Question

Posting a Lagrange Multiplier question shortly.

Answer 1

\[\text{Use the method of Lagrange multipliers to find the extrema of}\]\[f(x,y)=(x+1)^2+(y-\frac{1}{5})^{2}\]

Answer 2

\[\triangledown f = 2(x-1)i+2(y-\frac{1}{5})j\]

Answer 3

do you also have a constraint ?

Answer 4

On the ellipse \[x^2+4y^2+2x-16y=-13\]

Answer 5

Brb.

Answer 6

(Changing computers in the library)

Answer 7

Okay there is a sign mismatch in your gradient
check once

Answer 8

Alright, back online on the other PC.

Answer 9

\[\triangledown f=2(x+1)i+2(y-\frac{1}{5})j\]

Answer 10

\[\triangledown g = c = (2x+2)i+(8y-16)j\]

Answer 11

Whoops, equals c, not zero.

Answer 12

Alright, now I set the two up and solve a system of equations for points. This is, again, the thing that consistently messes me up the most, I suck at finding all the points, so help here above all is greatly welcomed. Going to take a shot now.

Answer 13

gradient vectors are parallel at extrema points : \[\nabla f = \lambda \nabla g\]

Answer 14

\[2(x+1)=2(x+1)\lambda \ \ \ (1)\]\[2(y-\frac{1}{5})=(8y-16)\lambda \ \ \ (2)\]

Answer 15

Alright, the first one seems super easy, at least finding some easy values. \[\lambda=1\]

Answer 16

maybe eliminate \(\lambda\) by dividing both equations

Answer 17

oh right didnt notice that haha!

Answer 18

But yeah, it's just getting every value necessary, I can always get a few, but I screw up properly grouping my x and y values into the appropriate points and such.

Alright, I'm guessing I should maybe plug that lambda value into both eqns and then do something?

Answer 19

Plugging it into the second one, \[2(y-\frac{1}{5})=(8y-16) \]Can now just straightforward solve for y I think.

Answer 20

\[2y-\frac{2}{5}=8y-16; \ \ \ 6y=\frac{78}{80}\]

Answer 21

we missed a value in the start, you keep going
we can go back in the end

Answer 22

^And that's the other thing, missing values. I need to figure out how to determine whether I've exhaustively checked for all values. Going to continue now, one sec.

Answer 23

how did u get 80 in the bottom ?

Answer 24

Oh, whoops, what was I thinking, my bad, supposed to be 5

Answer 25

\[6y=\frac{78}{5}; \ \ \ 3y=\frac{39}{5}\]

Answer 26

\[y=\frac{13}{5}\]

Answer 27

i am caught up hi guys

Answer 28

Hullo

Answer 29

yes use the constraint equation (ellipse) to find x value(s)

Answer 30

\[x^2+4y^2+2x-16y=-13\]

Answer 31

ah i see

Answer 32

\[x^2+4\bigg(\frac{13}{5}\bigg)^{2}+2x-16\bigg(\frac{13}{5}\bigg)=-13\]

Answer 33

standard equation

Answer 34

basically my friend you want to see how your function is changing along this new axis system which is your ellipse

Answer 35

you wan to find the maximum and minimum or extema of that slice place

Answer 36

lemme draw it, sec! i think it will all make sense why we would be parametrizing and seeing the directional gradient along it then

Answer 37

I dont mean to disturb you ganeshie just carry on ill draw around lol sorry!

Answer 38

i wanted to use some art in the start but i saw Mendicant already setup the equations... so i thought he/she already knows the geometric interpretation

Answer 39

dan is high ignore him

Answer 40

\[x^2+4\bigg(\frac{169}{25}\bigg)+2x-\frac{208}{5}=-13\]

Answer 41

Lol, well this looks atrocious.

Answer 42

wolfram it