310 Higher Engineering Mathematics
h
O
P
Q
r
h
2
R
5
1 2 c m
Figure 28.11
Volume of cylinder,
V = πr
2 h
(1)
Using the right-angled triangle OPQ shown in
Fig. 28.11,
r 2 +
h
2
2
= R 2 by Pythagoras’ theorem,
i.e.
r 2 +
h 2
4
= 144
(2)
Since the maximum volume is required, a formula for
the volume V is needed in terms of one variable only.
From equation (2),
r
2
= 144 −
h 2
4
Substituting into equation (1) gives:
V = π
144 −
h 2
4
h = 144πh −
πh 3
4
dV
dh
= 144π −
3πh 2
4
= 0,
for a maximum or minimum value.
Hence
144π =
3πh 2
4
from which,
h =
(144)(4)
3
= 13.86 cm
d 2 V
dh 2 =
−6πh
4
When h = 13.86,
d 2 V
dh 2 is negative, giving a maximum
value.
From equation (2),
r
2
= 144 −
h 2
4
= 144 −
13.86 2
4
from which, radius r = 9.80 cm
Diameter of cylinder = 2r = 2(9.80) = 19.60 cm.
Hence the cylinder having the maximum volume that
can be cut from a sphere of radius 12 cm is one in
which the diameter is 19.60 cm and the height is
13.86 cm.
Now try the following exercise
Exercise 123 Further problems on
practical maximum and minimum problems
1. The speed, v, of a car (in m/s) is related to
time t s by the equation v = 3 +12t − 3t 2 .
Determine the maximum speed of the car
in km/h.
[54 km/h]
2. Determine the maximum area of a rectangular piece of land that can be enclosed by
1200 m of fencing.
[90000 m 2 ]
3. A shell is fired vertically upwards and
its vertical height, x metres, is given by
x = 24t − 3t 2 , where t is the time in seconds.
Determine the maximum height reached.
[48 m]
4. A lidless box with square ends is to be made
from a thin sheet of metal. Determine the
least area of the metal for which the volume
of the box is 3.5 m 3 .
[11.42 m 2 ]
5. A closed cylindrical container has a surface
area of 400 cm 2 . Determine the dimensions
for maximum volume.
radius = 4.607 cm;
height = 9.212 cm
6. Calculate the height of a cylinder of maximum volume which can be cut from a cone
of height 20 cm and base radius 80 cm.
[6.67 cm]
h
O
P
Q
r
h
2
R
5
1 2 c m
Figure 28.11
Volume of cylinder,
V = πr
2 h
(1)
Using the right-angled triangle OPQ shown in
Fig. 28.11,
r 2 +
h
2
2
= R 2 by Pythagoras’ theorem,
i.e.
r 2 +
h 2
4
= 144
(2)
Since the maximum volume is required, a formula for
the volume V is needed in terms of one variable only.
From equation (2),
r
2
= 144 −
h 2
4
Substituting into equation (1) gives:
V = π
144 −
h 2
4
h = 144πh −
πh 3
4
dV
dh
= 144π −
3πh 2
4
= 0,
for a maximum or minimum value.
Hence
144π =
3πh 2
4
from which,
h =
(144)(4)
3
= 13.86 cm
d 2 V
dh 2 =
−6πh
4
When h = 13.86,
d 2 V
dh 2 is negative, giving a maximum
value.
From equation (2),
r
2
= 144 −
h 2
4
= 144 −
13.86 2
4
from which, radius r = 9.80 cm
Diameter of cylinder = 2r = 2(9.80) = 19.60 cm.
Hence the cylinder having the maximum volume that
can be cut from a sphere of radius 12 cm is one in
which the diameter is 19.60 cm and the height is
13.86 cm.
Now try the following exercise
Exercise 123 Further problems on
practical maximum and minimum problems
1. The speed, v, of a car (in m/s) is related to
time t s by the equation v = 3 +12t − 3t 2 .
Determine the maximum speed of the car
in km/h.
[54 km/h]
2. Determine the maximum area of a rectangular piece of land that can be enclosed by
1200 m of fencing.
[90000 m 2 ]
3. A shell is fired vertically upwards and
its vertical height, x metres, is given by
x = 24t − 3t 2 , where t is the time in seconds.
Determine the maximum height reached.
[48 m]
4. A lidless box with square ends is to be made
from a thin sheet of metal. Determine the
least area of the metal for which the volume
of the box is 3.5 m 3 .
[11.42 m 2 ]
5. A closed cylindrical container has a surface
area of 400 cm 2 . Determine the dimensions
for maximum volume.
radius = 4.607 cm;
height = 9.212 cm
6. Calculate the height of a cylinder of maximum volume which can be cut from a cone
of height 20 cm and base radius 80 cm.
[6.67 cm]