# 11 - Triangular gutter of maximum carrying capacity

**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**

$A = \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}L) \sin \theta$

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

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

$\cos \theta = 0$

$\theta = 90^\circ$ *answer*

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