a reservoir is in the form of a frustrum of a cone with upper base of radius 9 ft and lower base of radius 4ft and altitute of 10 ft. the water in the reservoir is x ft deep. if the level of the water is increasing at 4 ft/min, how fast is the volume of the water in the reservoir increasing when its depth is 2 ft?
NOTE: the volume of a frustrum of a cone of upper base raduis R and lower base raduis r and height h. V = 1/3 pi h (R^2 +r^2 + Rr)

Question

a reservoir is in the form of a frustrum of a cone with upper base of radius 9 ft and lower base of radius 4ft and altitute of 10 ft. the water in the reservoir is x ft deep. if the level of the water is increasing at 4 ft/min, how fast is the volume of the water in the reservoir increasing when its depth is 2 ft? 
NOTE: the volume of a frustrum of a cone of upper base raduis R and lower base raduis r and height h. V = 1/3 pi h (R^2 +r^2 + Rr)

ganeshie8 · Answer

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ganeshie8 · Answer

you're given 
$\dfrac{dx}{dt}=4$

jack1 · Answer

u lost me at "frustrum"... sorry  ;P

ganeshie8 · Answer

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can you express $R$ in terms of $x$ ?