Question

calc 3 help plz

use lagrange multipliers to find the max and min values of f(x,y,z)=2x+y-2z subject to the constraint x^2+y^2+z^2=4 and find the points at which these extreme values occur.

Answer 1

this isnt too bad, check on pauls online notes

Answer 2

would you like to go through it?

Answer 3

uhm sure.

Answer 4

Should I just say what I did?

Answer 5

yeah

Answer 6

found gradient of F and lamda gradient of G

Answer 7

put like terms together form the 2 functions, the i components together, j components and z components

Answer 8

http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

Answer 9

ok lets set it up

Answer 10

uhm I did gradient F = lamda gradient G

Answer 11

solve the system
<fx, fy,fz> = lambda <gx,gy,gz>
g(x,y,z) = k

Answer 12

found x=1/lamda, y=1/2lamda and z = -1/lamda

Answer 13

one sec

Answer 14

<2,1,-2> = L <2x, 2y, 2z>
2 = 2Lx, 1 = 2Ly , -2 = 2Lz

Answer 15

Yes I got that, then I solved for x, y and z

Answer 16

ok so we need to solve the system of equations
2 = 2Lx, 1 = 2Ly , -2 = 2Lz , x^2+y^2+z^2=4

Answer 17

so I get
1/x = 1/(2y)= -1/z ,

Answer 18

Yes, I plugged in the values I got, and solved for lamda

Answer 19

yes

Answer 20

so x/2 = y, and z = -x , we can plug that into g(x)
x^2 + (x/2)^2 + (-x)^2 = 4

Answer 21

Would it be right if I just plugged in the x, y, and z values?

Answer 22

x^2 +(x/2)^2 + (-x)^2 = 4 , solve that

Answer 23

Like rather than subbing it in terms of X, would I be able to just put in 1/L for x and 1/2L for y etc?

Answer 24

you want to solve for x, not for L

Answer 25

Ok, but before I solve it

Answer 26

L is just an auxiliary thing

Answer 27

why isn't it x^2+(2x)^2+(-x)^2=4?

Answer 28

do you agree that
2 = 2Lx, 1 = 2Ly , -2 = 2Lz , x^2+y^2+z^2=4

Answer 29

Yes

Answer 30

L = 1/x , L = 1/(2y ) , L = -1/z
now set them equal to each other, since they all equal to L

Answer 31

oohhh, that changes everything, I solved for the variables, not L

Answer 32

basically we want to get rid of the L. it is just a helper, we dont care about L actually

Answer 33

it helps to turn this into a one variable problem,

Answer 34

So after I solve for L

Answer 35

actually you can solve for L if you want

Answer 36

I just erased my work =[[

Answer 37

no problem

Answer 38

in this problem, either way works actually

Answer 39

after you solved L = 1/x, 1/2y, -1/z what do you do?

Answer 40

so I got x = positive or negative 4/3

Answer 41

hhhmm

Answer 42

do you agree x^2 +(x/2)^2 + (-x)^2 = 4

Answer 43

yeah that what i got

Answer 44

x^2+(x^2/4)+x^2=4

Answer 45

x^2 + x^2/4 + x^2 = 4
9/4 *x^2 = 4

Answer 46

its 16/9

Answer 47

x^2= 16/9

Answer 48

right, woops

Answer 49

ok , i agree x = + - 4/3

Answer 50

Yes, I almost erased it again aahhaha

Answer 51

from there you plug x back into L = 1/L correct?

Answer 52

then solve for L, and plug L back into the other ones?

Answer 53

hmmm, we plug this x into f(x,y,z)
first find y and z though

Answer 54

we have two solutions for x ,
(4/3 , , )
(-4/3 , , )

Answer 55

for my Y i got positive and negative 2/3

Answer 56

and x = -z so positive and negative 4/3?

Answer 57

yes

Answer 58

Alright, then plug in EVERY combination of those points, back into the F function, find which one gives you the highest and lowest value, and that's the max/min and the points are the coordinates for which they occur?

Answer 59

so we test
(4/3, 2/3, -4/3) and (-4/3, -2/3, 4/3)

Answer 60

right

Answer 61

I got 8 different points

Answer 62

but the points you just said.... are the max/min points ahah, how'd you figure them out so quick?

Answer 63

8 different points?

Answer 64

since x = 4/3, -4/3
that means there will be two points, remember y and z depend on x here because of our work above

Answer 65

two points (x,y,z) to plug into f(x,y,z) , one will be greater than the other, so that tells you which is min which is max

Answer 66

oh you're right, x = -z

Answer 67

you could solve this by using L, you were right earlier