A large industrial water tank is shaped like an inverted right circular cone with a semi-vertical…

Let $r$ be the radius and $h$ be the height of the water in the conical tank. Given $\tan\alpha = \frac{r}{h} = 0.5 \Rightarrow r = 0.5h$. Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (0.5h)^2 h = \frac{\pi}{12} h^3$. Differentiating with respect to $t$: $\frac{dV}{dt} = \frac{\pi}{12}(3h^2)\frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$. Given $\frac{dV}{dt} = -5\text{ m}^3/\text{min}$ and $h = 4\text{ m}$: $-5 = \frac{\pi}{4}(4^2)\frac{dh}{dt} = 4\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = -\frac{5}{4\pi}\text{ m/min}$. Thus, the water level is dropping at a rate of $\frac{5}{4\pi}\text{ m/min}$.

The slant height is $l = \sqrt{r^2 + h^2} = \sqrt{(0.5h)^2 + h^2} = \sqrt{1.25h^2} = \frac{\sqrt{5}}{2}h$. The wetted surface area is $S(h) = \pi r l = \pi (0.5h)\left(\frac{\sqrt{5}}{2}h\right) = \frac{\sqrt{5}}{4}\pi h^2$. Differentiating with respect to $h$: $\frac{dS}{dh} = \frac{\sqrt{5}}{2}\pi h$. Since $h > 0$ for a physical tank, $\frac{dS}{dh} > 0$ for all $h \in (0, \infty)$. Therefore, the wetted surface area $S(h)$ is a strictly increasing function of the water depth $h$.

Given cost $C = KS^2 = K\left(\frac{\sqrt{5}}{4}\pi h^2\right)^2 = K\frac{5\pi^2}{16}h^4$. The rate of change of cost with respect to time is: $\frac{dC}{dt} = \frac{dC}{dh} \cdot \left|\frac{dh}{dt}\right| = \left(K\frac{5\pi^2}{4}h^3\right) \left(\frac{5}{\frac{\pi}{4}h^2}\right) = K\frac{5\pi^2}{4}h^3 \cdot \frac{20}{\pi h^2} = 25K\pi h$. Since $\frac{dC}{dt}$ is a strictly increasing linear function of $h$, the rate of cost change increases as $h$ increases. Therefore, the cost is increasing most rapidly at the maximum depth allowed by the tank (or maximum $h$).