# Thomas Simpson and Maxima and Minima

## Introduction

Thomas Simpson (1710-1761) was a self-taught English mathematician who started his working life as a weaver, his father’s trade. Quite early he showed a keen interest in mathematics and later in life became an accomplished writer of textbooks on algebra, geometry, the calculus, and other mathematical subjects. His life was quite remarkable, from being a weaver to becoming a fellow of the Royal Society in 1745 (Clarke 1929 ).  Nowadays, Simpson is best remembered for the numerical integration technique that bears his name.

Simpson’s most widely known book appeared in print in 1750 under the title The Doctrine and Application of Fluxions. The fact that it was reprinted as late as 1823 (Simpson 1823 ) attests to its wide popularity. By modern standards it is an unusual work in the sense that applications of the calculus appear rather early and pervade all of the book.

After a first section on the nature of fluxions, and how to calculate with them, Simpson discusses with great care a collection of twenty two examples about maxima and minima. Fifteen of these examples are of a geometrical nature, three are applications to kinematics, and only four are strictly mathematical. We will discuss in detail eight of them, none commonly found in contemporary Calculus textbooks, replacing the word "fluxion" with "derivative" whenever the former appears in Simpson’s book. As expected, a certain amount of editing has been necessary, but we have kept the core of Simpson’s approach and explanations. We share the belief that it is a fruitful endeavor to engage students in the solution of mathematical problems from the past (Swetz 1995 ).

It is to be noted that we will use the concept of function, an idea that took almost two hundred years to mature since Leibniz and Johan Bernoulli introduced it at the end of the seventeenth century. Euler, the greatest eighteenth century mathematician, used the symbol f(x) starting in 1734 (Siu 1995 ). The notation or definition of function is nowhere to be found in Doctrine although one might surmise that it is implicitly employed, one way or another, in the work.

Furthermore, Simpson does not apply the second derivative test for extrema; it is not even stated in his work. Neither does he use the first derivative test except when discussing example XXII, as we will see later on. Despite this fact, no errors are to be found throughout the section on maxima and minima; the very nature of the problems, mostly applications to geometry and kinematics, helped Simpson avoid any pitfalls. For him it was enough to take the first derivative of the pertinent expression and then find the critical point. Of course, we can check, through the first or second derivative test, that things work well in all the examples of Simpson’s that we will discuss.

After these preliminary considerations, let us discuss in detail some of the examples from section II of Doctrine. We will state them almost verbatim, then we will provide a solution patterned on Simpson’s solution, and finally we will make some remarks pertinent to each problem.

## Maximizing area of right triangles

Of all right-angled plane triangles having the same given hypotenuse, to find that whose area is the greatest (Example IV, page 17)

Let a be the hypotenuse and x, y the legs. We have
 y = Ö a2 - x2 ,       so        area(x) = x 2 Ö a2 - x2 .
Let f(x) = (x2/4)(a2 - x2 ), the square of the area. Then 0 = f¢(x) = (2a2/4)x - x3, consequently x = a/Ö2. It follows that
 y = Ö a2 - a2/2 = a Ö2 .
Thus, the best we can do is to choose the isosceles right triangle with given hypotenuse a.

Simpson provides an alternative way of solving the problem at hand (he says "the same otherwise"): Assuming that y is a function of x, implicit differentiation - a term not used by the author - leads to 2x + 2yy¢ = 0 since x2 + y2 = a2, thus y¢ = - x/y . On the other hand area(x) = (1/2)xy(x), so area¢(x) = (1/2)(y + xy¢). We are looking for the maximum of the area function, thus we have to make area¢(x) = 0. Therefore (1/2)(y + xy¢) = 0, which in turn leads to y¢ = -y/x. Consequently -x/y = -y/x, that is to say x = y. Again we reach the conclusion that the right isosceles triangle, of given hypotenuse a, encloses the largest area.

