. Differential and integral calculus. cone. 14. Find the right cylinder of greatest convex surface that can be inscribed in a given right cone. irab Ans. S = 2 15. Show that the cone of greatest volume and greatest con-vex surface that can be inscribed in a sphere of radius (a) has-| a for its altitude. 16. Show that the altitude of the cone of least volume thatcircumscribes the sphere is 4 a. 17. Of all cones of a given slant height show that the one, the ratio of whose altitude to the radius of its base is —-, is a V2maximum. 18. Of all circular sectors of a given perimeter show that theone
. Differential and integral calculus. cone. 14. Find the right cylinder of greatest convex surface that can be inscribed in a given right cone. irab Ans. S = 2 15. Show that the cone of greatest volume and greatest con-vex surface that can be inscribed in a sphere of radius (a) has-| a for its altitude. 16. Show that the altitude of the cone of least volume thatcircumscribes the sphere is 4 a. 17. Of all cones of a given slant height show that the one, the ratio of whose altitude to the radius of its base is —-, is a V2maximum. 18. Of all circular sectors of a given perimeter show that theone whose arc = twice the radius is greatest. Let x = radius ; then, 2 x -f- arc = p, a constant, and Area = \ x (p — 2 x). 19. A Norman window, consisting of a semicircle surmountinga rectangle, is to be of a given perimeterand so constructed that the light ad-mitted shall be a maximum; required,the height and breadth of the window. With the notation of the figure, wehave, Area = 2 xy -\ , 2 and Perimeter = p = 2 y -{- 2 x -)- 156 Differential Calculus .*. Area =f(x) = px — 2 x2 .*. fix) = p — — irx = o. Hence, y = 4 + 7T / 4 + * Therefore the height of rectangle = radius of semicircle. We have heretofore, where two variables occurred in the ex-pression we desired to investigate for maximum and minimumvalues, substituted for one of the variables its value in terms ofthe other, as derived from given conditions. Thus, in A = 7TX2 2 xy -\ above we substituted for y its value in terms of x, as 2 determined by the conditionp =2y -\- 2 x -\- irx. It frequently happens that this substitution may be made more conveniently after differentiation. . ttxThus, f\x) = 2 xy -\ And from j>=2y-\-2X-\-Trx ; have, dy ^ j 0 = 2 — + 2+ 7T. dx dy 2 + itdx 2 Substituting now this value of -j- together with that of ydrawn from the value of/ in the value oif\x), we have, = p — 4 x — irx, as before. Maxima and Minima 157 20. A person in a boat 3 miles from shore wishes to reach
Size: 1707px × 1463px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, booksubjectcalculu, bookyear1918
