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19901
THE TEACHING OF MATHEMATICS 421
Once (1) is established for R", we can extend it to more general inner product
spaces. For example, let w,, w,,. . . , w,, be n strictly positive real numbers and
consider the weighted Euclidean inner product 2 . y' = C:=,w,x,y,. Using the change
we obtain (1) for any weighted Euclidean
of variables (, = &xi and 7 , =
inner product space. More generally we can extend (1) to any finite dimensional
inner product space using coordinates of vectors with respect to any orthonormal
basis [I, p. 2251.
Finally. for any inner product space we can establish (1) by a slight variation of
the proof that projection_onto an rz-dimensional subspace is norm-decreasing [4, pp.
145-1471. Assume x'# 0 and let S E R be such that x'.y'= 8121 Iy'll. Let us
remark that equality holds in (1) if and only if I S = 1 or y' = 6. Let us consider the
scalar h such that (y'- Ax') . ,i? = 0. Hence A = x' . y'/11x'112 = Slly'll/ll2ll and
117- h,i?l12 = (1+- S 2 ) l y l / 2 It. follows that -1 < S < 1 and y'= hx'if and only if
61 = 1 ory'=O.
,/FYI
REFERENCES
1. H. Anton and C. Rorres, Elementary Linear Algebra with Applications, Wiley, New York, 1987.
2. A. Cauchy, Cours d'analyse de 1 ' ~ c o l eRoyale Polytechnique, Ire partie: analyse algtbrique, Paris,
Debure frPres 1821, in Oeuvres cotnplPtes d'Augustin Cauchy, IIe strie, tome 111, Paris. Gauthier-Villars, 1897.
3. I. Niven, Maxima and Minima without Calculus, Dolciani Mathematical Expositions, No. 6,
Mathematical Association of America, 1981.
4. G. Strang, Linear Algebra and Its Applications (3rd ed.), Harcourt, Brace, Jovanovich, San Diego,
1988.
An Old Max-Min Problem Revisited
MARYEMBRY-WARDROP
Dej~ur.tirlerrro / Mutherrzutrc.i. Cerltrui Mrchrgurz Unlve~sriy,M f . Pleu.sunt, M I 488.59
Recently, while teaching a beginning calculus class, I introduced the problem of
inscribing a rectangle of maximum area in a given right triangle and asked the class
how the rectangle should be oriented inside the triangle. I was slightly surprised
when a student suggested that one side of the rectangle should be on the hypotenuse
of the triangle. In previous classes the suggestions were to fit one corner of the
rectangle inio the right angle of the triangle. The class soon bogged down in trying
to solve the problem with one side of the rectangle on the hypotenuse, but
subsequently solved the problem using the other orientation.
I was sufficiently intrigued to solve the problem with both orientations and was
surprised to discover that in either case the maximum area was $ bh, where b and h
are the base and height of the triangle. Furthermore in each case corners of the
rectangle of maximum area lie on the midpoints of the two legs of the triangle.
At this point I was thoroughly intrigued: Could the rectangle be more advantageously oriented and for other orientations did a corner of the rectangle of
maximum area lie on the midpoint of a leg of the triangle?
The following assertions describe the results of my investigation. The basic
mathematical tools used were the Law of Sines and the technique of maximizing or
422
[May
MARY EMBRY-WARDROP
minimizing a function, usually discussed in a beginning calculus class. I omit the
solutions under the assumption that most mathematicians enjoy finding their own
solutions more than reading another's.
Let the rectangle be oriented in the triangle as pictured in FIGURE1.The angle a
satisfies 8 < a < 71/2. For a given angle a let A(a) be the area of the rectangle of
maximum area which can be so inscribed in the triangle.
Assertion I . Given a the rectangle of largest area is obtained when
h
b sin 8
X = - and y =
2 sin a
2cos(a - 8 )
'
In this case
A(a)
bh sin 8
=
4sinacos(a
-
8)
and P is the midpoint of the side of the triangle opposite the angle 8.
Assertion 2. The maximum area function A(a), 8
the two endpoints of the interval:
< a < :,
Assertion 3. The maximum area function A(a), 8
the midpoint of the interval:
<a <
A -8+-
has largest value at
has smallest value at
bh sin 8
=
2(1+sin8)'
Moreover in this case, the shaded triangle in Figure 1 is isosceles with equal sides of
length t h .
There are numerous problems here for a first-semester calculus class and each of
the solutions calls upon skills that the students should have developed in geometry,
trigonometry, and calculus. One of the interesting facets of Assertions 2 and 3 is
that the students would be invited to consider maximizing and minimizing A(a),
each value of which is itself a maximum. Few problems of this type seem to be
accessible for beginning students. Furthermore, there is a collection of geometric
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THE TEACHING OF MATHEMATICS
423
problems only touched upon in Assertions 1-3: Why is P a fixed point as a corner
of the rectangles of maximum area? Why is the shaded triangle isosceles when A ( a )
is minimum? Why is A ( a ) minimum at the midpoint of the domain of A ( a ) ? Is
there a geometric proof that A ( a ) is obtained with sides given by the equation in
Assertion 1 or a geometric proof that A ( B ) and A ( ; ) are the maxima of A ( a ) and
A($B g) is the minimum?
It is interesting to note that the problem of inscribing a rectangle of maximum
area in an acute triangle is no more difficult than the same problem for a right
triangle. Let the rectangle be oriented in an acute triangle, as pictured in FIGURE2.
The angle a satisfies 0, < a < 7/2.
+
Once again for a given angle a the rectangle of maximum area, A ( a ) , which can
be so inscribed is obtained when P is the midpoint of the side of the triangle
opposite 0,. The maximum of A ( a ) occurs at the endpoints 0, and ;and the
of the domain of A . Finally, if the
minimum occurs at the midpoint :8,
triangle is obtuse, 8, > 7/2 for example, the solution is the same except that A ( a )
would only be defined for ;- 0, < a < ;and in thls case the maximum of A ( a )
would be attained only at 7/2.
Surely the results of this note are known. However, not a single mathematician to
whom I have mentioned them admitted having considered the maximum area
problem with any orientation other than the most standard one.
+
Functional Equations and L'HGpital's Rule in an Exact Poisson Derivation J. A c z f ~
Centre for l i ~ f o r n ~ u t Ti oh~e ~o y und Quanrirative Econom~cs,University of Wuterloo, Wuterloo,
Ontario, Cunuda N 2 L 3G1
In [2], Cooke derived the Poisson formula
P,,( t ) = Prob[n arrivals in the interval (0, t )]