1819.) | and Minima of Quantities. 195 
Multiplying by 2 — a, and transposing, the equation becomes 
w—mr+am=o0-.2%1—m=o0andm=2x. 
But from the first equation m = wo = 22, which 
ct—a4 z-—@4 
reduced gives x= 2 a. 
_ When radical quantities enter the proposed function, it does 
not appear that the common rule for toe equations havi 
two equal roots is generally applicable. e shall, therefe 
investigate one by which examples of this kind may be resolved. 
Since the equation fx = y has two affirmative roots, let p = 
less = root and p + e = greater (e being their difference, and 
represented in the illustration of the lemma by the line P P’). 
Then p and p + e substituted for x in the equation fxr = y give 
the same result, viz. y. Therefore fp = y = f (p + e) (A). 
After developing the second member of this equation, and taking 
away the quantities that are common to each side, all the 
remaining terms will be divisible by e, and we shall have an 
equation containing p, e, and constant quantities, which will be 
true for every value of the function. But when y = m, e= 0; 
therefore all the terms containing e and its powers will vanish, 
and we shall have an equation expressing the relation between p 
and given quantities, from the resolution of which, p or its equal 
x will be known. 
Let us apply this method to problem 2, where we have given 
r— 2x = y, to find r when y = m. This equation has two 
affirmative roots (by the lemma). Let p = less root and p + e 
= greater.. Then these substituted for x in the proposed 
equation give the same result, viz. y. Therefore, p — p®? = y 
=p he— (p + ef, orp = p= p+ ie — p— 3 pe — 
3pe—e. Taking away p — p° from each side, and dividing 
the remaining terms by e, we have 1 — 3 p>? — 3pe —e =o. 
Now make e = o (because when y = m, e vanishes), and the 
last equation becomes 1 — 3 p? = 0..p = WL =z. 
From this example, it is evident that in developing the func- 
tions of (p + e) only two terms of the series need be taken, 
because the succeding ones contain e*, e°, &c. and, therefore, 
ultimately disappear. 
Problem 4.—Given in position two points A, B, and the line 
CK; it is required to determine the point E ia this line, so 
that (A E + B E) may be the least possible. 
From the points A and B, let 
fall the perpendiculars A C, B D, 
upon C K, and let E be the re- 
ee point. Put AC = a, 
D=b,CD=candCE= , 
x; thnED=c—2,AE= - pa ha 
Va+x,BE= SB 
J (ce — rf + B. 
Fig. 3. 
Cc E D K 
