# Drawing the region which the pedestrian can cover in 1 hour.

A straight path separates a meadow from a field.

A pedestrian travels

• along the path at a speed of 5 km/hr,
• through the meadow at a speed of 4 km/hr,
• and through the field at a speed of 3 km/hr.

Initially, the pedestrian is on the path.

Draw the region which the pedestrian can cover in 1 hour.

I think this is equivalent to drawing an inequality through it seems to be difficult since I didn't really understand the problem well. If anyone can clarify it to me it will be greatly helpful

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I think you are to assume that the pedestrian moves as quickly as possible on the path for a while, then turns into the meadow or field and moves as quickly as possible in a straight line until the hour is up. This includes the cases where he stays on the path or gets off the path immediately.

In the diagram below, the $x$-axis is the straight path, the half-plane $y>0$ above that is the meadow, the half-plane $y<0$ below that is the field, the origin is the starting position of the pedestrian, and the scale in both $x$ and $y$ is in kilometers.

If the pedestrian stays on the path, he can go as far as $-5$ or $5$ on the $x$-axis. If however he immediately moves into the meadow he can go as far as the semicircle of radius $4$ centered at the origin, and if he immediately moves into the field he can go as far as the semicircle of radius $3$ centered at the origin.

This is more tricky, but if he stays on the path for a while then moves off into the meadow or field, he moves to a semicircle off-center from the origin. The envelope of these semicircles is a line segment from the end of the path at $(\pm 5, 0)$ tangent to the appropriate semicircle from the origin. (I showed this not with pure math but by using the Trace feature in Geogebra.)

If the pedestrian does not move in straight paths, he stays somewhere within the boundaries I have outlined. Therefore, the total region the pedestrian can cover is within the red boundary in the diagram above.

If you are to "draw" the region rather than sketch it, you probably need to find the points on the semicircles at the endpoints of the tangent segments. Ask if you need help with that calculation or with drawing the diagram in Geogebra.

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Here I just complement Rory Daulton's answer with formulas.

Let's focus on top part of image: path & meadow.

And consider $x\ge 0$ due to symmetry.

For each $x\le 5$ we need to find maximal value of $y$ that can be attained by pedestrian.

By intuition, optimal way is built of $2$ straight parts: $OA$ on path, and $AC$ on meadow ($|OA|\le |OB|$).

If pedestrian goes $t$ hours by path, then $(1-t)$ hours by meadow. $(0\le t \le 1)$.

$|OA| = 5t$, $~~~|AC| = 4(1-t)$, $~~~ |AB|=x-5t$.

We need to find such $t\in[0,x/5]$, that maximizes value $y=|BC|=\sqrt{AC^2-AB^2}$, or (simplier) maximizes value $$y^2 = AC^2-AB^2= 16(t-1)^2-(5t-x)^2.$$

So, for each (fixed) value $x\in[0,5]$ we consider function $$f(t) = 16(t-1)^2-(5t-x)^2 \rightarrow \max, \qquad (t\in[0,x/5]).$$

There exists one point of extrema: point $t$, where $f'(t)=0$.

$$f'(t) = 32(t-1)-10(5t-x) =0,$$ $$t_0 = \dfrac{5x-16}{9}.$$

To find maximal value, we compare $3$ values of $f(t)$ (on the bounds, and in $t_0$, if it exists here):

• when $t=0$;
• when $t=t_0$, if $t_0\in[0,x/5]$;
• when $t=x/5$.

A). if $t=0$, then $f(t) = f(0) = 16-x^2$ (red part on the image below);

B). if $t=t_0 = \dfrac{5x-16}{9}$, possible only for $x\in[16/5,~ 5]$, then $f(t) = 16\cdot \dfrac{25}{81}(x-5)^2 - \dfrac{256}{81}(x-5)^2 = \dfrac{16}{9}(x-5)^2$; (blue part on the image below);

C). if $t=x/5$, then $f(t)=f(x) = 16(1-x/5)^2 = \dfrac{16}{25}(5-x)^2$; (green part).

To get correct region bounds, we need to choose higher of $3$ lines. After that, we'll get image exactly as one drawn by Rory Daulton before.

(First, I had some doubts to Rory's image, so decided to check it via formulas. Formulas show that all were drawn correctly).

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Nicely done, and more rigorous than my answer! Before I posted my answer, I worked briefly on a proof that the region was the convex hull of the ending points of leaving the path immediately (the upper semicircle of radius 4 and the lower semicircle of radius 3) and staying on the path ($\pm 5$ on the $x$-axis). I have little experience with convex hulls and less with envelopes, so I didn't finish the proof. Your answer teaches me about envelopes. I still wonder about the convex hull, though. –  Rory Daulton Aug 18 '14 at 0:12