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Calculus Optimization: Cone of Maximal Volume Inscribed in a Sphere

Updated on August 01, 2014
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TR Smith is a product designer and former teacher who uses math in her work every day.

Lots of classical calculus problems involve finding the optimal dimensions of a shape inscribed in a sphere so that the shape has the largest possible volume. One of the easiest examples of such a problem is finding the dimensions of a cylinder inscribed within a sphere. Here we tackle a very similar problem, finding the maximum volume cone that can fit inside a sphere. Though a cone has two independent measurements, base radius and height, the constraint that it must fit inside a unit sphere (sphere of radius 1) lets us transform a two-variable calculus problem into a simpler single-variable calculus problem.

Step 1: Transforming a 2-Variable Equation Into a 1-Variable Equation

The formula for the volume of a cone is V(r, h) = (π/3)hr^2. Here, the function V is a function of r and h. But when a cone must fit exactly inside a unit sphere, we can discover a relation between r and h that allows us to rewrite r as a function of h, or h as a function of r. The figure below shows how we can relate r and h for a cone inscribed in a unit sphere.

When a cone is inscribed in a unit sphere, the relation between h and r is

(h-1)^2 + r^2 = 1

which can be simplified to

h^2 - 2h + 1 + r^2 = 1
h^2 - 2h + r^2 = 0
r^2 = 2h - h^2
r = sqrt(2h - h^2)

Now we can rewrite the cone volume formula solely in terms of h:

V(r, h) = (π/3)hr^2
V(h) = (π/3)h*[sqrt(2h - h^2)]^2
= (π/3)h(2h - h^2)
= (π/3)(2h^2 - h^3)

Step 2: Taking the Derivative, Finding h and r

The last step is to take the derivative of the volume function V(h), set the derivative expression equal to zero, solve for h, and then use our value of h to back solve for r. Doing so gives us

V'(h) = (π/3)(4h - 3h^2)


(π/3)(4h - 3h^2) = 0
4h - 3h^2 = 0
4h = 3h^2
4 = 3h
h = 4/3

So the cone with a maximal volume that can be inscribed within a unit circle has a height of 4/3. To figure out the value of r, we use the expression r = sqrt(2h - h^2) that we found in Step 1. Plugging h = 4/3 into this expression gives us

r = sqrt(8/3 - 16/9)
= sqrt(8/9)
= 2*sqrt(2)/3

Finally, we can write the solution as h = 4/3 and r = sqrt(8/9). The volume of this cone works out to be

V = (π/3)(4/3)(sqrt(8/9))^2
= 32π/81
≈ 1.241123

The volume of a unit sphere is 4π/3, which means the cone of maximal volume takes up 8/27 or about 29.63% of the unit sphere's volume.

Graph of V(h) = (π/3)(2h^2 - h^3).  As you can see from the volume graph, the value h = 4/3 yields the maximum volume.
Graph of V(h) = (π/3)(2h^2 - h^3). As you can see from the volume graph, the value h = 4/3 yields the maximum volume.

Comparison to the Cylinder Problem

In the closely related problem of a finding a maximal volume cylinder inscribed within a unit sphere, the optimal dimensions turned out to be h = 2/sqrt(3) and r = sqrt(2/3), with a total cylindrical volume of V = 4π/[3*sqrt(3)] ≈ 2.4184.

Solving as a 2-Variable Calculus Problem

Instead of transforming this into a single-variable problem, you can solve it as a 2-variable problem using the technique of Lagrange multipliers. This is a technique covered in multivariable calculus or Calc III. With Lagrange multipliers, we call the constraint equation g(r, h) = r^2 - 2h + h^2 = 0 and solve the system of equations

∂V/∂r = λ*∂g/∂r
∂V/∂h = λ*∂g/∂h
g = 0

which is the system

(2π/3)hr = 2λr
(π/3)r^2 = 2λ(1-h)
r^2 - 2h + h^2 = 0

Since it is a system of non-linear equations, you cannot use matrices to solve it, but instead substitution.


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