method of Lagrange Multipliers

Question

find the points on the ellipse x^2+16y^2=16 that are furthest away from the point(0,-1)using the method of Lagrange Multipliers

Answer 1

Square of distance D(x,y) from point (x, y) to point (0, -1) is D(x,y) = x² + (y + 1)². We have to find max of this function subject to the consrain: f(x,y) = x² + 16y²- 16 = 0. We create Lagrange function H(x, y,λ) = D(x,y) +λf(x); H(x,y,λ) = x² + y² + 2y + 1 + λ(x² + 16y² - 16)

∂H/∂x = 2x + 2λx = 0 (1)

∂H/∂y = 2y + 2 + 32λy = 0 (2)

∂H/∂λ = x² + 16y² - 16 = 0 (3)

From (1) x = 0 or λ = -1.

If x = 0 then from (3) 16y² - 16 = 0, 16y² = 16, y² = 1, y = ± 1;

If λ = - 1 then from (2) we have 2y + 2 - 32y = 0, y = 1/15

Then from (3) x² + 16(1/15)² = 16; x² = 16(1 - 1/225); x² = 16(224/225); x = ±16(14)^1/2/15

We get 2 points: (±16(14)^1/2/15, 1/15)

Answer 2

The setup for this problem is the objective function Q(x ,y)

Q(x,y) = x² +(y + 1)² + λ ( x² + 16 y² -16)

The first two terms are the square of the distance we are trying to maximize, The expression multiplying λ is the constraint. The method of Lagrange multipliers is to obtain two equations by setting the partial derivatives of Q with respect to x and with respect to y both equal to zero. A third equation is x² + 16 y² = 16

This gives three equations with three unknowns ( x,y, λ)

The first equation is 2 x + λ 2 x = 0 There are two solutions: either x = 0 or λ = -1/2

The first leads to y = ±1 and so to the points (0, 1) and (0, -1). It turns out that the points obtained by following λ = -1/2 are further away from ( 0, -1)

The second equation is 2(y + 1) + λ 32 y ,= 0 Substituting λ = -1/2 leads to y = 1/7

Substituting y = 1/7 into x² + 16 y² = 16 , gives x = ± 16 sqrt(3) / 7

Thus two points are (16 sqrt(3) /7, 1/7) and ( -16 sqrt(3) /7, 1/7)

These two points are equidistant from ( 0 , -1) and further from it than is (0,1)