Problem 11 A gutter having a triangular cross-section is to be made by bending a strip of tin in the middle. Find the angle between the sides when the carrying capacity is to a maximum.

Solution 11

$A = \frac{1}{8}L^2 \sin \theta$

$\dfrac{dA}{d\theta} = \frac{1}{8}L^2 \cos \theta = 0$

$\cos \theta = 0$

$\theta = 90^\circ$ answer

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