Optimizing an Inscribed Cone

A Cone Inside a Sphere

A right circular cone can be inscribed in a sphere by placing its apex on the sphere and its base in a plane that cuts the sphere. The base is then a circle on the sphere’s surface. The cone’s height and base radius cannot vary independently: the sphere fixes their relationship.

That geometric constraint turns the cone’s volume into a function of one variable. We can then use calculus to identify the dimensions that produce the greatest volume, while checking that the candidate is feasible and truly maximizes the volume.

Relating the Cone to the Sphere

Place the sphere’s center at the origin, and orient the vertical axis so the cone’s apex is at height \(R\). The base plane is \(h\) units below the apex, so its signed vertical coordinate is \(R-h\).

A cross-section through the sphere’s center, the cone’s axis, and a radius of the base circle forms a right triangle. Its horizontal leg has length \(r\), its hypotenuse has length \(R\), and its vertical leg has length \(|R-h|\). The absolute value matters: if the base plane passes below the sphere’s center, the signed displacement is negative, but a leg’s length cannot be.

Expanding and simplifying gives the required relationship between the cone’s dimensions:

For a cone with positive height and positive base radius, both \(h>0\) and \(2R-h>0\) must hold. Thus the feasible heights for a nondegenerate cone are \(0<h<2R\). At \(h=0\) or \(h=2R\), the base radius is zero and the cone is degenerate. Those boundary cases are useful when comparing volumes, even though they are not ordinary cones.

Worked Example: A Base Below the Sphere’s Center

Building the One-Variable Volume Function

The volume of a right circular cone is one third of the base area times the height: \(V=\frac{1}{3}\pi r^2h\). The sphere relation expresses \(r^2\) in terms of \(h\), so substitute \(r^2=h(2R-h)\):

The function includes the two degenerate boundary cases, where its value is zero. For actual cones, the domain is the open interval \(0<h<2R\). The closed interval is convenient for comparing the interior candidate with both boundary values.

Worked Example: Evaluating a Feasible Cone

Finding and Verifying the Maximum

To find the greatest volume, differentiate the one-variable function. Here \(R\) is fixed, and only \(h\) varies:

Inside the feasible interval, \(h>0\), so an interior critical point must satisfy \(4R-3h=0\). This gives \(h=\frac{4R}{3}\), which is feasible because \(0<\frac{4R}{3}<2R\) when \(R>0\).

A critical point alone does not establish an absolute maximum. We check the derivative’s sign and the boundary values. For \(0<h<\frac{4R}{3}\), both factors \(h\) and \(4R-3h\) are positive, so \(V'(h)>0\). For \(\frac{4R}{3}<h<2R\), \(h>0\) and \(4R-3h<0\), so \(V'(h)<0\). The volume increases up to the critical height and decreases afterward. Also, \(V(0)=V(2R)=0\), whereas \(V(h)>0\) for every \(0<h<2R\). Therefore this interior candidate gives the absolute maximum on the feasible interval.

Why the Maximum Occurs at Those Dimensions

The result follows from the geometry and the derivative sign, not from guessing a proportion. At the maximizing height, the sphere relation determines the radius:

Since a radius is positive, \(r=\frac{2\sqrt{2}R}{3}\). Substituting the maximizing height and radius into the cone-volume formula gives

These dimensions satisfy the sphere constraint: \(r^2+(R-h)^2=\frac{8R^2}{9}+(-\frac{R}{3})^2=R^2\). Thus the maximizing candidate is not only feasible; its dimensions also reproduce the required maximum volume.

Worked Example: Maximum Volume for a Specific Sphere

Common Pitfall: Calling a Signed Displacement a Length

A frequent error is to describe the vertical leg as having length \(R-h\). That statement fails whenever \(h>R\). For example, if \(h=\frac{3R}{2}\), then \(R-h=-\frac{R}{2}\). A negative number cannot be a leg length.

The correct description is that the signed vertical displacement is \(R-h\), while the vertical leg’s length is \(|R-h|\). The Pythagorean equation uses the square of that length, so it can equivalently be written using the signed displacement squared:

Another mistake is to assume the base must lie above the sphere’s center, restricting \(h\) to \(h\le R\). That omits feasible cones. A base below the center is allowed: the full nondegenerate range is \(0<h<2R\). The absolute value handles the change in direction of the displacement without changing the geometry.

Why the Model Is Useful

This example shows a standard strategy for geometry optimization. First, identify a geometric constraint; then use it to eliminate one dimension from the objective. The resulting one-variable function makes calculus possible, but the model is only complete when its feasible domain and boundary cases are also considered.

Here the sphere constraint determines the base radius for each height. The volume increases while the cone is relatively shallow, then decreases as the height approaches \(2R\), where the base circle shrinks to a point. The maximum occurs between those extremes. Checking feasibility, derivative signs, and boundary values is what turns a critical-point calculation into a justified absolute maximum.