Show that its volume is a maximum

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Homework Help Overview

The problem involves a right circular cone with a variable base radius \( r \) and height \( h \), while maintaining a constant curved surface area. The objective is to demonstrate that the volume of the cone reaches a maximum under a specific relationship between \( r \) and \( h \).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship \( r = r\sqrt{2}h \) and question its implications, particularly whether it is a typographical error in the original statement. There is uncertainty regarding the interpretation of this equation and its impact on the values of \( r \) and \( h \).

Discussion Status

The discussion is ongoing, with participants seeking clarification on the equation presented. Some have expressed confusion about potential typographical errors and are looking for confirmation from others regarding the validity of the relationship.

Contextual Notes

There is mention of a constant curved surface area, but the specifics of this constraint and its implications on the variables are not fully explored. The discussion reflects uncertainty about the original problem statement.

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Homework Statement



A right circular cone has base radius , r and height h . As r and h vary its curved surface area is kept constant . Show that its volume is a maximum when [tex]r=r\sqrt{2}h[/tex]

Homework Equations





The Attempt at a Solution



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

V=1/3 pi r^2 h
 
Last edited:
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what is m?
[tex] r=r\sqrt{2}h[/tex]
means: [tex] 1=h\sqrt{2}[/tex] ? and h=0.707 ?
 
Last edited:


Rajini said:
what is m?

Its a typo , the m shouldn't be there . Edited
 


Rajini said:
what is m?
[tex] r=r\sqrt{2}h[/tex]
means: [tex] 1=h\sqrt{2}[/tex] ? and h=0.707 ?

thats what confuses me , i am not sure whether that's a typo in the book , i need someone to confirm that for me . What do you think ?
 

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