Question

Find the dimensions of the rectangle of perimeter $36$cm which will sweep out a volume as large as possible when revolved about one of its sides.

Answer 1

Hint: In these type of questions, we make use of the formula of the perimeter of the rectangle, and the volume of the cylinder, and from the formula of the volume of the cylinder, the maximum volume of the cylinder can be calculated by using derivatives.

Complete step-by-step answer:
Given the perimeter of the rectangle is $36$cm.
Let us consider the length of the rectangle be $l$cm and the breadth of the rectangle be $b$cm,
Now using the formula for perimeter of rectangle i.e.,
Perimeter of rectangle, $ = 2\left( {l + b} \right)$
Given perimeter of rectangle is 36cm,
$ \Rightarrow 2\left( {l + b} \right) = 36$,
Now taking 2 to other side we get,
$ \Rightarrow l + b = \dfrac{{36}}{2}$,
Now applying division we get,
$ \Rightarrow b = 18 - l$,
Now taking $b$ to the other side, we get,
$b = 18 - l$,
Step 2:
From the given data, the rectangle of perimeter $36$cm which will sweep out a volume as large as possible when revolved about one of its sides,
Now let us consider that the rectangle is being resolved about its length $b$ and then the Volume of the cylinder $V$ is given by the formula of the volume of the cylinder i.e.,
$V = \pi {l^2}b$,
Now substituting the value of \[2\pi r\left( {h + r} \right)\] in the volume formula, we get,
$V = \pi {l^2}\left( {18 - l} \right)$,
Now multiplying we get,
$V = \pi \left( {18{l^2} - {l^3}} \right)$,
As the rectangle sweep out a volume as large as possible when revolved about one of its sides, we have to find the maximum value using differentiation, then it will be the maximum volume of the cylinder,
Now differentiating both sides we get,
$\dfrac{{dV}}{{dl}} = \dfrac{d}{{dl}}\pi \left( {18{l^2} - {l^3}} \right)$
Now differentiating on both sides we get,
$ \Rightarrow \dfrac{{dV}}{{dl}} = \pi \left( {\dfrac{d}{{dx}}18{l^2} - \dfrac{d}{{dx}}{l^3}} \right)$,
Now differentiating each term we get,
\[ \Rightarrow \dfrac{{dV}}{{dl}} = 18\pi \dfrac{d}{{dl}}{l^2} - \pi \dfrac{d}{{dl}}{l^3}\],
Now using derivatives formula we get,
\[ \Rightarrow \dfrac{{dV}}{{dl}} = \pi \left( {36l - 3{l^2}} \right)\]
Now equate\[\dfrac{{dV}}{{dl}} = 0\], then we get,
\[ \Rightarrow 0 = \pi \left( {36l - 3{l^2}} \right)\]
Now taking constant terms to L.H.S we get,
\[ \Rightarrow 36l - 3{l^2} = 0\]
Now taking 3 common in L.H.S we get,
\[ \Rightarrow 3l\left( {12 - l} \right) = 0\]
So now taking the term\[3l\] to \[0\] we get,
\[ \Rightarrow 12 - l = 0\],
Now taking the constant term to L.H.S we get,
\[ \Rightarrow l = 12\],
Now substituting the value of\[l\] in the equation\[l + b = 18\], we get,
\[ \Rightarrow 12 + b = 18\],
Now taking constant term to one side we get,
\[ \Rightarrow b = 18 - 12\],
Now applying subtraction we get,
\[ \Rightarrow b = 6\],
Now again to find the maximum value of the volume then we have to differentiate the volume for the second time we get,
\[\dfrac{{{d^2}V}}{{d{l^2}}} = \dfrac{d}{{dl}}\pi \left( {36l - 3{l^2}} \right)\]
Now differentiating we get,
\[ \Rightarrow \dfrac{{{d^2}V}}{{d{l^2}}} = \pi \left( {\dfrac{d}{{dl}}36l - \dfrac{d}{{dx}}3{l^2}} \right)\]
Now differentiating each part we get,
\[ \Rightarrow \dfrac{{{d^2}V}}{{d{l^2}}} = \pi \left( {36 - 6l} \right)\],
As we know the value of\[l = 12\], substituting the value in the above equation we get,
\[ \Rightarrow \dfrac{{{d^2}V}}{{d{l^2}}} = \pi \left( {36 - 6(12)} \right)\]
Now simplifying we get,
\[ \Rightarrow \dfrac{{{d^2}V}}{{d{l^2}}} = \pi \left( {36 - 72} \right)\],
Now subtracting the terms we get,
\[ \Rightarrow \dfrac{{{d^2}V}}{{d{l^2}}} = - 36\pi \],
So, as it is a negative number it is always smaller than zero, the volume of the cylinder will be maximum when\[l = 12\],
Now calculating the maximum volume of the cylinder using volume formula of the cylinder,\[V = \pi {l^2}b\],
Now substituting the values of \[l\] and\[b\] in the volume formula we get,
\[V = \pi {\left( {12} \right)^2}\left( 6 \right)\]
Now simplifying we get,
\[V = \pi \left( {144} \right)\left( 6 \right)\],
Now multiplying we get,
\[V = \pi \left( {144} \right)\left( 6 \right)\],
Now multiplying we get,
\[V = 864\pi \],
SO, the maximum volume of the cylinder is\[864\pi c{m^3}\].
\[\therefore \]the dimensions of the rectangle are length is equal to 12 cm and breadth is equal to 6 cm.

The dimensions of the rectangle given length is equal to 12 cm and breadth is equal to 6 cm.

Note:
In this type of questions the students must know the formulas for the perimeter and area of the rectangle and volume, surface area of the cylinder, students must be careful when using the formulas, as there are many formulas related to the area, perimeter, volume, surface area of the shapes. Here are some formula which are useful:
Area of square of side \[x = {x^2}\],
Perimeter of square of side \[x = 4x\],
Area of the rectangle=\[lb\],
Perimeter of rectangle\[ = 2\left( {l + b} \right)\],
Volume of the cylinder \[ = \pi {r^2}h\]
Surface area of cylinder\[ = 2\pi r\left( {h + r} \right)\]