Question

Alaina wants to get to the bus stop as quickly as possible. The bus stop is across a grassy park, 2,000 feet east and 600 feet north of her starting position. Alaina can walk along the edge of the park on the sidewalk at a rate of 6 feet/sec. She can also travel through the grass in the park, but only at a rate of 4 feet/sec (dogs are walked there, so she must move with care or get a surprise on her shoes). What path will get her to the bus stop the fastest

Answer 1

Answer:

She should walk approximately 1,463.34369 feet along the sidewalk going to the east, then walk the remaining straight line distance along the grass.

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Explanation:

Refer to the diagram below.

Alaina starts at point A. If she stays on the sidewalk the entire time, then she can go from A to C to D in that order.

If she takes the direct line route, then she would go from A to D across the grass entirely.

The two paths have pros and cons. The optimal solution is a mix of the two routes. This means she'll be on the sidewalk for some amount of time, and then cut along the grass once reaching a certain point.

Let's say point B is the point where she decides to cut along the grass. In the diagram below, I made $\text{BC} = x$ so that I could find $\text{BD} = \sqrt{x^2+600^2}$ through the pythagorean theorem fairly quickly.

If $\text{BC} = x$, then the remaining bit of the east/west side walk is $\text{AB} = 2000-x$ since the two portions must add to 2000 feet.

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If she travels along the side walk at a rate of 6 ft/sec, and travels 2000-x feet (from A to B), then it will take her

time = distance/speed = $\frac{2000-x}{6}$ seconds

Then if she cuts across the grass to follow path BD, then she'll take another

time = distance/speed = $\frac{\sqrt{x^2+600^2}}{4}$ seconds since she's traveling 4 ft/sec here.

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If Alaina goes from A to B to D in that order, then she takes a total of

$\text{(time from A to B)}+\text{(time from B to D)}\\ = \frac{2000-x}{6}+\frac{\sqrt{x^2+600^2}}{4}$

That represents the total time taken when following this path at those specified speeds mentioned.

The goal is to find the minimum of that function. So we'll need to compute the derivative.

$f(x) = \frac{2000-x}{6}+\frac{\sqrt{x^2+600^2}}{4}\\\\ f \ '(x) = -\frac{1}{6}+2x*\frac{1}{2*4\sqrt{x^2+600^2}}\\\\ f \ '(x) = -\frac{1}{6}+\frac{x}{4\sqrt{x^2+600^2}}$

Set the derivative equal to zero and solve for x

$f \ '(x) = 0\\\\ -\frac{1}{6}+\frac{x}{4\sqrt{x^2+600^2}} = 0\\\\ \frac{x}{4\sqrt{x^2+600^2}} = \frac{1}{6}\\\\ 6x = 4\sqrt{x^2+600^2}\\\\ (6x)^2 = \left(4\sqrt{x^2+600^2}\right)^2\\\\ 36x^2 = 16(x^2+600^2)\\\\ 36x^2 = 16x^2+16*600^2$

$36x^2 = 16x^2+5,760,000\\\\ 36x^2-16x^2 = 5,760,000\\\\ 20x^2 = 5,760,000\\\\ x^2 = 5,760,000/20\\\\ x^2 = 288,000\\\\ x = \sqrt{288,000}\\\\x \approx 536.65631$

If you were to perform the first derivative test, you should find that the local min occurs at roughly x = 536.65631

This means she must walk a approximately 2000-x = 2000-536.65631 = 1,463.34369 feet when going from A to B such that to minimize the walking time.

Side note: you can use the minimum feature on your calculator to confirm the answer