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A solid consists of a circular cylinder of radius r and height h, joined to a hemisphere whose radius is r cm, the flat base of the hemisphere being in contact with one of the circular ends of the cylinder. The total surface area of the solid is s.

(i) Prove that its volume is r/6 (3s - 5pr2) cm3.

(ii) Hence show that, for a given total surface area, the volume of the solid is greatest when h = r.

(iii) If the total surface area is 125p, find its maximum volume.

(iv) If h = 7cm, and the radius increases at a rate of 0.2 cm/s, find the rate of change of the volume when r = 3 cm.

Given that y = 2x3 - 3x2 + 1, use calculus to find, in terms of p, the approximate percentage increase in y when x increases from 2 by p% when p is small.

Answers

(i) Total surface area, s = 1/2 .4pr2 + pr2 + 2prh
                             = 3pr2 + 2prh
                                       ....................(1)
    Volume 
                
 
(ii) dV/dr = s/2 - 5pr2/2
 
     At maximum volume, dV/dr = 0
                 s/2 = 5pr2/2
                    s = 5pr2
 
     Substitute into equation (1),
                           
                      = r
 
     d2V/dr2 = -5pr
          which is always negative for all values of r.
 
     The volume is greatest when h = r.
 
(iii) At maximum volume,
                s = 5pr2 = 125p
                          r = 5
 
      Maximum volume = 5/6 [5p(5)3 - 5p(5)2]
                               = 625p/3
 
(iv) When h = 7,
               s - 3pr2 = 14pr
               s = 3pr2 + 14pr
 
     dV/dr = 1/2 (3pr2 + 14pr) - 5pr2/2
             = 1/2 [3p(3)2 + 14p(3)] - 5p(3)2/2
             = 12p
 
     dV/dt = dV/dr . dr/dt
             = 12p . 0.2
             = 2.4p cm3/s
 
 
     y = 2x3 - 3x2 + 1
 
     dy/dx = 6x2 - 6x
 
when x = 2,
     dy/dx = 12
     y = 5
 
     dy = 12(2)p% = 24p%
 
Percentage change in y = 24p% / 5 = 4.8p%