The right circular cone of largest volume that can be enclosed by

SOLUTION

Concept:

\({\rm{Volume\;of\;Right\;circular\;cone\;}} = \frac{1}{3}{\rm{π }}{{\rm{r}}^2}{\rm{h}}\)

The method of finding Maxima and Minima of Y = f(x)

1) Find f’(x) and f”(x) for given function Y = f(x)

2) Equate f’(x) to zero to obtain stationary points x = a

3) Calculate f”(x) at each stationary points x = a (i.e f”(a))

4) We obtain following three conditions:

(i) If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)

(ii) If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)

(iii) If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a 

Calculation:

By hypoteneous theorem in ΔOAB

(h-1)² + r² = 1² 

∴ r² = 2h - h²

\({\rm{Volume\;of\;Right\;circular\;cone\;}} = \frac{1}{3}{\rm{π }}{{\rm{r}}^2}{\rm{h}}\)

\({\rm{V}} = \frac{1}{3}{\rm{π }}{{\rm{(2h-h^2)}}}{\rm{h}}\)

\({\rm{V}} = \frac{π}{3}{\rm{}}{{\rm{(2h^2-h^3)}}}{\rm{}}\)

Let V = f(h)

∴ \({\rm{f(h)}} = \frac{π}{3}{\rm{}}{{\rm{(2h^2-h^3)}}}{\rm{}}\)

1) \({\rm{f'(h)}} = \frac{π}{3}{\rm{}}{{\rm{(4h-3h^2)}}}{\rm{}}\), \({\rm{f''(h)}} = \frac{π}{3}{\rm{}}{{\rm{(4-6h)}}}{\rm{}}\)

2) f’(h) = 0, ∴ \(\frac{π}{3}{\rm{}}{{\rm{(4h-3h^2)}}}{\rm{}}=0\), h = 0, 4/3(stationary point)

3) f"(h) at stationary point h = 0, f”(h) = 4π/3 and at h = 4/3, f”(h) = -4π/3 

4) f"(h) < 0 at h = 4/3

∴ f(h) is maximum at h = 4/3

∴ The largest volume of right circular cone which will fit in a sphere of radius 1 m will have a height h = 4/3 m