2

Answer 37

just the opposite

Answer 38

correct...opposite sign

Answer 39

now look at x+y+z=1

Answer 40

and x = 1!

Answer 41

yep

Answer 42

awesome thank you so much!

Answer 43

np