The strenght of a rectangular beam is proportion to its width (w) and its square thickness (t) that is S=kwt^2, where k is a constant. Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius 20cm.
It's been a long time since we reviewed optimization, so I might be wrong.... Solve for S'=0 to get the max strength and use this: (w^2+t^2)/2=r
that should be w^2 + t^2 = (2r)^2 right?
thank you for replyin but where did you get the (w^2+t^2)/2=r from? I understand to find the max strength we set s' to zero, but the second part you mentioned confused me o.0
i dont think what he gave you is correct. draw a cross-section of the beam, a circle, a square inscribed in the circle, so the sides of the rectangle should be the width and thickness, and the diagonal of the rectangle is the diameter
Yeah my bad. sorry. half the diagonal of a enclosed rectangle is the radius of the outside circle is what I was aiming for.
hey INT are you sure you got that series question?
okay no i understand, thank you guys so much!!
*now
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