Higher Engineering Mathematics, Sixth Edition

Some applications of differentiation 309

Hence for the least surface area, a cylinder of vol-
ume 200cm^3 has a radius of 3.169cm and height of
6.339cm.

Problem 18. Determine the area of the largest

piece of rectangular ground that can be enclosed by

100m of fencing, if part of an existing straight wall

is used as one side.

Let the dimensions of the rectangle bexandyas shown
in Fig. 28.9, wherePQrepresents the straight wall.

P Q

yy

x

Figure 28.9

From Fig. 28.9,

x+ 2 y= 100 (1)

Area of rectangle,

A=xy (2)

Since the maximum area is required, a formula for area
Ais needed in terms of one variable only.
From equation (1),x= 100 − 2 y
Hence areaA=xy=(100− 2 y)y= 100 y− 2 y^2

dA

dy

= 100 − 4 y= 0 ,

for a turning point, from which,y=25m

d^2 A

dy^2

=− 4 ,

which is negative, giving a maximum value.
Wheny=25m,x=50m from equation (1).
Hence themaximum possible area=xy=( 50 )( 25 )=
1250m^2.

Problem 19. An open rectangular box with

square ends is fitted with an overlapping lid which

covers the top and the front face. Determine the

maximum volume of the box if 6m^2 of metal are

used in its construction.

A rectangular box having square ends of sidexand
lengthyis shown in Fig. 28.10.

x

x y

Figure 28.10

Surface area of box,A, consists of two ends and five

faces (since the lid also covers the front face.)

Hence

A= 2 x^2 + 5 xy=6(1)

Since it is the maximum volume required, a formula

for the volume in terms of one variable only is needed.

Volume of box,V=x^2 y.

From equation (1),

y=

6 − 2 x^2

5 x

=

6

5 x

−

2 x

5

(2)

Hence volume

V=x^2 y=x^2

(

6

5 x

−

2 x

5

)

=

6 x

5

−

2 x^3

5

dV

dx

=

6

5

−

6 x^2

5

= 0

for a maximum or minimum value.

Hence 6= 6 x^2 ,givingx=1m (x=−1 is not possible,

and is thus neglected).

d^2 V

dx^2

=

− 12 x

5

Whenx= 1 ,

d^2 V

dx^2

is negative, giving a maximum value.

From equation (2), whenx=1,

y=

6

5 ( 1 )

−

2 ( 1 )

5

=

4

5

Hence the maximum volume of the box is given by

V=x^2 y=( 1 )^2

( 4

5

)

=^45 m^3

Problem 20. Find the diameter and height of a

cylinder of maximum volume which can be cut

from a sphere of radius 12cm.

A cylinder of radiusrand heighthis shown enclosed

in a sphere of radiusR=12cm in Fig. 28.11.