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.
Let breadth and length of the rectangle be x and y respectively.
∵Perimeter of the rectangle=36cm⇒2x+2y=36⇒x+y=18⇒y=18−x
The rectangle is being revolved about its length y.
Then, volume (V) of resultant cylinder =πx2.y⇒V=πx2.(18−x)=18πx2−πx3=π[18x2−x3]
On differentiating both sides w.r.t.x, we get
dVdx=π(36(3x2))Now,dVdx=0⇒36x=3x2⇒3x2−36x=0⇒3(x2−12x)=2⇒3x(x−12)=0⇒x=0,x=12∴x=12
Again, differentiating w.r.t. x we get
d2Vdx2=π(36−6x)⇒(d2Vdx2)x=12=π(36−6×12)=−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(18−12)]=π×144×6=864πcm3