TITLE: Solving an equation with fractional powers
QUESTION [6 upvotes]: I was trying to find the maximum value for a function. I took the first derivative and arrived at this horrible expression:
$$ (x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
How can I find the extrema by hand?

REPLY [2 votes]: Define a new variable:
$$ r = \sqrt{x^2 + y^2} $$
Then it follows:
$$ \begin{array}{rcl}
 (x^2 + y^2)^\frac{3}{2} - y \frac{3}{2}(x^2 + y^2)^{\frac{1}{2}}2y & = & 0 \\
 (x^2 + y^2)^\frac{3}{2} - 3(x^2 + y^2)^{\frac{1}{2}}y^2 & = & 0 \\
   r^3 - 3y^2r & = & 0 \\
   r^2 - 3y^2 & = & 0 \\
   r^2 & = & 3y^2 \\
   x^2 + y^2 & = & 3y^2 \\
   x^2 & = & 2y^2 \\
   x & = & \pm \sqrt{2}y \\
\end{array}
$$