The strength of a rectangular beam is directly proportional to the product of its width w and the square of its depth d by the formula: S = kwd^2, where k is a constant. Use trigonometric functions to find the dimensions of the strongest beam that can be cut from a circular log 16.0 cm in diameter.
If you can show a step by step answer that'd be great! =)
If you can show a step by step answer that'd be great! =)
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Let R = radius of the log
w = width , d = depth
With Pythagora's theorem:
R^2 = (w/2)^2 + (d/2)^2
d^2 = 4R^2 - w^2
w^2 = 4R^2 - d^2
put 1 of them into
S = kwd^2
S = kw(4R^2 - w^2)
S = k(4R^2w - w^3)
set S' = 0 to find te maximum
S' = k(4R^2 - 3w^2) = 0
4R^2 = 3w^2
w = 2R/√3
set into
d^2 = 4R^2 - w^2
d^2 = 4R^2 - 4R^2/3 = 2R^2/3
sub R = 8 cm
w = 16/√3 cm
d = 13.06 cm
OG
w = width , d = depth
With Pythagora's theorem:
R^2 = (w/2)^2 + (d/2)^2
d^2 = 4R^2 - w^2
w^2 = 4R^2 - d^2
put 1 of them into
S = kwd^2
S = kw(4R^2 - w^2)
S = k(4R^2w - w^3)
set S' = 0 to find te maximum
S' = k(4R^2 - 3w^2) = 0
4R^2 = 3w^2
w = 2R/√3
set into
d^2 = 4R^2 - w^2
d^2 = 4R^2 - 4R^2/3 = 2R^2/3
sub R = 8 cm
w = 16/√3 cm
d = 13.06 cm
OG