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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.

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The perimeter of the rectangle with length L and breadth b is 2(l + b) 

Therefore, 

2(L + b) = 36 

L + b = 18 

b = 18 - L 

Let the rectangle be rotated about its breadth. 

Then the resulting cylinder formed will be of radius L and height b. 

Volume of cylinder formed V = πL2b = π(18L2 - L3

To find the dimensions that will result in the maximum volume :

\(\frac{dV}{dL}\) = π(18 x 2 x L - 3 x (L2) = 0

36L = 3 x (L2)

L = 12,0

L cannot be 0. L is taken as 12 cm.

Therefore,

b = 24.

\(\frac{d^2V}{dL^2}\) = π(18 x 2 - 3 x (L))

At L = 12,

\(\frac{d^2V}{dL^2}\) = - 36π

= a -ve value

Therefore a maxima exists at L = 12, meaning the volume of the constructed cylinder will be maximum at L = 12 cm.

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