13-14 Water flowing into trapezoidal trough

$V = 20y + 10xy$
 

From the figure:
$\dfrac{x}{y} = \dfrac{1}{2}$

$x = \frac{1}{2}y$
 

$V = 20y + 5y^2$

$\dfrac{dV}{dt} = 20\dfrac{dy}{dt} + 10y \dfrac{dy}{dt}$
 

when y = 6 in or 0.5 ft
$10 = 20\dfrac{dy}{dt} + 10(0.5) \dfrac{dy}{dt}$

$\dfrac{dy}{dt} = 0.4 \, \text{ ft/min}$           answer

When y = 1 ft
$\dfrac{dV}{dt} = 20\dfrac{dy}{dt} + 10y \dfrac{dy}{dt}$

$10 = 20\dfrac{dy}{dt} + 10(1) \dfrac{dy}{dt}$

$\dfrac{dy}{dt} = \frac{1}{3} \, \text{ ft/min}$           answer