1. **State the problem:** We have a triangular trough 10 feet long, 6 feet across the top, and 3 feet deep. Water flows in at 12 cubic feet per minute. We want to find how fast the water surface is rising (in inches per minute) when the water is 6 inches (0.5 feet) deep. 2. **Define variables:** Let $h$ = water depth in feet, $V$ = volume of water in cubic feet, $\frac{dh}{dt}$ = rate of change of the water depth (ft/min). Given $\frac{dV}{dt} = 12$ ft$^3$/min. 3. **Volume formula:** The trough is triangular in cross section, length $L=10$ ft. The cross-sectional area $A$ depends on water depth $h$. The full triangle has base $b=6$ ft and height $H=3$ ft. Since the water surface forms a similar triangle: $$ b_h = 6 \times \frac{h}{3} = 2h $$ Cross-sectional area of water surface: $$ A = \frac{1}{2} b_h h = \frac{1}{2} (2h)(h) = h^2 $$ 4. **Volume as function of $h$:** $$ V = A \times L = h^2 \times 10 = 10 h^2 $$ 5. **Differentiate volume wrt time:** $$ \frac{dV}{dt} = 10 \times 2h \frac{dh}{dt} = 20h \frac{dh}{dt} $$ 6. **Solve for $\frac{dh}{dt}$ when $h=0.5$ ft:** $$ 12 = 20 \times 0.5 \times \frac{dh}{dt} $$ $$ 12 = 10 \frac{dh}{dt} $$ $$ \frac{dh}{dt} = \frac{12}{10} = 1.2 \text{ ft/min} $$ 7. **Convert to inches per minute:** Since 1 ft = 12 inches, $$ \frac{dh}{dt} = 1.2 \times 12 = 14.4 \text{ inches per minute} $$ **Final answer:** The surface is rising at \boxed{14.4} inches per minute when the water is 6 inches deep.