A water-trough is 10m long and has a cross-section which is the shape of an isosceles trapezoid
that is 30cm wide at the bottom, 80cm wide at the top, and has height 50cm. If the trough
is being filled with water at the rate of 0.2 m^3/min, how fast is the water level rising when
the water is 30cm deep?
The area of a cross-section is A = 1/2(a + b)h where a = width at the top, b = width at the bottom.

Firstly, I converted everything to meters, so it's 0.3m at the bottom, 0.8m at the top, height 0.5m.

Then, I multiplied the area by the depth of the trough to get
V = 5(a + b)h

So b = 0.3, a = (0.3 + h) since the water surface expands to the left and right at a 1:1 ratio with depth.

Substituting: V = 5(0.3 + h + 0.3)h = 5h(h + 0.6) = 5h^2 + 3h

Then I differentiate to find an equation for V' and sub in the current depth of 0.3m:

V' = 10hh' + 3h'
V' = 10(0.3)h' + 3h'
V' = 3h' + 3h' = 6h'
0.2 = 6h'

so the water level is rising at a rate of h' = 0.033... meters/min, which is equal to 3+1/3 cm/min, yet my answer key states 1/3 cm/min.

Just wondering if anyone can catch a mistake a made? Thanks.