Application of related rates. Problem below

Question

A horizontal trough is 15m long and its ends are isosceles triangles with an altitude of 5m and a base of 5m. Water is being drained from the trough at a rate of 12 m^3/min. How fast is the water level lowering when the water is 1m deep?

Answer 1

When the depth of the water in the trough is H, then the volume of water is

V = (Base)x(Height) = [ (1/2) H·H] ·15 = (15/2)H²

Then dV/dt = 15H·dH/dt

12 = 15·1·dH/dt

dH/dt = 4/5 m/min

Answer 2

Volume of the trough = V = bhl/2 = 5(5)(15)/2

= 375/2 = 187.5 m^3/min

V'= 12 m^3/min

V(h)= 5(15)h/2 = 75h/2 = 34.5h when h=5 m

V(1) = 1(15)h/2 = 7.5h when h= 1m

V(1) = 7.5(1) = 7.5 m^3

V'(1) = 7.5(hb'+bh')= 7.5(b'+h')= 7.5(2h') =12

h' = 12/7.5(2)= 12/15 = 0.8 m/min rate of decrease in water level when water level

it may help to sketch a diagram of the trough