Lagrange Multipliers: X y X y
Lagrange Multipliers: X y X y
Lagrange Multipliers: X y X y
In a rst calculus course, we learn that the extrema of z (t) over [a; b] must exist
and occur either at the critical points or the endpoints of [a; b]. Since the curve
is closed, we only need consider the critical points of z (t) in [a; b], which are
solutions to
dz
= rf v = 0
dt
where v is the velocity of r (t) : That is, the critical points of z (t) occur when
rf ? v: Since also rg ? v, it follows that the extrema of f (x; y) subject to
g (x; y) = k occur when rf is parallel to rg:
If rf is parallel to rg; then there is a number for which
rf = rg
Thus, the extrema of f (x; y) subject to g (x; y) = k must occur at the points
which are the solution to the system of equations
hfx ; fy i =
hgx ; gy i ;
1
g (x; y) = k
(1)
and
rg = h2x; 2yi
x = 2y
y
2x
and
x
2y
2
=)
y
x
=
2x
2y
x2 = 9;
x=
x; so that the
(3; 3) ; ( 3; 3) ; (3; 3) ; ( 3; 3)
However, f (3; 3) = f ( 3; 3) = 23; while f ( 3; 3) = f (3; 3) = 5:
Thus, the maxima of f (x; y) = xy + 4 over x2 + y 2 = 18 occur at
(3; 3) and ( 3; 3) ; while the minima of f (x; y) = xy + 4 occur at
( 3; 3) and (3; 3) :
(g (x; y)
k)
@L
= 0;
@y
@L
=0
@
and Lx = fx
gx ; Ly = fy
gy , and L = g (x; y) k: That is, the critical
points of L (x; y; ) are solutions of the system of equations
fx = gx ;
fy = gy ;
3
g (x; y) = k
(2)
x2 + y 2
(2x) ; Ly = y
x2 + y 2
18
(2y) ; and
18
Check your Reading: Can you identify the maxima and minima on the graph
shown above.
It follows that the highest level curve of f (x; y) intersecting a section of g (x; y) =
k must be tangent to the curve g (x; y) = k; which is possible only if their
gradients rf and rg are parallel.
Consequently, if g (x; y) = k is smooth and compact, then nding the extrema of f (x; y) subject to g (x; y) = k is equivalent to nding the critical points
onf the function L (x; y; ) ; and then evaluating f (x; y) at those critical points
(and if applicable the boundary points of g (x; y) = k):
EXAMPLE 3 Find the point(s) on the curve y = 1:5 x2 closest
to the origin both visually and via the Lagrange Multiplier method.
problem is to
Minimize f (x; y) = x2 + y 2
Subject to x2 + y = 1:5
We will thus let g (x; y) = x2 + y: Graphically, we can nd the point
on y = 1:5 x2 closest to the origin by drawing concentric circles
centered at the origin with ever greater radii until they interesect
the curve. The rst intersection of a circle with the curve will
correspond to a circle tangent to the curve i.e., a point where rf
is parallel to rg:
Points with jxj > 1:5 are more distant than any of the points for
jxj
1:5; so the point (0:5; 1) and ( 0:5; 1) are the closest to the
origin. Alternatively, the Lagrangian for this problem is
L (x; y; ) = x2 + y 2
1:5
(1) ; and L = x2 + y
Since Lx = 2x
(2x) ; Ly = 2y
critical points of L occur when
2x = 2x;
x2 + y
2y = ;
x2 + y = 1:5
1:5; the
= 2y into the
x = 2xy
If x = 0; then y = 1:5: Thus, (0; 1:5) is the critical point corresponding to = 3: If x 6= 0; then 1 = 2y so that y = 0:5 and
x2 + 0:5 = 1:5;
6
x=
Thus, the critical points are (0; 1) ; (1; 0:5) and ( 1; 0:5). However,
f (1; 0:5) = f ( 1; 0:5) = 1:25;
and
( 1; 0:5)
Similarly, an equation of the form g (x; y; z) = k denes a level surface in 3dimensions, and nding the extrema of a function f (x; y; z) subject to a constraint g (x; y; z) = k is equivalent to nding a level surface of f (x; y; z) that is
tangent to a constraint surface g (x; y; z) = k: It follows that rf is parallel to
rg at this point, so that if g (x; y; z) = k is a closed, bounded surface (i.e,. a
surface with nite extent that contains its boundary points), then solving
Optimize w = f (x; y; z)
Subject to g (x; y; z) = k
is equivalent to nding the critical points of the Lagrangian in 4 variables given
by
L (x; y; z; ) = f (x; y; z)
(g (x; y; z) k)
and then evaluating f (x; y; z) at those critical points (and if applicable the
boundary points of g (x; y; z) = k): Lets revisit a problem from the previous
section to see this idea at work.
