Optimization problems: Finding the optimal path I'm still trying to get the hang of optimization problems in calculus and I'm looking for a little help. I'm having trouble finding equations to model the following problem: I'm fairly sure I need to combine two variables into a single variable and set up a related rates problem, but I'm not sure how to get started.

The problem is as follows: Alaina wants to get to the bus stop as quickly as possible. The bus stop is across a grassy park, 2000 feet west and 600 feet north of her starting position. She can walk west along the edge of the park on the sidewalk at a speed of 6 ft/s. She can also travel through the grass in the park, but only at a rate of 4 ft/s. What path will get her to the bus stop the fastest?

Could someone help set it up for me? Thanks!
EDIT

To solve your question:

The path you should take is a combination of both traveling on the sidewalk and through the grass.

Let's start:

$$T(x) = \frac{\text{Distance through grass}}{\text{Rate through grass}} + \frac{\text{Distance on sidewalk}}{\text{Rate on sidewalk}}$$

$$T(x) = \frac{2000-x}{6} + \frac{\text{Distance through grass}}{4}$$

And 

$$600^2 + x^2 = (\text{Distance through grass})^2$$

$$\text{Distance through grass} = \sqrt{360000 + x^2}$$

$$T(x) = \frac{2000-x}{6} + \frac{\sqrt{360000 + x^2}}{4}$$

$$T'(x) = \frac{-1}{6} + \frac{x}{4\sqrt{360000 + x^2}} = 0$$

Solving, we get $x = 240\sqrt{5}$

And our distances are:

$$\text{Through Grass} = 360\sqrt{5}$$
$$\text{On sidewalk} = 2000 - 240\sqrt{5}$$

Comment if you need further explanation.

Also take a look at this, it might help.
Compare the travel time along the two sides of the right triangle (2000 ft+600 ft @ speed of 6 ft/s) versus the travel time along the hypotenuse of the right triangle (2088.06 ft @speed of 4 ft/s).

It is faster to travel along the edge of the park on the sidewalk, besides being grass-friendly!