Maximizing the volume of a beam cut from a cylindrical trunk

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Discussion Overview

The discussion revolves around determining the dimensions of a rectangular beam that can be cut from a cylindrical trunk with a given diameter "D" and length "L". The focus is on maximizing the volume of the beam, considering the geometric constraints of the circular cross-section of the trunk.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant inquires about the dimensions of a rectangular beam that maximizes volume from a cylindrical trunk.
  • Another participant suggests that the problem involves maximizing the area of a rectangle inscribed within a circle and prompts for the constraints and objective function.
  • A subsequent reply proposes that the optimal shape of the beam is a square with a side length of D/2, maintaining that the length of the beam equals the length of the trunk.
  • Further contributions clarify the objective function as the area of the rectangle, expressed as A(x,y) = xy, and introduce the constraint based on the Pythagorean theorem: x² + y² = D².
  • One participant indicates they derived a relationship leading to y = x = D/sqrt(2) as part of the maximization process.

Areas of Agreement / Disagreement

Participants present multiple approaches to the problem, with some agreeing on the use of inscribed rectangles and others proposing specific dimensions. The discussion remains unresolved regarding the optimal dimensions and the derivation process.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in maximizing the area or volume, and assumptions about the relationships between the dimensions and the constraints are still being explored.

leprofece
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what are the dimensions of rectangular beam of volume maximum that can be cut from a trunk in diameter "D" and length "L", assuming that the trunk has the shaped of a straight circular cylinder shape?

Answer Width =lenght
 
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Re: max and min 295

What have you tried?

Do you recognize that this problem boils down to maximizing the area of a rectangle inscribed within a circle?

As always, what is your constraint? What is your objective function?

Hint: look at the diagonal of the rectangle and how it relates to the diameter of the circle. And then how do the sides of the rectangle relate to its diagonal?
 
Re: max and min 295

MarkFL said:
What have you tried?

Do you recognize that this problem boils down to maximizing the area of a rectangle inscribed within a circle?

As always, what is your constraint? What is your objective function?

Hint: look at the diagonal of the rectangle and how it relates to the diameter of the circle. And then how do the sides of the rectangle relate to its diagonal?

It shall be a square beam of side equal to D/2.
Length equal to length of trunk.

If it is the area of a rectangle inscribed within a circle?

a = pir
and R2= x2-r2
r= sqrt(R2-x2)
A=2pisqrt(R2-x2)
I derive A and I must get the answer
 
Re: max and min 295

Here is a cross-section:

View attachment 2086

Our objective function is the area of the rectangle:

$$A(x,y)=xy$$

Subject to the constraint (by Pythagoras):

$$x^2+y^2=D^2$$

So, solve the constraint for either variable, and substitute for that variable into the objective function so that you only have one variable, and then maximize.
 

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Re: max and min 295

MarkFL said:
Here is a cross-section:

View attachment 2086

Our objective function is the area of the rectangle:

$$A(x,y)=xy$$

Subject to the constraint (by Pythagoras):

$$x^2+y^2=D^2$$

So, solve the constraint for either variable, and substitute for that variable into the objective function so that you only have one variable, and then maximize.

Ok I got y = x = D/sqrt(2)
 

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