Optimization problem: find strongest beam

[zebra]

The strength of a rectangular beam is directly proportional to the product of its width and the square of the depth of a cross section. Find the dimensions of the strongest beam that can be cut from a cylindrical log of a radius a.

I interpreted this as the Strength = (the Width of the beam or log)x(the length x of the square inside of the beam)x(the width y of the square)

so S=Wxy

in looking for another equation, i found that x^2+y^2=W^2 although i'm totally stuck at this point. i could not find another way to connect Strength to x and y, and i am not even sure if the equation that i originally thought of is correct or sufficient in obtaining the answer.

 

[galactus]

The variable of interest is the strength of the beam.

The strength of the beam varies as the product of the width and the square of the height:

w=width
h=heighth
S=strength
k=constant(type of wood, etc)
It doesn't matter, though, because k is a constant in this problem.

\(\displaystyle \L\\S=kwh^{2}\).....[1]

We need to express S as a function of a single variable.

We must relate w and h. Draw a line form the center of the log to a corner of the beam. Now, you have a right triangle and we can use ol' Pythagoras.

\(\displaystyle \L\\(\frac{w}{2})^{2}+(\frac{h}{2})^{2}=a^{2}\)

Solve the above for \(\displaystyle h^{2}\) in terms of w:

\(\displaystyle \L\\w^{2}+h^{2}=4a^{2}\)

\(\displaystyle \L\\h^{2}=4a^{2}-w^{2}\)

Sub into [1]:

\(\displaystyle \L\\S=kw(4a^{2}-w^{2})\)

\(\displaystyle \L\\S=k(4a^{2}w-w^{3})\)

Now, we have as a function of w alone:

\(\displaystyle \L\\S(w)=k(4a^{2}w-w^{3}), \;\ 0\leq{w}\leq2a\)

Differentiate:

\(\displaystyle \L\\S'(w)=4ka^{2}-3kw^{2}=k(4a^{2}-3w^{2})\)

Set to 0 and solve for w. Then find h.