I have worked out how to find the optimum surface area dimensions for a full cone with a set volume, but am unsure how to go about this when the cone is truncated (bottom pointed end cut off)
How do I find the optimum dimensions (radius and height) of a frustum cone with a set volume? I've derived the below volume and surface area formulas for a frustum cone.
$$V = \frac{1}{3}\pi h(R^2 + Rr + r^2)$$
$$SA = \pi(R + r)[(R-r)^2 + h^2]^{1/2} $$
I then found what h was using the first volume formula:
$$h = \frac{3V}{\pi(R^2 + Rr + r^2)}$$
Then I subbed h into the SA formula such that:
$$SA = \pi(R + r)\left[(R-r)^2 + \frac{9V^2}{\pi^2(R^2 + Rr + r^2)^2}\right]^{1/2} $$
I think that I am supposed to use differential calculus now to find the critical values/local minimum but am not sure how to do this because there are two different radii values?
Best Answer
For $R=r$ you have $SA=2V/R$, which has no minimum as it tends to $0$ for $R\to\infty$.
Notice however that your formula does NOT include the area of the bases.