A cone is inscribed in a sphere of radius a, centred at the origin. The

height of the cone is x and the radius of the base of the cone is r, as
shown in the diagram opposite.
Find the height, x, for which the volume of the cone is a maximum.
(HINT: show that the volume of the cone is given by
V=1/3pi(2ax^2-x^3)

To find the height, x, for which the volume of the cone is a maximum, we can use calculus and find the critical points of the volume function.

The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height.

In this case, the radius of the base, r, is not given explicitly. However, we can use the information given in the problem to determine the relationship between r and x.

Notice that when the cone is inscribed in the sphere, the line from the center of the sphere (at the origin) to the top of the cone is also the height of the cone. This means that the radius of the base of the cone is also equal to the perpendicular distance from the origin to the edge of the cone.

Using the Pythagorean theorem, we have r^2 = a^2 - x^2.

Now we can substitute this relationship into the volume formula:
V = (1/3)π(2ax^2 - x^3).

To find the maximum volume, we need to find the critical points of the volume function. We can do this by taking the derivative of V with respect to x and setting it equal to zero:

dV/dx = (2/3)π(4ax - 3x^2).

Setting this derivative equal to zero, we get:
(2/3)π(4ax - 3x^2) = 0.

Simplifying, we have:
4ax - 3x^2 = 0.

Now we can solve this equation for x:
x(4a - 3x) = 0.

This equation gives us two solutions: x = 0 or 4a - 3x = 0.

Since the height, x, cannot be zero for a non-trivial cone, we set 4a - 3x = 0 and solve for x:
4a - 3x = 0,
3x = 4a,
x = (4/3)a.

So the height, x, for which the volume of the cone is a maximum, is x = (4/3)a.