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?

Answer 1

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

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

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#.

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Answer 2

To find the dimensions (radius r and height h) of the cone of maximum volume inscribed in a sphere of radius 2, the radius and height of the cone can be related using similar triangles. The volume of the cone can then be expressed in terms of one variable, which can be maximized using calculus. The resulting dimensions are ( r = \frac{4}{3}\sqrt{2} ) and ( h = \frac{8}{3} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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