Michael Corral: Vector Calculus

2.7 Constrained Optimization: Lagrange Multipliers 99
Substituting this into g(x,y)= x 2 +y 2 = 80 yields 5x 2 = 80,
so x=±4. So the two constrained critical points are (4,8)
and(−4,−8). Since f(4,8)=45and f(−4,−8)=125, andsince
there must be points on the circle closest to and farthest
from (1,2), then it must be the case that (4,8) is the point
on the circle closest to (1,2) and (−4,−8) is the farthest from
(1,2) (see Figure 2.7.1).
Notice that since the constraint equation x 2 +y 2 = 80 describes
a circle, which is a bounded set in 2 , then we were
guaranteed that the constrained critical points we found
were indeed the constrained maximum and minimum.
y
x 2 +y 2 = 80
(4,8)
(1,2)
0
(−4,−8)
Figure 2.7.1
x
The Lagrange multiplier method can be extended to functions of three variables.
Example 2.27.
Maximize (and minimize) : f(x,y,z)= x+z
Solution: Solve the equation∇f(x,y,z)=λ∇g(x,y,z):
given : g(x,y,z)= x 2 +y 2 +z 2 = 1
1=2λx
0=2λy
1=2λz
The first equation impliesλ0 (otherwise we would have 1=0), so we can divide
byλin the second equation to get y = 0 and we can divide byλin the first and
third equations to get x= 1
2λ
= z. Substituting these expressions into the constraint
( )
equation g(x,y,z)= x 2 + y 2 + z 2 1
= 1 yields the constrained critical points √
1
,0, √2
( ( ) (
2
−1
and √,0, −1 √
1
). Since f √,0, √2
1 −1
> f √,0, −1 √
), and since the constraint equation
2 2 2 2 2
( )
x 2 + y 2 + z 2 = 1 describes a sphere (which is bounded) in 3 1
, then √
1
,0, √2 is the
( )
2
−1
constrained maximum point and √,0, −1 √ is the constrained minimum point.
2 2
SofarwehavenotattachedanysignificancetothevalueoftheLagrangemultiplier
λ. Weneededλonlytofindtheconstrainedcriticalpoints,butmadenouseofitsvalue.
It turns out thatλgives an approximation of the change in the value of the function
f(x,y) that we wish to maximize or minimize, when the constant c in the constraint
equation g(x,y)=cis changed by 1.