1. (Calculus Solution)

Let the river bank be the x-axis with the origin placed directly below the fire. In other words, the campfire is at coordinates (0, 300) and the car is at coordinates (1200, 600). We are seeking the x-value which minimizes the path from the campfire to the river and back to the car. For sake of clarity, call the point (x, 0) where we pick up water from the river W. The distance from the campfire to W is Sqrt[x^2 + 90000] and the distance from W to the car is Sqrt[360000 + (1200-x)^2]. The following Mathematica code yields a formula for the total distance.

Input := 

distance[x_] = Sqrt[x^2 + 90000] + Sqrt[360000 + (1200-x)^2];

It's helpful to plot the distance function to see where it is minimized.

Input := 

Plot[distance[x], {x, 0, 1200},
	PlotLabel -> "Distance D vs. Water Pick-up Location X",
	AxesLabel -> {"X [feet]", "D [feet]"}]

Output =

-Graphics-

In order to minimize the distance, we look for critical points by differentiating the distance function and setting it equal to zero.

Input := 

Solve[distance'[x]==0,x]

Output =

{{x -> 400}}

Therefore, W is 400 feet down the river bank from the campfire or equivalently, 800 feet up the river bank from the car.