de Maxlmis & Mimffils,

ig

for the Area of the Triangle

V X 1 -—a

being a Minimum, its Square must be a Minimum , and

Consequently —st-f—, or its--

!, Therefore, which is

mum also *. Whose Fluxion.

3 xXx+al Xx

x X x -f a

the Whole divided by

being put = 0> an d

a 1 ... -'

.I , we alto. get 3 x x—a

•— x + a = o ; whence x— 2a : Therefore, OD being— 2.OS, and the Triangles ODS and BDC equiangular,it is evident that DC is likewise = 2BC = AC ; and sothe Triangle ACD, when the least possible, is equila-teral.

EXAMPLE VII.

30. To determine the greatejl Cylinder , dg, that tan beinscribed in ct .given Cone ADB.

Let <?—BC, ,the Altitude of the Cone j5 —AL, the Diameter of its Base ;

(dh) the Diameter of the Cylinder, con-

sidered as variable;

p = sh l6 r l Skiff s) the Area of the Circlev -4 /

whose Diameter is Unity.

Then, the Areas of Circles being to one another asthe Squares of their-Diameters, we have, T: x % : :p : (px i ) the Area of the Circle fsgr: Moreover, fromthe Similarity of the Triangles ABC and Ads, we have

ib (AC) : « (BG) :: ib— ix {Ad) : df .

b

which multiplied by the Area px % (found above) gives

C 2

pabx *