Question

Is this problem "Lagrangable"?

What is the minimum value of \(x^2 + y^2 + z^2\) given\[\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{z^2}{25}=1\]and \(z=x+y\).

Answer 1

Or just maybe...\[\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{(x+y)^2}{25}=1\]

Answer 2

yes looks like nice candidate for lagrange :)

Answer 3

But I'd like a solution through Lagrange Multipliers (if one exists).

Answer 4

OK, I'm just learning about them.\[x^2 + y^2 + z^2 +\lambda (x+y-z)\]Or will we use the second condition?

Answer 5

\[\dfrac{\partial f}{\partial x} =2x + \lambda \]Is this right?

Answer 6

Similarly,\[\dfrac{\partial f}{\partial y}=2y + \lambda\]\[\dfrac{\partial f}{\partial z}=2z-\lambda\]

Answer 7

OK, so \(y =z\)

Answer 8

hey constraint is this :

\(\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{(x+y)^2}{25}=1\)

right ?

Answer 9

Oh, right...

Answer 10

So will we just remove \(z\) from existence and plug in \(x+y\) in everything?

Answer 11

you're given two constraints, and one function to optimize

Answer 12

So i think we need to merge the constraint and take it for lagrange

Answer 13

Anyway\[x^2+y^2 + z^2+\lambda\left(\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{x^2 + 2xy + y^2}{25}\right)\]

Answer 14

Oh, forgot the \(-1\) there

Answer 15

^^

Answer 16

\[x^2+y^2 + z^2+\lambda\left(\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{x^2 + 2xy + y^2}{25}-1\right)\]OK.

Answer 17

it shouldnt matter as it becomes 0 when u take derivative..

Answer 18

Oh whoops, didn't plug in \(x+y \) for \(z\) either.

Answer 19

\[x^2+y^2 + (x+y)^2+\lambda\left(\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{x^2 + 2xy + y^2}{25}-1\right)\]Finally

Answer 20

Oh, you want to eliminate z and do it in two variables is it ?

Answer 21

Yes. :)

Answer 22

earlier one is also fine, but this gives less equations to solve... :)

Answer 23

Yes, OK - should I first combine the terms or go for it?

Answer 24

go for it may be...

Answer 25

Either should work. Bleh.\[\dfrac{\partial{f}}{\partial x}=2x +2x + 2y + \dfrac{\lambda}{2} + \dfrac{2\lambda}{25} + \dfrac{2\lambda y}{25}\]Phew.

Answer 26

looks good !

Answer 27

Nononono not that one... should be \(2\lambda x/4\)

Answer 28

I've confused myself. ?_?

Answer 29

Actually, both of them. Dagnabbit.

Answer 30

\(\large f(x,y) = x^2+y^2 + (x+y)^2+\lambda\left(\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{x^2 + 2xy + y^2}{25}-1\right) \)


\(\large \dfrac{\partial{f}}{\partial x}=2x +2(x + y) + \lambda\left( \frac{2x}{4}+ \frac{2x + 2y}{25}\right)\)

Answer 31

And\[\large \dfrac{\partial f}{\partial y}=2y + 2(x+y) +\lambda \left(\dfrac{2y}{5} + \dfrac{2y + 2x}{25}\right)\]Lesson of the day: use brackets wisely.

Answer 32

\(\large f(x,y) = x^2+y^2 + (x+y)^2+\lambda\left(\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{x^2 + 2xy + y^2}{25}-1\right) \)


\(\large \dfrac{\partial{f}}{\partial x}=2x +2(x + y) + \lambda\left( \frac{2x}{4}+ \frac{2x + 2y}{25}\right) \)

\(\large \dfrac{\partial{f}}{\partial y}=2y +2(x + y) + \lambda\left( \frac{2y}{5}+ \frac{2x + 2y}{25}\right) \)

Answer 33

And I was sitting there expanding every little thing. -_-

Answer 34

hahah more variables more fun :)

Answer 35

hey... whats the difference between \dfrac, and \frac ?

Answer 36

`\dfrac` doesn't reduce the numbers in fractions to half their original size.

Answer 37

oh !

\(\large \frac{1}{2}, \dfrac{1}{2}\)

gotcha :)

Answer 38

Like you can see how the fractions come out really small in `\frac`. By the way, fractions shouldn't be coming out small if you write \(\LaTeX\) within `\[ \]`. I see that you use `\( \)`

Answer 39

All right, let's subtract the two equations.

Answer 40

\[\dfrac{2x}{4} - \dfrac{2y}{5}=0\]

Answer 41

Oh wait.

Answer 42

\[\left( 2x -2y \right) +\left( \dfrac{x}2 - \dfrac{2y}{5}\right)=0\]

Answer 43

\(\LaTeX\), Y U NO WORK

Answer 44

Oh, looks like there should be a \(\lambda\) in there too?

Answer 45

yes lol

Answer 46

\[\left( 2x -2y \right) +\lambda \left( \dfrac{x}2 - \dfrac{2y}{5}\right)=0\]

Answer 47

the better way is not to subtract the equations

Answer 48

How do we get rid of the \(\lambda\)? ;_;

Answer 49

cuz, by subtracting the equations we lost some info

Answer 50

Lagrange works like this :

At local max/min points, the gradiants are proportional, so we get :
1) \(\dfrac{\partial f}{\partial x} = 0\)

2) \(\dfrac{\partial f}{\partial y} = 0\)

Answer 51

above both give u two equations in 3 variables : \(\large x, y, \lambda\)

Answer 52

but how can u solve 3 variables wid just 2 equations ?

Answer 53

Oh, so we should have included the \(z\) :O

Answer 54

use the constraint as 3rd equation :

3) \(\dfrac{x^2}{4} + \dfrac{y^2}{5} + \dfrac{(x+y)^2}{25}=1\)

Answer 55

Oh, OK.

Answer 56

3 equations and 3 unknowns, feed it to wolfram.
dont bother solving it manually lol

Answer 57

it gets messy

Answer 58

OK, thank you for helping! ^_^

The way I learnt the basics was this: https://www.quora.com/What-is-a-Lagrange-multiplier

Answer 59

http://www.wolframalpha.com/input/?i=2x%2B2%28x%2By%29+%2B+%5Clambda*%282x%2F4+%2B+2%28x%2By%29%2F25%29%3D0%2C+2y%2B2%28x%2By%29+%2B%5Clambda*%282y%2F5+%2B+2%28x%2By%29%2F25%29%3D0%2C+x%5E2%2F4+%2B+y%5E2%2F5+%2B+%28x%2By%29%5E2%2F25+%3D+1

Answer 60

Can you explain the "gradients are proportional" part?

Answer 61

once u have 3 equations, its just algebra to sovle them..

Answer 62

yeah sure :)

Answer 63

Aah, I'm beginning to see why \(\partial/\partial x\) is 0.

Answer 64

It's just like how any one-variable function works.

Answer 65

two different things :
1) finding max/min of a multivariable funciton : f(x, y, z)
2) finding max/min of a multivariable funciton : f(x, y, z) under given constraints