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