EXAMPLE 2 Find the extreme values of the function f(x,y)=2x2+4y2 on the circle x2+y2=1 SOLUTION We are asked for extreme values of f subject to the constraint g(x,y)=x2+y2=1. Using Lagrange multipliers, we solve the equations f=g and g(x,y)=1, which can be written as fx=gxfy=gyg(x,y)=1 or as (1) =2 (2) =2y (3) x2+y2=1. From (1) we have x= or =2. If x=, then (3) gives y=1. If =2, then y= from (2), so then (3) gives x=1. Therefore f has possible extreme values at the points (0,1),(0,1),(1,0), and (1,0). Evaluating f at these four points, we find that f(0,1)f(0,1)f(1,0)f(1,0)==4==2. Therefore the maximum value of f on the circle x2+y2=1 is f(0,1)= and the minimum value is f(1,0)= . Checking with the figure, we see that these values look reasonable