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

Question

The lateral edge of a regular rectangular pyramid is $a$ cm long.The lateral edge makes an angle $\alpha$ with the plane of the base.What is the value of $\alpha$ for which the volume of the regular rectangular pyramid is greatest?

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Since the volume of the rectangular pyramid is $V=\frac{lbh}{3}$,where $l$ is

the length of the base of the pyramid,$b=$width of the base of the pyramid,

$h=$ height of the pyramid.

and $\sin \alpha=\frac{h}{a}$,where $a$ is the lateral edge(as given in the question.)$\Rightarrow h=a\sin\alpha$

So $V=\frac{l\times b \times a\times\sin\alpha}{3}$


What should i do to maximize the volume?

Answer 1

In a constrained problem where surface area or some other item is an implied constraint , volume of a cuboid or regular rectangular pyramid that has a constant multiplier, a product of 3 numbers is maximum when they are all equal $= b= l= h $, it can be proved separately.

EDIT 1/2

When all three dimensions are equal,

$$ \tan \alpha = \dfrac{2h}{ \sqrt{l^2+b^2}} = \sqrt2. $$

Answer 2

To answer the question with respect to the $\alpha$ term.

The way your question reads, especially with the $\sin\alpha=\frac{h}{a}$ part, $l$ and $b$ must be equal as you are implicitly constraining a with l and b. In this case, $l=b=2a$.

Therefore, $$V=\frac{l\times b\times a\sin{\alpha}}{3}=\frac{4a^3\sin{\alpha}}{3}$$ Take the derivative of V w.r.t $\alpha$ to see how the volume changes. Set this equal to zero to get the turning point, then solve for $\alpha$.

$$\frac{d}{d\alpha}V=\frac{4a^3\cos{\alpha}}{3}=0$$

$${4a^3\cos{\alpha}}=0$$

The times which this will equal zero are $\alpha=\frac{\pi}{2},\frac{3}{4}\pi$. Well, you can constrain that $0<\alpha<\frac{\pi}{2}$. Unless you would like a 2d shape($\alpha=0$), a cube ($\alpha=\frac{\pi}{2}$) or something else which isn't a pyramid ($\alpha>\frac{\pi}{2}$) .

If we graph the original equation and arbitrarily select a value of $a=10$, then you can tell that the maximum volume will occur as $\alpha$ approaches but never equals $\frac{pi}{2}$. If $\alpha=\frac{pi}{2}$ you have vertical sides, so the sides will not converge to a single point at the top of the structure.

Summary:

$$V_{MAX}= \lim \limits_{\alpha \to \frac{pi}{2}} \frac{4a^3\sin{\alpha}}{3}$$ where $0<\alpha<\frac{\pi}{2}$.

You can reach this point without anything other than your initial, $\alpha$ dependent equation and some thought. You have a multiplier and a sinusoid factor. The sinusoid is at maximum when at certain values, but some of those values are not valid as they are outside of the permitted angles. The only one which is close to valid is the $\frac{\pi}{2}$ term, but that gives a cuboid, so the answer is to get as close as possible to it.

Answer 3

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: $$ \cos^2\alpha=2\sin^2\alpha, \quad\hbox{that is:}\quad \tan\alpha={1\over\sqrt2}, \quad \alpha\approx35.26°. $$

Answer 4

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.