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 11 miles down the shore from P in the least time?

She must row a distance of : $\displaystyle \sqrt{x^2 + 16}$

She must walk a distance of : $\displaystyle 11-x$

$\displaystyle t=\frac{d}{t}$

The time traveled over water plus the time traveled over land can be written:

$\displaystyle \frac{\sqrt{x^2 + 16}}{3} + \frac{11-x}{4}$

If this is correct, how would I find the derivative of the above function..?

Would this be the derivative?

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

Finding a common denomenator:

$\displaystyle \frac{4x-3\sqrt{x^2 + 16}}{12\sqrt{x^2 + 16}}$ ?

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I figured out the solution myself. If anyone is curious to how I solved it, here is what I did:

$\displaystyle 4x-3\sqrt{x^2 +16} = 0$

$\displaystyle x = \frac{12}{\sqrt{7}}$

From here I plugged in 0, 11, and $\displaystyle \frac{12}{\sqrt{7}}$ back into my original equation. I found that$\displaystyle \frac{12}{\sqrt{7}}$was the minimum time that occured at the stationary point. Therefore, she would land the boat$\displaystyle \frac{12}{\sqrt{7}}$miles down the shore from point P.