Question

Use the method of Lagrange multipliers to find the location (x, y) of the point on the
curve y^2-x+2= that is closest to the origin.

Answer 1

y^2-x+2=0

?

Answer 2

yeah =0

Answer 3

distance^2= x^2+y^2

F= x^2 +y^2
grad F= {2x,2y}

grad g={-1,2y}

\[ grad F= \lambda grad G\]

Answer 4

\[2x=-\lambda\]
\[2y=\lambda 2y\]
y^2-x+2=0

Answer 5

thanks dear

Answer 6

is it finished?

Answer 7

you have to solve for x and y

Answer 8

oh kul,thanks

Answer 9

2y=λ2y
can mean two thing

1) λ=1
2) y=0

let's assume second case
y=0

y^2-x+2=0
0^2- x+2
x=2

(2,0)

Answer 10

This also makes a lot of intuitive sense. Draw the graph of y^2 = x - 2 and you'll see.