Question

CAL 3 Problem

Answer 1

Solve using Lagrange Multipliers

Find the critical points of
\[f(x.y)= x^2+y^2\]
on the curve \[xy-x-y=1\]

Answer 2

\[f=x^2+y^2\\g=xy-x-y\]
we first set up the Lagrange equation system:
\[
f_x=\lambda g_x\\
f_y=\lambda g_y\\
g=1
\]

Answer 3

so,
\[
\rm(1)\qquad2x=\lambda(y-1)\\
\rm(2)\qquad2y=\lambda(x-1)\\
\rm(3)\qquad x(y-1)-y=1
\]

Answer 4

solve these equations using either matrix method or by eliminations

Answer 5

eq(1)-eq(2)
\[2(x-y)=\lambda[(y-1)-(x-1)]\\
2\cancel{(x-y)}=-\lambda\cancel{(x-y)}\\
\implies\lambda=-2
\]

Answer 6

so,
\[2x=-2(y-1)\qquad 2y=-2(x-1)\\
x=1-y\qquad\qquad\quad y=1-x\]

Answer 7

there are no stationary points!!!

Answer 8

I believe it was simplified. if you distribute it you'll get the equation of the curve

Answer 9

the third equation is the derivative with respect to \(\lambda\)
\[L(x,y,\lambda)=f(x,y)-\lambda[g(x,y)-c]\]

Answer 10

so,
\[0=g(x,y)-c\\\implies g(x,y)=c\]

Answer 11

was lambda was made negative because you get \[2(x-y)=\lambda[(y-1)-(x-1)] \rightarrow 2(x-y)=\lambda[(y-x)]\]

Answer 12

you can have it either way.. you'd end up with the same thig..
lambda is just an arbitrary scalar

Answer 13

also, in the step where I cancel off the (x-y) term, I do that "understanding" that under the given conditions,
\(x\ne y\)

Answer 14

kapeesh all?

Answer 15

@camoJAX the extracting "-" sign, yes thats why I did that

Answer 16

ohh ok now i gotcha

Answer 17

@hoa i was very confused when i was given this problem

Answer 18

ok,
so my frist post declares the three differntial equations.

Answer 19

1) partial diff with x
2) partial diff with y
3) partial diff with lambda -> gives you the criterion equation...

Answer 20

these equations can be readily obtained using the expression for "L(x,y,lambda)" I gave earlier

Answer 21

is (3) a general equation?

Answer 22

no (3) is the constraint for optimization.. specific for each problem

Answer 23

L(x,y,lambda) is the Lagrangian function.
this function combines (linearly) the cost function (f) with the constraint (g=c)

Answer 24

ok, from top,..
cost function: \(f(x,y)\)
constraint:\(g(x,y)=c\)

Answer 25

@Hoa perfect...
so, now, do you get any values for "x" and "y"?

Answer 26

no.. so, no solution exists

Answer 27

back
we then combine the cost function and the constraint using a linear relationship
\[L(x,y,\lambda)=f(x,y)-\lambda\left[g(x,y)-c\right]\]

Answer 28

ohh okay i understand now

Answer 29

now, this function represents the "function with the constyraint"

Answer 30

@Hoa "x=y" can be satisfied by infinite values.. and is not possible so, no solution

Answer 31

this Lagrangian manifests as a Hamiltonian in the applications esp., Quantum mechanics

Answer 32

please elaborate

Answer 33

@Hoa lol. then ignore that piece

Answer 34

well, if you want, try to replace all "x" in g(x,y)=c with "y"
and what do you get?

Answer 35

so the critical points are
x=1-y
y=1-x

Answer 36

@electrokid

Answer 37

@camoJAX He ran away. just you and me for the topic , now. review. he said that L (x,y,lamda) = f(x,y ) - lamda[g (x,y)-c] and then take derivative to get (3). I never know about that, but temporarily accept the formula to step up. But, when take derivative respect to lamda of that L, I got something with negative sign, like -[ (x(y-1)-y) -1] and cannot be like what he said.

Answer 38

what do you think? let him go, I feel embarrass when I waste other's time with my stupidity.

Answer 39

ok, after changing the side, I got what he said.

Answer 40

my bad i stepped away, i was lost with that one too but now i get it also

Answer 41

@Hoa and @camojax the derivatives should equal zero since, when f(x,y) is maxima or minima, L has an extrema too

Answer 42

@camoJAX check with the \(\mathcal{L}\) function and its partial derivative with respect to \(\lambda\)

Answer 43

that should give you the (3) equation.

Answer 44

\[L(x,y,\lambda)=x^2+y^2-\lambda xy+\lambda x+\lambda y+\lambda\]
\[\frac{ \partial L }{ \partial \lambda } L(x,y,\lambda)=-x(y-1)+y+1\]
@electrokid I did not get the same result as you did for (3)

Answer 45

and this should become =0 since we are looking for an extrema..

Answer 46

ok, thanks

Answer 47

surething

Answer 48

actually, the so-called L(x,y,lamda) is exactly the constraint. kid confused me,A ha!

Answer 49

"L" is a linear combination of the function to maximize/minimize and the constraints on the optimization.
so, L becomes your new cost function that implicitly optimizes cost "f" under constraint "g"

Answer 50

you are very good at foundation of math. good kid

Answer 51

@Hoa thank you. that is the only way I can stay "sane".

Answer 52

@Hoa @camoJAX this linear combination is performed along a new "dimension of \(\lambda\)". that is why we have to optimize along all the dimensions (dimensions of "f" and "g" and the new one, lambda)

Answer 53

got it. sir