Question

Lagrange's - Understanding how;
Find the distance between the curve; x^2*y = 128 and origo.

Now, I already have the "answer sheet", and it says you have to find the distance from origo (formula?). The "amounts to minimization" is f(x,y) = x^2 + y^2.
I do not understand how you get to f(x,y) = x^2 + y^2 from this. Help understanding would be much appreciated! (SOLUTION ATTACHED BELOW)

Answer 1

I suspect it will always be f(x,y) = x^2 + y^2 regardless of curve, as long as we want to find the distance between curve and origin.

Answer 2

You have a set of points \(x^2y=128\), distance of every point on it like \((x, y)\) from the origin is \(\sqrt{x^2+y^2}\). right?

Answer 3

Why is the root removed then? Why f(x,y) = x^2 + y^2? Hmm...

Answer 4

because minimum of \(x^2+y^2\) gives the minimum of \(\sqrt{x^2+y^2}\). sqrt is an increasing function

Answer 5

Okay, so regardless of curve, I can use (note down) that the minimum function will be f(x,y) = x^2 + y^2 ?
Would this still be true, if say, instead from origo and curve, we want (1,1) and curve?
Or would that be (x+1)^2+(y+1)^2 or something like that?


I may not understand it right away, but knowing how to solve it will go a long way in me doing it over and over until I understand it :)

Answer 6

I think it should be x-1 and y-1 (not +), but Im not sure

Answer 7

Ah, right :)

Answer 8

Thanks a lot :)
Just to double check; it would be x-1 and y-1 instead of x and y when the position moves to (1,1)? And not +, since I recall something about it being x-x_0, y-y_0 :)

Answer 9

@mukushla

Answer 10

yeah, that's right, minimum function for distance from \((a, b)\) will be\[f(x, y)=(x-a)^2+(y-b)^2\]

Answer 11

Thanks! Appreciate the help :)

Answer 12

anytime ;)