Remarks: It might surprise the reader that, in the first approach, Simpson does not take the derivative of the area but of its square. This is a clever choice since the latter is a simple polynomial. As the author writes right before example IV: "It will be proper to observe here, that the value of a quantity, when a maximum or minimum, may oftentimes be determined with more facility by taking the fluxion of some given part, multiple, or power thereof, than from the fluxion of the quantity itself." Simpson does not, however, mention that a simple algebraic alternative exists, due to the fact that f(x) happens to be a biquadratic function. Indeed -(1/4)x4 + (a2/4)x2 can be transformed into - (1/4)z2 + (a2/4)z by defining z = x2. Completing squares, we can tell that the function in the variable z adopts its minimum at

 z = - a2/4 2( - 1/4) = a2 2 .
So the original equation attains its minimum at
 x = Ö a2 /2 = a Ö2 .

## Least isosceles triangle circumscribing a circle

To determine the dimensions of the least isosceles triangle ACD that can circumscribe a given circle (Example VI, p. 18).

Let O be the center of the circle, with radius a, inscribed in the isosceles triangle ACD. Draw the altitude DB and let x = OD. We have DDSO is similar to DDBC. Thus DS/DB = SO/BC. Hence

BC = (x + a)a

 Ö x2 - a2
so        area(x) = (x + a)2 a

 Ö x2 - a2
.

Define

 f(x) = (x + a)4 x2 - a2 ,

which is the square of the area multiplied by 1/a2. Note that

 f(x) = (x + a)3 x - a so        f¢(x) = 3(x + a)2 (x - a) - (x + a)3 (x - a)2 .

The numerator will be zero when x = 2a. Therefore, the area attains its minimum at x = 2a. From the similarity of triangles DDSO and DDBC, we get DC/OD = BC/SO. Thus DC = x ·BC/a = 2a ·BC/a = 2BC = AC. In other words, the solution is to construct the equilateral triangle that circumscribes the given circle.

Remark: It is interesting to observe that Simpson implicitly assumes that the altitude BD passes through the center of the circle. Undoubtedly, he believes that his readers are knowledgeable in geometry and will realize that the center of the inscribed circle is at the point of intersection of the angle bisectors of the given triangle. Because the latter is isosceles, BD is both an altitude and an angle bisector; thus BD goes through O.

## Maximizing the volume of a cone

Of all cones under the same given superficies to find that (ABD) whose solidity is the greatest (Example IX, p. 21).

It should be noted that by "superficies" Simpson means "lateral surface", while "solidity" is an archaic synonym of "volume". We have s = px2 + pxy, where s is the lateral surface, y is the length of the slant side, and x the radius of the base. Then y = s/px - x. The height h is found by using the Pythagorean proposition. Thus

 h = Ö (s/px - x)2 - x2 = Ö s2/p2 x2 - 2s/p ,
and consequently
 V(x) = 1 3 px2 Ö s2/p2 x2 - 2s/p .

Define f(x) = (s2/9)x2 - (2ps/9)x4, the square of the volume. Then 0 = f¢(x) = (2s2/9)x - (8/9)psx3 implies x = Ö{s/4p}, which is the value of the radius at which the volume is the greatest. Under these circumstances

y = s - px2

px
=
 s - ps 4p

 p Ö s/4p
= 3
Ö

s/4p

= 3x.

Remark: As in example IV, Simpson does not mention that the square of the volume is a biquadratic expression. There is a simple algebraic alternative to determine its maximum: define z = x2, thus the problem is reduced to finding the point where the parabola defined by -(2ps/9)z2 + (s2/9)z attains its maximum. Given that this is a quadratic polynomial, we know that this takes place at

 z = -s2/9 -4ps/9 = s 4p ,

thus x = Ös/4p.

## Motion of bodies I

Two bodies move at the same time, from two given places A and B, and proceed uniformly from thence in given directions, AP and BQ, with celerities in a given ratio; it is proposed to find their position, and how far each has gone, when they are nearest possible to each other (Example XIII, page 28).

Let a = AC, b = BC, c = DC in the adjacent figure. Assume m is the velocity ("celerity") of the body that moves in the direction AP while n is the velocity of the body that moves in the direction BN.

