Solving an equation with fractional powers 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?
 A: After some simplifying steps,
$$(x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0, \\
(x^2 + y^2)^\frac{3}{2} - 3y^2(x^2 + y^2)^{\frac{1}{2}} = 0, \\
(x^2 + y^2)^{\frac{1}{2}}\left(x^2+y^2-3y^2 \right)=0, \\
x^2+y^2=0 \Rightarrow x=0,\;y=0, \\
\text{or}\\
x^2-2y^2=0 \Rightarrow |x|=\sqrt{2}|y|.
$$
A: After factoring out $(x^2 + y^2)^{\frac{1}{2}}$ and setting the second term $=0$:
$$2y^{2} = x^2$$
$$ y = \frac{x}{\sqrt{2}} $$
A: $$ (x^2 + y^2)^{3/2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
$$\iff (x^2 + y^2)^{1/2}\left((x^2 + y^2 - 3y^2)\right) = 0$$
$$\iff (x^2 + y^2)^{1/2}\left(x^2 - 2y^2\right) = 0$$
$$\iff (x^2 + y^2) = 0 \;\;\text{ or }\;\;x^2 = 2y^2$$
$$\iff \quad ?$$
A: Define a new variable:
$$ r = \sqrt{x^2 + y^2} $$
Then it follows:
$$ \begin{array}{rcl}
 (x^2 + y^2)^\frac{3}{2} - y \frac{3}{2}(x^2 + y^2)^{\frac{1}{2}}2y & = & 0 \\
 (x^2 + y^2)^\frac{3}{2} - 3(x^2 + y^2)^{\frac{1}{2}}y^2 & = & 0 \\
   r^3 - 3y^2r & = & 0 \\
   r^2 - 3y^2 & = & 0 \\
   r^2 & = & 3y^2 \\
   x^2 + y^2 & = & 3y^2 \\
   x^2 & = & 2y^2 \\
   x & = & \pm \sqrt{2}y \\
\end{array}
$$
A: Hint: factor out $(x^2+y^2)^\frac{1}{2}$
