Question

minimize f(x,y)=x²+4y² subject to the constraint xy=1

(we are using Langrangian formula) minimize f(x,y)=x²+4y² subject to the constraint xy=1

(we are using Langrangian formula) @Mathematics

Answer 1

this is quite similar to the other ones right?

Answer 2

yes but i couldnt solve for all the variables :/

Answer 3

F(x,y)=x^2+4y^2 ; xy-1=0

F(x,y,L) = x^2+4y^2 + L(xy-1)

Answer 4

right

Answer 5

Fx = 2x + Ly
2x + Ly = 0
2x = -Ly
x = -Ly/2

Fy = 8y + Lx
8y + Lx = 0
8y + L(-Ly/2) = 0
8y -L^2y/2 = 0
(16y -L^2y)/2 = 0
y(16 -L^2)/2 = 0
y = 2/(16 -L^2)

looks messy, but does the math look ok?

Answer 6

one sec!

Answer 7

now what?

Answer 8

im thinking the answer is 4; but that is without going thru the Lagrange mess

Answer 9

since the constraint is that xy = 1; y = 1/x can be subbed into the original f(x,y)

Answer 10

other than that, i cant get a good bead on it

Answer 11

i dont understand :(

Answer 12

f(x,y)=x²+4y² subject to the constraint xy=1

xy=1 means y = 1/x

f(x,y(x))=x^2+4/x^2

Answer 13

if we do it the other way:

f(x,y)=x²+4y² ; xy=1

xy=1 can also mean x = 1/y

f(x(y),y)=1/y^2+4y^2

Answer 14

either way they minimize at 4