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!
 A: Use Schwarz inequality
$$(x+y)^2\le \left((\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{3}})^2\right)\left( (\sqrt{2}x)^2+(\sqrt{3}y)^2\right)\le 5/6.$$
A: 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 $\sqrt{\frac3{10}}+\sqrt{\frac2{15}}$.
A: Since it's a linear functional, use the fact that it's level sets form a family of parallel straight lines to see why the maxima must occur at a boundary point.Now at the boundary, use the parametrization $x= a cos \theta$ and $y= b sin \theta$, where $a$ and $b$ are lengths of semi-major and semi-minor axes respectively, and the fact that $a\hspace{0.5mm}sin \theta + b \hspace{0.5mm} cos \theta \leq \sqrt{ a^2 + b^2}$ with the bound being actually attained, to get your answer when $a = \frac{1}{\sqrt{2}}$ and $b= \frac{1}{\sqrt{3}}$ respectively.
Edit - as an exercise,you might try showing that a linear functional on $\mathbb{R}^n$ can attain it's maxima on a compact subset $K$ only at its boundary points, and that the conclusion can further be strengthened to extreme points if $K$ is convex.
A: Hint.
This is a maximum/minimum problem involving homogeneous functions. The objective function being linear, the maxima/minima should be located at the boundary so this problem, making $y = \lambda x$ can be read as the unconstrained.
$$
\max_{\lambda}\frac{1+\lambda}{\sqrt{2+3\lambda^2}}
$$
A: Alternatively, the objective function $z=x+y$ will achieve its maximum when the contour line $y=-x+z$ will touch the ellipse $2x^2+3y^2=1$ from above. So, the slope of the tangent line must be equated to the slope of the contour line:
$$4x_0^2+6y_0^2y'=0\Rightarrow y'=-\frac{2x_0}{3y_0}=-1\Rightarrow x_0=\frac32y_0$$
Now plug this into the equation of ellipse:
$$2\cdot \left(\frac32y_0\right)^2+3y_0^2=1\Rightarrow y_0=\sqrt{\frac2{15}}\Rightarrow x_0=\sqrt{\frac3{10}}.$$
Hence, the maximum is $z(x_0,y_0)=\sqrt{\frac3{10}}+\sqrt{\frac2{15}}$.
