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Minimizing travel time via critical point analysis

A person is in a boat at a point on a lake, located 2 miles from the nearest point AA on a straight shoreline. A cabin is located on the shoreline 8 miles from AA. The person plans to row in a straight line to some point on the shore located a distance xx from AA (with 0≤x≤80 \le x \le 8), and then walk the rest of the way to the cabin along the shore. The person rows at 3 mph and walks at 5 mph. The total travel time (in hours) as a function of xx is given by T(x)=x2+43+8−x5.T(x) = \frac{\sqrt{x^2+4}}{3} + \frac{8-x}{5}.

Find the landing point xx (in miles from AA) that minimizes the total travel time T(x)=x2+43+8−x5T(x) = \dfrac{\sqrt{x^2+4}}{3} + \dfrac{8-x}{5}, where 0≤x≤80 \le x \le 8. State the value of xx that gives the minimum travel time.

Hint

Differentiate T(x)T(x), set T′(x)=0T'(x) = 0, and solve for xx. Then confirm it's a minimum by comparing to the endpoint values T(0)T(0) and T(8)T(8), or by using the first derivative test.

Step-by-step solution

Differentiate: T′(x)=x3x2+4−15T'(x) = \dfrac{x}{3\sqrt{x^2+4}} - \dfrac{1}{5}. Setting T′(x)=0T'(x)=0 gives x3x2+4=15\dfrac{x}{3\sqrt{x^2+4}} = \dfrac{1}{5}, so 5x=3x2+45x = 3\sqrt{x^2+4}.

Squaring both sides: 25x2=9(x2+4)25x^2 = 9(x^2+4), giving 16x2=3616x^2 = 36, so x2=9/4x^2 = 9/4 and x=3/2x = 3/2 (taking the positive root since x≥0x \ge 0). Checking T′T' changes from negative to positive around x=3/2x=3/2 confirms this is a minimum, and it lies within [0,8][0,8], so it is the answer.

Minimizing travel time via critical point analysis — solution graph

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