If $T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$. Determine $||T||$. Define $T : \mathbb{R}^2 → \mathbb{R}$ ( $\mathbb{R}^2$ & $\mathbb{R}$ being
equipped with the Euclidean norm) by
$T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$.
Determine $||T||$.

My thoughts:- 
We know that
$ \qquad \left\|{T}\right\| = \sup \left\{{\left|{Th}\right|: \left\|{h}\right\| = 1}\right\}$
here $h=(x,y)$ and $||h||=\sqrt{x^2+y^2}=1$
So we need to find out the maximum value of $2x+y$ with the condition that $x^2+y^2=1$.
Am I right?
 A: Yes, you are right. To do so, note that 
$$ T'(x,y) = \begin{pmatrix} 2 & 1 \end{pmatrix} $$
and for $F(x,y) = x^2 + y^2 - 1$ we have 
$$ F'(x,y) = \begin{pmatrix} 2x & 2y \end{pmatrix} $$
We hence have to solve
\begin{align*}
  2 &= 2x \lambda\\
  1 &= 2y\lambda\\
  x^2 + y^2 &= 1
\end{align*} 
So $x = 1/\lambda$, $y = 1/2\lambda$, hence $y = \frac x2$. Plugging this into the third equation gives
$$ x^2 + \frac{x^2}4 = 1 \iff x^2 = \frac 45 \iff x = \pm \frac 2{\sqrt 5} $$
This gives $y = \pm \frac 1{\sqrt 5}$. Now evaluate $T$ at the two points.
A: There's a nice method that maximizes linear functions subject to quadratic constraints: Cauchy-Schwarz gives
$$
|2x+y| = \big|(2,1)\cdot(x,y)\big| \le \|(2,1)\|\, \|(x,y)\| = \sqrt5 \sqrt{x^2+y^2}.
$$
So under the constraints $\sqrt{x^2+y^2}=1$, we have $|2x+y|\le\sqrt5$. And in Cauchy-Schwarz, equality is attained when the two vectors are parallel, so that $(x,y) = \alpha(2,1)$ (and it's easy to calculate then that $\alpha=1/\sqrt5$).
A: You need to use something called Lagrange Method to solve this :
Consider $M(x,y)=f(x,y)+\lambda g(x,y)=(2x+y)+\lambda(x^2+y^2-1)$


*

*$\frac{\partial M}{\partial x}=2+2\lambda x =0$

*$\frac{\partial M}{\partial y}=1+2\lambda y =0$

*$\frac{\partial M}{\partial \lambda}=x^2+y^2 -1=0$
All you need to is solve for $\lambda,x,y$ and you have to evaluate $f(x,y)$ at those points.
