optimization: find best width, height for beam w/ radius R

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erikalava

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Jun 28, 2007
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This is an easy question...

You have a cylindrical log to be cut into a rectangular beam. The strength of the beam is directly proportional to the product of the width and the square of the height. Find the width and the height for the best beam possible from a log with radius R.

humm??
 
The max-min variable is the strength of the beam. Since it varies as the product of the width and the square of the height gives us:

\(\displaystyle \L\\S=kwh^{2}\)

S= strength and k is a constant that depends on the type of wood and the units used.

We must find an equation which relates w and h. Draw a radius to one corner of the rectangle and make a right triangle with hypoteneuse R and sides of length w/2 and h/2.

We get: \(\displaystyle \L\\(\frac{w}{2})^{2}+(\frac{h}{2})^{2}=R^{2}\)

Now, it may be best to avoid radicals and solve for h^2.

This gives: \(\displaystyle \L\\h^{2}=4R^{2}-w^{2}\)

Now, sub this into S, differentiate, set to 0 and solve for w. You're on your way.

Does that help?.

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erikalava said:
This is an easy question...
Click to expand...
So you're saying you were able to make a lot of progress, but you're stuck near the end...? Or that you got the whole thing done, but your answer doesn't match the back of the book...? :shock:

Please reply with clarification, including a clear listing of all of your work and reasoning. Thank you! :D

Eliz.
 
no I finished the problem to the end. I am not sure because a log is a Cylinder with the formula V=Pi*r^2H. The answer I came up with and the answer that is shown on here is fine if you are using a one dimentional circle. Once you add in the fact that you have the area of the beam (L *W*H) and the area of a cylinder to work with.. the S=K * W*H^2 all of a sudden becomes a little more complicated.
 
You're certainly making this harder than need be. You do not have to take the length of the log into consideration. Besides, that isn't part of the problem statement. I, too, would like to see your method.
 
That is what i wasn't sure about. I used the same basic method that galactus did... except the i used the equation (2r)^2= w^2 + H^2 and solved for H. basically getting the same answer. I was just wondering IF i was making it out harder than what it needed to be... thanks for your imput
 
What did you get for an answer?. Show me yours and I'll show you mine.
 
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