# Get Answers to all your Questions

#### 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. Also, find the maximum volume.

Given: rectangle of perimeter 36cm

To find: to estimate the dimensions of a rectangle in a way that it can sweep out maximum amount of volume when resolved to about one of its sides. Also, to find the maximum volume

Explanation: x and y can be the length and the breadth of the rectangle

The known perimeter of the rectangle is 36cm

$\\ { \Rightarrow 2x+2y = 36}\\ { \Rightarrow x+y = 18}\\ { \Rightarrow y = 18-x \ldots \ldots \ldots (i)}\\$

Now when the rectangle revolve about side y it will form a cylinder with y as the height and x as the radius, then if the volume of the cylinder is V, then we know

$V = \pi x\textsuperscript{2}y$

Applying the value from equation (i) in above equation we get

$V = \pi x\textsuperscript{2}(18-x)$

$\Rightarrow V = \pi (18x\textsuperscript{2}-x\textsuperscript{3})$

Then find out the first derivative of the given equation,
$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\pi\left(18 \mathrm{x}^{2}-\mathrm{x}^{3}\right)\right)}{\mathrm{dx}}$
Taking out the constant terms from equation followed by using the sum rule of differentiation,
$\\ \mathrm{V}^{\prime}=\pi\left[\frac{\mathrm{d}\left(18 \mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{d} \mathrm{x}}\right] \\\mathrm{v}^{\prime}=\pi\left[18(2 \mathrm{x})-3 \mathrm{x}^{2}\right] \\\mathrm{V}^{\prime}=\pi\left[36 \mathrm{x}-3 \mathrm{x}^{2}\right] \ldots \ldots (ii)$

To calculate critical point, we will equate the first derivative to 0, i.e.,

$\\ {V' = 0}\\ { \Rightarrow \pi (36x-3x\textsuperscript{2}) = 0}\\ { \Rightarrow 36x-3x\textsuperscript{2} = 0}\\ { \Rightarrow 36x = 3x\textsuperscript{2}}\\ { \Rightarrow 3x = 36}\\ { \Rightarrow x = 12 \ldots \ldots ..(iii)}\\$

By differentiating the second equation, second derivative of the volume equations can be easily calculated,

$\mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(\pi\left(36 \mathrm{x}-3 \mathrm{x}^{2}\right)\right)}{\mathrm{dx}}$
Taking out the constant terms from equation followed by using the sum rule of differentiation,
$\\\mathrm{V}^{\prime \prime}=\pi\left[\frac{\mathrm{d}(36 \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}\right] \\\mathrm{V}^{\prime \prime}=\pi[36-3(2 \mathrm{x})]\\V^{\prime \prime}=\pi[36-6 x]$

Now substituting x = 12 (from equation (iii)), we get

$V''_{x} = 12 = \pi [36 - 6(12)] \\ V''_{x} = 12 = \pi [36 - 72] \\ V''_{x} = 12 = -36\pi < 0$

Hence at x = 12, V will have maximum value.

The maximum value of V can be found by substituting x = 12 in

$V = \pi (18x\textsuperscript{2}-x^3),$ i. e $V _x = 12 = \pi (18(12)^2-(12)^3)$

$\\ V_{x} = 12 = \pi (18(144)-(1728))\\$

$V_{x} = 12 = \pi (2592-(1728))\\$

$V_{x} = 12 = 864 \pi cm \textsuperscript{3}$

Therefore, the dimensions of the rectangle which will sweep out a volume as large as possible, when revolved about one of its sides equal to 12cm.

And the maximum volume is $864 \pi cm \textsuperscript{3}$.