Posted by Megan on Wednesday, November 23, 2011 at 7:04pm.
since h = 2r
v = 1/3 pi r^2 (2r) = 2/3 pi r^3
dv/dt = 2 pi r^2 dr/dt = 2*pi*25*2 = 314.16
This is an unrealistic problem. If sand is running out, it's likely that the flow gradually slows down. But this setup has the radius of the pile steadily growing, meaning that the volume increases at an increasing rate. They must be piling more sand into the tank, so it comes out faster and faster to keep the radius steadily growing.
Find the absolute max and min of f(x,y)=4xy^2-x^2y^2-xy^2 on the region D
where D is bounded by x=0, y=0 and y=-x+6
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