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.