Pre- Calc Help!?

Question

A cone is inscribed in a sphere of radius 6.
a. Express the function of x. ( Hint V = 1/3 pi times r^2 times h)
Then give the domain of this function.
b. Find the approximate value of x that maximizes the volume. Then give the approximate maximum volume.

Answer 1

Hi,

Let x = the radius of the base of the cone. Assuming the cone has a height longer than the radius of 6, then the radius of the sphere is the hypotenuse of a right triangle formed by the radius of the cone's base, x, and the portion of its height that is below the sphere's centerpoint, which we will call z. Since 6² = x² + z², we can solve for z and get z = √(36 - x²).

Since the radius of the cone has to be positive, but can't be bigger than the radius of the sphere, the domain is 0 < x < 6.

That means the entire height of the cone would be 6 + √(36 - x²). So the formula for the volume of the cone would be:

V = 1/3Pi*r²h

y = 1/3Pi*x²(6 + √(36 - x²))

If this is entered in your calculator with the appropriate window, you can use the MAXIMUM command under 2nd [TRACE], CALC to find where the maximum volume would occur.

It occurs when the radius of the cone, x, is 5.657. The maximum volume is 268.08.

To make sure this is correct, you also need to check the maximum volume if the cone had a height shorter than 6. In that case, the height of the cone would be 6 + √(36 - x²). So the volume formula would be:

V = 1/3Pi*r²h

y = 1/3Pi*x²(6 - √(36 - x²))

If this is entered in your calculator with the appropriate window, you can use the MAXIMUM command under 2nd [TRACE], CALC to find where the maximum volume would occur.

It occurs when the radius of the cone, x, is 6. Then, the maximum volume is 226.19. Since this volume is less than the previous maximum volume, the answer remains x = 5.657 to form a cone with a maximum volume is 268.08.

I hope that helps!! :-)

Answer 2

Let me see if my approach would be easier.
You did not specify which variable (V,h,r) should be x. Thus I can choose an appropriate x.
First of all, I declair that corresponding to a single r (r <> radius 6), there are two cones can be inscribed in this sphere of radius 6, one h < the radius and the other h > the radius. since we are interested in the maximum volume, we are deducing a function V = V(h, r), such that h > 6, in order to have V maximized.
Very clearly, there is a relation between r and h with Pathagoran's theorem:
r^2 = 6^2 - (h - 6)^2 = 12h - h^2
As you see from the hint, that:
V = (1/3)pi*r^2*h = (pi/3)(12h^2 - h^3)
The domain of the function is 6 <= h <= 12, even though any h value from 0 to 12 would not make V negative.
I am at this point unable to find the max without calculus or approximation. With calculus, the max of V occurs when dV/dh = 0. That is:
24h - 3h^2 = 0 or h = 8.
max V = (pi/3)8^2(12 - 8) = (256/3)*pi ≈ 268.08257

Answer 3

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