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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 side. 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 side. Also, find the maximum volume.

 

Answers (1)

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}

Posted by

Safeer PP

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