Water is leaking out of an inverted conical tank at a rate of _ cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has helght 8 meters and the diameter at the top is 3 meters. If the water level is rising at a rate of 19 centimeters per minute when the height of the water is 4.5 meters, find the rate at which water is being pumped into the tank in eubic centimeters per mitiute. _ \frac{\mathrm{cm}^{3}}{\mathrm{~min}}

We begin by letting • \( h \) be the height of the water (in centimeters), • \( r \) be the radius of the water’s surface (in centimeters), and • \( V \) be the volume of water in the tank (in cubic centimeters). The tank is an inverted cone with a height of 8 meters and a top diameter of 3 meters. First we convert these dimensions to centimeters: \[ \text{Height: } 8\,\text{m} = 800\,\text{cm},\quad \text{Top radius: } \frac{3\,\text{m}}{2} = 1.5\,\text{m} = 150\,\text{cm}. \] Because the water in the tank also forms a cone, the radius \( r \) and the water height \( h \) are related by similar triangles. In the full tank, \[ \frac{r}{h}=\frac{150}{800}=\frac{3}{16}\quad \Longrightarrow\quad r=\frac{3}{16}\,h. \] The volume of a cone is \[ V=\frac{1}{3}\pi r^2h. \] Substitute \( r=\frac{3}{16}h \) into the volume formula: \[ V=\frac{1}{3}\pi \left(\frac{3}{16}h\right)^2h=\frac{1}{3}\pi\frac{9}{256}h^3=\frac{3\pi}{256}h^3. \] Differentiate both sides with respect to time \( t \): \[ \frac{dV}{dt}=\frac{d}{dt}\left(\frac{3\pi}{256}h^3\right) =\frac{9\pi}{256}\,h^2\,\frac{dh}{dt}. \] We are told that when the water depth is \( h=4.5\,\text{m} \) (which is \( 450\,\text{cm} \)) the water level is rising at a rate of \( \frac{dh}{dt}=19\,\text{cm/min} \). Plug these values into the differentiated equation: \[ \frac{dV}{dt}=\frac{9\pi}{256}\,(450)^2\,(19). \] Now, calculate step by step: 1. Compute \( 450^2 \): \[ 450^2=202500. \] 2. Multiply by 9 and 19: \[ 9\times202500=1\,822\,500, \] \[ 1\,822\,500\times19=34\,627\,500. \] So, \[ \frac{dV}{dt}=\frac{34\,627\,500\,\pi}{256}\quad \text{(in cm}^3/\text{min)}. \] This \( \frac{dV}{dt} \) is the net rate of change of volume in the tank. In our situation water is both being pumped in and leaking out. Let • \( Q \) be the rate at which water is being pumped into the tank (in cm\(^3\)/min), and • \( L \) be the rate at which water is leaking out (in cm\(^3\)/min). Since water is being added and simultaneously lost by leakage, we have \[ \frac{dV}{dt}=Q-L. \] The problem states that water is leaking out at a rate of \[ L=10\,000\,\frac{\text{cm}^3}{\text{min}}. \] (If another numerical leak rate were given you would substitute that value instead.) Thus, solving for \( Q \): \[ Q=\frac{dV}{dt}+L =\frac{34\,627\,500\,\pi}{256}+10\,000. \] To get a numerical approximation, take \(\pi\approx 3.1416\): 1. Multiply numerator by \(\pi\): \[ 34\,627\,500\times3.1416\approx108\,786\,000. \] 2. Divide by 256: \[ \frac{108\,786\,000}{256}\approx425\,289\,\frac{\text{cm}^3}{\text{min}}. \] Now add the leak rate: \[ Q\approx425\,289+10\,000\approx435\,289\,\frac{\text{cm}^3}{\text{min}}. \] Thus, water is being pumped into the tank at approximately \[ 435\,289\,\frac{\text{cm}^3}{\text{min}}. \]