Chapter 2, Problem 30P

Minimizing Time A man stands at a point A on the bank of a straight river, 2 mi wide. To reach point B, 7 mi downstream on the opposite bank, he first rows his boat to point P on the opposite bank and then walks the remaining distance x to B, as shown in the figure. He can row at a speed of 2 mi/h and walk at a speed of 5 mi/h.

1. (a) Find a function that models the time needed for the trip.
2. (b) Where should he land so that he reaches B as soon as possible?

To determine

To find: The function that models the time needed by the man for the trip.

Answer to Problem 30P

The function that models the time taken by the man is $T=\frac{\sqrt{x^2-14x+53}}{2}+\frac{x}{5}$.

Explanation of Solution

Given:

Width of river is $2\text{mi}$, speed of boat $\left(s\right)$ is $2\text{mi}/\text{hr}$ and the speed $\left(s_m\right)$ at which the man walk is $5\text{mi}/\text{hr}$.

Calculation:

Draw the figure according to given situation in the question as shown below.

Figure (1)

In the Figure (1) $A$ is the starting point of the trip and $B$ is the end point of the trip.

The man reaches at point $P$ from A in boat and walks the distance $PB$ as shown in the Figure (1).

Width of river $\left(OA\right)$ is $2\text{mi}$ and distance between point $O$ and $B$ is $7\text{mi}$.

Total time of the trip is sum of time taken to cross the river and time taken to walk the remaining distance.

Let time taken to cross the river be $t_1$ and time taken to walk is $t_2$.

Total time $\left(T\right)$ is,

$T=t_1+t_2$

The formula to calculate $t_1$ is,

$\begin{array}{c}{t}_1=\frac{\text{Distance}\text{travelled}\text{by}\text{the}\text{boat}}{\text{Speed}\text{of}\text{boat}}\\ =\frac{y}{s}\end{array}$ (1)

Distance travelled by the boat is $y$.

Apply Pythagorean Theorem in triangle $OAP$ shown in Figure (1).

${\left(AP\right)}^2={\left(OA\right)}^2+{\left(OP\right)}^2$

Substitute $y$ for $AP$, 2 for $OA$ and $7-x$ for $OP$ in above equation.

$\begin{array}{c}{\left(y\right)}^2={\left(2\right)}^2+{\left(7-x\right)}^2\\ y^2=4+49-14x+x^2\left[\because {\left(a-b\right)}^2=a^2-2ab+b^2\right]\\ =53-14x+x^2\end{array}$

Take square root of above equation.

$y=\sqrt{x^2-14x+53}$

Substitute $\sqrt{x^2-14x+53}$ for y and 2 for $s$ in equation (1) and calculate $t_1$.

$t_1=\frac{\sqrt{x^2-14x+53}}{2}$

The formula to calculate $t_2$ is,

$\begin{array}{c}{t}_2=\frac{\text{Walking}\text{distance}}{\text{Speed}\text{of}\text{man}}\\ =\frac{x}{s_m}\end{array}$

Substitute 5 for $s_m$ in above equation

$t_2=\frac{x}{5}$

Summarize all the information in the table as shown below.

In Words	In Algebra
Total time for the trip	$T$
Time taken to cross the river	$\frac{\sqrt{x^2-14x+53}}{2}$
Time taken in walking	$\frac{x}{5}$

Use the information in the table and model the function.

$\begin{array}{c}T=t_1+t_2\\ T=\frac{\sqrt{x^2-14x+53}}{2}+\frac{x}{5}\end{array}$

Thus, the function that models the time taken by the man is $T=\frac{\sqrt{x^2-14x+53}}{2}+\frac{x}{5}$.

To determine

To find: The point at which the man should reach to minimize the time of trip.

Answer to Problem 30P

The distance is $6.127\text{mi}$ from point $B$ to minimize the time of trip.

Explanation of Solution

The function that models the time as calculated in part(a) is $T=\frac{\sqrt{x^2-14x+53}}{2}+\frac{x}{5}$

Sketch the graph of the function as shown below.

Figure (2)

Observe from the graph that function attains a minimum value when x is $6.127$.

Thus, man should land at distance of $6.127\text{mi}$ from point $B$.