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Problem

The stiffness of a wooden beam of rectangular cross section is proportional to the product of the width and the cube of the depth of the cross section. Find the width and depth of the stiffest beam that can be cut out of a circular log of radius R. stiffness is $cxy^3$ where $c>0$ and is constant.

Attempt to solve

Since stiffness is proportion to the product of the width and the cube of the depth of the cross section we know that. We can mark stiffness with variable $s$

$$ s=cxy^3 $$ where $c>0$.

Since i want to find out what is stiffest possible beam. Which is when $s$ is in it's maximum value. I want to find out a formula that describes the relation between $x$ ,$y$ and the radius of a circle. $s=cxy^3$ only describes relation between $x$ and $y$.

I drew image of the problem.

enter image description here

Now based on the image we can see that number of things are true based on the geometry of this.

$$ \sin(\alpha)=\frac{\frac{1}{2}x}{R} $$ $$ \sin(\beta)=\frac{\frac{1}{2}y}{R} $$

We can solve these equations for $x$ and $y$ since we wanted to know maximum height and depth.

$$ x=2R\sin\alpha $$ $$ y=2R\sin\beta $$

We can also notice from the image that $$ \alpha + \beta = \frac{1}{2}\pi\quad(radian) $$ $$ \alpha=\frac{1}{2}\pi-\beta $$ $$ \beta=\frac{1}{2}\pi-\alpha $$ is true. However we don't know values for $\alpha$ and $\beta$.

$$ cxy^3=c(2R\sin(\alpha))(2R\sin(\beta))^3 $$ division both sides by c. $$ xy^3=(2R\sin(\alpha))(2R\sin(\beta))^3$$ since $\beta=\frac{1}{2}\pi-\alpha$ $$ xy^3=(2R\sin(\alpha))(2R\sin(\frac{1}{2}\pi-\beta)) $$ since $\sin(\frac{1}{2}pi-\alpha)=\cos(\alpha)$ $$ xy^3=16R^4\sin(\alpha)\cdot \cos(\alpha) $$ Now i could solve for y and x but problem is i don't know angles for $\alpha$ and $\beta$ ? I cannot define $x$ or $y$ when these angles are unknown.

Now if someone can point out what went wrong or if i am missing something. That would be highly appreciated.

Tuki
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  • Along with the supplied answer you could have also written things in terms of the bounding circle which relates x and y with each other as $(x/2)^2 + (y/2)^2= R^2$ – Triatticus Nov 20 '17 at 18:09
  • I just noticed a severe error in the post: You dumped the exponent in $\sin^3(\beta)$. My answer has been edited to reflect that, assuming you meant $$xy^3 = 16R^4 \sin\alpha \cos^3\alpha$$ – AlexR Nov 20 '17 at 18:11

1 Answers1

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After fixing the mentioned error you have turned the original problem into

$$\max s = \max c \cdot 16 R^4 \sin\alpha \cos^3\alpha \simeq \max_{\alpha\in[0,\frac\pi2)} \sin\alpha \cdot \cos^3 \alpha$$

(omiting all things constant). This is proportional to $s$, so you've successfully reduced a 2D problem to 1D.

What remains is to find this maximum with standard analysis techniques ($\frac{\mathrm d}{\mathrm d\alpha}$ will help)

Note that the boundary values ($\alpha\in\{0,\frac\pi2\}$) give a stiffness of $0$ so the maximum will be within the interval. Solve

$$\frac{\mathrm d}{\mathrm d\alpha} (\sin\alpha \cdot \cos^3\alpha) = 0\qquad \alpha\in (0,\frac\pi2)$$

Doing this you will wind up with $\alpha = \frac\pi6$ which you can use to compute $x$ and $y$ respectively with $x = 2R\sin\alpha, y = 2R\cos\alpha$

AlexR
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