The volume of water in a right circular frustum glass is 1176 pi mm^3, the height of the water reaches 18mm. find the radius of the lower base if the product of its radii is 60mm^2. Note that the lower base is smaller than the upper base.

r = ______ millimeters
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Volume of a cone=⅓AH where A is the base area and H the height.

Volume of a frustum is the difference in the volumes of two similar cones. If the larger cone has a radius of R  and height H, and the smaller a radius of r, then the height of the smaller cone=Hr/R.

The height of the frustum, h, is H-Hr/R=(H/R)(R-r)=

A=πr² for the smaller cone and πR² for the larger one. 

Volume of frustum=⅓(πR²H-πr²Hr/R)=(πH/3)(R²-r³/R)

⅓(πH/R)(R³-r³)=⅓(πH/R)(R-r)(R²+Rr+r²)=⅓πh(R²+Rr+r²)=1176π.

We know Rr=60, h=18. 6(R²+Rr+r²)=1176, (R²+Rr+r²)=1176/6=196.

R²+2Rr+r²=(R+r)²=196+60=256, so R+r=16. R=60/r so 60/r+r=16, 60+r²=16r, r²-16r+60=0=(r-10)(r-6).

So r=6mm and R=10mm. The base radius is smaller, so it’s 6mm.

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