Given a surface area of $2m^2$, what is the maximum volume of an open-top cone?
h=height of cone r=radius of base L=slant height=$√(h^2+r^2)$
$2=πr√(h^2+r^2)$ -> $h=√(4-π^2r^4)/πr$
Plugging the height formula into the Volume Formula: $(πr^2h)/3$
Solving for $r$, I get $0.606m$, giving a max. volume of $0.33m^3$.
Could someone verify this or tell me where I went wrong?