Question 25 - CBSE Class 12 Sample Paper for 2019 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

Question 25

A cuboidal shaped godown with square base is to be constructed. Three times as much cost per square meter is incurred for constructing the roof as compared to the walls. Find the dimensions of the godown if it is to enclose a given volume and minimize the cost of constructing the roof and the walls.

Question 25 A cuboidal shaped godown with square base is to be constructed. Three times as much cost per square meter is incurred for constructing the roof as compared to the walls. Find the dimensions of the godown if it is to enclose a given volume and minimize the cost of constructing the roof and the walls.
Given a godown with square base
So,
Length of godown = Breadth of godown
Now,
Volume of godown = Length × Breadth × Height
Volume of godown = l × l × h
V = l2h
h = 𝑉/𝑙^2
Here, volume of given.
So, it is constant.
Thus, we write height in terms of V and l
Given that,
it costs three times as much cost per square meter is incurred for constructing the roof as compared to the walls.
Let
Rate of constructing the wall = Rs k per m2
So, Rate of constructing the roof = Rs 3k per m2
Cost of constructing the walls
= Rate of construction of wall × Area of walls
= k × 4lh
Putting value of h from (1)
= k × 4l × 𝑉/𝑙^2
= 4𝑘𝑉/𝑙
Cost of constructing the roof
= Rate of construction of roof × Area of roof
= 3k × l × l
= 3kl2
Now,
Cost of constructing roof and wall
= Cost of constructing roof + Cost of constructing the wall
= 3kl2 + 4𝑘𝑉/𝑙
C = 3kl2 + 4𝑘𝑉/𝑙
Now, we need to minimize Cost
i.e. we need to minimize C
Since length (l) is variable, we differentiate w.r.t l
𝑑𝐶/𝑑𝑙 = 6kl + 4kv × ((−1)/𝑙^2 )
𝑑𝐶/𝑑𝑙 = 6kl – 4𝑘𝑉/𝑙^2
Putting 𝑑𝐶/𝑑𝑙 = 0
0 = 6kl – 4𝑘𝑉/𝑙^2
6kl = 4𝑘𝑉/𝑙^2
6k𝑙^3 = 4𝑘𝑉
𝑙^3 = 4𝑘𝑉/6𝑘
𝑙^3 = 2𝑉/3
𝑙 = (2𝑉/3)^(1/3)
Now,
Finding (𝑑^2 𝐶)/(𝑑𝑙^2 )
𝑑𝐶/𝑑𝑙 = 6kl – 4𝑘𝑉/𝑙^2
Differentiating w.r.t l
(𝑑^2 𝐶)/(𝑑𝑙^2 ) = 6k – 4kV((−2)/𝑙^3 )
(𝑑^2 𝐶)/(𝑑𝑙^2 ) = 6k + 8𝑘𝑉/𝑙^3
Putting 𝑙 = (2𝑉/3)^(1/3)
(𝑑^2 𝐶)/(𝑑𝑙^2 ) = 6k + 8𝑘𝑉/(2𝑉/3 )
(𝑑^2 𝐶)/(𝑑𝑙^2 ) = 6k + 12k
(𝑑^2 𝐶)/(𝑑𝑙^2 ) = 18k
Since k is positive,
(𝑑^2 𝐶)/(𝑑𝑙^2 ) > 0
Therefore,
(𝑑^2 𝐶)/(𝑑𝑙^2 ) > 0 when 𝑙 = (2𝑉/3)^(1/3)
C is minimum when 𝑙 = (2𝑉/3)^(1/3)
Now, Finding h
From (1)
h = 𝑉/𝑙^2
h = 𝑉/(2𝑉/3)^(2/3)
h = (3/2)^(2/3) 𝑉/𝑉^(2/3)
h = (3/2)^(2/3) 𝑉^(1 − 2/3)
h = (3/2)^(2/3) 𝑉^(1/3)
h = (3^2/2^2 )^(1/3) 𝑉^(1/3)
h = (9/4)^(1/3) 𝑉^(1/3)
h = (9𝑉/4)^(1/3)
h = (9𝑉/4)^(1/3)
Thus,
Length of godown = 𝑙 = (2𝑉/3)^(1/3)
Breadth of godown = 𝑙 = (2𝑉/3)^(1/3)
Height of godown = h = (9𝑉/4)^(1/3)

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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