**Problem**

A gutter whose cross-section is trapezoidal is to be made of galvanized iron sheet of 24 m wide. If its carrying capacity is maximum, find the dimension of the base.

A. 4 m

B. 6 m

C. 8 m

D. 10 m

**Solution**

$A = \frac{1}{2} y^2 (2 \cos \theta \sin \theta) + 24y \, \sin \theta - 2y^2 \, \sin \theta$

$A = \frac{1}{2} y^2 \sin 2\theta + 24y \, \sin \theta - 2y^2 \, \sin \theta$ ← eq. (1)

From eq. (1)

$\dfrac{\partial A}{\partial y} = y \, \sin 2\theta + 24 \sin \theta - 4y \, \sin \theta = 0$

$y \, (2 \sin \theta \cos \theta) + 24 \sin \theta - 4y \, \sin \theta = 0$

$y \, \cos \theta + 12 - 2y = 0$

$\cos \theta = \dfrac{2y - 12}{y}$

$\cos \theta = \dfrac{2(y - 6)}{y}$ ← eq. (2)

From eq. (1)

$\dfrac{\partial A}{\partial \theta} = y^2 \cos 2\theta + 24y \, \cos \theta - 2y^2 \, \cos \theta = 0$

$y (2 \cos^2 \theta - 1) + 24 \, \cos \theta - 2y \, \cos \theta = 0$

$2y \, \cos^2 \theta - y + 24 \, \cos \theta - 2y \, \cos \theta = 0$

Substitute cos θ of eq. (2)

$2y \left[ \dfrac{2(y - 6)}{y} \right]^2 - y + 24 \left[ \dfrac{2(y - 6)}{y} \right] - 2y \left[ \dfrac{2(y - 6)}{y} \right] = 0$

$2y \left[ \dfrac{4(y - 6)^2}{y^2} \right] - y + \dfrac{48(y - 6)}{y} - 4(y - 6) = 0$

$\dfrac{8(y - 6)^2}{y} - y + \dfrac{48(y - 6)}{y} - 4(y - 6) = 0$

$8(y - 6)^2 - y^2 + 48(y - 6) - 4y(y - 6) = 0$

$8(y^2 - 12y + 36) - y^2 + (48y - 288) - (4y^2 - 24y) = 0$

$8y^2 - 96y + 288 - y^2 + 48y - 288 - 4y^2 + 24y = 0$

$3y^2 - 24y = 0$

$y = 8 ~ \text{m}$

Width of base = 24 - 2(8) = 8 m Answer: [ C ]