blueEXAMPLE 4 blackFind the point(s) on the plane x+y z = 3
that are closest to the origin.
= x2 + y 2 + z 2
x+y z =3
(x + y
2z = ; and x + y
3)
(x + y
z=3
z
(3)
( x) = 3; 3x = 3; x = 1
3) ;
= 0 in example 4?
Applications
Many of the optimization word problems in a rst calculus course are, in fact,
constrained optimization problems of the form
Optimize :
Subject to :
f (x; y)
g (x; y) = k
In such problems, the Lagrange multiplier method produces a family of "extrema" of f (x; y) parameterized by ; so that eliminating and combining the
result with the constraint is essentially equivalent to nding which member of
the family satises the constraint.
EXAMPLE 5 John happens to acquire 420 feet of fencing and decides to use it to start a kennel by building 5 identical adjacent
rectangular runs (see diagram below). Find the dimensions of each
run that maximizes its area.
(10x + 6y
420)
x=6 ;
10x + 6y = 420
9
420: Thus,
5x
3
= 420;
x = 21 f eet
=)
fx
gx
=
fy
gy
10
r2 h
0:25
(2 rh) ; 2 r =
r2
(2 rh)
( r2 )
=)
2r + h
2h
=
r
r
or h = 2r
since r cannot be 0. That is, all cans with minimal surface area
will have h = 2r; which means a height equal to the diameter. To
determine which such can satises the constraint, we substitute to
obtain
r
3 0:25
2
3
r (2r) = 0:25; r =
2
which leads to r = 0:3414 feet, with h = 0:6818 feet. To see that
a minimum must occur, we notice that the constraint implies that
h = 0:25= r2 ; which leads to S as a function of r in the form
S = 2 r2 +
0:5
r2
Check Your Reading: What is the value of the Lagrange multiplier in example 5?
Multiple Constraints
Typically, if given a constraint of the form g (x; y) = k; we instead let g1 (x; y) =
g (x; y) k and use the constraint g1 (x; y) = 0: Thus, Lagrangians are usually
of the form
L (x; y; z; ) = f (x; y; z)
g1 (x; y; z)
11
g1 (x; y; z)
h1 (x; y; z)
l: As before, the
If we assume that the carry-on is to have (at least roughly) the shape
of a rectangular box and one dimension is no more than half of one
of the other dimensions (to insure "slide under seat" is possible),
then what dimensions of the carryon lead to maximum storage (i.e.,
maximum volume)?
= f (x; y; z; )
g1 (x; y; z; )
= xyz
(x + y + z 45)
45 and
h1 (x; y; z; )
(y 2x)
= yz
= xy
( 2) ; Ly = xz
and L = x + y + z
must satisfy
45; L = y
yz =
2 ; xz =
xy =
along with the constraints. Combining the last two equations yields
xz = xy + ; so that the rst equation becomes
yz = xy
2 (xz
xy) or yz = 3xy
2xz
Exercises
Use the method of Lagrange Multipliers to nd the extrema of the following
functions subject to the given constraints. (Notice that in each problem below
the constraint is a closed curve).
1.
3.
5.
7.
9.
f (x; y) = 3x + 2y
subject to: x2 + y 2
f (x; y) = x 2y
subject to: x2 + y 2
f (x; y) = x2 + 2y 2
subject to: x2 + y 2
f (x; y) = x4 + y 2
subject to: x2 + y 2
f (x; y) = x sin (y)
subject to: x2 + y 2
2.
