Question 1085987
.
 a sphere of {{{highlight(the_radius)}}} 10 m and a right circular cone of base radius 10 m and height 15 m stands on a table. 
at what height from the table should the two solids be cut in order to have equal circular sections 
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<pre>
Let "z" be the vertical coordinate/axis starting (z=0) at the table and directed vertically up.


Then the horizontal section area at elevation z is:

    for the sphere S(z) = {{{pi*(10^2-(10-z)^2)}}},   and

    for the cone   C(Z) = {{{pi*(10*((15-z)/15))^2}}}.


They want you find z such that S(z) = C(z),  or,  which is the same

{{{pi*(10^2-(10-z)^2)}}} = {{{pi*(10*((15-z)/15))^2}}},


I will leave the solution of this equation to you.


Instead, I'll give you the plots of the two functions S(z) and C(z).
</pre>


{{{graph( 330, 330, -1.5, 16.5, -40.5, 400.5,
          3.14*(10^2-(10-x)^2), 3.14*(10*((15-x)/15))^2
)}}}


Plots S(z) = {{{pi*(10^2-(10-z)^2)}}} (red)  and C(z) = {{{pi*(10*((15-z)/15))^2}}} (green)



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<U>Final notice</U>. It is difficult to imagine the two solids of this size standing on a table.