James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

976 Chapter 14 Partial Derivatives
h=c
FIGURE 5
C
g=k
±g
P
±f
±h
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MasterID: 01639
Two Constraints
Suppose now that we want to find the maximum and minimum values of a function
f sx, y, zd subject to two constraints (side conditions) of the form tsx, y, zd − k and
hsx, y, zd − c. Geometrically, this means that we are looking for the extreme values of f
when sx, y, zd is restricted to lie on the curve of intersection C of the level surfaces
tsx, y, zd − k and hsx, y, zd − c. (See Figure 5.) Suppose f has such an extreme value at
a point Psx 0 , y 0 , z 0 d. We know from the beginning of this section that =f is orthogonal to
C at P. But we also know that =t is orthogonal to tsx, y, zd − k and =h is orthogonal
to hsx, y, zd − c, so =t and =h are both orthogonal to C. This means that the gradient
vector =f sx 0 , y 0 , z 0 d is in the plane determined by =tsx 0 , y 0 , z 0 d and =hsx 0 , y 0 , z 0 d. (We
assume that these gradient vectors are not zero and not parallel.) So there are numbers
and (called Lagrange multi pliers) such that
16 =f sx 0 , y 0 , z 0 d − =tsx 0 , y 0 , z 0 d 1 =hsx 0 , y 0 , z 0 d
In this case Lagrange’s method is to look for extreme values by solving five equations in
the five unknowns x, y, z, , and . These equations are obtained by writing Equa tion 16
in terms of its components and using the constraint equations:
f x − t x 1 h x
f y − t y 1 h y
f z − t z 1 h z
tsx, y, zd − k
hsx, y, zd − c
The cylinder x 2 1 y 2 − 1 intersects
the plane x 2 y 1 z − 1 in an ellipse
(Figure 6). Example 5 asks for the
maximum value of f when sx, y, zd is
restricted to lie on the ellipse.
4
EXAMPLE 5 Find the maximum value of the function f sx, y, zd − x 1 2y 1 3z on the
curve of intersection of the plane x 2 y 1 z − 1 and the cylinder x 2 1 y 2 − 1.
SOLUtion We maximize the function f sx, y, zd − x 1 2y 1 3z subject to the constraints
tsx, y, zd − x 2 y 1 z − 1 and hsx, y, zd − x 2 1 y 2 − 1. The Lagrange condition
is =f − =t 1 =h, so we solve the equations
17 1 − 1 2x
z
3
2
1
0
_1
_2
_1
FIGURE 66
0
y
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1
18 2 − 2 1 2y
19 3 −
20 x 2 y 1 z − 1
21 x 2 1 y 2 − 1
Putting − 3 [from (19) ] in (17), we get 2x − 22, so x − 21y. Similarly, (18)
gives y − 5ys2d. Substitution in (21) then gives
1
1 25
2 4 − 1 2
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