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Question

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.

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Solution

Let breadth and length of the rectangle be x and y respectively.

Perimeter of the rectangle=36cm2x+2y=36x+y=18y=18x
The rectangle is being revolved about its length y.
Then, volume (V) of resultant cylinder =πx2.yV=πx2.(18x)=18πx2πx3=π[18x2x3]
On differentiating both sides w.r.t.x, we get
dVdx=π(36(3x2))Now,dVdx=036x=3x23x236x=03(x212x)=23x(x12)=0x=0,x=12x=12
Again, differentiating w.r.t. x we get
d2Vdx2=π(366x)(d2Vdx2)x=12=π(366×12)=36π<0
At x = 12, volume of the resultant cylinder is the maximum.
So, the dimensions of rectangles are 12 cm and 6 cm, respectively.

Maximum volume of resultant cylinder,
(V)x=12=π[18.(12)2(12)3]=π[122(1812)]=π×144×6=864πcm3


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