1. ## Rectangular Wooden Beam

The strength of a rectangular wooden beam is proportional to the product of the width and the cube if its depth. If the beam is to be cut from a log in the shape of a cylinder of radius 3 feet, express the strength S of the beam as a function of the width x. Also, what is the domain of S?

2. Originally Posted by symmetry
The strength of a rectangular wooden beam is proportional to the product of the width and the cube if its depth. If the beam is to be cut from a log in the shape of a cylinder of radius 3 feet, express the strength S of the beam as a function of the width x. Also, what is the domain of S?
Hello,

I've attached a rough sketch of the situation.

According to the problem

$\displaystyle S=x \cdot d^3$

From the diagram you know:

$\displaystyle x^2+d^2=(2r)^2 \Longleftrightarrow d=\sqrt{4r^2-x^2}$

Now plug in the term for d into the equation with S:

$\displaystyle S(x)=x \cdot \left(\sqrt{4r^2-x^2}\right)^3$.

To get real results it is necessary that the radicand is equal or greater than zero. Thus the value of x must be greater than zero and smaller than 2r:

$\displaystyle domain=\text{]}0, 2r\text{[}_{\mathbb R}$

EB

3. ## ok

Great reply as always but how do you make such diagrams?

4. ## earboth

You said:

"Now plug in the term for d into the equation with S."

What exactly do you mean by this statement?

5. Originally Posted by symmetry
You said:

"Now plug in the term for d into the equation with S."

What exactly do you mean by this statement?
Hello,

from my previous post:

$\displaystyle d=\underbrace{\sqrt{4r^2-x^2}}_{\text{this is a term for d}}$

Now I took the RHS of this equation which contains the term for d and put it into the equation for S:

$\displaystyle S=x \cdot d^3$

$\displaystyle S=x \cdot \left(\underbrace{\sqrt{4r^2-x^2}}_{\text{this is a term for d}} \right)^3$

EB

6. ## ok

I finally get it.

Thanks!

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### the strength, s, of a wood

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