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You are given a straight stick of length 21.97 cm. You break the stick at a position chosen uniformly at random along its length. Each of the two stick portions you break in half and make a rectangle with the four bits of the stick. What is the expected area of the rectangle?

If I call the first point the stick is broken at $X$ then I have obtained an expression for the area of the rectangle: $\frac{21.97X-X^2}4$. But I am not able to understand how to work out the value of $X$. I have no idea how to go about finding the area of the rectangle made by the four sticks.

Any help would be much appreciated.

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    $\begingroup$ You cannot find the value of $X$, you will only find it's expected value. So, your expression seems correct. Just take the expectation. $\endgroup$
    – Jimmy R.
    Commented Oct 17, 2016 at 13:10

2 Answers 2

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For the sake of generality, call the stick length $L$ units instead of fixing it at 21.97. Then the area of the rectangle if the stick is first broken at a point $x$ units from an end is $\frac14x(L-x)$.

The expected area of the rectangle is then the average of this function over $[0,L]$: $$\begin{align} E[A]&=\frac1{L-0}\int_0^L\frac14x(L-x)\ dx\\ &=\frac1{4L}\int_0^L(Lx-x^2)\ dx\\ &=\frac1{4L}\left[\frac12Lx^2-\frac13x^3\right]_0^L\\ &=\frac1{4L}\left(\frac12L^3-\frac13L^3\right)\\ &=\frac1{4L}\cdot\frac{L^3}6\\ &=\frac{L^2}{24} \end{align}$$ In particular, if $L=21.97\text{ cm}$ as in the question, the expected area is $\frac{21.97^2}{24}=20.1117\dots\text{ cm}^2$.

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Suppose the rod is long 1. Let $X$ be the length of the initial stick when you break the rod at random in position $X$.When you divide each of the resulting sticks again, you get a rectangle with sides $X/2$ and $(1-X)/2$. The area will be $A = X(1-X)/2$. Since $X$ has a uniform distribution, the expected value of the area is $$E(A) = E \left[ \frac{X-X^2}{4} \right] = \frac{1}{8} - \frac{1}{12} = \frac{1}{24}$$ If the rod is long $1 \times 21.97$, the same reasoning gives you $A = 21.97^2 X (1-X)/2$ (this is slightly different from the formula in your question) and the expected value of the area is $21.97^2 \cdot (1/24)$.

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