Although this might seem like a homework question, it isn't; I was curious of how to optimize volumes given a contraint. If we are given a certain amount of surface area, what is the largest possible square pyramid (volume-wise) that can be constructed? I have worked out the following:

let surface area=$k$, a constant

length of base=$b$, height of pyramid=$h$, and slant height=$s$

$k=b^2+4(\frac12bs)$ since $b^2$ is the base area and $\frac12bs$ is the triangle's area

$s=\sqrt{(\frac{b}2)^2+h^2}$ in terms of b and h, using Pythagorean Theorem

$k=b^2+2b\sqrt{(\frac{b}2)^2+h^2}$ (substituting s)

$\sqrt{(\frac{k-b^2}{2b})^2-(\frac{b}2)^2}=h$ (solving for h)

$\frac{\sqrt{k^2-2kb^2}}{2b}=h$ (simplified)

Since volume $V=\frac13b^2h$ and we are finding the maximum, do we differentiate $V$ to find a value where $\frac{dV}{db}=0$, or is there another way to do this?

  • $\begingroup$ @RossMillikan Yeah, I just put that in the question (although it was already in the title). $\endgroup$ – ayane_m May 6 '14 at 23:36

You should substitute for $h$ to get: $$V=\dfrac{1}{3}\dfrac{b\sqrt{k^2-2kb^2}}{2}$$ Then differentiate w.r.t to $b$: $$\dfrac{dV}{db}=\dfrac{1}{6}\left(\dfrac{1}{2\sqrt{b^2k^2-2kb^4}}\times(2bk^2-8kb^3)\right)=\dfrac{1}{6}\left(\dfrac{bk}{\sqrt{b^2k^2-2kb^4}}\times(k-4b^2)\right)=0\\ \implies b=\dfrac{\sqrt k}{2}$$ Of course, another way to do this is by trial and error, which is not very effective. The above method is the most effective, and works very well.

| cite | improve this answer | |
  • $\begingroup$ @RossMillikan Thanks - I saw where my mistake was, and I corrected the post. $\endgroup$ – user122283 May 6 '14 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.