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

Section 14.8 Lagrange Multipliers 973
This is a system of four equations in the four unknowns x, y, z, and , but it is not necessary
to find explicit values for .
For functions of two variables the method of Lagrange multipliers is similar to the
method just described. To find the extreme values of f sx, yd subject to the constraint
tsx, yd − k, we look for values of x, y, and such that
=f sx, yd − =tsx, yd and tsx, yd − k
This amounts to solving three equations in three unknowns:
f x − t x f y − t y tsx, yd − k
Our first illustration of Lagrange’s method is to reconsider the problem given in
Example 14.7.6.
EXAMPLE 1 A rectangular box without a lid is to be made from 12 m 2 of cardboard.
Find the maximum volume of such a box.
SOLUtion As in Example 14.7.6, we let x, y, and z be the length, width, and height,
respectively, of the box in meters. Then we wish to maximize
subject to the constraint
V − xyz
tsx, y, zd − 2xz 1 2yz 1 xy − 12
Using the method of Lagrange multipliers, we look for values of x, y, z, and such that
=V − =t and tsx, y, zd − 12. This gives the equations
V x − t x
V y − t y
V z − t z
which become
2xz 1 2yz 1 xy − 12
2 yz − s2z 1 yd
3 xz − s2z 1 xd
4 xy − s2x 1 2yd
5 2xz 1 2yz 1 xy − 12
There are no general rules for solving systems of equations. Sometimes some ingenuity
is required. In the present example you might notice that if we multiply (2) by x, (3) by y,
and (4) by z, then the left sides of these equations will be identical. Doing this, we have
Another method for solving the system
of equations (2 –5) is to solve each of
Equations 2, 3, and 4 for and then to
equate the resulting expressions.
6 xyz − s2xz 1 xyd
7 xyz − s2yz 1 xyd
8 xyz − s2xz 1 2yzd
We observe that ± 0 because − 0 would imply yz − xz − xy − 0 from (2), (3),
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