Question

Help please! Need help with Lagrange multipliers.
Calc 3.

f(x,y)=4x^3+y^2 ; constraint 2x^2 +y^2=1

Answer 1

where are you stuck?

Answer 2

I've found F_x, F_y, and F_lambda, but in order to solve for x or y you have to divide, which is where I am stuck. x or y can be zeros, so I;m not sure what to do with that. Also, the Ys seem to cancel when solving F_y. Hopefully that makes sense.

Answer 3

what did you get for \(f_x,f_y\)

Answer 4

Let Lambda = L.
F_x=12x^-4Lx
F_y=2y-2Ly

Answer 5

12x^2**

Answer 6

What are you using \(F(x,y,\lambda)\)?

Answer 7

whoops

Answer 8

Yes I am using F(x,y,L)

Answer 9

I found the partials for each from F(x,y,L); then I set them equal to zero.

Answer 10

ok..so you have one equation being


\[2y=\lambda 2y\] correct?

Answer 11

yes

Answer 12

what does that tell you?

Answer 13

L=1

Answer 14

is that all?

Answer 15

well, y=2Ly/2
can't really see much else

Answer 16

does y=0 satisfy the equation too?

Answer 17

I need to do something...be back in a few mins

Answer 18

Okay

Answer 19

back...you you see what I am talking about?

Answer 20

No, I'm not quite sure what you mean.

Answer 21

if \(2y=\lambda y\) then either \(y=0\) or \(y\ne0\) and therefroe \(\lambda=1\)

Answer 22

forgot a 2 \(2y=\lambda 2y\)

Answer 23

\[2\cdot 0=\lambda 2\cdot 0\]

\[0=0\]

Answer 24

Right, but we don't know what y actually equals, but if it did equal zero, does that not mean that we can not divide by it?

Answer 25

if y=0 we cannot divide by it

Answer 26

Right, that's where I am stuck. In order to solve for anything in the equation 2y=2Ly we have to divide.

Answer 27

i'm just looking at cases...either y=0 or it doesn't

Answer 28

if y=0 then it satisfies the equation \(2y=\lambda 2y\)

if \(y\neq 0\) then we can divide by it and get \(\lambda=1\)

Answer 29

What happens if y is 0?

Answer 30

if y=0 what does x have to be?

Answer 31

+- 1/sqrt2 ?

Answer 32

correct...so you now have 2 critical points...are there others?

Answer 33

what happens then if y is not zero and lambda=1

Answer 34

3x=1?

Answer 35

yep

Answer 36

so what do I do for that?

Answer 37

so what is x...and therefore what is y?

Answer 38

x=1/3, y=sqrt7/3?

Answer 39

so now you have two more critical numbers

Answer 40

you forgot the \(\pm\)

Answer 41

Hm, I just plug in to original and see which is biggest at this point, right?

Answer 42

yes and smallest (i guess it depends on what the original question asked for )

Answer 43

Max and min. Okay, I think I get it better now. Check the constraint if x or y is 0...that's where I was lost. Thanks for your help!!

Answer 44

no problem

Answer 45

you do have more critical points though

Answer 46

one of your equations was

\[12x^2=\lambda 4x\]

\(x=0\) satisfies this equation

Answer 47

So here x=L/3

Answer 48

if \(x=0\) then \(y=\pm 1\)

Answer 49

Oh okay. So that would be the last one right

Answer 50

yes

Answer 51

So CPs: (0,+-1), (+- 1/sqrt2,0), (1/3, sqrt7/3)

Right?

Answer 52

scratch the 0 on the second one

Answer 53

yes

Answer 54

the last one is +-

Answer 55

Oh yeah. So quick question: if my constraint were to be something like xy=1, then x and y both could not equal 0 right?

Answer 56

correct

Answer 57

Okay, then I fully understand it now! Appreciate the man.

Answer 58

help*

Answer 59

no problem