Cups A and B are cone shaped and have heights of `32 cm` and `14 cm` and openings with radii of `15 cm` and `12 cm`, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

Volume cone = `(pi r^2 h)/3`

`V_A = (pi xxcolor(red)(15^2xx32))/3 and V_B = (pixxcolor(blue)(12^2 xx14))/3`

We only need to know which one is bigger, not the actual volumes.
We can therefore ignore `pi and 3` as they are common to both.

By inspection we can see that :`color(red)(V_A) >color(blue)(V_B)`

So A will not overflow, but how high will the water reach?

The water in A and the whole cup A are similar figures.
Therefore we can compare the ratio of the cubes of the heights with the ratio of their volumes.

`(h/H)^3 =V_B/V_A`

`(h/H)^3 = h^3/H^3 = h^3/32^3 = (color(blue)(12^2xx14))/(color(red)(15^2xx32)) = V_B/V_A`

`h^3 = (32^3xx12^2xx14)/(15^2xx32) = 9,1750.4`

`h = root3(9,1750.4) = 20.93cm`

Volume of cone`=1/3pir^2h`
A`=1/3 pi^2*15^2*32`
`=1/3*3.141592654*225*32`
`=1/3*22619.467`
`=22619.467/3`
Vol. A`=7539.822cm^3`
Vol.B`=1/3pi12^2*12`
`=1/3*3.41592654*144*12`
`=1/3*5428.672105`
`=5428.672105/3`
`Vol.B=1809.557cm^3`
New vol cup A=`1/3*pi15^2*x=1809.557cm^3`
`235.619x=1809.557`
`x=1809.557/235.619`
`x=7.68cm`

`Volume of con = 1/3*pi*r^2*h`.
where r = radius and h = height

`Va=1/3*pi*15^2*32`
`Va = 2400pi`

`Vb=1/3*pi*12^2*14`
`Vb=672pi`

so, cup A wont overflow because it is bigger than cup B.

let say T is the height of cup A when content in cup B poured to cup A.

Va = Vb

`cancel(1/3)*cancelpi*15^2*T = cancel(1/3)*cancelpi*12^2*14 =(672*pi)`

`15^2*T = 12^2*14`

`T= (12^2*14)/(15^2)`

`T = 8.96 cm`