Use the lagrange's multipliers method to find a points on an ellipse

Question

Question:

Using the Lagrange's Multipliers method, find the points on the ellipse $x^2+2y^2=1$, that are situated in the longest and shortest distance from the line $x+y=2$.

I know how to use Lagrange's Method but I do not know how to restrict this question to: 'find a functions max and min on a given set', thus you need to help me solely with that.

Answer 1

If you realize that the distance between the lines $x+y=k$ and $x+y=2$ is just $\frac{|k-2|}{\sqrt{2}}$ the problem boils down to computing the minimum and maximum of $x+y$ over $x^2+2y^2=1$. Lagrange's method gives $x=2y$, hence $(x,y)=\pm\frac{1}{\sqrt{6}}(2,1)$ and $k=\pm\sqrt{\frac{3}{2}}$.

Answer 2

If $f(x,y) = x + y$ then $\nabla f (x,y) = (1,1)$ orthogonal to the line $x + y = 2$. Considering $\varphi (x,y) = x^2 + 2y^2$ we have that $\nabla \varphi (x,y) = (2x , 4x)$.

Notice that $1$ is a regular point of $\varphi$ and $M = \varphi^{-1} (1)$ is an ellipse, therefore compact. The restriction $f|_M$ has then at least two critical points (Weiestrass Theorem) in which assumes its maximum and minimum value.

The critical points of $f|_M$ are the solutions $(x,y)$ of $$\nabla f(x,y) = \lambda \cdot \nabla \varphi (x,y)$$