Question 35 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at May 29, 2023 by Teachoo

A given quantity of metal is to be cast into a solid half circular cylinder with a rectangular base and semi-circular ends. Show that in order that total surface area is minimum, the ratio of length of cylinder to the diameter of semi-circular ends is π ∶ π + 2.

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Question 35 (OR 1st Question) A given quantity of metal is to be cast into a solid half circular cylinder with a rectangular base and semi-circular ends. Show that in order that total surface area is minimum, the ratio of length of cylinder to the diameter of semi-circular ends is π ∶ π + 2.
Let r be the radius and h be the height of the half cylinder
The Half circular cylinder with look like
Let V & S be the volume & total surface area of the half cylinder respectively
Since volume of half cylinder is constant
Volume of half cylinder = 1/2×𝜋(𝑟𝑎𝑑𝑖𝑢𝑠)^2 (ℎ𝑒𝑖𝑔ℎ𝑡)
V = 1/2 𝜋𝑟^2 ℎ
2𝑉/𝜋=𝑟^2 ℎ
ℎ=(2𝑉/𝜋) 1/𝑟^2
ℎ= 𝑘/𝑟^2
where k = 2𝑉/𝜋
Since V is constant
⇒ (2𝑉/𝜋) is also constant
Let k = 2𝑉/𝜋
We need to minimize
Total Surface of Half Cylinder
Now,
Total Surface of Half Cylinder
= 1/2 × Curved Surface Area of Cylinder
+ 2 × Area of Semcircle
+ Area of bottom rectangle
= 1/2×2𝜋𝑟ℎ+2×(1/2 𝜋𝑟^2 )+2𝑟×ℎ
= 𝜋𝑟ℎ+𝜋𝑟^2+2𝑟ℎ
Putting h = 𝒌/𝒓^𝟐
= 𝜋𝑟×𝑘/𝑟^2 +𝜋𝑟^2+2𝑟×𝑘/𝑟^2
= 𝜋×𝑘/𝑟+𝜋𝑟^2+2𝑘/𝑟
= 𝜋𝑘×1/𝑟+𝜋𝑟^2+2𝑘×1/𝑟
Diff w.r.t r
𝑑𝑆/𝑑𝑟=𝜋𝑘×(1/𝑟)^′+𝜋(𝑟^2 )^′+2𝑘×(1/𝑟)^′
=𝜋𝑘×((−1)/𝑟^2 )+𝜋×2𝑟+2𝑘×((−1)/𝑟^2 )
=−𝜋𝑘×1/𝑟^2 +2𝜋×𝑟−2𝑘×(1/𝑟^2 )
Putting 𝑑𝑆/𝑑𝑟=0
−𝜋𝑘×1/𝑟^2 +2𝜋×𝑟−2𝑘×(1/𝑟^2 )=0
2𝜋𝑟=2𝑘×(1/𝑟^2 )+𝜋𝑘×1/𝑟^2
2𝜋𝑟=𝑘(2/𝑟^2 + 𝜋/𝑟^2 )
2𝜋𝑟=𝑘((2 + 𝜋)/𝑟^2 )
(2𝜋𝑟^3)/((2 + 𝜋))=𝑘
𝑘=(2𝜋𝑟^3)/((2 + 𝜋))
Now,
Finding (𝑑^2 𝑆)/(𝑑𝑟^2 ) for this value of k
𝑑𝑆/𝑑𝑟 =−𝜋𝑘×1/𝑟^2 +2𝜋×𝑟−2𝑘×(1/𝑟^2 )
𝑑𝑆/𝑑𝑟 =𝑘((−𝜋)/𝑟^2 −2/𝑟^2 )+2𝜋𝑟
𝑑𝑆/𝑑𝑟 =𝑘((−𝜋 − 2)/(2𝑟^2 ))+2𝜋𝑟
𝑑𝑆/𝑑𝑟 =−𝑘((𝜋 + 2)/(2𝑟^2 ))+2𝜋𝑟
𝑑𝑆/𝑑𝑟 =−𝑘((𝜋 + 2)/2)×1/𝑟^2 +2𝜋𝑟
Diff. again w.r.t x
(𝑑^2 𝑆)/(𝑑𝑟^2 )=−𝑘((𝜋 + 2)/2)×(−2)/𝑟^3 +2𝜋
(𝑑^2 𝑆)/(𝑑𝑟^2 )=𝑘(𝜋+2)×1/𝑟^3 +2𝜋
Since everything is positive on the right side
Therefore,
(𝑑^2 𝑆)/(𝑑𝑟^2 )>0 when 𝑘=(2𝜋𝑟^3)/((2 + 𝜋))
i.e. S is minimum when 𝑘=(2𝜋𝑟^3)/((2 + 𝜋))
Now,
k = 2𝑉/𝜋
(2𝜋𝑟^3)/((2 + 𝜋)) = 2𝑉/𝜋
Putting Volume V = 𝟏/𝟐 𝝅𝒓^𝟐 𝒉
(2𝜋𝑟^3)/((2 + 𝜋)) = (2 × 1/2 𝜋𝑟^2 ℎ)/𝜋
(2𝜋𝑟^3)/((2 + 𝜋)) = (𝜋𝑟^2 ℎ)/𝜋
(2𝜋𝑟^3)/((2 + 𝜋)) = 𝑟^2 ℎ
𝑟^3/(𝑟^2 ℎ) = ((2 + 𝜋))/2𝜋
𝑟/ℎ = ((2 + 𝜋))/2𝜋
ℎ/𝑟 = 2𝜋/((2 + 𝜋))
We need to prove ratio of Height and Diameter
Therefore,
ℎ/2𝑟 = 2𝜋/(2(2 + 𝜋))
ℎ/2𝑟 = 𝜋/((𝜋 + 2))
Hence proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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