A cylinder Area is 60cm^2, What`s the largest and the smallest possible volume.?

Question

Using Calculus: If the surface area of a cylinder including both ends is 60cm^2, what is the largest and smallest volume possible?

Answer 1

Write an equation for the volume of the cylinder in terms of the radius.

A = 2(pi)r^2 + 2pi(r)(h)

60 = 2(pi)r^2 + 2pi(r)(h)

30 = (pi)r^2 + pi(r)(h)

Solve for h

30 - (pi)r^2 = pi(r)(h)

30/(pi(r)) - r = h

Substitute into the Volume formula for h.

V = pi(r^2)(h)

V = pi(r^2)[ (30/(pi(r)) - r ]

Take the derivative of this function and find the critical points to locate the minimum(s) or maximum (don't forget to find zeros in this case because r can't be negative)

Or, graph it and locate the minimum(s) or maximum.

The maximum is 35.682 cm^3 which occurs when r = 1.784 cm.

The minimum is zero (or has a limit of zero, some will argue that it doesn't exist at zero.. and in that case there is no 'smallest' volume)... and these occur at r = 0 cm and r = 3.090 cm.

Answer 2

permit the radius and height of the inscribed cylinder be R and H. Now, R and H are related to r and h as follows: seem at how the inscribed cylinder ameliorations as a function of H: As H will enhance, R decreases -- it somewhat is, the cylinder gets narrower because it gets taller until, while H is as tall via fact the cone, its radius R is 0. while R = r, H = 0, and while H = h, then R = 0. In the two of those severe cases, the quantity of the cylinder is 0. we'd desire to discover the fee for H and R such that pi R^2 H is at a optimal. H is a function of R, and for 0 < R < r, H(R) = h - R(h/r) So, the quantity of the cylinder V(H, R) = pi R^2 H = -pi R^2 h (R - r)/r V' = -pi R h (3R-2r)/r R = 0, 2/3 r. via fact R = 0 is the degenerate case, our optimal happens the place R = 2/3 r. H(2/3 r) = h - (2/3 r)(h/r) = h / 3. So, our optimal happens at pi (2/3 r)^2 (h / 3), or at 4/27 pi h r^2

Answer 3

The smallest volume is zero. Then the cylinder ends would each be 30 cm^2, and the height would be zero. This happens at R^2 = 60 / (2pi).

The area is 2 pi R^2 + pi R H = 60
The volume is pi R^2 H
H = ( 60 - 2 pi R^2) / (pi R )
V = pi R^2 [ ( 60 - 2 pi R^2) / (pi R ) ]
V = 60 R - 2 pi R^3
V' = 60 - 6 pi R^2
set V' = 0
then
60 = 6 pi R^2
10 = pi R^2
10 / pi = R^2
3.183098 = R^2
R = 1.784124
V = 71.364965

Answer 4

surface of cylinder is: 2 x 3.14 x r x r + h x 2 x 3,14 x r= 60 cm2
Therefore: 2x3.14 xr ( r+ h) - 60= 0
and ; h= (60 - 2x3,14rxr)/2x3.14xr

volume is: 3,14 x r x r x h



Small h give big r and oposite
Smallest volume is 0 with infinith length or radius.

Biggest volume i will be found by traditionel estremal calkulation of the volume formula with h substractet as the funktion of r.

Answer 5

as h (height) goes to zero, volume becomes null.
i.e. choose h=.1, then V is small
try radius equal to square root of thirty over pi...find height
so, smallest volume possible is zero.

or, you could have a very tall cylinder, with a radius of (almost) zero, then the volume is still zero!

Answer 6

2*3.14*r*h+2*3.14*r^2=60 so

r*h +r^2=10 so to speak.

you have V= 3.14*r^2*h

h=(10-r^2)/r

You want to find the minimum and maximum of r^2*(10-r^2)/r
that is r*(10-r^2).

I think you need to find the first derivative which is: 10- 3r^2.
Second derivative is: -6r. Since it is negative the original equation only has a maximum.

The maximum is for r=sqrt(10/3). So Vmax=3.14*10/3*20/3*sqrt(3/10). Don't forget I took pi=3.