Question #183696

 Find the minimum value of the function 

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


1
Expert's answer
2021-05-07T12:16:37-0400

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


Constraint: g(x,y)=x2+y2=1       (11)g(x, y)= x^2+y^2=1~~~~~~~-(11)


Using Lagrange multipliers,


fx=λgx,fy=λgy,g(x,y)=1f_x= λg_x,f_y= λg_y,g(x,y) = 1


which become

2x=2xλ          (1)4y=2yλ           (2)x2+y2=1           (3)2x= 2xλ ~~~~~~~~~~-(1) \\ 4y= 2yλ ~~~~~~~~~~~-(2) \\ x^2+y^2= 1 ~~~~~~~~~~~-(3)


 From (1) we have x=0 or λ=1x=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).(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

x2+y2=1 is f(0,±1)=2x^2+y^2=1 \text{ is } f(0,±1) =2 and the minimum value is f(±1,0)=1.f(±1,0) =1.  


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