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I have no clue how to do this textbook question: An environmental artist is planning to construct a rectangle with 36m of fencing as part of an outdoor installation. If the length of the rectangle is a randomly chosen integral number of $L$ metres, what is the expected area of this enclosure?

imranfat
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Noah Woodworth
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    I'm guessing the possible side lengths are all equally likely? Also, is the rectangle with dimensions $18 \times 0$ allowed? – Theo Bendit Jun 05 '18 at 23:12

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Let $L$ denote the length of the rectangle, and $W$ the width.

We know that $2L+2W=36$, so $L+W=18$, i.e. $W=18-L$. The rectangle's area is $A=LW=L(18-L)$.

Since $L$ is known to be an integer, and $W,L$ must both be nonnegative, then $L$ is chosen from $$\{0,1,2,3,\ldots, 18\}$$

Now, computed the expected value of $A$, as $L$ varies among its possible values. Do you need help with this part?

Adding more details per request. The desired value is $$\sum_{L}APr(A)=\sum_L L(18-L)\frac{1}{19}=\frac{1}{19}\sum_{L=0}^{18} 18L - L^2=$$ $$=\frac{18}{19}\sum_{L=0}^{18}L - \frac{1}{19}\sum_{L=0}^{18}L^2$$

These last sums can be computed using Faulhaber's formulas.

vadim123
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  • I understand what you have written, but do I use a binomial distribution now? If so, how do I apply it? – Noah Woodworth Jun 05 '18 at 23:18
  • Do you know how to calculate $E(L^2)$? Either in principle, or actually doing it by hand? – ConMan Jun 05 '18 at 23:34
  • @ConMan Unfortunately, I do not. – Noah Woodworth Jun 05 '18 at 23:39
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    The distribution of $L$ is not stated in the question, but it seems fair to assume that each possible value of $L$ is equally likely (so $P(L) = 1/19$ for each possible $L$ if $L$ can be any value in ${0,\ldots,18}$).

    For computing the expected value you can just apply the definition directly.

     – arkeet Jun 05 '18 at 23:47
  • Great answer (+1). It is also useful to point out that an $18 \times 0$ "rectangle" might not be considered a genuine rectangle, and might therefore be counted out of consideration. Under that interpretation, the expected area will be $19 / 17$ times the above calculation. – Ben Jun 06 '18 at 00:59