# Optimizing a Conical Frustum Using Partial Differentiation

I am working on minimizing the surface area of a frustum with a known volume, however, the equations for the volume and surface area (listed below) have three variables. I am trying to use partial differentiation to optimize surface area with volume as a constraint, yet, differentiating with respect to any variable results in very messy work that can't be solved for a variable. Is there a better way to do it, or can anyone please show me how to do it using partial differentiation?

$$V=1/3(πh)(r^2+rR+R^2)$$

$$S.A=π(r+R)(\sqrt{(r-R)^2+h^2}+πr^2+πR^2$$

Thank you.

Define the Lagrangian function

$$\Lambda(r,R,h;\lambda):=\pi(r+R) \sqrt{(r-R)^2+h^2}+\pi r^2+\pi R^2 -\lambda(V_0-(\pi h) (r^2+rR+R^2)/3),$$

with $$V_0$$ denoting the allocated volume.

The partial derivatives are

\begin{align}\Lambda_r&=\pi \sqrt{h^2+(r-R)^2}+\frac{\pi (r-R) (r+R)}{\sqrt{h^2+(r-R)^2}}+\frac{1}{3} \pi h \lambda (2 r+R)+2 \pi r, \\ \Lambda_R&=\frac{\pi \left(R^2-r^2\right)}{\sqrt{h^2+(r-R)^2}}+\pi \sqrt{h^2+(r-R)^2}+\frac{1}{3} \pi h \lambda (r+2 R)+2 \pi R, \\ \Lambda_h &=\frac{\pi h (r+R)}{\sqrt{h^2+(r-R)^2}}+\frac{1}{3} \pi \lambda \left(r^2+r R+R^2\right), \\ \Lambda_\lambda&=\frac{1}{3} \pi h \left(r^2+r R+R^2\right)-V_0. \end{align}

Proceed with a change of variables, motivated by the square root in the above \begin{align} h &= \rho \cos \theta, \\ r &= \frac{s+\rho \sin \theta}{2}, \\ R &= \frac{s-\rho \sin \theta}{2}. \end{align}

Setting the partial derivates above equal to zero, and applying the substitution gives

\begin{align} 0&=\frac{1}{6} \pi \lambda \rho \cos (\theta ) (\rho \sin (\theta )+3 s)+\pi (\sin (\theta )+1) (\rho +s),\\ 0&=-\frac{1}{6} \pi \lambda \rho \cos (\theta ) (\rho \sin (\theta )-3 s)-\pi (\sin (\theta )-1) (\rho +s), \\ 0&=\frac{1}{24} \pi \lambda \left(-\rho ^2 \cos (2 \theta )+\rho ^2+6 s^2\right)+\pi s \cos (\theta ),\\ 0 &=\frac{1}{12} \pi \rho \cos (\theta ) \left(\rho ^2 \sin ^2(\theta )+3 s^2\right)-V_0. \end{align}

The first three equations can be written in linear form as

$$\left( \begin{array}{ccc} \pi (\sin (\theta )+1) & \pi (\sin (\theta )+1) & \frac{1}{6} \pi \rho \cos (\theta ) (\rho \sin (\theta )+3 s) \\ \pi (1-\sin (\theta )) & \pi (1-\sin (\theta )) & \frac{1}{6} \pi \rho \cos (\theta ) (3 s-\rho \sin (\theta )) \\ 0 & \pi \cos (\theta ) & \frac{1}{24} \pi \left(-\rho ^2 \cos (2 \theta )+\rho ^2+6 s^2\right) \\ \end{array} \right) \begin{pmatrix} \rho \\ s \\ \lambda \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}.$$

If the matrix in the above is invertible, we must have $$\rho=0$$, which implies a planar frustum ($$h=0$$). Hence, we can assume that the matrix is singular, meaning its determinant $$\frac{1}{3} \pi ^3 \rho \sin (\theta ) \cos ^2(\theta ) (\rho -3 s)=0.$$

This vanishing determinant condition reduces the search to two seemingly admissible cases:

1. $$\rho=3s \implies h =2 \sqrt{(2 r+R) (r+2 R)}$$
2. $$\rho \sin \theta =0 \implies r=R$$

Case 1:

The first three equations, upon divisions allowed by $$s>0, \cos \theta \neq 0$$ become $$\frac{1}{2} \pi (3 \lambda s \cos (\theta )+8) =0, \\\frac{1}{2} \pi (3 \lambda s \cos (\theta )+8)=0, \\\pi \cos (\theta )+\frac{1}{8} \pi \lambda s (5-3 \cos (2 \theta ))=0.$$ It takes a little work to deduce from this that $$\sin^2 \theta = \frac{1}{9}$$, which gives $$r=0$$ or $$R=0$$. This renders Case 1 inadmissible.

Case 2:

Let us return to the Cartesian form of the system, assuming that $$r=R$$:

$$\pi (h \lambda r+h+2 r)=0, \\ \pi (h \lambda r+h+2 r)=0, \\\pi r (\lambda r+2)=0, \\\pi h r^2-V_0 .$$

Since $$r>0$$ the third equation immediately gives $$\lambda = -2/r$$. It follows from there that $$h=2r$$ and finally that $$r=\sqrt[3]{\frac{V_0}{2 \pi }}.$$

Since there is a unique critical point, it has to be the minimum. That is, the frustum of volume $$V_0$$, of least surface area is a cylinder with radii $$R=r=\sqrt[3]{\frac{V_0}{2 \pi }}$$ and height $$h=2\sqrt[3]{\frac{V_0}{2 \pi }}$$.

Remark: This is somewhat expected based on the iso-perimetric inequality.

• @GlennBoll That difference is equivalent to taking $\lambda \mapsto - \lambda$. It doesn't affect the solution. Feb 14, 2021 at 1:28