Question

Let f(x,y) be xy^2 and let the constraint be g(x,y):x+4y=4
Apply the mothod of lagrange and interpret the result

Answer 1

So lagrange method is:
\[\Delta f = \lambda \Delta g\]

Answer 2

Only those deltas are upside down (gradients)

Answer 3

lol, they are called nabla

Answer 4

For this problem it looks like this:
\[\Delta f = <x,2xy>\]

Answer 5

The gradient symbols?

Answer 6

yes

Answer 7

Im not seeing an option for them here lol

Answer 8

\nabla

Answer 9

So i'll just keep using delta

Answer 10

\[\nabla \]

Answer 11

Magic!

Answer 12

\[\nabla F=<y^2,2y x>\]

Answer 13

yep

Answer 14

and \[\lambda /\nabla g = <\lambda,4\lambda>\]

Answer 15

lambda * , not lambda /

Answer 16

yep, just set up equal to each other and solve for x and y

Answer 17

So since those are equal we can write the following equations:
\[x=\lambda\]
and
\[2xy=4\lambda\]

Answer 18

solving the bottom equation gives:\[y=2\lambda/x\]

Answer 19

from the first equation we know lambda is x

Answer 20

so we know y = 2

Answer 21

plugging that into our constraint equation gives:\[x+4(2)-4=0-> x=4\]

Answer 22

so \[f(4,2)\]
is some minimum or maximum of the function

Answer 23

Right?

Answer 24

plug that into function to see if it works out

Answer 25

constraint function , I mean

Answer 26

no, it doesnt

Answer 27

then , there is mistake somewhere

Answer 28

(4)+4(2)=4

Answer 29

is supposed to =4 i mean

Answer 30

it equals 12

Answer 31

\[y^2=\lambda 1\]

\[2xy=\lambda 4\]

x+4y=4

Answer 32

ohh, doy

Answer 33

For fx i had x, not y^2

Answer 34

4y^2=2xy

x+4y=4

Answer 35

ok

Answer 36

So pretty much

Answer 37

http://www.wolframalpha.com/input/?i=Solve%5B4%20y%5E2%3D%3D2xy%2Cx%2B4y%3D%3D4%5D&t=crmtb01

Answer 38

after doing the lagrange multiplier its free game to do whatever you have to to find x,y?

Answer 39

y=0 , x=4

x=4/3 , y=2/3

plug these two points into function, one will be max other will be min

Answer 40

how are you getting 4y^2?

Answer 41

\[y^2=\lambda \]

while

\[2xy=4 \lambda \]

Answer 42

ah ok

Answer 43

This is really easy

Answer 44

Im thinking into too much and its becoming hard haha

Answer 45

Thanks for the help

Answer 46

When in doubt, don't think too much

Answer 47

you won't always get a/the max and min when you do Lagrange multipliers.

Answer 48

saddle point?

Answer 49

sort of

Answer 50

on this constraint the function f has no min

Answer 51

at both of your critical values \(f(x,y)\ge 0\)

Answer 52

try x=-4 and y=2

Answer 53

hmm

Answer 54

There is another formula that is really useful for finding extrema

Answer 55

the only way we are guaranteed to find the max and min is if f is continuous and the constraint forms a compact set. (then we can use the extreme value theorem)

Answer 56

http://imgur.com/CAy0U

Answer 57

Lagrange Multipliers is the best way to do it...you just need to be careful when checking if you really have maxs and mins

Answer 58

Yea

Answer 59

Like here I am still trying to solve those systems

Answer 60

Ive found x=4/3

Answer 61

But thats it

Answer 62

what happened?

Answer 63

I cant figure out how wolfram gets x=4,y=0 for that system

Answer 64

I see if you multiply the top system by y and subtract you get the fractional answers for x and y

Answer 65

4y^2=2xy

well , this works when y=0

0=0

assuming y=0



x+4(0)=4
x=4

Answer 66

Hmm. ok

Answer 67

So interpreting these results, would it be correct to say: The function f(x,y) has a minimum or maximum at f(4,0) or f(4/3,2/3)