# What is the maximum possible volume of a cone that has a slant height of 10 cm?

The volume of a cone with perpendicular height h and radius of the base r is given by `V = (1/3)*pi*r^2*h`

If a cone has a slant height (S), perpendicular height (h) and radius (r), the three are related by `S^2 = h^2 + r^2`

Here S = 10

If...

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The volume of a cone with perpendicular height h and radius of the base r is given by `V = (1/3)*pi*r^2*h`

If a cone has a slant height (S), perpendicular height (h) and radius (r), the three are related by `S^2 = h^2 + r^2`

Here S = 10

If the height is x, the radius is: `sqrt(100 - x^2)`

The volume of the cone is `V = (1/3)*pi*(100 - x^2)*x`

To maximize V, solve `(dV)/(dx) = 0`

=> `(1/3)*pi*100 - (1/3)*pi*3*x^2 = 0`

=> `x^2 = 100/3`

=> `x = 10/sqrt 3`

The height of the cone is `10/sqrt 3` and the radius is `sqrt(200/3)` . The volume of the cone is: `(1/3)*pi*(200/3)(10/sqrt 3) = 403.06`

The maximum possible volume of a cone with slant height 10 cm is 403.06 cm^3.

Approved by eNotes Editorial Team