Christopher F. answered • 04/13/20

M.S. in Mathematics with 500+ hours in Calculus tutoring

What we're trying to maximize is the distance from a point (x,y) to the point (0, -1) with the constraint x^{2} + 16y^{2 }= 16.

So - using the distance formula for two points - our f(x, y) is √((x-0)^{2} + (y-(-1))^{2}) (what we're maximizing)

And g(x,y) = x^{2} + 16y^{2 } (our constraint)

Maximizing the distance between two points is equivalent to maximizing the distance SQUARED between two points. Therefore, for the sake of simplicity, f(x, y) = (x-0)^{2} + (y+1)^{2 }= x^{2} + (y+1)^{2}

By Lagrange, we set up the equations ∇ f(x, y) = λ ∇ g(x, y).

First, ∇ f(x, y) = < 2x, 2(y+1) > and ∇ g(x, y) = < 2x, 32y >

Equating components, we have 2x = λ2x and 2(y+1)=λ32y

The first equation tells us λ=1, which we plug into the second to get 2y + 2 = 32y

So we get y = 1/15 ≈ .0667

We plug this y-value into the constraint equation (where the point (x,y) is on!) and solve for x:

x = ± 3.991

Therefore, the points on the ellipse which are furthest away from (0, -1) are:

(3.991, 0.0667) and (-3.991, 0.0667)