Some applications of differentiation 309
Hence for the least surface area, a cylinder of volume 200 cm 3 has a radius of 3.169 cm and height of
6.339 cm.
Problem 18. Determine the area of the largest
piece of rectangular ground that can be enclosed by
100 m of fencing, if part of an existing straight wall
is used as one side.
Let the dimensions of the rectangle be x and y as shown
in Fig. 28.9, where P Q represents the straight wall.
P
Q
y
y
x
Figure 28.9
From Fig. 28.9,
x + 2y = 100
(1)
Area of rectangle,
A = x y
(2)
Since the maximum area is required, a formula for area
A is needed in terms of one variable only.
From equation (1), x = 100 −2y
Hence area A =xy = (100 −2y)y = 100y −2y 2
d A
d y
= 100 − 4y = 0,
for a turning point, from which, y = 25 m
d 2 A
d y 2 = −4,
which is negative, giving a maximum value.
When y = 25 m, x = 50 m from equation (1).
Hence the maximum possible area = x y = (50)(25) =
1250 m
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 6 m 2 of metal are
used in its construction.
A rectangular box having square ends of side x and
length y is 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 = 2x
2
+ 5x y = 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 − 2x 2
5x
=
6
5x
−
2x
5
(2)
Hence volume
V = x
2 y = x
2
6
5x
−
2x
5
=
6x
5
−
2x 3
5
dV
dx
=
6
5
−
6x 2
5
= 0
for a maximum or minimum value.
Hence 6 =6x 2 , giving x = 1 m (x =−1 is not possible,
and is thus neglected).
d 2 V
dx 2 =
−12x
5
When x = 1,
d 2 V
dx 2 is negative, giving a maximum value.
From equation (2), when x = 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
=
4
5 m
3
Problem 20. Find the diameter and height of a
cylinder of maximum volume which can be cut
from a sphere of radius 12 cm.
A cylinder of radius r and height h is shown enclosed
in a sphere of radius R = 12 cm in Fig. 28.11.
Hence for the least surface area, a cylinder of volume 200 cm 3 has a radius of 3.169 cm and height of
6.339 cm.
Problem 18. Determine the area of the largest
piece of rectangular ground that can be enclosed by
100 m of fencing, if part of an existing straight wall
is used as one side.
Let the dimensions of the rectangle be x and y as shown
in Fig. 28.9, where P Q represents the straight wall.
P
Q
y
y
x
Figure 28.9
From Fig. 28.9,
x + 2y = 100
(1)
Area of rectangle,
A = x y
(2)
Since the maximum area is required, a formula for area
A is needed in terms of one variable only.
From equation (1), x = 100 −2y
Hence area A =xy = (100 −2y)y = 100y −2y 2
d A
d y
= 100 − 4y = 0,
for a turning point, from which, y = 25 m
d 2 A
d y 2 = −4,
which is negative, giving a maximum value.
When y = 25 m, x = 50 m from equation (1).
Hence the maximum possible area = x y = (50)(25) =
1250 m
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 6 m 2 of metal are
used in its construction.
A rectangular box having square ends of side x and
length y is 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 = 2x
2
+ 5x y = 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 − 2x 2
5x
=
6
5x
−
2x
5
(2)
Hence volume
V = x
2 y = x
2
6
5x
−
2x
5
=
6x
5
−
2x 3
5
dV
dx
=
6
5
−
6x 2
5
= 0
for a maximum or minimum value.
Hence 6 =6x 2 , giving x = 1 m (x =−1 is not possible,
and is thus neglected).
d 2 V
dx 2 =
−12x
5
When x = 1,
d 2 V
dx 2 is negative, giving a maximum value.
From equation (2), when x = 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
=
4
5 m
3
Problem 20. Find the diameter and height of a
cylinder of maximum volume which can be cut
from a sphere of radius 12 cm.
A cylinder of radius r and height h is shown enclosed
in a sphere of radius R = 12 cm in Fig. 28.11.