At time t the first body will be at M while the second body will be at N (Simpson calls these two points "cotemporary"). Next draw perpendiculars NE and BD to AP, and let x = CN. Since DECN is similar to DDCB, we can conclude that b/x = c/CE. Because M and N are cotemporary points it follows that AM/m = BN/n, therefore AM = m/n(x - b). So CM = AC - AM = a - m/n(x - b) = d - m/nx, where d = a + m/nb. By the law of cosines we have MN2 = CM2 + CN2 - 2(CM)(CN)cosC. But cosC = EC/CN. Therefore

 MN2
 = CM2 + CN2 - 2(CM)(CE) = æ è d - m n x ö ø 2 + x2 - 2 æ è d - m n x ö ø cx b
 = d2 - æ è 2dm n + 2cd b ö ø x + æ è m2 n2 + 2cm nb +1 ö ø x2.

Taking the derivative of MN2, which is obviously a function of x, and making it equal to zero we get

 x = mnbd + cdn2 bm2 + 2cmn + bn2 .
This is the value where MN2, and consequently MN, attains its minimum.

Remark: Maybe Simpson should have mentioned that MN2 is a second degree polynomial in the variable x, thus on the Cartesian plane it represents a vertical parabola that opens upwards with minimum at -r/2s where r and s are the coefficients of x and x2 respectively. We would get exactly the same value obtained before, namely

 x = mnbd + cdn2 bm2 + 2cmn + bn2 .
So, in this problem there is an alternative to the use of derivatives.

## Motion of bodies II

Let a body move uniformly from A towards Q, with the celerity m, and let another body proceed from B, at the same time, with the celerity n. Now it is proposed to find the direction BP of the latter, so that the distance MN of the two bodies, when the latter arrives in the way of direction AQ of the former, may be the greatest possible (Example XIV, page 31).

Draw the perpendicular BC and let b = BC, a = AC. Let N be the point where the second body crosses the path from A to Q, and define x = BN, a = mÐCBN. Let M be the position of the first body when the other is at N.

Since M and N are cotemporary we will have AM/m = BN/n. Thus AM = (m/n)x, and consequently CM = AM - AC = (m/n)x - a. Then MN(x) = CN - CM = Ö(x2 - b2) - (m/n)x + a. Taking the derivative and making it equal to zero we get

0 = MN¢(x) = x

 Ö x2 - b2
- m

n
,

so MN adopts its maximum at x = mb/Ö(m2 - n2). Therefore

cosa = b

x
= b / æ
ç
è
mb

 Ö m2 - n2
ö
÷
ø
=
 Ö m2 - n2

m
.

Consequently a = arccos(Ö(m2 - n2 )/m) is the direction to be adopted by the second body if the distance MN is to be maximized.

Remarks: It is to be noted that

MN æ
ç
è
mb

 Ö m2 - n2
ö
÷
ø
=
 na - b Ö m2 - n2

n
,
so we must have na - bÖ(m2 - n2 ) > 0 in order to have a solution. Furthermore, one can check that
 MN"(x) = - b2 (x2 - b2 )3/2 ;
therefore
MN" æ
ç
è
mb

 Ö m2 - n2
ö
÷
ø
< 0,

so indeed we are dealing with a maximum! Simpson assumes that m > n. What happens if n ³ m? Under these circumstances there is not a mathematical solution. From a practical point of view, the body that starts at B would have to follow an angle a = 90 - e where e is a very small positive number. On the other hand, Simpson only discusses the case when M is between C and N. It can happen that M is between A and C, in which case

 MN = AN - AM = a + Ö x2 - b2 - m n x

(the very same expression for MN found before). If M is to the right of N, then

 MN = AM - AN = m n x - æ è a + Ö x2 - b2 ö ø .

We note that

MN¢(x) = m

n
- x

 Ö x2 - b2
.
Thus
0 = m

n
- x

 Ö x2 - b2

leads, as expected, to x = mb/Ö(m2 - n2).

