A right cylinder is inscribed in a sphere of radius 93 inches. Find the largest possible volume of such as cylinder.

4 years ago

Answered By Jason R

To solve this, we should maximize a function that gives the volume of a cylinder that can fit inside a sphere of radius 93 inches.

For a right cylinder, Volume = Area of the circular base x distance between bases.

Summarized the usual way, this looks like V = [pi]r^{2}h.

We can find a relation between r and h so we can write the volume function with just one independant variable.

My drawing isn't very good so I will describe it.

We can make a right angle triangle with the radius of the sphere, the radius of the base of the cylinder, and the distance from the centre of the sphere to the centre of the base of the cylinder (which is half the height of the cylinder).

Using Pythagorus, 93^{2} = r^{2} + (h/2)^{2}.

We can replace r^{2} in our volume function with 93^{2} - (h/2)^{2}.

So V = [pi][93^{2} - (h/2)^{2}]h

To find the maximum value of this, we should look at the first derivative. Where the derivative has a zero is where the local maximum or minimums will be in a polynomial funcition like this one.

Let's rewrite it and then take the derivative.

V = [pi][93^{2}h - h^{3}/4] .... V = [pi/4][186^{2}h - h^{3}] (I made 93^{2}h into a fraction over 4 so I could bring that outside the brackets)

Now the derivative, dV/dh = [pi/4][186^{2} - 3h^{2}].

Set it equal to zero and solve ... this simplifies to 0 = [186^{2} - 3h^{2}].

So 186^{2} = 3h^{2} and h^{2} = 186^{2}/3 ... h = 62 sqrt(3).

We can plug this h value into our volume function to get a final result for the maximum volume of the cylinder that fits in the sphere with radius 93 inches.

To the nearest unit, I got 1,945,256 in^{3}. (don't forget to use cubic units for your volume answer).

4 years ago

## Answered By Jason R

To solve this, we should maximize a function that gives the volume of a cylinder that can fit inside a sphere of radius 93 inches.

For a right cylinder, Volume = Area of the circular base x distance between bases.

Summarized the usual way, this looks like V = [pi]r

^{2}h.We can find a relation between r and h so we can write the volume function with just one independant variable.

My drawing isn't very good so I will describe it.

We can make a right angle triangle with the radius of the sphere, the radius of the base of the cylinder, and the distance from the centre of the sphere to the centre of the base of the cylinder (which is half the height of the cylinder).

Using Pythagorus, 93

^{2}= r^{2}+ (h/2)^{2}.We can replace r

^{2}in our volume function with 93^{2}- (h/2)^{2}.So V = [pi][93

^{2}- (h/2)^{2}]hTo find the maximum value of this, we should look at the first derivative. Where the derivative has a zero is where the local maximum or minimums will be in a polynomial funcition like this one.

Let's rewrite it and then take the derivative.

V = [pi][93

^{2}h - h^{3}/4] .... V = [pi/4][186^{2}h - h^{3}] (I made 93^{2}h into a fraction over 4 so I could bring that outside the brackets)Now the derivative, dV/dh = [pi/4][186

^{2}- 3h^{2}].Set it equal to zero and solve ... this simplifies to 0 = [186

^{2}- 3h^{2}].So 186

^{2}= 3h^{2}and h^{2}= 186^{2}/3 ... h = 62 sqrt(3).We can plug this h value into our volume function to get a final result for the maximum volume of the cylinder that fits in the sphere with radius 93 inches.

To the nearest unit, I got 1,945,256 in

^{3}. (don't forget to use cubic units for your volume answer).I hope this helps.

Jason

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