Cylindrical object question

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mares09

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A silo is to hold 10,000 cubic feet of grain. the silo will be cylindrical in shape and have a flat top. the floor of the silo will be the earth. what dimensions of the silo will use the least material for construction?

I need help with this problem, i have been trying to resolved but I havent been able to. This is what I got so far.

The equation for the Surface area is 2 pi r^2 + 2 pi r h. And I know the volume of the silo is 10000 cubic ft, but I dont know how to relate those quantities to each other.

Any help would be apprecciated. :shock:
 
mares09 said:
A silo is to hold 10,000 cubic feet of grain. the silo will be cylindrical in shape and have a flat top. the floor of the silo will be the earth. what dimensions of the silo will use the least material for construction?

I need help with this problem, i have been trying to resolved but I havent been able to. This is what I got so far.

The equation for the Surface area is 2 pi r^2 + 2 pi r h. And I know the volume of the silo is 10000 cubic ft, but I dont know how to relate those quantities to each other.

Any help would be apprecciated. :shock:
Click to expand...

M(aterial) = pi * r^2 + 2*pi*r*h

v = pi*r^2*h

h = v/(pi*r^2)

M = pi*r^2 + 2*v/r

Now minimize 'M' for onstant 'v'.
 
Hello Subhotosh Khan, I just wanted to know if you could tell me if what I did is correct.

I took the equation A = pi*r^2 + 20000r^-1.
and I got the derivative: 2pi*r-(20000/ (r^2)).

Then I substracted and got (2pi*r^3 - 20000) / (r^2)
Then I set everything to zero and solved for r.
So r = (20000/ 2pi )^1/3 = 14.7

Since r= 14.7 I replaced to know what h would be:

h = 10000/ (pi*(14.7^2)
h= 14.7

So the dimensions would need to be r and h both = 14.7.

Somehow that doesnt make any sense for them to be the same value though.
somebody else replied to my message and told me that this was the way it was done:

As = 2*[pi*r^2] + 2*pi*r*h Your Surface Area (As) equation

V = [pi*r^2] * h = 10,000 Your Volume equation (V)
This is a constraint on h and r, solve this eqn for h to eliminate one variable:

h = 10,000 / [pi*r^2]
Now substitute the eqn for h back into the As equation

As = 2*[pi*r^2] + 2*pi*r*[10,000/(pi*r^2)]

As = 2*pi^2*r^4 + 20,000*pi*r
Now we have As strictly as a function of one variable!

d(As) / dr = 2*pi^2 * 4*r^3 + 20,000*pi
To minimize any quantity, take the first derivative and set it equal to zero.

d(As) / dr = 8*pi^2*r^3 + 20,000*pi = 0

r^3 = 20,000*pi / (8*pi^2) = 2500 / pi
Solve this resulting eqn for r

r = [2500 / pi] ^ 1/3

r = 9.266

h = 10,000 / [pi * r^2] = 10,000 / [pi * 9.266^2] = 37.067
Substitute back into the Volume equation to obtain h

To minimize Surface Area (As) and presumably building costs, while holding V = 10,000, the dimensions need to be:
r = 9.266
h = 37.067

I know is a long message but I would really appreciate if you could help me.

Thank you!!!
 
mares09 said:
Hello Subhotosh Khan, I just wanted to know if you could tell me if what I did is correct.

I took the equation A = pi*r^2 + 20000r^-1.
and I got the derivative: 2pi*r-(20000/ (r^2)).

Then I substracted and got (2pi*r^3 - 20000) / (r^2)
Then I set everything to zero and solved for r.
So r = (20000/ 2pi )^1/3 = 14.7

Since r= 14.7 I replaced to know what h would be:

h = 10000/ (pi*(14.7^2)
h= 14.7

So the dimensions would need to be r and h both = 14.7.

Somehow that doesnt make any sense for them to be the same value though.
somebody else replied to my message and told me that this was the way it was done:

As = 2*[pi*r^2] + 2*pi*r*h Your Surface Area (As) equation .... This is incorrect

since earth is used as bottom floor - you do not need any material for floor.

V = [pi*r^2] * h = 10,000 Your Volume equation (V)
This is a constraint on h and r, solve this eqn for h to eliminate one variable:

h = 10,000 / [pi*r^2]
Now substitute the eqn for h back into the As equation

As = 2*[pi*r^2] + 2*pi*r*[10,000/(pi*r^2)]

As = 2*pi^2*r^4 + 20,000*pi*r
Now we have As strictly as a function of one variable!

d(As) / dr = 2*pi^2 * 4*r^3 + 20,000*pi
To minimize any quantity, take the first derivative and set it equal to zero.

d(As) / dr = 8*pi^2*r^3 + 20,000*pi = 0

r^3 = 20,000*pi / (8*pi^2) = 2500 / pi
Solve this resulting eqn for r

r = [2500 / pi] ^ 1/3

r = 9.266

h = 10,000 / [pi * r^2] = 10,000 / [pi * 9.266^2] = 37.067
Substitute back into the Volume equation to obtain h

To minimize Surface Area (As) and presumably building costs, while holding V = 10,000, the dimensions need to be:
r = 9.266
h = 37.067

I know is a long message but I would really appreciate if you could help me.

Thank you!!!
Click to expand...
 
Did I get the right answer thought? 14.7 for both height and radious?

Thank you.
 
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