2.2. THE EXTREME VALUE THEOREM REDUX 53
Solution:
In this problem we are cutting a parabolic segment out of the plane s.x; y/ D 2x C y 4. We
get that the ends of the domain of optimization are .2; 1/ and .2; 1/ by plugging in the ends of
the interval in x to the formula for the parabolic segment. Substituting the parabolic segment
into the plane yields
s.x; x
2
3/ D 2x C x
2
3 4 D x
2
C 2x 7
is means our critical point appears at 2x C 2 D 0 or x D 1, making the candidate point
.1; 2/. Plugging the candidate points into s.x; y/ yields:
s.2; 1/ D 7
s.1; 2/ D 8
s.2; 1/ D 1
is means that the maximum value is 1 at .2; 1/, and the minimum is 8 at .1; 2/.
˙
Some sets of boundaries are simple enough that we can use geometric reasoning to avoid needing
to use calculus on the boundaries.
Example 2.26 Find the maximum value of
h.x; y/ D x
2
C y
2
for 2 x; y 3.
Solution:
First of all, we know that this surface has a single critical point at .0; 0/ – this surface is
an old friend (see graph in Figure 1.2).
e domain of optimization is a square with corners .2; 2/ and .3; 3/. Along each side
of the square, we see that the boundary is a line – and so has extreme values at its ends. is
means that we need only check the corners of the square; the interior of the edges – viewed as
line segments – cannot attain maximum or minimum values by the extreme value theorem. is
means we need only add the points .2; 2/, .2; 3/, .3; 2/, and .3; 3/ to our candidates.
54 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
Plugging in the candidate points we obtain:
h.2 2/ D 8
h.0; 0/ D 0
h.2; 3/ D 13
h.3; 2/ D 13
h.3; 3/ D 18
So the maximum is 18 at .3; 3/, and the minimum is 0 at .0; 0/.
˙
e goal for this section was to set up LaGrange Multipliers – the topic of Section 2.3. e
take-home message from this section is that the extreme value theorem implies that optimizing
on a boundary is the difficult added portion of optimizing on a bounded domain.
e techniques that we will develop in the next section require that the boundary itself be a
differentiable curve. Some of the boundaries in this section are made of several differentiable
curves, meaning that the techniques in this section may be easier for those problems. If we have
a boundary that is not differentiable, then the techniques in this section are the only option.
PROBLEMS
Problem 2.27 If we look at the points on
h.x; y/ D x
2
C y
2
C 4x C4
that fall on a line, then there is a minimum somewhere on the line. Find that minimum value
for the following lines.
1. y D 3x C 1
2. y D 4 x
3. x C y D 6
4. 2x 7y D 24
5. x C y D
p
3=2
6. 3x C 5x D 7
2.2. THE EXTREME VALUE THEOREM REDUX 55
Problem 2.28 Find the maximum of
f .x; y/ D .x
2
C y
2
/ e
xy
with .x; y/ in the first quadrant where 0 < x; y.
Hint: this function is symmetric. Use this fact.
Problem 2.29 Find the maximum and minimum of
q.x; y/ D
1
x
2
C y
2
2x 4y C 6
in the first quadrant: 0 < x; y.
Problem 2.30 Find the maximum of
g.x; y/ D .x
4
C y
4
/
on the set of points
f.x; y/ W x
2
C y
2
25g:
Problem 2.31 Maximize
h.x/ D x
4
C y
4
on the set of points f.x; y/ W x
2
C y
2
25g.
Problem 2.32 If
g.x; y/ D 2x
2
C 3y
2
;
find the minimum of g.x; y/ for those points .x; y/ on the line y D x 1.
56 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
Problem 2.33 For the function
s.x; y/ D x
2
C y
2
2x C 4y C 1
with .x; y/ on the following line segments, find the minimum and maximum values.
1. y D 2x 2 on 2 x 2
2. y D x C 7 on 1 x 4
3. y D 6 5x on 0 x 6
4. x C y D 10 on 4 x 10
5. 2x 7y D 8 on 10 x 10
6. 2x y D 13 on 1 x 5
Problem 2.34 Suppose that
p.x; y/ D 3x 5y C 2
Find the maximum and minimum values on the following parametric curves.
1.
.
3t C1; 5 t
/
I 5 t 5
2.
.
cos.t/; sin.t/
/
3.
.
sin.t/; 3 cos.t/
/
4.
.
sin.2t/; cos.t/
/
5.
.
t cos.t/; t sin.t /
/
I 0 t 2
6.
.
3 sin.t/; 2 sin.t/
/
Problem 2.35 Find the maximum and minimum of
q.x; y/ D
1
x
2
C y
2
C 4x 12y C 45
in the first quadrant: 0 < x; y.
Problem 2.36 Suppose that P D f .x; y/ is a plane and that we are considering the points
where x
2
C y
2
D r
2
for some constant r. If f .x; y/ is not equal to a constant, explain why
there is a unique minimum and a unique maximum value.
Problem 2.37 If
f .x; y/ D x
2
C y
2
;
find the minimum of f .x; y/ for those points .x; y/ on the line y D mx C b.
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