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I recently came up to this exercise where the requested thing to do was to model the problem as a minimization problem:

"A person is at point $0$ and needs to get to point $D$ and it is raining. The point $D$ is $20$ meters away. The person needs to decide their speed $u$, their height $h$ and width $w$ (one can decide if they bow or stand on their toes so that's how height and width of the person changes) to get to the point $D$ so that he/she gets wet as little as possible. Due to physical restrictions the height of the person must be between $0.5$ and $2$ meters. There must also hold that $wh = 0.3$ and the person can't run faster than $10m/s$. The speed of the rain is $u_r=20m/s$."

The solution given was:

$\min Dh+w\frac{Du_r}{u}+\frac{1}{2}w^2\frac{u_r}{u}$

such that:

$wh=0.3$

$0.5 \leq h \leq 2$

$0 \leq u \leq 10$

My question is what exactly is $\frac{1}{2}w^2\frac{u_r}{u}$ in the solution. I think that $Dh$ is the water that you hit with the front side of your body (and face) while walking/running and that $w\frac{Du_r}{u}$ is the water that falls to the top of your head while you are in the rain. Am I correct in these? If yes, what does $\frac{1}{2}w^2\frac{u_r}{u}$ stand for? If no, what is the mistake in my point of view?

Thanks in advance. Maybe this question helps someone trying to answer me.

michalis vazaios
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1 Answers1

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First you need to have a good understanding of the 'width' of a person. It does not mean the shoulder-to-shoulder distance, but it is measured in the direction of running.

At the start and end of the trajectory, part of the body is already (or still) exposed while another part is still (or already) covered. The last term captures that. The last term is valid as long as $w \leq D$.

LinAlg
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