A man in a boat rows 3 miles off along straight shore and wants to reach a point 4 miles up the shore on the opposite side. If he can row 2 miles/hour and walk 4 miles/hour, describe his fastest route.

Answer

To solve the problem step by step, we will use the information given about the distances and speeds of the man rowing and walking.

### Step 1: Understand the Problem
The man is 3 miles off the shore and wants to reach a point 4 miles up the shore. He can row at a speed of 2 miles/hour and walk at a speed of 4 miles/hour.

### Step 2: Set Up the Coordinates
Let’s denote:
- The starting point of the man in the boat as point A (0, 0).
- The point directly across the shore (the point he rows to) as point B (3, 0).
- The destination point up the shore as point C (3, 4).

### Step 3: Define the Variables
Let \( x \) be the distance he rows to a point D on the shore, which is directly across from point B. The coordinates of point D will be (3, y) where \( y \) is the distance he rows. The distance he walks from point D to point C will be \( 4 - y \).

### Step 4: Calculate the Distances
Using the Pythagorean theorem, the distance he rows (AD) can be expressed as:
\[ AD = \sqrt{(3)^2 + (y)^2} = \sqrt{9 + y^2} \]

### Step 5: Calculate the Time Taken
The time taken to row to point D (t1) is:
\[ t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{\sqrt{9 + y^2}}{2} \]
The time taken to walk from point D to point C (t2) is:
\[ t_2 = \frac{4 - y}{4} \]

### Step 6: Total Time Function
The total time \( T \) taken is the sum of the rowing time and walking time:
\[ T = t_1 + t_2 = \frac{\sqrt{9 + y^2}}{2} + \frac{4 - y}{4} \]

### Step 7: Differentiate the Total Time
To find the minimum time, we differentiate \( T \) with respect to \( y \) and set the derivative equal to zero:
\[ \frac{dT}{dy} = \frac{1}{2} \cdot \frac{1}{2\sqrt{9 + y^2}} \cdot 2y - \frac{1}{4} = \frac{y}{2\sqrt{9 + y^2}} - \frac{1}{4} \]
Setting \( \frac{dT}{dy} = 0 \):
\[ \frac{y}{2\sqrt{9 + y^2}} = \frac{1}{4} \]

### Step 8: Solve for y
Cross-multiplying gives:
\[ 4y = 2\sqrt{9 + y^2} \]
Squaring both sides:
\[ 16y^2 = 4(9 + y^2) \implies 16y^2 = 36 + 4y^2 \implies 12y^2 = 36 \implies y^2 = 3 \implies y = \sqrt{3} \]

### Step 9: Calculate the Distance y
Now substituting \( y \) back into the equations:
- The distance he rows is \( \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \).
- The distance he walks is \( 4 - \sqrt{3} \).

### Final Answer
The fastest route for the man is to row \( 2\sqrt{3} \) miles to the shore and then walk \( 4 - \sqrt{3} \) miles to his destination.