Solve equations ∇f= λ ∇g and g(x,y)=1 using Lagrange multipliers Constraint:
g(x,y)=x2+y2=1Using Lagrange multipliers,
fx=λgx,fy=λgy,g(x,y)=1which become
2x=2xλ1=2yλx2+y2=1If x=0, then y2=1=>y=±1.
If λ=1, then y=21, and x=±23
Therefore f has possible extreme values at the points (0,−1),(0,1),(−23,21), and (23,21). Evaluating fat these four points, we find that
f(0,−1)=−1f(0,1)=1f(−23,21)=45f(23,21)=45Therefore the maximum value of f on the circle x2+y2=1 is
f(±23,21)=45,the minimum value of f on the circle x2+y2=1 is
f(0,−1)=−1
Comments
Leave a comment