Probem39+Lucas,+Ian

 //(ms sweeney): very good. the only problem is that since we cannot assume that the value you found, Ian, yields the absolute maximum. Only plugging into the objective function and comparing this outcome with that at the endpoints will tell for sure!// Ian R. Wilson:

39. Show that, when a cone is inscribed within another cone such that the vertex of the inner cone is tangent to the base of the outer cone at the base's center, the largest volume possible for the inner cone as we are varying the height of said inner cone is 1/3 that of the outer cone.

Working under the assumption that the largest possible value for the inscribed cone's volume exists when the height of that cone (h) is 1/3 that of the larger cone (the height of which shall be H), we are given the following. //(ms sweeney) You cannot assume this! This is what you are trying to prove!! //

the volume of the inner cone (v) is described by: v=1/3π(r^2)h the volume of the outer cone (V) is described by:V=1/3 π(R^ 2)H There is another cone for which the base is that of the cone the volume of which is V, but the height of which is H-h. Since this third cone is similar to the cone the volume of which is V it follows that:

H/R=(H-h)/r rH/R-H=-h h=H-rH/R

Therefore:

v=1/3π(r^2)(H-rH/R)

let v=f(x)

f(x)= 1/3π(r^2)(H-rH/R) f(x)=(1/3) (πr^2)H-(1/3R) π(r^3R)H f'(x)=(2/3) πHr- (πH/3)r^2 0= (2/3) πHr- (πH/R)r^2 (2/3) πHr=( πH/R)r^2 <span style="color: rgb(12, 14, 12);">(2/3) <span style="color: rgb(12, 14, 12);">πH= <span style="color: rgb(12, 14, 12);">( <span style="color: rgb(12, 14, 12);">πH/R)r r=(2/3)R at maximum

if the maximum value for r is 2/3R then the maximum value of (H-h) is 2/3H, therefore the value of h which shall yield the highest value v, being largest when r too is largest, is 1/3H. <span style="color: rgb(12, 14, 12);"> <span style="color: rgb(12, 14, 12);"> <span style="color: rgb(12, 14, 12);">