Median of Square's Area

Shaded areas represent possible values for (L, W). If you can find a curve LW = A bisects the area of this square, then A is the required median. Calculate the shadow area of the upper half, we come to this formula. The final solution A = 99.6773

We will determine the probability function $F_A(a) = \mathbb P(A < a)$ . The median is then the value $m$ for which $F_A(m) = 0.5$ .

The distributions for $W$ and $L$ have probability function $F_{W,L}(x) = \begin{cases} 0 & x < 9 \\ \tfrac12(x-9) & 9 \leq x \leq 11 \\ 1 & x > 11\end{cases} \ \ \ \ \ f_{W,L}(x) = F'_{W,L}(x) = \begin{cases} 0 & x < 9 \\ \tfrac12 & 9 \leq x \leq 11 \\ 0 & x > 11 \end{cases}$ In principle, $F_A(a) = \mathbb P(A < a) = \iint_{\ell w = a} f_W(w)f_L(\ell)\:dw\:d\ell = \int_9^{11} \tfrac12 F_L(a/w)\:dw.$ If $a < 81$ , we have $a/w < 9$ on the entire integration interval, so that $F_A(a) = 0$ .

If $a > 121$ , we have $a/w > 11$ on the entire integration interval, so that $F_A(a) = \int_9^{11} \tfrac12\cdot 1\:dw = 1$ .

If $81 \leq a \leq 99$ , then $a/w < 11$ on the entire integration interval, but $a/w < 9$ for $w > a/9$ . Therefore we have $F_A(a) = \int_9^{a/9} \tfrac14\left(\frac a w - 9\right)\:dw = \tfrac14\left[a\ln w - 9w\right]_9^{a/9} = \tfrac14\left(a\ln a - 2a \ln 9 + 81 - a \right).$

If $99 \leq a \leq 121$ , then $a/w > 9$ on the entire integration interval, but $a/w > 11$ for $w < a/11$ . Therefore we have $F_A(a) = 1 - \int_{a/11}^{11} \tfrac14\left(11 - \frac a w\right)\:dw = 1 - \tfrac14\left[11w - a\ln w\right]_{a/11}^{11} = \tfrac14\left(2a\ln 11 - a\ln a + 117 - a \right).$

Thus the probability function for the area is $F_A(a) = \begin{cases} 0 & a < 81 \\ \tfrac14\left(a\ln a - 2a \ln 9 + 81 - a\right) & 81 \leq a \leq 99 \\ \tfrac14\left(2a\ln 11 - a\ln a + 117 - a\right) & 99 \leq a \leq 121 \\ 1 & a > 121\end{cases}$ At the transition point $F_A(99) \approx 0.466600$ ; the median is therefore greater than 99, and we solve numerically $\tfrac14\left(2a\ln 11 - a\ln a + 117 - a \right) = 0.5\ \ \ 2a\ln 11 - a\ln a + a = 119\ \ \ \therefore\ \ \ m = a \approx \boxed{99.6773}.$

Graphs of the probability function $F_A$ and probability density function $f_A = F'_A$ :

Mean (81,121) = 101 this is wrong answer .. but the app allows it

For each value, we have 3 possible values 9, 10 and 11 As such, the possible outcomes for A are 81 90 99 100 110 121. everything but the squares occur 2 times {90, 99, 110} while the squares occur 1 time {81, 100, 121}.

Therefore the complete set looks like this:

{81, 90, 90, 99, 99, 100, 110, 110, 121} which includes

1 2 3 4 5 6 7 8 9

The result should, therefore, be 99. Even if I made a mistake, the only possible answers could ever be 99 or 99,5 as in (99+100)/2. Whatever was calculated was NOT the mean! It probably was the average, but I'm too lazy to check. Just as I am too lazy to format this here, even if it wouldn't be hard in Latex...

In a spreadsheet, 21 rows labeled 9.025 to 10.975 in .05 steps. Same labels for 21 columns. In the grid between them, value equals row label times column label. In a separate cell, put the median function over then entire grid. Answer is 99.68, within the acceptable accuracy of the answer. (If more accuracy is needed, use more steps.)

You may have noticed, my answers tend to be more practical than theoretical.