2022-07-23

Probability with elements of geometry
We have a wire of length 20. We bend this wire in random point. And then bend again to get rectangual frame. What's the probability that area of this rectangle is less than 21.

Reinfarktq6

Expert

Step 1
Let $X\sim \mathsf{U}\left(0,20\right)$ be the point at which we first bend the pole. Then the height of the rectangle is $H=X\wedge 20-X$ and the width of the rectangle is $W=2\left(10-H\right)$, so the area of the rectangle is $A=WH=2H\left(10-H\right)$. The distribution of H is U(0,10) by symmetry. To compute $\mathbb{P}\left(A⩽21\right)=\mathbb{P}\left(2H\left(10-H\right)⩽21\right)=\mathbb{P}\left(H\left(10-H\right)⩽\frac{21}{2}\right),$ consider the function $a\left(h\right)=h\left(10-h\right)$. We have ${a}^{\prime }\left(h\right)=10-2h$, so the maximum of a is 25, attained at $h=5$, with a′ negative on $\left(0,5\right)\cup \left(5,10\right)$. The equation $h\left(10-h\right)=\frac{21}{2}$ has solutions $h=5±\frac{\sqrt{58}}{2}$, so
$h\left(10-h\right)⩽\frac{21}{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}h\in \left[0,5-\frac{\sqrt{58}}{2}\right]\cup \left[5+\frac{\sqrt{58}}{2},10\right].$
Step 2
By symmetry we have $\begin{array}{rl}\mathbb{P}\left(H\left(10-H\right)⩽\frac{21}{2}& =2\mathbb{P}\left(H⩽5-\frac{\sqrt{58}}{2}\right)\\ & =2\cdot \frac{1}{10}\left(5-\frac{\sqrt{58}}{2}\right)\\ & =1-\frac{\sqrt{58}}{10}.\end{array}$

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