Minimum and maximum points using Lagrange's method

Question

F(x,y)=x+2y, G(x,y)=x^2+y^2-1 I want to find the minimum and maximum points using lagrange's method, I tried but I always get lost, I could use a little bit help. Thank you.

Answer 1

Define,

$L(x,y, \lambda) = F(x,y) + \lambda G(x,y) = x + 2 y + \lambda (x^2 + y^2 - 1) $

Then, at the extreme points,

$\dfrac{\partial L}{\partial x} = 1 + \lambda (2 x) = 0 $

$\dfrac{\partial L}{\partial y} = 2 + \lambda (2 y) = 0 $

$\dfrac{\partial L}{\partial \lambda} = x^2 + y^2 - 1 = 0$

From the first two equations, it follows that,

$ \lambda = \dfrac{-1}{2x} = \dfrac{-2}{2y} $

Hence,

$2 x = y$

Plug this into the third equation,

$ x^2 + (2 x)^2 = 1 $

$ x = \pm \dfrac{1}{\sqrt{5}} $

$ y = \pm \dfrac{2}{\sqrt{5}} $

Plugging these values into the function $F(x,y)$ we deduce that the maximum is $\sqrt{5}$ and is achieved at $(\dfrac{1}{\sqrt{5}}, \dfrac{2}{\sqrt{5}})$ and the minimum is $-\sqrt{5}$ achieved at $(-\dfrac{1}{\sqrt{5}}, -\dfrac{2}{\sqrt{5}})$.