Question

Use Lagrange multipliers to find the points on a given curve that are nearest the origin. (You are not given the function f but it will be the distance formula between the point(x,y) and the point given.)

Need a worked example please

Answer 1

Let g(x, y, z) is a given curve and d = f(x, y, z) is the distance between any point (x, y, z) on the curve and the origin.

Then, according to the method of Langrange's multipliers, the distance d = f(x, y, z) is minimum when

$\nabla f=\lambda\nabla g \text{ subject to } g=0$

Example:

Let us find the point on the plane on the plane X-y+32 = 1 which is closet to the oregion. Distance of an arbitrary point (2,4, 2) on the plane curve g(x,y,z) = 2-4 +32-1=0 from the origin is given by de 5(2-0)2 + (4-0)2 + (2-0)2 =) d=5a2 +42 +22 It is clear that. absolute minimum of this function for (2,4, 2) lying on the plane. To find it, we instead minimize the function d²= f(x, y, z) = ² + y² +22 Subject to the constraint g(x, y, z) = 0 whese, g(x,y,z) = x-y+32-1. 2

According to the method of langrangers multipliers, the absolute minimum of the distance function d²= f(x, y, z) will occur at a point where If=X&g g=0 Now, f = x2 + y2 +22 Groadient of f, xf = 2(x2), 2(42), 2(22) Әу in o and ax az 5) Of = (2x, 24, 22) And, g x-y+ 3z-1 Gradient of g, vg = Cocos, en ), &(32))

бе - Вд - (1,-1,3) from o Vf c Ayg Э) (22, 2y, 27) - A(-1,3) 2 (23, 2y, 22) = (2,-2, 35) *, 2х - -) x, 2y =-2 ) y=- 27 - 3, 5 - 3% and - О 2 x-y+3z - | ЕО (2

Putting the values of 2, 4,2 in ③. 출 - (출) + 3(2) -) = 0 2) +2 +92 - 2 2. 02 ) 112 -2 O 코 2 || 류 키 출해 JI Hence, R 즐 =급 2 다를템 E 22 려를 그 규, ad려류, Thus, (ti ) (뉴) & the , 3, 4) is the point closed to oregin. ast