Question

Prove that a conical tent of given capacity will require the least amount of canvas when the height is \[\sqrt 2 \] times the radius of the base.

Answer 1

Hint:
Here, we will apply Pythagoras theorem in the cone such that the square of the slant height is equal to the sum of squares of the height and the radius respectively. Now, we will differentiate the surface area and equate it zero to find the value of height. We will then double differentiate the surface area to find the maxima and minima that will help us to verify the given statement.

Formula Used:
We will use the following formulas:
1) Volume of a conical tent, \[V = \dfrac{1}{3}\pi {r^2}h\], where, \[r\] is the radius of the base of the conical tent and \[h\]is the height of the conical tent.
2) Surface area of a conical tent, \[S = \pi rl\], where \[l\]is its slant height.
By Pythagoras theorem, in a cone, \[{l^2} = {h^2} + {r^2}\]

Complete step by step solution:
According to the question, we are given a conical tent of given capacity or volume.
Now, also, we know that, in the right triangle formed inside the cone, by Pythagoras theorem,
\[{l^2} = {h^2} + {r^2}\]
Now, we know that Surface area of a conical tent, \[S = \pi rl\].
Squaring both sides of the above equation, we get
\[{S^2} = {\pi ^2}{r^2}{l^2}\]
Let \[{S^2} = Z\] and substituting \[{l^2} = {h^2} + {r^2}\], we get,
\[ \Rightarrow Z = {\pi ^2}{r^2}\left( {{h^2} + {r^2}} \right)\]……………………………….\[\left( 1 \right)\]
Now, we know that, \[V = \dfrac{1}{3}\pi {r^2}h\].
Squaring both sides of the above equation, we get
\[ \Rightarrow {V^2} = \dfrac{1}{9}{\pi ^2}{r^4}{h^2}\]
Hence, we get the value of \[{h^2}\] as:
\[ \Rightarrow {h^2} = \dfrac{{9{V^2}}}{{{\pi ^2}{r^4}}}\]
Substituting this value in equation \[\left( 1 \right)\], we get
\[Z = {\pi ^2}{r^2}\left( {\dfrac{{9{V^2}}}{{{\pi ^2}{r^4}}} + {r^2}} \right)\]
Multiplying the terms, we get
\[ \Rightarrow Z = \dfrac{{9{V^2}}}{{{r^2}}} + {\pi ^2}{r^4}\]
Now, differentiating both sides with respect to \[r\] using the formula \[\dfrac{{dy}}{{dx}}{x^n} = n{x^{n - 1}}\], we get
\[ \Rightarrow \dfrac{{dZ}}{{dr}} = \dfrac{{\left( { - 2} \right)9{V^2}}}{{{r^3}}} + 4{\pi ^2}{r^3}\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{{dZ}}{{dr}} = 4{\pi ^2}{r^3} - \dfrac{{18{V^2}}}{{{r^3}}}\]
Now, substituting \[\dfrac{{dZ}}{{dr}} = 0\] in the above equation, we get
\[ \Rightarrow \dfrac{{dZ}}{{dr}} = 4{\pi ^2}{r^3} - \dfrac{{18{V^2}}}{{{r^3}}} = 0\]
Taking LCM in the LHS, we get
\[ \Rightarrow 4{\pi ^2}{r^6} - 18{V^2} = 0\]
\[ \Rightarrow 4{\pi ^2}{r^6} = 18{V^2}\]
Dividing both sides by 2, we get
\[ \Rightarrow 2{\pi ^2}{r^6} = 9{V^2}\]
But, we know that \[{h^2} = \dfrac{{9{V^2}}}{{{\pi ^2}{r^4}}}\] or \[9{V^2} = {\pi ^2}{r^4}{h^2}\].
Hence, we get,
\[ \Rightarrow 2{\pi ^2}{r^6} = {\pi ^2}{r^4}{h^2}\]
Cancelling the similar terms, we get
\[ \Rightarrow 2{r^2} = {h^2}\]
Taking square root on both sides, we get
\[ \Rightarrow h = \sqrt 2 r\]
Now, differentiating again \[\dfrac{{dZ}}{{dr}} = 4{\pi ^2}{r^3} - \dfrac{{18{V^2}}}{{{r^3}}}\] with respect to \[r\], we get
\[ \Rightarrow \dfrac{{{d^2}Z}}{{d{r^2}}} = 4{\pi ^2}\left( 3 \right){r^2} + 3 \times \dfrac{{18{V^2}}}{{{r^4}}}\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{{{d^2}Z}}{{d{r^2}}} = 12{\pi ^2}{r^2} + \dfrac{{54{V^2}}}{{{r^4}}}\]
Substituting \[h = \sqrt 2 r\] in the above equation, we get
Clearly, \[\dfrac{{{d^2}Z}}{{d{r^2}}} = 12{\pi ^2}{r^2} + \dfrac{{54{V^2}}}{{{r^4}}} > 0\] for all values of \[r\]and \[V\]
Hence, we have proved that \[Z = {S^2}\] is minimum when \[h = \sqrt 2 r\].
Therefore, we can say that a conical tent of given capacity will require the least amount of canvas when the height is \[\sqrt 2 \] times the radius of the base.
Hence, proved

Note:
There are two types of surface areas. One is lateral surface area in which we include the area of the lateral or side surface only. Hence, in case of cones, when we take its lateral surface area, the area of the circular base of the cone is neglected, thus we consider the formula as \[\pi rl\]. But, in case of total surface area, we take the overall area of the surface of the given shape. Hence, in case of a cone, we consider the area of the circular base, i.e. \[\pi {r^2}\] as well. Thus, the total surface area of a cone becomes: \[\pi rl + \pi {r^2}\]. Thus, the difference between the two must be kept in mind while solving the questions.