You are to maximise $f(d,w)=wd^2$ subject to the constraint $(d/2)^2+(w/2)^2=r^2$ .. which is the requirement that the beam fit in the log.
Is this problem even correct? First of all I would think that depth is the dimension that goes inside the log; then if depth^2 times width = strength then since depth isn't bounded by anything then it follows that I should just take the biggest possible width.The strength of a rectangular beam is proportional to the product of its width w times the square of its depth d. Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius r
Answers: $w=\dfrac{2r}{\sqrt{3}}, h=\dfrac{2\sqrt{2}r}{\sqrt{3}}$
And if the depth is this vertical side of the beam as shown in the picture, then why does the answer they give is in terms of w and h? What is h (height I assume?) then? Is it the dimension that "goes inside" the log? But then it isn't bounded by anything, so I can make it arbitrarily big and it wouldn't matter.
Can someone clear this up?
Draw a line segment from the centre to the top right corner. Label the centre O and the top right corner A. Draw a normal from the centre to the right hand side of the beam and label the point where it meets the right hand edge B. Now OBA is a right triangle with right angle at B, and sides r, the radius, and w/2 and d/2.
Alternatively draw in the diagonal of the beam. This passes through the centre of the circle, and so is a diameter of length 2r. So now we have a right triangle of sides 2r, w, d.
Now apply Pythagoras' theorem to these sides.
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