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# Calculus/application of derivite

Question
show that of right circular cone of maximum volume that can be inscribed in a sphere radius is 4r/3

The equation of a sphere is x²+y²+z²=r².

Let's put the top point of the cone in the sphere at (0,0,r),

The base of the sphere is in the plane z = -a where a<r.
This makes the equation of the base is x²+y²=r²-a².
This means the radius of the base is sqrt(r²-a²).

Since the base is a units under the origin and the tip is r units up, the height is r+a.

The volume of a cone is V = πs²h/3 where s is the radius of the base and h is the height.
It is known that base radius is given by s = sqrt(r²-a²), so s² = r²-a².
It is known that the height is given by r+a.
This gives us V = π(r²-a²)(r+a)/3.

Since r is a constant, the only variable is a, making it really be V(a).
Since we know that V(a) = π(r²-a²)(r+a)/3, we can multiply it out to get
V(a) = π(r³+ar²-a²r-a³)/3.

The derivative with respect to a, since that's the variable, is V'(a) =  π(r²-2ar-3a²)/3.
This is zero when r² - 2ar - 3a² = 0.

Since a is the variable, putting it with a² first gives -3a² - 2ra + r² = 0.
The quadratic equation gives a = (2r ± sqrt(4r²+12r²))/6.
The r²'s in the square-root add to 16r², and the sqrt(16r²) is 4r.

That leaves us with (2r±4r)/6.
Not that is we look at the plus, we get 6r/6, so a=r, and this means the cone is a single point.  This gives it a volume of 0, and that's the minimum.

If we take the minus sign, we get a = (2r-4r)/6 = -2r/6 = -r/3.
Putting a=-r/3 in V(a) gives V(-r/3) = π(r³ - r³/3 - r³/9 + r³/27)/3.
Simplifying gives V(-r/3) = π(27r³/27 - 9r³/27 - 3r³/27 + r³/27)/3.
This reduces to V(-r/3) = π(16r³/81), and that is the maximum volume.

Note that 16/81 is really (2/3)^4 for 2^4 = 16 and 3^4 = 81.

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