Maximum value of equation $ab+bc+ca$ if $a+2b+c=4$

Question

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!

Answer 1

$$a+b+c=4-b$$ $$a^2+b^2+c^2+2(ab+ac+bc)=16-8b+b^2$$ $$ab+bc+ca=\frac{16-8b-a^2-c^2}2$$

Since $b=\frac{4-a-c}2$, $$ab+bc+ca=\frac{16-16+4a+4c-a^2-c^2}2=\frac{a(4-a)+c(4-c)}2\le\frac{4+4}2=4$$

Answer 2

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 $4$.

Answer 3

$$f(a)=ab+c(a+b)=ab+(4-a-2b)(a+b)=-a^2+a(4-2b)+b(4-2b)$$

Once we got a quadratic function, the maximum is given by

$$f_{max}(a)=-\frac{\Delta}{4(-1)}=\frac{(4-2b)^2+4b(4-2b)}{4}=(2-b)(2+b)$$

Calling $g(b)=(2-b)(2+b)$ we see that the maximum is $4$ then $f_{max}(a)=4$.