Question

a small island is 4 miles from the nearest point p on the straight shoreline of a large lake. if a woman on the island can row a boat 3 miles per hour and can walk 4 miles per hour, where should the boat be landed in order to arrive at a town 12 miles down the shore from p in the least time?
the boat should be landed \\(\square\\) miles down the shore from p.
(type an exact answer.)

Explanation:

Step1: Define Variables

Let \( x \) be the distance (in miles) down the shore from point \( P \) where the boat is landed. The distance rowed is \( \sqrt{4^2 + x^2} \) (by Pythagoras), and the distance walked is \( 12 - x \). The time taken to row is \( \frac{\sqrt{16 + x^2}}{3} \) (since rowing speed is 3 mph) and the time taken to walk is \( \frac{12 - x}{4} \) (walking speed is 4 mph). The total time \( T(x) \) is \( T(x)=\frac{\sqrt{x^2 + 16}}{3}+\frac{12 - x}{4} \).

Step2: Take Derivative

To minimize \( T(x) \), take the derivative \( T'(x) \). The derivative of \( \frac{\sqrt{x^2 + 16}}{3} \) is \( \frac{x}{3\sqrt{x^2 + 16}} \), and the derivative of \( \frac{12 - x}{4} \) is \( -\frac{1}{4} \). So \( T'(x)=\frac{x}{3\sqrt{x^2 + 16}}-\frac{1}{4} \).

Step3: Set Derivative to Zero

Set \( T'(x) = 0 \): \( \frac{x}{3\sqrt{x^2 + 16}}=\frac{1}{4} \). Cross - multiply: \( 4x = 3\sqrt{x^2 + 16} \). Square both sides: \( 16x^2 = 9(x^2 + 16) \).

Step4: Solve for x

Expand: \( 16x^2=9x^2 + 144 \). Subtract \( 9x^2 \): \( 7x^2 = 144 \). Then \( x^2=\frac{144}{7} \), but wait, this is wrong. Wait, let's re - check the problem. Wait, maybe the rowing speed is 3 mph? Wait, the original problem (probably) has rowing speed 3 mph and walking speed 4 mph, and the island is 4 miles from P, town is 12 miles from P along the shore.

Wait, let's redo the derivative step. Let's assume rowing speed \( v_r = 3 \) mph, walking speed \( v_w = 4 \) mph. Distance rowed: \( d_r=\sqrt{4^2 + x^2}=\sqrt{x^2 + 16} \), time rowed \( t_r=\frac{\sqrt{x^2 + 16}}{3} \). Distance walked: \( d_w = 12 - x \), time walked \( t_w=\frac{12 - x}{4} \). Total time \( T(x)=\frac{\sqrt{x^2 + 16}}{3}+\frac{12 - x}{4} \).

Derivative \( T'(x)=\frac{x}{3\sqrt{x^2 + 16}}-\frac{1}{4} \). Set to zero: \( \frac{x}{3\sqrt{x^2 + 16}}=\frac{1}{4} \). Cross - multiply: \( 4x = 3\sqrt{x^2 + 16} \). Square both sides: \( 16x^2 = 9(x^2 + 16)\Rightarrow16x^2=9x^2 + 144\Rightarrow7x^2 = 144\Rightarrow x^2=\frac{144}{7}\Rightarrow x=\frac{12}{\sqrt{7}} \). But this seems odd. Wait, maybe the rowing speed is 5 mph? Wait, no, maybe I misread the problem. Wait, the user's problem: "A small island is 4 miles from the nearest point P on the straight shoreline of a large lake. If a woman on the island can row a boat 3 miles per hour and can walk 4 miles per hour, where should the boat be landed in order to arrive at a town 12 miles down the shore from P in the least time?"

Wait, let's check the standard problem. The standard problem is similar, and the correct approach is to set up the time function and take derivative. Wait, maybe I made a mistake in the derivative. Let's re - express:

Let \( x \) be the distance from P to landing point. Then rowing distance is \( \sqrt{4^2 + x^2} \), rowing time \( \frac{\sqrt{16 + x^2}}{r} \), walking time \( \frac{12 - x}{w} \), where \( r = 3 \), \( w = 4 \).

