# Mr B... you lost me ...I'll do the full problem below

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Posted by Subhotosh Khan on October 09, 2002 at 10:22:54:

In Reply to: stepping in... posted by Denis Borris on October 08, 2002 at 22:48:24:

: : : I understand what needs to be done, Mr K, but can't finish it.
: : : Too many things I haven't seen before.
: : : :mAOB = T
: : : I understand "AOB", but what is T? degrees, like 75 degrees? Yes
: : : What does "m" mean? It is short of "measure of" or "value of" angle AOB
: : : :From that you can get the area of the base(A_1)
: : : Why do I need to get the area? Look below
: : : :volume of small cone (V_1)=1/3*R*A
: : : where does that come from? V = (pi r^2 h) / 3, is it not? In this case A = pi * r^2 .... and h = R (the radius of the circle you started with) so we are talking about the same thing
: : ***********************************
: : Assume mAOB = T (in radians)
: : So chord AB = R * T (definition of angle)
: : So the circumference of the base of the small cone
: : 2* pi * r = R * T ? (r = radius of the circular base of the small cone)
: : r = R * T/(2*pi)
: : V_1 = (pi)* r^2 * h / 3
: : = (pi)* [R * T/(2*pi)]^2 * R / 3
: : = R^3/(12*pi) * T^2
: : Similarly find V_2

: was fooling around with this; interim question, Mr K (I'm rusty with cones!):

: height of smaller cone = Rsqrt(360^2 - T^2) / 360

: height of bigger cone = Rsqrt[T(720 - T)] / 360

: are those correct?

: if Mr K says "yes", hope I've helped you Kurt!!
************************************
V_1 = R^3/(12*pi) * T^2

V_2 = R^3/(12*pi) *(2pi - T)^2

V = V_1 + V_2 = V_2 = R^3/(12*pi) *(T)^2
+ R^3/(12*pi) *(2pi - T)^2

dV/dT = R^3/(12*pi) * [2T - 2*(2pi -T)] = 0 ... for maximum

[2T - 2*(2pi -T)] = 0

T = pi ......[or 180°]

Mr. B as the problem is presented, the heights of the cones are equal - the radius of the original circle.

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