Question

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

Answer 1

Hint: Here, we have to find the dimensions of the rectangle of a given perimeter which covers maximum volume when revolving about one of its lengths. For that, we will first assume the dimensions of the rectangle and then we will find the perimeter of the assumed dimensions. We will equate the perimeter with the given perimeter, we will get the relation between x and y. Then to find the value of x and y, we need another equation which we will get after equating the derivative of the volume of the rectangle with zero as it is given in the question that the volume covered is maximum. Then we will solve these two equations to get the value of x and y. We will check whether the volume covered is maximum or not. For that we will differentiate the volume equation twice and we will put all the values obtained. If the value is less than zero than the obtained dimensions are the required dimensions.

Complete step-by-step answer:
Let the length of the rectangle be $x$ and breadth be $y$
Now let’s find the perimeter of the rectangle length and of breadth
Perimeter = $2(x+y)$ …………..$(1)$
Given perimeter of the rectangle $=36cm$
Now, we will equate the perimeter obtained in the equation $(1)$with the given perimeter.
Therefore,
$2(x+y)=36$
Now, we will divide 36 by 2.
$\Rightarrow$ $(x+y)=\dfrac{36}{2}$
After dividing 36 by 2, we get
$\Rightarrow$ $(x+y)=18$
Therefore,
$\Rightarrow$ $y=18-x$…………….$(2)$
Now, we will find the volume covered by the rectangle while rotating about one of its lengths.
$\Rightarrow$ $V=\pi {{x}^{2}}y$
On putting value of y in terms of x obtained in equation$(2)$, we get
$\Rightarrow$ $V=\pi {{x}^{2}}(18-x)$
On applying the distributive law of multiplication, we get
$\Rightarrow$ $V=\pi (18{{x}^{2}}-{{x}^{3}})$………….$(3)$
Now, we will differentiate the equation with respect to x.
$\Rightarrow$ $\dfrac{dV}{dx}=\dfrac{d\pi (18{{x}^{2}}-{{x}^{3}})}{dx}$
We will differentiate each term with respect to x.
$\Rightarrow$ $ \dfrac{dV}{dx}=\pi \left( 36x-3{{x}^{2}} \right)$……………$(4)$
We will equate $\dfrac{dV}{dx}$with zero.
$\Rightarrow$ $\Rightarrow \pi \left( 36x-3{{x}^{2}} \right)=0$
Now we will factorize the left hand side equation.
$\Rightarrow$ $\Rightarrow 3\pi x\left( 12-x \right)=0$
From here, we get
$\Rightarrow$ $\left( 12-x \right)=0$and $x=0$
$\Rightarrow$ $x=12$but $x\ne 0$ as the length of the rectangle can’t be equal to zero.
 We will put the value of x in equation $(2)$
$\Rightarrow$ $y=18-12=6$
$\Rightarrow$ $y=6$
Now, we will again differentiate the equation $(4)$with respect to x.
$\Rightarrow$ $\dfrac{d}{dx}\left( \dfrac{dV}{dx} \right)=\dfrac{d}{dx}\left[ \pi \left( 36x-3{{x}^{2}} \right) \right]$
$\Rightarrow$ $\dfrac{{{d}^{2}}V}{d{{x}^{2}}}=\pi \left( 36-6x \right)$
Now, we will put the value of x here.
$\Rightarrow$ $\dfrac{{{d}^{2}}V}{d{{x}^{2}}}=\pi \left( 36-6\times 12 \right)$
On further simplifying the terms, we get
$\Rightarrow$ $\dfrac{{{d}^{2}}V}{d{{x}^{2}}}=-36\pi <0$
Thus, for $x=12$ , the volume of the cylinder which we have obtained on the revolution of the rectangle about its length is maximum.
So the required dimensions of the rectangle are 12cm and 6cm.

Note: We have used distributive law of multiplication which says that the multiplication of a number with the addition of two numbers will be equal to sum of the multiplication of a number by each of these two numbers i.e. if a, b and c are three numbers then according to the distributive law of multiplication over addition is $a(b+c)=a.b+a.c$