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(x^{2}  b^{2})  (m/n)x + a. Taking the derivative and making it equal to zero we get
so MN adopts its maximum at x = mb/Ö(m^{2}  n^{2}). Therefore
cosa = 
b
x

= b / 
æ
ç
è 
mb

ö
÷
ø 
= 
m

. 

Consequently a = arccos(Ö(m^{2}  n^{2} )/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
so we must have
na  bÖ(
m^{2}  n^{2} ) > 0 in order to have a solution. Furthermore, one can check that
MN"(x) = 
 b^{2}
(x^{2}  b^{2} )^{3/2}

; 

therefore
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

x^{2}  b^{2}

 
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

x^{2}  b^{2}

ö
ø 
. 

We note that
Thus
leads, as expected, to x = mb/Ö(m^{2}  n^{2}).
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(x^{2}  1) . Hence (5/4)x + MN = 20 + Ö(x^{2}  1), i.e. MN(x) = 20 + Ö(x^{2}  1)  (5/4)x. Taking the derivative and making it equal to zero we get x^{2}/(x^{2}  1) = 25/16, thus x = 5/3. Finally, since cosa = 1/(5/3) = 3/5, we get a = arccos(3/5) = 53.13^{°}.