Thank you for your understanding.

In summary, the conversation is about finding the maximum volume of a beam that can be cut from a tree trunk with a truncated cone shape. The trunk has a length of 2 meters and its bases have diameters of 10 cm and 20 cm. The conversation also discusses different equations and asks for clarification on the problem, as well as the possibility of errors in the given answer. A diagram is requested to better understand the problem.
  • #1
leprofece
241
0
A tree trunk is shaped like a truncated cone it has 2 m of length and diameters of their bases are 10 cm and 20 cm. Cut a square straight section so that the axis of the beam coincides with the axis of the truncated cone. find the beam volume maximum that can be drawn from this form.

answer 13,3 cm 13,3 cm y 133 cm

I have no idea maybe the equations are

truncated cone
V = Pir^2H/3

a = piR^2

r= D/2
 
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  • #2
Re: max and min 289

I get completely different answers, which leads me to believe I am not interpreting your intentions correctly. Could you provide a diagram? You can draw a crude sketch using a graphics editing program (MS Paint if you run Windows) and upload it as an attachment. Unless we know what the problem is asking, we are at a loss to help you. Help us to help you. :D
 
  • #3
Re: max and min 289

MarkFL said:
I get completely different answers, which leads me to believe I am not interpreting your intentions correctly. Could you provide a diagram? You can draw a crude sketch using a graphics editing program (MS Paint if you run Windows) and upload it as an attachment. Unless we know what the problem is asking, we are at a loss to help you. Help us to help you. :D
volume cone truncated=[ (R²+r²+ R.r).¶.H]/ 3, with R=radio major bass
r radio bass minor
The tree trunk has two circles that are equals
its diaMEters are 10 and 20 cm so radius are 5 and 10
The two axes are equals when I cut the trunk
What is the trunk of max volume
If it is your interpretacion
this may be the answer because the books sometimes has mistakes


It is very easy suposse the graph so the book may be it is wrong
 
  • #4
volume cone truncated=[ (R²+r²+ R.r).¶.H]/ 3, with R=radio major bass
r radio bass minor
The tree trunk has two circles that are equals
its diaMEters are 10 and 20 cm so radius are 5 and 10
The two axes are equals when I cut the trunk
What is the trunk of max volume
If it is your interpretacion
this may be the answer because the books sometimes has mistakesIt is very easy suposse the graph so the book may be it is wrong
 
  • #5
leprofece said:
volume cone truncated=[ (R²+r²+ R.r).¶.H]/ 3, with R=radio major bass
r radio bass minor
The tree trunk has two circles that are equals
its diaMEters are 10 and 20 cm so radius are 5 and 10
The two axes are equals when I cut the trunk
What is the trunk of max volume
If it is your interpretacion
this may be the answer because the books sometimes has mistakesIt is very easy suposse the graph so the book may be it is wrong

Hello leprofece,

Repeating a post previously made or otherwise posting without adding anything new is what's called "bumping." We ask that you do not do this, as given in our first rule:

No bumping. Bumping a thread is posting a reply to that thread solely to raise its profile and return it to the top of the active threads list. This is forbidden at MHB. If you want to draw attention to an unanswered thread, then post something of value such as further progress. It is also forbidden to bump one thread by drawing attention to it in a different thread.

You are always welcome to post your progress or further thoughts, etc., but please do not simply repeat a previous post to draw further attention to a thread. This draws the attention of our helpers, and when we see that nothing new has been added, it has wasted our time. I hope you can understand why this rule has been established and how it makes MHB more efficient for everyone involved.

Now, I am waiting for you to provide a diagram, as I simply am not clear from your description what the problem actually is. A diagram would tell me what is actually being asked. Please help us to help you. :D
 

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