Atrough is 14 ft long and its ends have the shape of isosceles triangles that are 2 ft across at the top and have a height of 1 ft. if the trough is being filled with water at a rate of 8 ft3/min, how fast is the water level rising when the water is 9 inches deep?

The filling water forms a triangle which is geometrically similar to the triangle of the trough. Let the height of the water be h and its upper length be x. Using a ratio:
h : 1 = x : 2
h/1 = x/2
x = 2h
The volume of the water can be found by multiplying the are of this triangle by the length of the trough. Thus,
V = 1/2 * x * h * L
V = 1/2 * 2h * h * 14
V = 14h²
dV/dh = 28h
We are given dV/dt as 8 ft³/min and need to find dh/dt
Using the chain rule:
dh/dt = dV/dt x dh/dV
dh/dt = 8 x (1/28h); h = 9/12 feet
dh/dt = 8/21 ft/min

1.3 and 1 1/3 are both closest to 1. hope this ; )