A numerical example can further clarify the solution to the problem. Suppose m = 5 mph, n = 4 mph, b = 1 mile, a = 20 miles. Let t be the time it takes body B to reach the ray AQ at a certain point N. After t hours body A will be at M, so AM/5 = BN/4. Therefore AM = (5/4)x, where x = BN. But AM + MN = AC + CN = 20 + Ö(x2 - 1) . Hence (5/4)x + MN = 20 + Ö(x2 - 1), i.e. MN(x) = 20 + Ö(x2 - 1) - (5/4)x. Taking the derivative and making it equal to zero we get x2/(x2 - 1) = 25/16, thus x = 5/3. Finally, since cosa = 1/(5/3) = 3/5, we get a = arccos(3/5) = 53.13°.

## Ship and boat

Suppose a ship to sail from a given place A, in a given direction AQ, at the same time that a boat, from another given place B, sets out in order (if possible) to come up with her, and supposing the rate at which each vessel runs to be given; it is required to find in what direction the latter must proceed, so that if it cannot come up with the former, it may, however, approach it as near as possible (problem XV, page 32).

Let m be the constant velocity of the ship and n the constant velocity of the boat. If n ³ m, the boat will eventually meet the ship, so we may assume that m > n. After a certain time t, the ship is at D while the cotemporary point for the boat is at F. It is to be noted that all points on the circumference centered at B and radius BF are cotemporary with F, but the boat should follow the path BFD since any other path BF¢D, with F¢ on the circumference, is longer.

We need to minimize DF. Let BC be the perpendicular to AQ. Define AC = a, BC = b, CD = x, and DF = d. Since D and F are cotemporary and time = distance/velocity we can conclude that AD/m = BF/n.

Thus BF = (n/m)(a + x), and consequently

 d(x) = BD - BF = Ö b2 + x2 - na + nx m .

Therefore

d¢(x) = x

 Ö b2 + x2
- n

m
.
Making d¢(x) = 0, we get x = nb/Ö(m2 - n2), the value at which d(x) attains its minimum.

Remarks: We note that

 a = arctan æ è n/ Ö (m2 - n2) ö ø
while
 d æ è nb/ Ö (m2-n2) ö ø = (b Ö (m2 - n2) - na)/m.

So, the problem can be solved provided bÖ(m2 - n2) - na ³ 0. What happens if bÖ(m2 - n2) - na = 0? This equality implies DF = 0. In other words, the boat meets the ship if it takes the direction determined by the angle

 a = arctan æ è n/ Ö (m2-n2) ö ø

and bÖ(m2 - n2) - na = 0. Whenever bÖ(m2 - n2) - na > 0, we can be certain that

(bÖ(m2 - n2) - na)/m is the minimum distance that can be achieved.

## Maximizing y as an implicit function of x

To find the greatest value of y in the equation a4 x2 = (x2 + y2 )3 (Example XX, p. 42).

Assuming that y is a function of x, implicit differentiation ( Simpson says: "by putting the whole equation into fluxions") leads to 2a4 x = 3(x2 + y2 )2 (2x + 2yy¢). But we have to make y¢ = 0, so 2a4 x = = 3(x2 + y2 )2 (2x). Thus a2/Ö3 = x2 + y2, which in turn becomes a6/3Ö3 = (x2 + y2)3. Since (x2 + y2 ) = a4 x2, we can conclude that a6/3Ö3 = a4 x2, whence

 x = a/ Ö 3Ö3

.

Replacing this value in the original expression we get

 y = a Ö 2/3Ö3 .

An alternative approach, also presented in Doctrine, goes as follows: Taking the cube root we arrive at a4/3 x2/3 = x2 + y2, thus y2 = a4/3 x2/3 - x2. Consequently, 2yy¢ = (2/3)a4/3x-1/3 - 2x, i.e.

y¢ =
 2 3 a4/3 x-1/3 - 2x

2y
.
Making y¢ = 0, it follows that (2/3)a4/3x-1/3- 2x = 0, hence
 x = a/ 4 Ö 27 = a / Ö 3Ö3 .

