48 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
Problem 2.17 For each of the following functions either demonstrate it is not monotone in-
creasing on the given interval or show that it is.
1. y D e
x
on .1; 1/
2. y
D
x
3
x
on
.
1
;
1
/
3. y D
x
x
2
C 1
on .1; 1/
4. y D tan
1
.x/ on .1; 1/
5. y D
x
2
x
2
C 1
on .0; 1/
6. y D
2x
x
2
C 1
on .1; 1/
7. y D
e
x
e
x
C 1
on .1; 1/
Problem 2.18 Suppose that we are going to cut a line of length L into three pieces. Find the
division into pieces that maximizes the sum of the squares of the lengths. is can be phrased
as a multivariate optimization problem.
Problem 2.19 Find the largest interval on which the function
x
x
2
C 16
is monotone increasing.
Problem 2.20 Find the largest interval on which the function
x
2
1
x
2
C 4
is monotone increasing.
2.2 THE EXTREME VALUE THEOREM REDUX
e extreme value theorem, last seen in Fast Start Differential Calculus, says that e global max-
imum and minimum of a continuous, differentiable function must occur at critical points or at the
boundaries of the domain where optimization is taking place. is is just as true for optimizing a
function z D f .x; y/. So what’s changed? e largest change is that the boundary can now have
a very complex shape.
In the initial section of this chapter we carefully avoided optimizing functions with boundaries,
thus avoiding the issue with boundaries. In the next couple of examples we will demonstrate
techniques for dealing with boundaries.
2.2. THE EXTREME VALUE THEOREM REDUX 49
Example 2.21 Find the global maximum of
f .x; y/ D .xy C 1/e
xy
for x; y 0.
Solution:
is function is to be optimized only over the first quadrant. We begin by finding criti-
cal points in the usual fashion
f
x
.x; y/ D ye
xy
C .xy C1/e
xy
.1/ D .xy C y 1/e
xy
f
y
.x; y/ D xe
xy
C .xy C1/e
xy
.1/ D .xy C x 1/e
xy
Since the extreme value theorem tells us that the optima occur at boundaries or critical points,
we won’t need a second derivative test – we just compare values. is means out next step is to
solve for any critical points. Remember that powers of e cannot be zero, giving us the system of
equations:
xy C y 1 D 0
xy C x 1 D 0
xy D x 1 D y 1 so x D y
x
2
C x 1 D 0
x
2
x C 1 D 0
x D
1 ˙
p
1 4
2
ere are no critical points!
is means that the extreme value occurs on the boundaries – the positive x and y axes where
either x D 0 or y D 0. So, we need the largest value of f .0; y/ D e
y
for y 0 and f .x; 0/ D
e
x
for x 0. Since e
x
is largest (for non-negative x) at x D 0, we get that the global maximum
is f .0; 0/ D 1.
˙
While the extreme value theorem lets us make decisions without the second derivative test, it
forces us to examine the boundaries, which can be hard. It is sometimes possible to solve the
50 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
problem by adopting a different point of view. e next example will use a transformation to a
parametric curve to solve the problem.
Example 2.22 Maximize
g.x; y/ D x
4
C y
4
for those points f.x; y/ W x
2
C y
2
4g.
Solution:
is curve has a single critical point at .x; y/ D .0; 0/ which is extremely easy to find.
e boundary for the optimization domain is the circle x
2
C y
2
D 4, a circle of radius 2
centered at the origin. is boundary is also the parametric curve .2 cos.t/; 2 sin.t//. is means
that on the boundary the function is
g.2 cos.t/; 2 sin.t// D 16 cos
4
.t/ C 16 sin
4
.t/:
is means we can treat the location of optima on the boundary as a single-variable optimization
task of the function g.t/ D 16 cos
4
.t/ C 16 sin
4
.t/. We see that
g
0
.t/ D 64 cos
3
.t/ .sin.t // C64 sin
3
.t/ cos.t/ D 64 sin.t/ cos.t/.sin
2
.t/ cos
2
.t//
Solve:
sin.t/ D 0 t D .2n C1/
2
cos.t/ D 0 t D n
sin
2
.t/ cos
2
.t/ D 0
sin
2
.t/ D cos
2
.t/ t D ˙.2n C1/
4
If we plug these values of t back into the parametric curve, the points on the boundary
that may be optima are: .0; ˙2/, .˙2; 0/, and .˙
p
2; ˙
p
2/. Plugging the first four of these
into the curve we get that g.x; y/ D 16. Plugging the last four into the function we get
g.x; y/ D 2 .
p
2/
4
D 2 4 D 8. At the critical point (0,0) we see g.x; y/ D 0. is means that
the maximum value of g.x; y/ on the optimization domain is 16 at any of .0; ˙2/; .˙2; 0/.
˙
2.2. THE EXTREME VALUE THEOREM REDUX 51
Notice that the solutions to the parametric version of the boundary gave us an infinite number
of solutions. But, when we returned to the .x; y; z/ domain, this infinite collection of solutions
were just the eight candidate points repeated over and over. is means that our failure to put a
bound on the parameter t was not a problem; bounding t was not necessary.
A problem with the techniques developed in this section is that all of them are special purpose.
We will develop general purpose techniques in Section 2.3 – the understanding of which is
substantially aided by the practice we got in this section.
Example 2.23 If
h.x; y/ D x
2
C y
2
find the minimum value of h.x; y/ among those points .x; y/ on the line y D 2x 4.
Solution:
e function h.x; y/ is a paraboloid that opens upward. If we look at the points .x; y; z/
on h.x; y/ that happen to lie on a line, that line will slice a parabolic shape out of the surface
defined by h.x; y/. First note that h.x; y/ has a single critical point at .0; 0/ which is not in
the domain of optimization. is means we may consider only those points on y D 2x 4, in
other words on the function
h.x; 2x 4/ D x
2
C .2x 4/
2
D 5x
2
16x C 16
is means that the problem consists of finding the minimum of f .x/ D 5x
2
16x C 16.
f
0
.x/ D 10x 16 D 0
10x D 16
x D 8=5
y D 16=5 20=5 D 4=5
Since f .x/ opens upward, we see that this point is a minimum; there are no boundaries to the
line that is constraining the values of .x; y/ so the minimum value of h.x; y/ on the line is
h.8=5; 4=5/ D
64
25
C
16
25
D
80
25
D
16
5
˙
52 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
Example 2.23 did not really use the extreme value theorem. For that to happen we would need
to optimize over a line segment instead of a full line.
Example 2.24 If
h.x; y/ D x
2
C y
2
find the minimum and maximum value of h.x; y/ among those points .x; y/ on the line
y D x C 1 for 4 x 4.
Solution:
is problem is very similar to Example 2.23. e ends of the line segment are (-4,-3)
and (4,5), found by substituting into the formula for the line. On the line h.x; y/ becomes
h.x; x C 1/ D x
2
C .x C 1/
2
D 2x
2
C 2x C 1
So we get a critical point at 4x C 2 D 0 or x D 1=2 which is the point .1=2; 1=2/.
Plug in and we get
h.4; 3/ D 25
h.1=2; 1=2/ D 1=2
h.4; 5/ D 41
is means the minimum value is 1/2 at the critical point and that the maximum value is 41 at
one of the endpoints of the domain of optimization.
˙
Making the domain a line segment is one of the simplest possible options. Let’s look at another
example with a more complex domain of optimization.
Example 2.25 If
s.x; y/ D 2x C y 4
find the minimum and maximum value of s.x; y/ among those points .x; y/ on the curve
y D x
2
3 for 2 x 2.
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