Calculus 2nd Edition Rogawski

854 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES
In Theorem 1, the assumption ∇g P ̸= 0
guarantees (by the Implicit Function
Theorem of advanced calculus) that we can
parametrize the curve g(x, y) = 0 near P
by a path c such that c(0) = P and
c ′ (0) ̸= 0.
REMINDER Eq. (1) states that if a local
min or max of f (x, y) subject to a
constraint g(x, y) = 0 occurs at
P = (a, b), then
∇f P = λ∇g P
provided that ∇g P ̸= 0.
Proof Let c(t) be a parametrization of the constraint curve g(x, y) = 0 near P , chosen
so that c(0) = P and c ′ (0) ̸= 0. Then f(c(0)) = f (P ), and by assumption, f(c(t)) has
a local min or max at t = 0. Thus, t = 0 is a critical point of f(c(t)) and
∣
d ∣∣∣t=0
dt f(c(t)) = ∇f P · c ′ (0) = 0
} {{ }
Chain Rule
This shows that ∇f P is orthogonal to the tangent vector c ′ (0) to the curve g(x, y) = 0.
The gradient ∇g P is also orthogonal to c ′ (0) (because ∇g P is orthogonal to the level
curve g(x, y) = 0atP ). We conclude that ∇f P and ∇g P are parallel, and hence ∇f P is
a multiple of ∇g P as claimed.
We refer to Eq. (1) as the Lagrange condition. When we write this condition in terms
of components, we obtain the Lagrange equations:
f x (a, b) = λg x (a, b)
f y (a, b) = λg y (a, b)
Apoint P = (a, b) satisfying these equations is called a critical point for the optimization
problem with constraint and f (a, b) is called a critical value.
EXAMPLE 1 Find the extreme values of f (x, y) = 2x + 5y on the ellipse
( x
4
) 2
+
( y
3
) 2
= 1
Solution
Step 1. Write out the Lagrange equations.
The constraint curve is g(x, y) = 0, where g(x, y) = (x/4) 2 + (y/3) 2 − 1. We have
〈 x
∇f = ⟨2, 5⟩ , ∇g =
8 , 2y 〉
9
The Lagrange equations ∇f P = λ∇g P are:
〈 x
⟨2, 5⟩ = λ
8 , 2y 〉
⇒ 2 = λx 9
8 , 5 = λ(2y)
9
Step 2. Solve for λ in terms of x and y.
Eq. (2) gives us two equations for λ:
λ = 16
x , λ = 45
2y
To justify dividing by x and y, note that x and y must be nonzero, because x = 0or
y = 0 would violate Eq. (2).
Step 3. Solve for x and y using the constraint.
The two expressions for λ must be equal, so we obtain 16
x = 45 45
or y = x. Now
2y 32
substitute this in the constraint equation and solve for x:
(
( x
) 2 45
+
32 x
) 2
= 1
4 3
( 1
x 2 16 + 225 ) ( ) 289
= x 2 = 1
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