A container in shape of a right circular cylinder with no top has surface area 3*pi ft^2. What height h & base radius r will maximize the volume of the cylinder

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

Mitasvil P. | Math,Test Prep (SAT,GRE,ACT,SSAT)-Expert Tutoring at one goMath,Test Prep (SAT,GRE,ACT,SSAT)-Expert...

4.84.8(4 lesson ratings)(4)

1

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.

Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...

5.05.0(3 lesson ratings)(3)

0

B = area of a base of the cylinder = pi*r^2
C = circumference of a base of the cylinder = 2*pi*r

SA = surface area of the cylinder = 3*pi; given.
= B + C*h = 3*pi
= pi*r^2 + 2*pi*r*h = 3*pi
Solving for h:
h = (3*pi - pi*r^2)/(2*pi*r) = (3 - r^2)/(2*r)

V = volume of the cylinder = B*h
= pi*r^2*(3 - r^2)/(2*r)
= (pi/2)*r*(3 - r^2); which is a cubic polynomial function of r.
As r -> +infinity, V is negative, so end behavior is down to the right.
Since V is an odd function, end behavior is up to the left.
The zeros of V are r = 0 and r = +-sqrt(3).
V, r, and h all have to be positive; so
the realistic domain of V is 0 < r < sqrt(3).

Differentiating V(r) using constant-multiple, product, chain, and power rules:
Vā = (pi/2)*[ 1*((3 - r^2) + r*(-2*r) ] = (pi/2)*[ 3 - 3*r^2 ].
Maximum occurs when Vā = 0 = 3 - 3*r^2 and r = +1 foot.
Then h = (3 - 1^2)/(2*1) = 1 foot.