<s>[INST] Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$

My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{2 x}{3 y}$
$2 x^{2}+3 y^{2}=1$
What to do next? Any suggestion or Hint would be greatly appreciated!
 [/INST] You can use the method of Lagrange multipliers. If $f(x,y)=x+y$, then, for every $(x,y)\in\Bbb R^2$, $\nabla f(x,y)=(1,1)\ne(0,0)$. Therefore, the maximum of $f$ can only be attained at the boundary of the disk, that is, when $2x^2+3y^2=1$. So, solve the system$$\left\{\begin{array}{l}1=4\lambda x\\1=6\lambda y\\2x^2+3y^2=0.\end{array}\right.$$The only solutions are $(x,y)=\pm\left(\sqrt{\frac3{10}},\sqrt{\frac2{15}}\right)$. So, the maximum is attained at $\left(\sqrt{\frac3{10}},\sqrt{\frac2{15}}\right)$ and that maximum is $\sqrt{\frac3{10}}+\sqrt{\frac2{15}}$.
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