Question 35 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards

Last updated at Oct. 1, 2019 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.

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
= 𝜋×𝑘/𝑟+𝜋𝑟^2+2𝑘/𝑟
= 𝜋𝑘×1/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
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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.