Volume question, please, please HELP?

Question

Determine the dimensions of a closed cylindrical metal can that will minimize the amount of metal used, if the volume must be 1024x cm^3.

Answer 1

volume of cylinder = pi r^2 h = 1024
h= 1024/pi r^2
a = 2*pi*r*h + 2*pi*r^2
a= 2pi(r^2 + 1024r/(pi*r^2))
da/dr = 0 for max or min
da/dr = 2pi(2r - 1024/(pir^2)) = 0
4 pi r - 2048/r^2 = 0
taking lcm
4 pi r ^3 - 2048 = 0
r ^3 = 2048/4 pi = 163.057
r = 5.463(ans)
h= 1024/pi r^2
h = 10.927(ans)

Answer 2

Minimizing the amount of metal used is the same as minimizing the surface area of the cylinder.

A = 2*Pi*(r^2) + 2*Pi*r*h, where A = area, r = radius of the circle at either end of the cylinder, h = length of the cylinder

To find the miminum, take the derivative of A and set equal to zero. Right now A is expressed as a function of two variables, r and h. We are given a fixed volume, so we can express one of the variables as a function of the other.

V = Pi*(r^2)*h, where V = volume
h = V/(Pi*(r^2))

Now let's substitute this expression into our original expression for A.

A = 2*Pi*(r^2) + 2*Pi*r*h --> original expression
A = 2*Pi*(r^2) + 2*Pi*r*V/(Pi*(r^2)) --> substitution for h
A = 2*Pi*(r^2) + 2*V/r --> simplification

Find the derivative of A with respect to r.

dA/dr = 4*Pi*r - 2*V/(r^2)

Set the derivative equal to 0 and solve for r.

dA/dr = 4*Pi*r - 2*V/(r^2) = 0
4*Pi*(r^3) - 2*V = 0 --> multiply through by r^2
4*Pi*(r^3) = 2*V --> add 2*V to both sides
r^3 = 2*V/(4*Pi) --> divide through by 4*Pi
r = (2*V/(4*Pi))^(1/3)

We've found one of the dimensions, the radius of the cylinder. Now just plug this into the expression for h.

h = V/(Pi*(r^2))
h = V/(Pi*(((2*V/(4*Pi))^(1/3))^2))
h = V/(Pi*((2*V/(4*Pi))^(2/3)))

We've just found the other dimension, the height of the cylinder. I'll leave it to you to plug in the numbers and come up with the values.

Answer 3

The volume = 1024 cm^3

Volume of the cylindrical metal = pi*r^2*H

r = radius

H = altitude

1024 = pi*r^2*h

h = 1024 / pi*r^2

the area = 2pi*r*(r + h)

Area = 2pi*r^2 + 2pi*r*h

2pi*r^2 + 2pi*r*1024 / pi*r^2 = Area(r)

2pi*r^2 + 2048 / r = Area(r)

Let's find the minimum Area, using the first derivative :

Area'(r) = 4pi*r - 2048 / r^2 = 0

4pi*r = 2048 / r^2

r = 5.46 cm

So, well, to get the minum area, radius = 5.46 cm

and to find the altitude, let's replace the radius on the given volume :

1024 = pi*r^2*h

h = 10.9 cm

Hope that helps

Answer 4

Let radius = r cm and height = h cm
V = π r² h
1024 = π r² h
h = 1024 / πr²
Let Surface Area = S
S = 2π r² + 2 π.r.h
S = 2π r² + 2048 r^(-1)
dS / dr = 4π r - 2048 r^(-2) = 0 for minimum.
4 π r ³ = 2048
r ³ = 512 / π
r = 5.46 cm and h = 10.9 cm

Answer 5

if the volume is fixed, you have to use that much of metal regardless of the dimensions. I guess there is some mistake in problem