Answer:
0.75 ft/min
Explanation:
Given;
Length of trough l = 8 ft
base length of trough b = 3 ft
height of trough h = 1 ft
Change in volume dV/dt = 12 ft^3/min
At height hi = 8 in = 2/3 ft
Ratio of base to height;
b/h = 3/1
b = 3h .....1
The volume of the trough;
V = area of triangle × length
V = 1/2 × b×h × l
V = 1/2(bhl) .....2
Substituting l = 8ft and equation 1 (b = 3h) into equation 2;
V = 1/2(3h × h × 8)
V = 12h^2
differentiating V, change in volume per time is;
dV/dt = 12 × 2h .dh/dt
dV/dt = 24 hi .dh/dt
Substituting the values of dV/dt and hi;
12 = 24(2/3) .dh/dt
dh/dt = 12 ÷ 24(2/3) = 12/16
dh/dt = 3/4 ft/min = 0.75 ft/min