What is the value of $\alpha$ for which the volume of the regular rectangular pyramid is greatest?

Solution 1

You have not only $h=a\sin\alpha$, but also $b=\sqrt2 a\cos\alpha$, where $b$ is the side of the base (a regular rectangular pyramid has a square base). Plugging these into the volume formula, we get: $$ V={2\over3}a^3\sin\alpha\cos^2\alpha. $$ To find the maximum $V$ we just need to find the solutions of equation $dV/d\alpha=0$, which boils down to: $$ \cos^2\alpha=2\sin^2\alpha, \quad\hbox{that is:}\quad \tan\alpha={1\over\sqrt2}, \quad \alpha\approx35.26°. $$

Solution 2

The main reason this question looks confusing to me is that choosing $\alpha$ alone does not determine the volume, you also need to optimize $l$ and $b$. As half the diagonal of the base rectangle sits in a right-angeled triangle with $a$ and $h$, the constraint there is $$\frac{1}{4}l^2 + \frac{1}{4}b^2 + h^2 = a^2$$ or in other words $$l^2 + b^2 = 4(1-\sin^2(\alpha))\cdot a^2 = 4\cos^2(\alpha)\cdot a^2.$$ Now you can actually break the optimization problem up into two parts: For fixed $\alpha$, you can choose $l,b$ such that their product $P = P(\alpha)$ is maximal, then you optimize $V =\frac{1}{3} a \cdot P(\alpha)\cdot \sin(\alpha)$.

For the first part, note that $l\cdot b \leq \frac{1}{2} (l^2 + b^2)$ as $l^2 +b^2 - 2l\cdot b = (l-b)^2 \geq 0$; so $l\cdot b$ is maximal if $l = b$, and in this case we have $$l \cdot b = \frac{1}{2} (l^2 + b^2) = 2\cos^2(\alpha)\cdot a =: P(\alpha)$$

So your pyramid has quadratic base and Volume $$V = \frac{2}{3}a^2 \cdot(\cos^2(\alpha)\sin(\alpha)).$$

So you're left with optimizing $$\cos^2(\alpha)\sin(\alpha) = \sin(\alpha) - \sin^3(\alpha)$$ which you can do by taking the derivative w.r.t $\alpha$ and search for zeros.

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user1442

Updated on August 01, 2022