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

Section 14.8 Lagrange Multipliers 977
and so 2 − 29
4 , − 6s29 y2. Then x − 72ys29 , y − 65ys29 , and, from (20),
z − 1 2 x 1 y − 1 6 7ys29 . The corresponding values of f are
7 2
s29
1 2S 6 5
s29
D 1 3S1 6 7
s29
D − 3 6 s29
Therefore the maximum value of f on the given curve is 3 1 s29 .
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;
1. Pictured are a contour map of f and a curve with equation
tsx, yd − 8. Estimate the maximum and minimum values
of f subject to the constraint that tsx, yd − 8. Explain your
reasoning.
y
0
g(x, y)=8
70 60 50 40
30
20
10
2. (a) Use a graphing calculator or computer to graph the
7et14708x01
circle x 2 1 y 2 − 1. On the same screen, graph several
curves 05/12/10
of the form x 2 1 y − c until you find two that
just touch MasterID: the circle. What 01641 is the significance of the
values of c for these two curves?
(b) Use Lagrange multipliers to find the extreme values of
f sx, yd − x 2 1 y subject to the constraint x 2 1 y 2 − 1.
Compare your answers with those in part (a).
3–14 Each of these extreme value problems has a solution with
both a maximum value and a minimum value. Use Lagrange
multipliers to find the extreme values of the function subject to
the given constraint.
3. f sx, yd − x 2 2 y 2 ; x 2 1 y 2 − 1
4. f sx, yd − 3x 1 y; x 2 1 y 2 − 10
5. f sx, yd − xy; 4x 2 1 y 2 − 8
6. f sx, yd − xe y ; x 2 1 y 2 − 2
7. f sx, y, zd − 2x 1 2y 1 z; x 2 1 y 2 1 z 2 − 9
8. f sx, y, zd − e xyz ; 2x 2 1 y 2 1 z 2 − 24
9. f sx, y, zd − xy 2 z; x 2 1 y 2 1 z 2 − 4
10. f sx, y, zd − lnsx 2 1 1d 1 lnsy 2 1 1d 1 lnsz 2 1 1d;
x 2 1 y 2 1 z 2 − 12
x
;
11. f sx, y, zd − x 2 1 y 2 1 z 2 ; x 4 1 y 4 1 z 4 − 1
12. f sx, y, zd − x 4 1 y 4 1 z 4 ; x 2 1 y 2 1 z 2 − 1
13. f sx, y, z, td − x 1 y 1 z 1 t; x 2 1 y 2 1 z 2 1 t 2 − 1
14. f sx 1, x 2, . . . , x nd − x 1 1 x 2 1 ∙ ∙ ∙ 1 x n;
x 2
1 1 x 2 2 1 ∙∙∙ 1 x 2 n − 1
15. The method of Lagrange multipliers assumes that the
extreme values exist, but that is not always the case.
Show that the problem of finding the minimum value of
f sx, yd − x 2 1 y 2 subject to the constraint xy − 1 can be
solved using Lagrange multipliers, but f does not have a
maximum value with that constraint.
16. Find the minimum value of f sx, y, zd − x 2 1 2y 2 1 3z 2
subject to the constraint x 1 2y 1 3z − 10. Show that f
has no maximum value with this constraint.
17–20 Find the extreme values of f subject to both constraints.
17. f sx, y, zd − x 1 y 1 z; x 2 1 z 2 − 2, x 1 y − 1
18. f sx, y, zd − z; x 2 1 y 2 − z 2 , x 1 y 1 z − 24
19. f sx, y, zd − yz 1 xy; xy − 1, y 2 1 z 2 − 1
20. f sx, y, zd − x 2 1 y 2 1 z 2 ; x 2 y − 1, y 2 2 z 2 − 1
21–23 Find the extreme values of f on the region described by
the inequality.
21. f sx, yd − x 2 1 y 2 1 4x 2 4y, x 2 1 y 2 < 9
22. f sx, yd − 2x 2 1 3y 2 2 4x 2 5, x 2 1 y 2 < 16
23. f sx, yd − e 2xy , x 2 1 4y 2 < 1
24. Consider the problem of maximizing the function
f sx, yd − 2x 1 3y subject to the constraint sx 1 sy − 5.
(a) Try using Lagrange multipliers to solve the problem.
(b) Does f s25, 0d give a larger value than the one in part (a)?
(c) Solve the problem by graphing the constraint equation
and several level curves of f.
(d) Explain why the method of Lagrange multipliers fails to
solve the problem.
(e) What is the significance of f s9, 4d?
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