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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?

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  • $\begingroup$ @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
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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.

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

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