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Question : 28
Total: 29
Find the volume of the largest cylinder that can be inscribed in a sphere of radius r.
OR
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for sides, what is the cost of least expensive tank?
OR
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for sides, what is the cost of least expensive tank?
Solution:
The given sphere is of radius R. Let h be the height and r be the radius of the cylinder inscribed in the sphere.
Volume of cylinder
V =π R 2 h
In right angled triangle ΔOBA
A B 2 + O B 2 O A 2
R 2 +
r 2
So,R 2 r 2 −
Putting the value ofR 2
V = π( r 2 −
)
V = π( r 2 h −
)
∴
( r 2 −
)
For stationary point,
π( r 2 −
)
r 2
h 2
Now ,
( −
h )
∴[
] at h =
( −
,
)
∴ Volume is maximum at h =
Maximum volume is = π( r 2 ×
.
×
)
= π(
)
= π(
)
=
OR
Let l, b, and h denote the length breadth and depth of the open rectangular tank.
Given h = 2m
V =8 m 3
i.e. 2/b = 8
⇒ lb = 4 or b =
Surface area, S, of the open rectangular tank of depth 'h' = lb + 2 (l + b) × h
In this problem , b =
∴ S = 4 + 2 (l + 4/l) × 2
⇒ S = 4 + 4 (l + 4/l)
For maxima or minima, differentiating with respect to l we get,
( 1 −
)
l = 2m for minimum or maximum
Now,
So l = 2m is a point of minima and minimum surface area is
S = lb + 2 (l + b) × h
= 4 + 2 × 8 = 4 + 16 = 20 square meters
Base Area = 4 square metres; Lateral surface area = 16 square metres
cost = 4 × 70 + 16 × 45
= 280 + 720 = Rs. 1000
Volume of cylinder
V =
In right angled triangle ΔOBA
So,
Putting the value of
V = π
V = π
∴
For stationary point,
π
Now ,
∴
∴ Volume is maximum at h =
Maximum volume is = π
= π
= π
=
OR
Let l, b, and h denote the length breadth and depth of the open rectangular tank.
Given h = 2m
V =
i.e. 2/b = 8
⇒ lb = 4 or b =
Surface area, S, of the open rectangular tank of depth 'h' = lb + 2 (l + b) × h
In this problem , b =
∴ S = 4 + 2 (l + 4/l) × 2
⇒ S = 4 + 4 (l + 4/l)
For maxima or minima, differentiating with respect to l we get,
l = 2m for minimum or maximum
Now,
So l = 2m is a point of minima and minimum surface area is
S = lb + 2 (l + b) × h
= 4 + 2 × 8 = 4 + 16 = 20 square meters
Base Area = 4 square metres; Lateral surface area = 16 square metres
cost = 4 × 70 + 16 × 45
= 280 + 720 = Rs. 1000
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