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Contents
Quiz
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Quiz Changes
(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.
Answers (i)
Total surface
area, s = 1/2 .4pr2
+ pr2
+ 2prh
= 3pr2
+ 2prh
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%
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