=1
4.
= 25
6.
=1
8.
=1
10.
=1
f (x; y) = 2x y
subject to: x2 + y 2 = 1
f (x; y) = x + y
subject to: x2 + 2y 2 = 1
f (x; y) = x2 y
subject to: x2 + y 2 = 1
f (x; y) = x4 + y 4
subject to: x2 + y 2 = 1
f (x; y) = sin (x) cos (y)
subject to: x2 + y 2 = 1
x+y =1
x = y2 1
2
y = e x =2+2
12.
14.
16.
x + 2y = 5
x2 + 2y 2 = 1
2
y = 2e x =2
20. A farmer has 400 feet of fence with which to fence in a rectangular eld
adjoining two existing fences which meet at a right angle. What dimensions
14
15
This means minimizing the surface area of the can as well as the "wasted
area" between the ends and the squares they are punched from. Use Lagrange
multipliers to nd r and h that minimize the metal used to make the can. What
is the relationship between the resulting r and h?
23. What dimensions of the carry on in example 7 yield maximum volume if
we drop the requirement of one dimension being no more than half of another.
24. What dimensions of the carry on in example 7 yield maximum volume if
we drop the requirement of one dimension being no more than half the other
and set z = 9 (i.e., nd x and y ).
25. Redo example 7 for those airlines that allow 51 linear inches for a carryon.
26. What dimensions of the carry on in example 7 yield maximum volume if
we replace the constraint y = 2x by the constraint xy = 300? What does this
new constraint represent?
27.
27. The intersection of the plane z = 4:5 + 0:5x and the cone x2 + y 2 = z 2
is an ellipse. What point(s) on the ellipse are furthest from or closest to the
origin? What is signicant about these points?
28. The intersection of the plane z = 1 + x and the cone x2 + y 2 = z 2 is
a parabola. What point on the parabola is closes to the origin? What is
signicant about these point?
29. A house with width w and length l is be 18 feet tall at its tallest and 10
feet tall at each corner.
What dimensions for a 2000 square foot house (i.e., wl = 2000) minimize the
area of the roof and sides of the house?
30. The moons orbit about the earth is well-approximated by the curve
x2 + y 2 = (238; 957
0:0549y)
where distance is in miles. How close is the moon to the earth at its closest
point? What is the greatest distance between the moon and the earth?
Optimization with constraints occurs frequently in business settings. For example, if L denotes the number of manhours and K denotes the number of units
of capital required to produce q units of a commodity, then q is often related to
L and K by a Cobb-Douglas function
q = AL K
16
(4)
(5)
and
xy + yz + zx = 1
37. Find the point P on the curve 4x2 + 3xy = 45 that is closest to the
origin, and then show that the line from the origin through that point P is
perpendicular to the tangent line to the curve at P: In light of the method of
Lagrange multipliers, why would we expect this result? (Hint: you may prefer
to eliminate using the method described between examples 5 and 6).
17
5
4
3
2
1
0
-5
-4
-3
-2
-1
5
x
-1
-2
-3
-4
-5
38. Find the point P on the curve y = x2 + x 1:5 that is closest to the
origin, and then show that the line from the origin through that point P is
perpendicular to the tangent line to the curve at p: In light of the method of
Lagrange multipliers, why would we expect this result?
39. Find the maximum of
f (x; y) = e
x2 y 2
x2 y 2
@g @f
=0
@x @y
(6)
then nding the extrema of f (x; y) subject to g (x; y) = k is the same as nding
the extrema of
z (t) = f (x (t) ; y (t))
when z (a) = z (b) : )
44. Write to Learn: A region in the xy-plane is said to be convex if the line
segment joining any two points in the region is also in the region.
In a short essay, use Lagrange multipliers to explain why that the angles
2 that minimize the time required to travel from P to Q must satisfy
sin ( 1 )
sin ( 2 )
=
v1
v2
and
(7)