# 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 side. Also, find the maximum volume.

Let the sides of the rectangle are x cm and y cm.

Perimeter = 36 cm

2x + 2y = 36

y = 18 -x

$\\ \text {Volume of ractnagle when it rotating around one axis }( V)=\pi x^{2} y$

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

$V=\pi\left(18 x^{2}-x^{3}\right)$

Diffentiate w.r.t. x

$\frac{d V}{d x}=\pi\left(36 x-3 x^{2}\right)$

For critical point

$\frac{d V}{d x}=\pi\left(36 x-3 x^{2}\right) =0$

$x(12- x) =0$

$x =0 \ cm , 1 2 \ cm$

For minima and maxima

$\frac{d^2 V}{d x^2}=\pi\left(36 -6 x\right)$

$\frac{d^2 V}{d x^2}_{at x =12}=\pi\left(36 -72\right) <0$

It has maximum volume at x = 12 cm.

Hence the dimension of the rectangle is 12cm x 6 cm.

$\text { Maximum volume }=\pi x^{2} (18-x)$

$V_{max}= \pi 12^{2} \times 6 = 864 \pi \mathrm{cm}^{3}$

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