Question

The perimeter of an isosceles triangle is 2s. What must be it's sides so that the volume of the solid generated by revolving the triangle about the base is greatest possible​

Answer 1

Let the lengths of the the base and the congruent sides be bb and ll respectively.

⇒b+2l=2s⇒l=2s−b2.⇒b+2l=2s⇒l=2s−b2.

Let the length of the perpendicular from the vertex to the base be hh.

⇒h2=l2−b24=(2s−b2)2−b24.⇒h2=l2−b24=(2s−b2)2−b24.

⇒h2=s2−sb=s(s−b)⇒h2=s2−sb=s(s−b)

When this isosceles triangle is revolved about its base, we get two cones, each of base radius hh and height b2.b2.

The sum of the volumes of two cones is,

V=23πh2×b2=23πs(s−b)×b2.V=23πh2×b2=23πs(s−b)×b2.

⇒V=13πbs(s−b).⇒V=13πbs(s−b).

At the extreme values of V,V, its first derivative with respect to bb would be 0.0.

⇒