Answer to Question #210331 in Calculus for Ash

Question

use Lagrange’s Multiplier method to find the maximum and minimum of f(x,y)= y^2 - x^2 subjected to the constraint of 1/4 x^2 + y^2 = 1

Answer 1

Given 

$$f(x,y) =y^2-x^2$$

Let  

$$g(x,y) =\dfrac{1}{4} x^{2}+y^{2}-1$$

 and let 

$$F = f(x,y)+\lambda g(x,y)$$

where $\lambda$  is the Lagrangian multiplier.

$$F = y^2-x^2 + \lambda(\dfrac{1}{4} x^{2}+y^{2}-1)$$

To find the maximum and minimum values, we have to solve the system

$$\dfrac{\partial F}{\partial x} = 0$$

$$\dfrac{\partial F}{\partial y} = 0$$

 

$$\dfrac{\partial F}{\partial \lambda} = 0$$

Hence

$$-2x + \dfrac{1}{2}x\lambda = 0$$

$$2y+2y\lambda=0$$

$$\dfrac{1}{4} x^{2}+y^{2}-1=0$$

If $x=0,$ we get

$$x=0$$

$$\lambda=-1$$

$$y^2=1$$

When $\lambda=-1,$

$x=0, y=-1$ or $x=0, y=1.$

If $y=0,$ we get

$$\lambda=4$$

$$y=0$$

$$x^2=4$$

When $\lambda=4,$

$x=-2, y=0$ or $x=2,y=0.$

If $x\not=0, y\not=0,$ we get

$$\lambda = 4$$

$$\lambda=-1$$

$$\dfrac{1}{4} x^{2}+y^{2}-1=0$$

The system has no solution.

$$f(0, -1)=(-1)^2-(0)^2=1$$

$$f(0, 1)=(1)^2-(0)^2=1$$

$$f(-2, 0)=(0)^2-(-2)^2=-4$$

$$f(2, 0)=(0)^2-(2)^2=-4$$

The maximum and minimum values of f(x,y) are respectively,

maximum value:

$$f(0, -1)=f(0, 1)=1$$

minimum value:

$$f(-2, 0)=f(2, 0)=-4$$