Answer
The water level is rising at a rate of $~0.00164~ft/min$
Work Step by Step
Let $W$ be the width of the pool. Then $W = 20~ft$
Let $b_1$ be the distance across the bottom of the pool. Then $b_1 = 6~ft$
We can use similar triangles to write an expression for the distance $b_2$ across the top of the water in terms of the water level $h$:
$b_2 = b_1+\frac{11h}{3} = 6~ft+\frac{11h}{3}$
We can differentiate both sides of the equation for the water volume with respect to $t$:
$V = \frac{h}{2} (b_1+b_2)~W$
$V = \frac{h}{2} (6+6+\frac{11h}{3})~(20)$
$V = 10h~ (\frac{11h}{3}+12)$
$V = \frac{110}{3}h^2+120h$
$\frac{dV}{dt} = (\frac{220}{3}h + 120)~\frac{dh}{dt}$
$\frac{dh}{dt} = [\frac{1}{\frac{220}{3}h + 120}](\frac{dV}{dt})$
$\frac{dh}{dt} = [\frac{1}{\frac{220}{3}(5) + 120}](0.8)$
$\frac{dh}{dt} = 0.00164~ft/min$
The water level is rising at a rate of $~0.00164~ft/min~~$
