Question

You are standing at the edge of a slow-moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2 mph and walk at 3 mph. You must first swim across the river to any point on the opposite bank. From there walk to the campground, which is one mile from the point directly across the river from where you start your swim. What route will take the least amount of time?

(Note: Answers should be in decimal form. Up to two decimal places only.)

Answer 1

Time=distance/speed. If T(x) is the time to reach the campground, then it consists of the time taken to swim across the river=√(1+x²)/2 plus the time to walk to the campground=(1-x)/3, so we can write:

T(x)=√(1+x²)/2+(1-x)/3.

To find out how T changes with x we need dT/dx=x/(2√(1+x²))-1/3

When the rate of change is zero we are at a maximum or minimum.

So x/(2√(1+x²))-1/3=0, 3x-2√(1+x²)=0 (the denominator 6√(1+x²) is always positive).

Therefore 2√(1+x²)=3x.

Square each side:

4+4x²=9x², 5x²=4, x=2/√5=2√5/5=0.8944 (0.89mi to 2 dec places).

T(2√5/5)=√1.8/2+0.1056/3=0.7060 (0.71hr to 2 dec places).

Since T(0)=⅚ (0.83hr approx) and T(1)=0.7071hr, we know that T=0.7060 is a minimum value.