Maximum value of equation $ab+bc+ca$ if $a+2b+c=4$ If $a,b,c$ ${\in}$ $\mathbb{R} $ such that
$a+2b+c=4$, then find the $max$ value of
$ab+bc+ca$.
I always get stuck with max, min questions. We cannot apply AM:GM here, I have not studied much  calculus yet.
Can you do this with graphs or by plane algebra?
I don't really know what tag to put for this!

Substitution of $b=\frac{4-a-c}{2}$ into $ab+bc+ca$ gives:
$$\begin{align}-\frac{a^2}{2} + 2 a -\frac{c^2}{2} + 2 c 
& = \tfrac{1}{2}\left( 4a-a^2+4c-c^2\right) \\[3pt]
& = \tfrac{1}{2}\left( 4-\left(2-a \right)^2+4-\left(2-c \right)^2 \right) \\[3pt]
& = 4-\tfrac{1}{2}\left(2-a \right)^2-\tfrac{1}{2}\left(2-c \right)^2 
\end{align}$$
And this is clearly maximal when $a=c=2$; yielding a maximal value of $4$.