\( T(x)=\frac{\sqrt{x^2 + 16}}{3}+\frac{12 - x}{4} \)

\( T'(x)=\frac{x}{3\sqrt{x^2 + 16}}-\frac{1}{4} \)

Set to zero: \( \frac{x}{3\sqrt{x^2 + 16}}=\frac{1}{4} \)

\( 4x = 3\sqrt{x^2 + 16} \)

Square: \( 16x^2 = 9x^2 + 144 \)

\( 7x^2 = 144 \)

\( x=\frac{12}{\sqrt{7}}\approx4.53 \). But this doesn't seem right. Wait, maybe the rowing speed is 5 mph? Wait, no, maybe the problem is with the numbers. Wait, maybe the island is 3 miles from P, and town is 12 miles? No, the user's problem says 4 miles from P.

Wait, maybe I misread the rowing speed. Let's check the original image again. The user's image: "can row a boat 3 miles per hour and can walk 4 miles per hour, where should the boat be landed in order to arrive at a town 12 miles down the shore from P in the least time? The boat should be landed \(\square\) miles down the shore from P."

Wait, let's do the calculation again.

\( 4x = 3\sqrt{x^2 + 16} \)

Square both sides: \( 16x^2 = 9x^2 + 144 \)

\( 7x^2 = 144 \)

\( x^2=\frac{144}{7} \)

\( x = \frac{12}{\sqrt{7}}=\frac{12\sqrt{7}}{7}\approx4.53 \)

But maybe the problem has rowing speed 5 mph? Wait, no, the user's problem says 3 mph rowing, 4 mph walking.

Wait, maybe the correct answer is \( \frac{12}{\sqrt{7}} \), but let's check with calculus.

Alternatively, maybe the problem is: rowing speed 5 mph, walking speed 4 mph, island 3 miles from P, town 12 miles. No, the user's problem is 4 miles from P.

Wait, perhaps the user made a typo, but assuming the problem is as stated, the solution is \( x = \frac{12}{\sqrt{7}}=\frac{12\sqrt{7}}{7}\approx4.53 \). But let's check the standard problem. The standard problem is: A man is in a boat 2 miles from the nearest point on a straight shoreline. He wants to reach a point 6 miles down the shore. He can row at 3 mph and walk at 5 mph. Where should he land? The solution is \( \frac{3}{\sqrt{2}} \). But in our case, with 4 miles from P, 12 miles to town, row 3, walk 4.

Wait, maybe I made a mistake in the time function. Let's re - express:

Time rowed: distance rowed / rowing speed = \( \sqrt{4^2 + x^2}/3 \)

Time walked: distance walked / walking speed = \( (12 - x)/4 \)

Total time \( T(x)=\frac{\sqrt{x^2 + 16}}{3}+\frac{12 - x}{4} \)

To find the minimum, take derivative, set to zero.

\( T'(x)=\frac{x}{3\sqrt{x^2 + 16}}-\frac{1}{4}=0 \)

\( \frac{x}{3\sqrt{x^2 + 16}}=\frac{1}{4} \)

\( 4x = 3\sqrt{x^2 + 16} \)

\( 16x^2 = 9x^2 + 144 \)

\( 7x^2 = 144 \)

\( x=\frac{12}{\sqrt{7}}=\frac{12\sqrt{7}}{7}\approx4.53 \)

So the boat should be landed \( \frac{12\sqrt{7}}{7} \) miles from P. But let's rationalize: \( \frac{12\sqrt{7}}{7}\approx4.53 \).

Wait, but maybe the problem is with the rowing speed. If the rowing speed is 5 mph, then:

\( T'(x)=\frac{x}{5\sqrt{x^2 + 16}}-\frac{1}{4}=0 \)

\( 4x = 5\sqrt{x^2 + 16} \)

\( 16x^2 = 25x^2 + 400 \)

\( - 9x^2 = 400 \), which is impossible. So rowing speed must be less than walking speed.

Since 3 < 4, it's possible. So the correct answer is \( \frac{12\sqrt{7}}{7} \) miles.

Answer:

\(\frac{12\sqrt{7}}{7}\)