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**gk99** Im having some issues understanding an example in my book, mainly due to the fact that there is some missing explanation of what was done. The idea is I need to eventually use something similar to this to for another question to solve a minimum cost optimization problem.

A man launches his boat at point A on a straight river, 3km wide and wants to reach point B, 8km downstream. He can row at 6km/h and run at 8km/h

Rowing distance is $\displaystyle \sqrt{x^2 + 9}$

Running distance is $\displaystyle 8 - x$

$\displaystyle time = \frac {distance}{rate}$

$\displaystyle \frac{\sqrt{x^2 + 9}}{6} + \frac{8 - x}{8}$

$\displaystyle T'(x) = \frac {x}{6\sqrt{x^2 + 9}} - \frac {1}{8}$

Find where x = 0

$\displaystyle T'(x) = 0 = \frac {x}{6\sqrt{x^2 + 9}} = \frac {1}{8}$

Im guessing in the example we want to land part to be on the other side which is why its moved over.

This is where I get confused, Im not sure what exactly has happened, but I think we have some how canceled the top of the left side fraction than halved both sides.

$\displaystyle 4x = 3\sqrt{x^2 + 9}$

From here Im guessing we square both sides to remove the square root.

$\displaystyle 16x^2 = 9(x^2 + 9)$

Than remove the square root from the right so subtract the left sided one

$\displaystyle 7x^2 = 81$

The square root both sides

$\displaystyle x = \frac {9}{\sqrt{7}}$