Questions: EXAMPLE 2 Find the extreme values of the function f(x, y)=2 x^2+5 y^2 on the circle x^2+y^2=1. SOLUTION We are asked for extreme values of f subject to the constraint g(x, y)=x^2+y^2=1. Using Lagrange multipliers, we solve the equations ∇f=λ ∇g and g(x, y)=1, which can be written as fx=λ gx fy=λ gy g(x, y)=1 or as (1) (2) =2 y λ (3) x^2+y^2=1. From (1) we have x= or λ=2. If x= , then (3) gives y= ± 1. If λ=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) =5 f(1,0) = f(-1,0) =2 Therefore the maximum value of f on the circle x^2+y^2=1 is f(0, ± 1)= f( ± 1,0)=

Transcript text: EXAMPLE 2 Find the extreme values of the function $f(x, y)=2 x^{2}+5 y^{2}$ on the circle $x^{2}+y^{2}=1$. SOLUTION We are asked for extreme values of $f$ subject to the constraint $g(x, y)=x^{2}+y^{2}=1$. Using Lagrange multipliers, we solve the equations $\nabla f=\lambda \nabla g$ and $g(x, y)=1$, which can be written as \[ f_{x}=\lambda g_{x} \quad f_{y}=\lambda g_{y} \quad g(x, y)=1 \] or as (1) $\square$ (2) $\square$ $=2 y \lambda$ (3) $x^{2}+y^{2}=1$. From (1) we have $x=$ $\square$ or $\lambda=2$. If $x=$ $\square$ , then (3) gives $y= \pm 1$. If $\lambda=2$, then $y=$ $\qquad$ from (2), so then (3) gives $x= \pm 1$. 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 \[ \begin{aligned} f(0,1) & =\square \\ f(0,-1) & =5 \\ f(1,0) & =\square \\ f(-1,0) & =2 \end{aligned} \] Therefore the maximum value of $f$ on the circle $x^{2}+y^{2}=1$ is \[ \begin{array}{l} f(0, \pm 1)=\square \\ f( \pm 1,0)=\square \end{array} \]