194 Mr. Slee on the Maxima - [Marew, 
ordinate possible. Then P M first increases till it becomes equal 
to C D, and then it decreases. Con- 
sequently a line drawn from M parallel Miko ts 
to the axis A B, will cut the curve > 
again in some point M’ on the other M M’ 
side of C D. Draw M’ P’ parallel to 
M P; then M’P’ = MP. Therefore, 
corresponding to any particular value 
P M of the ordinate, there are two 
different abscissa, viz. A P and 
A P’, which are both affirmative, 
being on the same side of the 
point A; that is, x has two different 
affirmative values in the equation 
fx =y. But conceive P M, and 
consequently P’ M’, to move up to 
CD, then A P and A P’ are each 
equal to A C, and, therefore, equal to each other, which shows 
that the two affirmative roots of the preceding equation become 
equal to each other when y = m. The same reasoning is 
evidently applicable to fig. 2, where the ordinate P M is supposed 
to admit of a mmmum C D. 
Corollary.—Hence if we deduce a value of x from the equa- 
tion f x = m, upon the supposition of its having two equal roots, 
that value will correspond to a maximum or a minimum. 
We shall now proceed to illustrate this theory, by applying it 
to a few examples ; but we must first remind the reader of the 
rule, which is investigated in most works on algebra, for reduc- 
ing equations having two equal roots. It is as follows : Multiply 
each term of the equation by the dex of the unknown quantity 
in that term, diminish the index by unity, and equate the result 
to nothing. Thus if the equation a 2* + ba*—' $ ca". .... 
+ m = 0, have two equal roots, it may be demonstrated that 
ir aaa ee, _90O$0c07O7Oo 
nmax*+n—1.ba°? Fn—2.cxe#®F=Ho. 
Problem \.—To divide a given line or number (a) into two 
such parts that the rectangle under them may be a maximum (m). 
Let x = one of the parts, then a —xv = the other, anda — 2 
xt = e@r—x2 =m, ora —ar+m=o. 
This equation has two equal roots by the lemma, 
Therefore, by the preceding rule, we have 
22—-a=0-.7 = 5. 
Problem 2.—-Find that fraction which, being diminished by its 
cube, shall give the greatest remainder possible. Let « = the 
required fraction, then ¢ — 2 = morai—a2+4+m=o..by 
the rule 82? — 1 = oanda = Vi. 
i, 2 * 
» Problem 3.—Find x when ——— = m 
; wt — &@ 
