How to find the greatest value of the function? $$ f(x, y)=x^2+y^2,\quad x,\;y \ge 0,\quad 3x+2y \le 6$$
The max value is in the point $(0, 3)$, but how do I prove it? I may be able to prove that the function decreases on the curve $x(t)=2t,\; y(t)=3-3t$, but I don't know how.
 A: As always, pictures provide a great way to see what is happening. I will let you fill in rigorous arguments to make this into a proof.

EDIT
There are many ways to make this into a rigorous proof. Below is one way. Say the maximum of $x^2+y^2$ is $a^2$. Then we have that $x = a \cos(\theta)$ and $y = a \sin(\theta)$. The constraint enforces that $\theta \in [0, \pi/2]$ and
$$3a \cos(\theta) + 2a \sin(\theta) \leq 6 \implies a (3 \cos(\theta) + 2 \sin(\theta)) \leq 6$$
Since, we want to maximize $a$, we want to minimize $ (3 \cos(\theta) + 2 \sin(\theta))$ in the domain $\theta \in [0, \pi/2]$. Now this is a simple calculus problem in one variable. Find the derivative and set it equal to zero and check the boundary points to figure out the minimum of $ (3 \cos(\theta) + 2 \sin(\theta))$. This will give you that the minimum occurs at $\theta = 0$. Hence, the maximum value of $a$ is $3$. Hence, the maximum value of $x^2 + y^2$ is $9$.
A: Since
$$
f(x,y) \ge 0=f(0,0) \quad \forall\ (x,y) \in \mathbb{R}^2,
$$ 
the function
$$
f: \mathbb{R}^2 \to \mathbb{R},\ f(x,y)=x^2+y^2
$$
has a global minimum at $(0,0)$, therefore the maximum of $f$ on the compact set
$$
K=\{(x,y) \in \mathbb{R}^2:\ x,y \ge 0,\ 3x+2y\le 6\}
$$ 
is on its boundary
$$
\partial K=\{0\}\times[0,3]\cup[0,2]\times\{0\}\cup\{(t,-\frac32t+3):\ 0 \le t\le 2\}.
$$
We have
\begin{eqnarray}
f(x,y)&\le&9=f(0,3) \quad \forall\ (x,y) \in \{0\}\times[0,3]\\
f(x,y)&\le&4=f(2,0) \quad \forall\ (x,y) \in [0,2]\times\{0\},
\end{eqnarray}
and for every $t \in [0,2]$ we have
\begin{eqnarray}
\phi(t)&:=&f(t,-\frac32t+3)=t^2+\Big(-\frac32t+3\Big)^2 \le \max\{\phi(0),\phi(2)\}=9=\phi(0).
\end{eqnarray}
Hence 
$$
\max_{K}f=9=f(0,3).
$$
A: $f$ is critical when the gradient is $(0,0)$; that is,
$$
\nabla(x^2+y^2)=(2x,2y)=(0,0)
$$
$(0,0)$ is on the boundary of the triangle $x\ge0$, $y\ge0$, and $3x+2y\le6$.
An extremum can also be attained on the boundary where the gradient is perpendicular to the boundary.
On $x=0$, where the boundary is parallel to $(0,1)$ so we are looking for a point where
$$
(2x,2y)\cdot(0,1)=0
$$
Again, we have $(x,y)=(0,0)$.
On $y=0$, where the boundary is parallel to $(1,0)$, we are looking for a point where
$$
(2x,2y)\cdot(1,0)=0
$$
Yet again, we have $(x,y)=(0,0)$.
On $3x+2y=6$, where the boundary is parallel to $(2,-3)$, we are looking for a point where
$$
(2x,2y)\cdot(2,-3)=0
$$
$3x+2y=6$ and $4x-6y=0$ intersect at $\left(\frac{18}{13},\frac{12}{13}\right)$.
The other points are the singular points of the boundary (the vertices of the triangle): $(0,0)$, $(2,0)$, and $(0,3)$.
Checking each of these possibilities, we get
$$
\begin{align}
f(0,0)&=0\\
f\left(\frac{18}{13},\frac{12}{13}\right)&=\frac{36}{13}\\
f(2,0)&=4\\
f(0,3)&=9
\end{align}
$$
Therefore, $f(0,3)=9$ is the maximum.
