How do you find the dimensions (radius r and height h) of the cone of maximum volume which can be inscribed in a sphere of radius 2?

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Answer 1 · 2015-02-26T19:57:48

The answer is: $r=4/3sqrt2$ , $h=8/3$

We can imagine a vertical section of the figure, that would appear:

Google image

Let $r$ be the radius of the cone, $R$ be the radius of the sphere and $h$ be the height of the cone.

Let's put $DhatOB=x$ with limitations $0<=x<=pi$ .

In the right-angled triangle $DOB$ :

$r=DB=Rsinx$ , $OD=Rcosx$ , than

$h=CD=OC+OD=R+Rcosx=R(1+cosx)$ .

So the volume of the cone is:

$V=1/3pir^2hrArrV=1/3pi(Rsinx)^2*R(1+cosx)rArr$

$V=1/3piR^3sin^2x(1+cosx)$

$V'=1/3piR^3*[2sinx*cosx*(1+cosx)+sin^2x(-sinx)]=$

$=1/3piR^3sinx[2cosx(1+cosx)-sin^2x]=$

$=1/3piR^3sinx(2cosx+2cos^2x-sin^2x)=$

$=1/3piR^3sinx(2cosx+2cos^2x-1+cos^2x)=$

$=1/3piR^3sinx(3cos^2x+2cosx-1)$

Now let's find the signum of the derivative, since $sinx>=0$ for every $x$ in the limitations, than:

$V'>=0rArr3cos^2x+2cosx-1>=0rArr$

$Delta/4=(b/2)^2-ac=1+3=4$

$cosx=((-b/2)+-sqrt(Delta/4))/a=(-1+-2)/3$ ,

So:

$cosx<=-1vvcosx>=1/3$

The first one has only the solution:

$x=pi$ ,

the second:

$-arccos(1/3)<=x<=arccos(1/3)$ , but for the limitations:

$0<=x<=arccos(1/3)$ .

The function growths from zero to $arccos(1/3)$ , than it decreases.

So $x=arccos(1/3)$ is the maximum requested.

Let's find now $r$ and $h$ :

$r=Rsinx=Rsqrt(1-cos^2x)=2sqrt(1-(1/3)^2)=2sqrt(1-1/9)=$

$=2sqrt((9-1)/9)=2sqrt(8/9)=2*2sqrt2/3=4/3sqrt2$

$h=R(1+cosx)=2(1+1/3)=2((3+1)/3)=8/3$ .