# Question #a50c7

Mar 25, 2016

Maximum is $\frac{2}{3 \sqrt{3}}$ at $\left(\pm \sqrt{\frac{2}{3}} , \frac{1}{\sqrt{3}}\right)$

#### Explanation:

$F \left(x , y , \lambda\right) = {x}^{2} y + \lambda \left({x}^{2} + {y}^{2} - 1\right)$

${F}_{x} = 2 x y + 2 \lambda x$
${F}_{y} = {x}^{2} + 2 \lambda y$

Setting each of these to $0$ and solving for $- \lambda$, we get

$- \lambda = y = {x}^{2} / \left(2 y\right)$.

This leads to ${x}^{2} = 2 {y}^{2}$

Substituting into ${F}_{\lambda} = {x}^{2} + {y}^{2} - 1 = 0$

We get $3 {y}^{2} = 1$, so $y = \pm \frac{1}{\sqrt{3}}$

Using ${x}^{2} = 2 {y}^{2}$, we get $x = \pm \sqrt{\frac{2}{3}}$.

Simple arithmetic gets the maximum value is $\frac{2}{3 \sqrt{3}}$