EXAMPLE 2 Find the extreme values of the function f(x, y) -2x2 +5y2 on the circl
ID: 2888343 • Letter: E
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EXAMPLE 2 Find the extreme values of the function f(x, y) -2x2 +5y2 on the circle x2+y 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 Vf-A79 and g(x, y) = 1, which can be written as or as Video Example From (1) we have x = , then (3) gives y= ±1. If = 2, then 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 fat these four points, we find that ro, 1) = (O,-1)5 1,0) = /(-1,0) = 2 Therefore the maximum value of f on the circle x2 + y2 = 1 is 0, ±1) = and the minimum value is Checking with the figure, we see that these values look reasonableExplanation / Answer
We have given f(x,y)=2x^2+5y^2 subject to the constraint g(x,y)=x^2+y^2=1
using Lagrange multifliers
fx=gx implies 4x=2x --(1)
fy=gy implies 10y=2y --(2)
x^2+y^2=1 --(3)
from equation (1) (4x-2x)=0 implies 2x(2-)=0 implies x=0 or =2
if x=0 then (3) gives y=1,-1
if =2 then (2) gives 10y=4y implies (10y-4y)=0 implies 6y=0 implies y=0
if y=0 then (3) gives x=1,-1
therefore f has possible extreme values at the points (0,1),(0,-1),(1,0),(-1,0)
f(0,1)=2(0)^2+5(1)^2=0+5=5
f(0,-1)=2(0)^2+5(-1)^2=0+5=5
f(1,0)=2(1)^2+5(0)^2=2
f(-1,0)=2(-1)^2+5(0)^2=2
Therefore the maximum value of f on the circle x^2+y^2=1 is f(0,1)=f(0,-1)=5 and the minimum value is
f(1,0)=f(-1,0)=2
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