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  • 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.

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.

Answers (1)

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}.

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