Answer to Question #183696 in Calculus for Ganesh Chandra Mondal

Question

 Find the minimum value of the function 

f(x, y) = x2 + 2y2 on the circle x2 + y2 = 1. 

Answer 1

Solve equations $∇f= λ ∇g \text{ and } g(x,y)=1$ using Lagrange multipliers

Constraint: $g(x, y)= x^2+y^2=1~~~~~~~-(11)$

Using Lagrange multipliers,

$f_x= λg_x,f_y= λg_y,g(x,y) = 1$

which become

$2x= 2xλ ~~~~~~~~~~-(1) \\ 4y= 2yλ ~~~~~~~~~~~-(2) \\ x^2+y^2= 1 ~~~~~~~~~~~-(3)$

 From (1) we have $x=0 \text{ or } λ=1$ . If x=0, then (11) gives y=±1. If

λ=1, then y=0 from 2, so then (11) gives x=±1. Therefore f

has possible extreme values at the points $(0,1), (0,-1), (1,0), (1,0).$

Evaluating f at these four points, we find that

• f(0,1)=2

• f(0,-1)=2

• f(1,0)=1

• f(-1,0)=1

Therefore the maximum value of f on the circle

$x^2+y^2=1 \text{ is } f(0,±1) =2$ and the minimum value is $f(±1,0) =1.$