More Optimization — Question 5

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Question 5

A person is standing at point AA on the shore of a straight river that is 3 km wide. Point BB lies on the opposite shore, 5 km downstream from the point directly opposite AA. The person can run at 10 km/h along the shore and swim at 5 km/h directly across the river or at an angle.

The person first swims from AA to PP, then runs along the opposite shore to BB. At what point PP between the point directly opposite AA and BB should the person land to minimize total time to reach BB?

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Original worksheet page 1: question and worked solution for 4-9-005
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Question 5 - Solution

Let the swimmer land at point PP, which is xx km from the point directly across from AA.

Step 1: Write distances:

Swim distance: x2+32=x2+9\sqrt{x^2 + 3^2} = \sqrt{x^2 + 9} , Run distance: 5−x5 - x

Step 2: Write time expression:

Swimming speed: 5 km/h , Running speed: 10 km/h

T(x)=x2+95+5−x10T(x) = \frac{\sqrt{x^2 + 9}}{5} + \frac{5 - x}{10}

Step 3: Minimize time — take derivative:

T′(x)=x5x2+9−110T'(x) = \frac{x}{5\sqrt{x^2 + 9}} - \frac{1}{10}

Set T′(x)=0T'(x) = 0:

x5x2+9=110⇒2xx2+9=1⇒4x2=x2+9⇒3x2=9⇒x2=3⇒x=3\frac{x}{5\sqrt{x^2 + 9}} = \frac{1}{10} \Rightarrow \frac{2x}{\sqrt{x^2 + 9}} = 1 \Rightarrow 4x^2 = x^2 + 9 \Rightarrow 3x^2 = 9 \Rightarrow x^2 = 3 \Rightarrow x = \sqrt{3}

Step 4: Answer

x=3≈1.732kmTmin=when the person lands 3km from the point directly across from A.\boxed{ \begin{aligned} x &= \sqrt{3} \approx 1.732\ \text{km} \\ T_{\min} &= \text{when the person lands } \sqrt{3}\ \text{km from the point directly across from } A. \end{aligned} }

Original worksheet page 2: question and worked solution for 4-9-005

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