Nitin wants to construct a rectangular plastic tank for his house that can hold
80 lt' of water. The top of the tank is open. The width of tank will be 5 ft but the
length and heights are variables. Building the tank cost 20 per sq. foot for the
base and 10 per sq. foot for the side.
Based on the above information, answer the following questions
(i) In order to make a least expensive water tank, Nitin need to minimize its
(a) Volume
(b) Base
(c) Curved surface area
(d) Cost
(ii) Total cost of tank as a function of h can be represented as
(a) c(h) = 100 h - 320 - 1600/h
(b) c(h) 100 | 320 h20
1600
(C) C(h) 100 + 320 h + 160 h
(d) c(h) = 100 h + 320 + 1600/h
(iii) Range of h is
(a) (3.5)
(b) (0, infinity)
(c) (0.8)
(d) (0, 3)
(iv) Value of h at which c(h) is minimum, is
(a) 4
(b) 5
(c) 6
(d) 6.7
(v) The cost of least expensive tank is
(a) 1020
(b) 1100
(c) 1120
(d) 1220
Answers
Given : The top of the tank is open. The width of tank will be 5 ft but the
length and heights are variables.
Building the tank cost 20 per sq. foot for the base and 10 per sq. foot for the side.
Volume of tank is 80 cubic ft ( not 80 litre)
To Find : In order to make a least expensive water tank, Nitin need to minimize what ?
Solution:
In order to make a least expensive water tank, Nitin need to minimize cost
The width of tank will be 5 ft
Let sat height = h
Volume = length * width * height
=> 80 = 5 * length * h
=> length = 16/h
Base area = 5 * 16/h
CSA = 2 ( 5 + 16/h) * h
Cost = 20 ( 5 * 16/h ) + 10 * 2 ( 5 + 16/h) * h
= 1600/h + 100h + 320
c(h) = 100h + 320 + 1600/h
Range of h (0 , ∞)
c(h) = 100h + 320 + 1600/h
=> c'(h) = 100 - 1600/h²
c'(h) = 0
=> 1600/h² = 100
=> h = 4
c''(h) = 3200/h³ > 0
Hence cost is minimum at h = 4
The cost of least expensive tank
c(h) = 100h + 320 + 1600/h
c(4) = 400 + 320 + 400
= 1120
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