# Find the dimensions (height and radius) of the cone that has the maximum volume?Given that a right circular cone is inscribed in a sphere of radius 15cm.

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You need to find the dimension of height of cone such that:

`h = 15 + sqrt(15^2 - r^2)` (r expresses the radius of cone)

You need to remember the formula that expresses the volume of cone such that:

`V = (pi*r^2*h)/3`

You need to substitute 15 + sqrt(15^2 - r^2) for h in formula of volume such that:

`V = (pi*r^2*(15 + sqrt(15^2 - r^2)))/3`

`V(r) = 5pi*r^2 + pi*r^2*sqrt(15^2 - r^2)/3`

The problem provides the information that the cone has maximum volume, hence you need to differentiate the volume with respect to r, such that:

`V'(r) = 10pi*r + (pi/3)*((15r^4 - r^6)')/(2sqrt(15r^4 - r^6))`

`V'(r) = 10pi*r + (pi)*(10r^3 - r^5)/(sqrt(15r^4 - r^6))`

You need to solve the equation `V'(r) = 0` such that:

`10pi*r + (pi)*(10r^3 - r^5)/(sqrt(15r^4 - r^6)) = 0`

You need to divide by `pi` such that:

`10r + (10r^3 - r^5)/(sqrt(15r^4 - r^6)) = 0`

`10r(sqrt(15r^4 - r^6)) = r^5 - 10r^3`

`100(15r^4 - r^6) = (r^4 - 10r^2)^2`

`100(15r^4 - r^6) = r^8 - 20r^6 + 100r^4`

`1400r^4 - 80r^6 - r^8 = 0`

You need to factor out `-r^4` such that:

`-r^4(r^4 + 80r^2 - 1400) = 0`

r = 0 contradiction (the radius of cone is not zero)

`r^4 + 80r^2 - 1400 = 0`

You should come up with the substitution `r^2 = y` such that:

`y^2 + 80y - 1400 = 0`

`y_(1,2) = (-80+-sqrt(6400 + 5600))/2`

`y_(1,2) = (-80+-20sqrt(30))/2`

`y_(1,2) = (-40+-10sqrt(30))`

`r^2 = 14.77 =gt r = sqrt(14.77) ~~ 4`

Substituting 4 for r in equation of height yields:

`h = 15 +- sqrt(15^2 - 4^2)`

You need to select the positive value `+14.456` such that:

`h = 15 + 14.456`

`h = 29.456`

**Hence, evaluating dimensions of the cone under given conditions yields `r~~4` and `h~~29` .**

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