Water is being pumped into an inverted conical tank at some constant rate. The tank has a height of 600 cm and the diameter of the tank across the top is 400 cm. If the water level is rising at a rate of 20 cm/min when the height of the water is 200 cm, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)cm3/min

Answer:The water is being pumped at a speed of [tex] \boxed{\bf \frac{10592000*\pi}{27}\;cm^3/min}[/tex] Step-by-step explanation:By congruence of triangles, the radius r of the cone base when its height is 200 cm satisfies the relation r/200 = 400/600 (See picture attached) So, r = 400/3 cm when the water is 200 cm high. The volume of a cone with radius of the base = R is given by [tex] \bf V=\frac{\pi R^2h}{3}[/tex] So, the volume of water when it is 200 cm high is [tex] \bf V_1=\frac{\pi* (400/3)^2*200}{3}=\frac{32000000*\pi}{27}\;cm^3[/tex] One minute later, the height of the water is 200 cm + 20 cm = 220 cm The radius now satisfies r/220 = 400/600 and now the radius of the base is r = 440/3 and the new volume of water is  [tex] \bf V_2=\frac{\pi* (440/3)^2*220}{3}=\frac{42592000*\pi}{27}\;cm^3[/tex] So, the water is raising (being pumped) at a rate (speed) of [tex] \bf V_2-V_1=(\frac{42592000*\pi}{27}-\frac{32000000*\pi}{27})=\frac{10592000*\pi}{27}\;cm^3/min[/tex]