Remark: An aspect not considered by Simpson is whether a maximum is actually attained by y at

 x = a/ Ö 3Ö3

First of all, let us note that he is undoubtedly working with the "positive part" of y defined by

 y = Ö (a4 x2 )1/3 - x2

x £ a. This function adopts the value zero at x = 0 and x = a, and is positive on (0,a). There is just one critical point, namely

 x = a/ Ö 3Ö3

Hence y attains its greatest value at this point.

## Maximizing a polynomial function

To determine the different values of x, when that of 3x4 - 28ax3 + 84a2 x2 - 96a3 x + 48b4 becomes a maximum or minimum (Example XXII, page 44).

Let f(x) be the given expression. Then f¢(x) = 12x3 - 84ax2 + 168a2 x - 96a3 . The critical points of f(x) stem from the equation 12x3 - 84ax2 + 168a2 x - 96a3 = 0, which is equivalent to x3 - 7ax2 + 14a2 x - 8a3 = 0. By simple inspection we can realize that the latter equation has solutions x = a,x = 2a,x = 4a. Therefore f¢(x) = (x - a)(x - 2a)(x - 4a). When x < a we have x < 2a and x < 4a, so f¢(x) < 0; when x > a but x < 2a we get x < 4a, thus f¢(x) > 0. By the first-derivative test we can conclude that f(x) adopts a local minimum at x = a. In a similar fashion one can prove that f(x) adopts a local maximum at x = 2a and a local minimum at x = 4a.

As a remark ("scholium") after example XXII (page 45, last paragraph), Simpson discusses the function g(x) = 24a3 x - 30a2 x2 + 16ax3 - 3x4. We have g¢(x) = 24a3 - 60a2 x + 48ax2 - 12x3. Factoring out 12, without much effort we get g¢(x)=- 12(x - a)2 (x - 2a). In a small neighborhood of x = a the derivative is positive, so g(x) does not adopt a local maximum or minimum at x = a. But it does adopt a local maximum at x = 2a because for x < 2a the derivative is positive, while for x > 2a the derivative is negative.

Remark: Both f(x) and g(x) are easy to analyze through the first-derivative test since their derivatives can be factored without much difficulty. The above-mentioned examples may seem rather ad hoc, nonetheless they illustrate quite well how an important test works.

## Final Considerations

In the same year 1823 that Simpson’s work was reprinted, Resume Des Lecons Donnees a L’Ecole Royale Polytechnique: Calcul Infinitesimal by Augustin-Louis Cauchy was published in France (Cauchy 1823 ). This book is what we would nowadays call a textbook of Analysis rather than an introduction to the calculus. In many respects it resembles contemporary textbooks; for instance, the first and second derivative tests are stated with care and precision. However, Resume does not deal with any applications, as is customary in a book on Analysis, making it a work quite different from Simpson’s. One of the strengths of  Doctrine is precisely the number and variety of applications, which help to motivate the reader and renders it a masterpiece of mathematical exposition.

In hindsight we can see that Simpson opted for the right course by avoiding extensive discussions on the foundations of Calculus, which no one was able to put on a solid base during Simpson’s lifetime, and instead concentrating on the usefulness of the techniques invented by Newton .

## Bibliography

Cauchy, A.-L. (1823). Resume des Lecons Donees a L’Ecole Royale Polytechnique: Calcul Infinitesimal .  Paris . Reprinted in 1994 by Aubin Imprimeur, Liguge, Poitiers , France .

Clarke, F.M. (1929). Thomas Simpson and His Times. New York

Simpson, T. (1823). Doctrine and Application of Fluxions. Edinburgh : Bell and Bradfute.

Siu, M. (1995). "Concept of Function -- Its History and Teaching," in Learn from the Masters (Swetz F. J. et al, Editors). Washington: Mathematical Association of America .

Swetz, F. J. (1995). "Using Problems from the History of Mathematics in Classroom Instruction," in Learn from the Masters (Swetz F.J. et al., Editors). Washington: Mathematical Association of America .