Given that perimeter of the rectangular sheet= 36cm, Height of cylinder =y base circumference of the cylinder = x = 2pir
So x+y = 36/2 = 18 , y = 18-x, so height of the cylinder = 18-x and radius of cylinder = x/2pi
So volume of cylinder = pir^2h = pi(x/2pi)^2* (18-x) = (x^2* (18-x))/(4pi)
Let f(x) = (x^2(18-x))/(4pi)
We need to maximise this volume function.
So f'(x) = (1/4pi) ( -x^2 + 2x(18-x))= (1/4pi)(x)(-x+36-2x)= (1/4pi)(x)(-3x+36)
So f'(x) = 0⇒ (1/4pi)(x)(-3x+36)=0⇒x=0 or x= 12
Since x+y = 18, so 0<x<18
f'(x) = (1/4pi)(-3x^2+36x)
So f''(x) = (1/4pi)(-6x+36)⇒ f''(12) = (1/4pi)(-6*12+36) = -9pi <0
So using second derivative test x=12 gives the maximum volume.
Since x+y =18 so 12+y=18 ⇒y = 6
So x=12cm and y = 6cm give the largest volume.
