06-09 Trapezoidal gutter of greatest capacity

$A_2 = \frac{1}{2}(4\cos \theta)(4\sin \theta) = 8 \cos \theta \sin \theta$
 

$A = A_1 + 2A_2$

$A = 56\sin \theta + 2(8\cos \theta \sin \theta)$

$\dfrac{dA}{d\theta} = 56\cos \theta + 16(\cos^2 \theta - \sin^2 \theta) = 0$

$7\cos \theta + 2[ \, \cos^2 \theta - (1 - \cos^2 \theta) \, ] = 0$

$4\cos^2 \theta + 7\cos \theta - 2 = 0$

$(4\cos \theta - 1)(\cos \theta + 2) = 0$
 

for
$4\cos \theta - 1 = 0$

$\cos \theta = 1/4$
 

for
$\cos \theta + 2 = 0$

$\cos \theta = -2$   →   (meaningless)
 

use
$\cos \theta = 1/4$
 

$a = 14 + 2(4\cos \theta) = 14 + 8(1/4)$

$a = 16 \, \text{ inches}$           answer

$A_2 = \frac{1}{2}(3\cos \theta)(3\sin \theta) = 4.5\cos \theta \sin \theta$
 

$A = A_1 + 2A_2$

$A = 21\sin \theta + 2(4.5\cos \theta \sin \theta)$

$\dfrac{dA}{d\theta} = 21\cos \theta + 9(\cos^2 \theta - \sin^2 \theta) = 0$

$7\cos \theta + 3 [ \, \cos^2 \theta - (1 - \cos^2 \theta) \, ] = 0$

$6\cos^2 \theta + 7\cos \theta - 3 = 0$

$(3\cos \theta - 1)(2\cos \theta + 3) = 0$
 

for
$3\cos \theta - 1 = 0$

$\cos \theta = 1/3$
 

for
$2\cos \theta + 3 = 0$

$\cos \theta = -3/2$   →   (meaningless)
 

use $\cos \theta = 1/3$
 

$a = 7 + 2(3\cos \theta) = 7 + 6(1/3)$

$a = 9 \, \text{ inches}$           answer

$A_2 = \frac{1}{2}(3\cos \theta)(3\sin \theta) = 4.5\cos \theta \sin \theta$
 
$A = A_1 + 2A_2$

$A = 9\sin \theta + 2(4.5\cos \theta \sin \theta)$

$\dfrac{dA}{d\theta} = 9\cos \theta + 9(\cos^2 \theta - \sin^2 \theta) = 0$

$\cos \theta + [ \, \cos^2 \theta - (1 - \cos^2 \theta) \, ] = 0$

$2\cos^2 \theta + \cos \theta - 1 = 0$

$(2\cos \theta - 1)(\cos \theta + 1) = 0$
 
for
$2\cos \theta - 1 = 0$

$\cos \theta = 1/2$
 

for
$\cos \theta + 1 = 0$

$\cos \theta = -1 \,\,$ → (meaningless)
 

use $\cos \theta = 1/2$
 
$a = 3 + 2(3\cos \theta) = 3 + 6(1/2)$

$a = 6 \,\, \text{ inches }$           answer

$x = \frac{1}{2}(w - b) \cos \theta$

$h = \frac{1}{2}(w - b) \sin \theta$
 

$A_1 = bh$

$A_1 = \frac{1}{2}b(w - b) \sin \theta$
 

$A_2 = \frac{1}{2} xh$

$A_2 = \frac{1}{2} [\, \frac{1}{2}(w - b) \cos \theta \,] [\, \frac{1}{2}(w - b) \sin \theta \,]$

$A_2 = \frac{1}{8}(w - b)^2 \sin \theta \cos \theta$
 

$A = A_1 + 2A_2$

$A = \frac{1}{2}b(w - b) \sin \theta + 2 [ \, \frac{1}{8}(w - b)^2 \sin \theta \cos \theta \, ]$

$\dfrac{dA}{d\theta} = \frac{1}{2}b(w - b) \cos \theta + \frac{1}{4}(w - b)^2 (\cos^2 \theta - \sin^2 \theta)$

$2b \cos \theta + (w - b) [ \, \cos^2 \theta - (1 - \cos^2 \theta) \, ] = 0$

$2(w - b) \cos^2 \theta + 2b \cos \theta - (w - b) = 0$
 

By quadratic formula:
A = 2(w - b); B = 2b; C = -(w - b)

$\cos \theta = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

$\cos \theta = \dfrac{-2b \pm \sqrt{(2b)^2 - 4[2(w - b)][-(w - b)]}}{2[2(w - b)]}$

$\cos \theta = \dfrac{-2b \pm \sqrt{4b^2 + 8(w - b)^2}}{4(w - b)}$

$\cos \theta = \dfrac{-2b \pm 2\sqrt{b^2 + 4(w - b)^2}}{4(w - b)}$

$\cos \theta = \dfrac{-b \pm \sqrt{b^2 + 4(w - b)^2}}{2(w - b)}$

$\cos \theta = \dfrac{-b + \sqrt{b^2 + 4(w - b)^2}}{2(w - b)}$   and   $\dfrac{-b - \sqrt{b^2 + 4(w - b)^2}}{2(w - b)}$
 

$\cos \theta = \dfrac{-b - \sqrt{b^2 + 4(w - b)^2}}{2(w - b)}$   →   meaningless
 

Use
$\cos \theta = \dfrac{-b + \sqrt{b^2 + 4(w - b)^2}}{2(w - b)}$
 

$x = \frac{1}{2}(w - b)\cos \theta$

$x = \frac{1}{2}(w - b) \times \dfrac{-b + \sqrt{b^2 + 4(w - b)^2}}{2(w - b)}$

$x = \frac{1}{4} \left[ -b + \sqrt{b^2 + 4(w - b)^2} \right]$
 

$a = b + 2x$

$a = b + 2 \Big\{ \frac{1}{4} \left[ -b + \sqrt{b^2 + 4(w - b)^2} \right] \Big\}$

$a = \frac{1}{2} [\, -b + \sqrt{b^2 + 4(w - b)^2} \,]$           answer