A complete and practical solution book for the common school teacher . — x, FH=jj/, and the vol-ume of the cylinder=V. (3) We have v=iry*x. . (1). (4) From the similar trian- gles ADBand AHF, wehave AD : BD :: AH :FH, or a : b :: a—x : y. (5) .\y=— (a—x), which in (1) gives»=«£(«-,xYx . (2). (6) Dropping constant factors, we have u=(a— x)2x=a2x— 2ax2+x*. (7) .*. ~—a2— ±ax+3x2=0, or x*—$ax=—\a2. (8) .*. x—a, or \a. dn2 d2 it (9) -r^ ——±a-\-6x; -j—2=2a, when x=a, .. minimum. d2u (10) -j—2=—2#, when x — \a, .. maximum. (11) Hence, the altitude of the maximum cylinder is J of the cone. (12) The se
A complete and practical solution book for the common school teacher . — x, FH=jj/, and the vol-ume of the cylinder=V. (3) We have v=iry*x. . (1). (4) From the similar trian- gles ADBand AHF, wehave AD : BD :: AH :FH, or a : b :: a—x : y. (5) .\y=— (a—x), which in (1) gives»=«£(«-,xYx . (2). (6) Dropping constant factors, we have u=(a— x)2x=a2x— 2ax2+x*. (7) .*. ~—a2— ±ax+3x2=0, or x*—$ax=—\a2. (8) .*. x—a, or \a. dn2 d2 it (9) -r^ ——±a-\-6x; -j—2=2a, when x=a, .. minimum. d2u (10) -j—2=—2#, when x — \a, .. maximum. (11) Hence, the altitude of the maximum cylinder is J of the cone. (12) The second value of x in (2) gives V = —-(a—£#)q= (13) Volume of the cone = ^7r#£2. (14) .*. Volume of cylinder=| volume of cone. (15) y — ~{a—Jtf)=f£=radius of base of cylinder. (16) From the above the required result can easily be found. PROBLEM 447. From a cone, altitude 30 in. and radius of base 5 in., a 6-inch cyl-inder is cut as long as possible; from the top of the cone remaining- 226 FAIRCHILDS SOLUTION another cylinder is cut, of the same length as the former and the thick-est possible: how much of the volume of the original cone is cut away? Sohition. (1) Let ABC represent the cone, and SKDE and FETP thecylinders. (2) AO = 30in., OC=5in., LD = 3 in. (3) COA and DLA are similar triangles. (4) We have CO: AO::DL:AL, or 5 : 30 :: 3 : AL. Fromwhich AL = 18 in. (5) LO = 30 in.—18 in. = 12 in., the length of the cylinderSE, also the length of thecylinder FT. (6) ThetrianglesALDandTED are similar. (7) WehaveAL:LD::TE:ED, or ED=2 in. (8) LE=LD—ED=1 in., or ra- dius of small cylinder. (9) Volume of the large cylinder is (32ttX 12) =108tt cu. in. (10) Volume of the small cylinder is (12ttX 12) =12tt cu. in. (11) Volume of both cylinders=108ir+12ir = 120ir cu. in. (12) Volume of the cone ABC = 527rXl0=2507r cu. in. (13) Volume cut away = 2507r—1207r=130ir, or cu. in. PROBLEM 448. What are the two dimensions of a con
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