Please give a detailed description in the answer using diagrams if possible.

Let variable r be the radius of the circular base and variable h the height of the cylinder.

The total surface area of the cylinder is given to be

3*pi = (area of base) + (area of the curved side)

= pi*r^2 + (2*p*r)h ,

so that

2*pi*r*h = 3*pi - pi*r^2

or

h = { (3*pi ā pi*r^2 )/2*pi*r }

= { 3 /2*r } - (1/2) r .

We wish to MAXIMIZE the total VOLUME of the cylinder

V = (area of base) (height) = pi*r^2*h .

However, before we differentiate the right-hand side, we will write it as a function of r only. Substitute for h getting

V = pi*r^2*h

= pi*r^2( { 3/2*r } - (1/2) r)

= (3/2)*pi*r - (1/2)pi*r^3 .

Now differentiate this equation, getting

V' = (3/2)pi - (1/2)pi*3*r^2

= (3/2)pi( 1 - r^2 )

= (3/2)pi( 1 - r ) ( 1 + r )

= 0

for

r=1 or r=-1 .

But r =/ -1 since variable r measures a distance and r > 0 . Since the base of the box is a circle and there are 3*pi ft^2 of material, it follows that 0 <r<= sqrt{ 3 } . See the adjoining sign chart for V' .

If

r=1 ft. and h=1 ft. ,

then

V = pi ft^3

is the largest possible volume of the cylinder.