Day 14: Return to the Factory

We can use a double integral to solve this: $\sqrt{\frac{\int^1_0 \int^1_0 (x^2+y^2)\ dx\ dy}{1 \times 1}} = \sqrt{\frac23} = 0.8164\dots$ but this requires knowledge of double integrals, so here I will present a solution that only requires single integrals.

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We only need to consider one quarter of the square as the situation is perfectly symmetric. Consider the top left quadrant, and let the factory be the origin $(0,0)$ .

Note that the quadratic mean is given by: $\sqrt{\dfrac{l_1^2+l_2^2+\cdots+l_n^2}{n}}$ Basically we want to find the square root of the (arithmetic) mean of the squares of the lengths. Note that $l^2 = x^2 + y^2$ by Pythagoras' Theorem. So the above expression is equal to: $\sqrt{\dfrac{x_1^2+x_2^2+\cdots+x_n^2+y_1^2+y_2^2+\cdots+y_n^2}{n}}$ Now what we want to do is to limit $n$ to infinity as we take all values of $x$ and $y$ from $0$ to $1$ .

As $x$ and $y$ are perfectly symmetric we can consider them separately.

Now the mean of the $x^2$ terms is precisely: $\frac{\int_0^1 x^2 dx}{1} = \left[ \frac{x^3}{3}\right] = \frac13$

In a similar way the mean of the $y^2$ terms is $\frac13$ too.

Substituting in shows us that the final answer is $\sqrt{\frac23} = 0.8164\dots$ and this gives the answer $\boxed{816}$ as required.