A complete and practical solution book for the common school teacher . FIG. 114. MENSURA TION. 225. FIG. 115. PROBLEM 446. Find the maximum cylinder which can be inscribed in a conicalcup 1 ft. deep and 10 in. in diameter. Solutio?i. (1) Let ABC be the cone, and MN the inscribed cyl-inder. (2) Let BD=£, AD=tf, HD — 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*.
A complete and practical solution book for the common school teacher . FIG. 114. MENSURA TION. 225. FIG. 115. PROBLEM 446. Find the maximum cylinder which can be inscribed in a conicalcup 1 ft. deep and 10 in. in diameter. Solutio?i. (1) Let ABC be the cone, and MN the inscribed cyl-inder. (2) Let BD=£, AD=tf, HD — 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 re
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