Higher Engineering Mathematics, Sixth Edition

310 Higher Engineering Mathematics

h O

PQ

r

h

2

R^5

12 cm

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 volumeVis 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= 144 πh−

πh^3

4

dV

dh

= 144 π−

3 πh^2

4

= 0 ,

for a maximum or minimum value.

Hence

144 π=

3 πh^2

4

from which, h=

√

( 144 )( 4 )

3

= 13 .86cm

d^2 V

dh^2

=

− 6 πh

4

Whenh= 13. 86 ,

d^2 V

dh^2

is negative, giving a maximum

value.

From equation (2),

r^2 = 144 −

h^2

4

= 144 −

13. 862

4

from which, radiusr= 9 .80cm

Diameter of cylinder= 2 r= 2 ( 9. 80 )= 19 .60cm.

Hence the cylinder having the maximum volume that

can be cut from a sphere of radius 12cm is one in

which the diameter is 19.60cm and the height is

13.86cm.

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
   timetsby the equationv= 3 + 12 t− 3 t^2.
   Determine the maximum speed of the car
   in km/h. [54km/h]


2. Determine the maximum area of a rectangu-
   lar piece of land that can be enclosed by
   1200m of fencing. [90000m^2 ]


3. A shell is fired vertically upwards and
   its vertical height, xmetres, is given by
   x= 24 t− 3 t^2 ,wheretis the time in seconds.
   Determine the maximum height reached.
   [48m]


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 3.5m^3. [11.42m^2 ]


5. A closed cylindrical container has a surface
   area of 400cm^2. Determine the dimensions
   for maximum volume.[
   radius= 4 .607cm;
   height= 9 .212cm

]

6. Calculate the height of a cylinder of max-
   imum volume which can be cut from a cone
   of height 20cm and base radius 80cm.
   [6.67cm]