Arectangular sheet of perimeter of 36 cm and dimensions x cm by y cm is to rolled into a cylinder. what values of x and y give the largest volume

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.

so the possibilities of the measurement of the side of the triangle are either it is longest side or not the longest side

when it is the longest side

s < 25 + 35s < 60

if it is not the longest side

35 < 25 + s10 < sso 10 < s < 60so the perimeter70 < p < 130

could u provide us some answer choices or hmm idk maybe a table for us to anser with ? ?