To solve the problem, we minimize the time taken by the swimmer to reach the cottage by considering the tradeoff between swimming and walking distances.

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Variables:

1. $x$: Distance from the closest point on the shoreline where the swimmer comes ashore.
2. Total time $T(x)$: Time required to reach the cottage.

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Key Observations:

• The swimming distance is the hypotenuse of a right triangle with one side as 500 m (the perpendicular distance to the shoreline) and the other as $x$. $\text{Swimming Distance} = \sqrt{500^2 + x^2}$
• The walking distance is the horizontal distance from the point of coming ashore to the cottage, which is $800 - x$.

Time Expressions:

• Swimming time: $T_{\text{swim}} = \frac{\sqrt{500^2 + x^2}}{4}$
• Walking time: $T_{\text{walk}} = \frac{800 - x}{6}$
• Total time: $T(x) = \frac{\sqrt{500^2 + x^2}}{4} + \frac{800 - x}{6}$

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Optimization:

Minimize $T(x)$ by finding $\frac{dT(x)}{dx} = 0$.

Compute $\frac{dT(x)}{dx}$:

1. Derivative of the swimming time: $\frac{d}{dx}\left(\frac{\sqrt{500^2 + x^2}}{4}\right) = \frac{1}{4} \cdot \frac{x}{\sqrt{500^2 + x^2}}$
2. Derivative of the walking time: $\frac{d}{dx}\left(\frac{800 - x}{6}\right) = -\frac{1}{6}$

Set the derivative of $T(x)$ to zero:

$\frac{1}{4} \cdot \frac{x}{\sqrt{500^2 + x^2}} - \frac{1}{6} = 0$

Simplify: $\frac{x}{4\sqrt{500^2 + x^2}} = \frac{1}{6}$

Multiply through by $6 \cdot 4\sqrt{500^2 + x^2}$: $6x = 4\sqrt{500^2 + x^2}$

Square both sides: $36x^2 = 16(500^2 + x^2)$

Expand and simplify: $36x^2 = 16 \cdot 500^2 + 16x^2$ $20x^2 = 16 \cdot 500^2$ $x^2 = \frac{16 \cdot 500^2}{20}$ $x^2 = \frac{16}{20} \cdot 500^2$ $x^2 = 0.8 \cdot 500^2$ $x = 0.8^{1/2} \cdot 500$

Numerical approximation: $x \approx 0.894 \cdot 500$ $x \approx 447 \, \text{m}$

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Answer:

The swimmer should come ashore approximately 447 meters from the closest point on the shoreline.

Would you like further details, a step-by-step breakdown, or graphical visualization?

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Related Questions:

1. How does swimming speed affect the optimal point of coming ashore?
2. What if the walking speed is slower than the swimming speed?
3. How does the perpendicular distance to the shore influence the result?
4. What if the cottage was closer (e.g., 500 m) downshore instead of 800 m?
5. Can this problem be solved geometrically instead of using calculus?

Tip:

Always double-check optimization problems by considering the physical meaning of the result to ensure it makes sense.