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. Solution:
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. Solution:

We should believe x and y to be the length and expansiveness of given square shape ABCD.

As per the inquiry, the square shape will be settled with regards to side AD which making a chamber with range x and stature y.

Along these lines, the volume of the chamber \[V\text{ }=\text{ }\pi r2h\text{ }=\text{ }\pi x2y\text{ }\ldots \text{ }.\text{ }\left( 1 \right)\]

NCERT Exemplar Solutions Class 12 Mathematics Chapter 6 - 42

Presently, border of square shape \[P\text{ }=\text{ }2\left( x\text{ }+\text{ }y \right)\]

\[36\text{ }=\text{ }2\left( x\text{ }+\text{ }y \right)\]

\[18\text{ }=\text{ }x\text{ }+\text{ }y\]

\[y\text{ }=\text{ }18\text{ }\text{ }x\text{ }\ldots \text{ }..\text{ }\left( ii \right)\]

Placing the worth of y in the situation (I), we get

\[V\text{ }=\text{ }\pi x2\left( 18\text{ }\text{ }x \right)\text{ }=\text{ }\pi \left( 18×2\text{ }\text{ }x3 \right)\]

Separating the two sides w.r.t. x, we get

\[dV/dx\text{ }=\text{ }\pi \left( 36x\text{ }\text{ }3×2 \right)\text{ }\ldots \text{ }.\text{ }\left( iii \right)\]

For neighborhood maxima and nearby minima \[dV/dx\text{ }=\text{ }0\]

\[\pi \left( 36x\text{ }\text{ }3×2 \right)\text{ }=\text{ }0\]

\[36x\text{ }\text{ }3×2\text{ }=\text{ }0\]

\[3x\left( 12\text{ }\text{ }x \right)\text{ }=\text{ }0\]

\[x\text{ }\ne \text{ }0\text{ }and\text{ }12\text{ }\text{ }x\text{ }=\text{ }0\Rightarrow x\text{ }=\text{ }12\]

From condition (ii), we have

\[y\text{ }=\text{ }18\text{ }\text{ }12\text{ }=\text{ }6\]

Separating condition (iii) w.r.t. x, we get

\[d2V/dx2\text{ }=\text{ }\pi \left( 36\text{ }\text{ }6x \right)\]

At \[x\text{ }=\text{ }12,\]

\[d2V/dx2\text{ }=\text{ }\pi \left( 36\text{ }\text{ }6\text{ }x\text{ }12 \right)\text{ }=\text{ }\pi \left( 36\text{ }\text{ }72 \right)\]

\[=\text{ }-\text{ }36\text{ }\pi \text{ }<\text{ }0\text{ }maxima\]

Presently, volume of the chamber so framed \[=\text{ }\pi x2y\]

\[=\text{ }\pi \text{ }x\text{ }\left( 12 \right)2\text{ }x\text{ }\left( 6 \right)\text{ }=\text{ }\pi \left( 144 \right)2\text{ }x\text{ }6\text{ }=\text{ }864\pi \text{ }cm3\]

In this way, the necessary measurement are \[12\text{ }cm\text{ }and\text{ }6\text{ }cm\] and the greatest volume is\[864\pi \text{ }cm3\] .