i don't understand this calculus problem plz help

Question

A water trough is 5 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 60 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.2 m3/min how fast is the water level rising when the water is 20 cm deep?

Answer 1

Volume of a trapezoidal prism = [(A + B)H/2]*L

In our problem:

H = y m

B = 20 cm = 0.2 m = (1/5) m

A = 20 + y cm = (1/5) + y m

L = 5 m

V = [(1/5) + y + (1/5)](y/2)(5)

V = [(2/5) + y](y/2)(5)

V = (2/5)(5y/2) + (y)(5y/2)

V = y + (5/2)y^2

Differentiating to find dV/dt:

dV/dt = dy/dt + 5y(dy/dt)

Rearranging to find dy/dt:

(dy/dt)(1 + 5y) = dV/dt

dy/dt = (dV/dt)/(1 + 5y)

Given values:

dV/dt = (1/5) m^3/min

y = 1/5 m

Substituting and solving:

dy/dt = (1/5)/[1 + 5(1/5)]

dy/dt = (1/5)/(2)

dy/dt = 1/10 m/min

dy/dt = 0.1 m/min