# Maximum value of the product ab

Problem 312. What is the maximum value of the product $ab$ if $a$ and $b$ are non-negative numbers such that $a+2b=3$?

Solution. It is easier to say when the product of two non-negative numbers $a$ and $2b$ (whose sum equals 3) is maximal. It is maximal when these numbers are equal, that is, $a=2b=3/2$. The product of $a$ and $b$ is half the product of $a$ and $2b$; its maximum value is $$\frac{3}{2}*\frac{3}{4}=\frac{9}{8}.$$

I solved this problem geometrically correctly before I saw this solution, and when I saw this one I didn't understand to be honest, can someone explain this solution to me? Especially why/how 3 became $3/2$. Thanks in advance.

-

It's essentially a parameter substitution. Let $c = 2b$. Then the question becomes

Maximise $\tfrac{1}{2}ac$ given that $a+c=3$.

By symmetry you can argue that the answer must be $a=c$ and thus $a=3/2$, $c=3/2$.

The reason for this is that if you had an asymmetrical solution, $a=x$ and $c=3-x$ then there would be another solution, $a=3-x$ and $c=x$. Therefore there is no solution at all (e.g. consider the same problem with "minimize" instead of "maximize") or there is a unique, symmetric solution.

Finally you undo the substitution, getting $b = c/2 = 3/4$.

-
Thanks for the explanation. –  Paul Dirac Jan 16 '13 at 11:48
The minimum value of 2ab is 0, and occurs when $a=0, b=3/2$ or $a=3, b=0$. There are 2 non-symmetric solutions. You have to check your boundary conditions. –  Calvin Lin Jan 16 '13 at 12:14
@CalvinLin Oops, I didn't read the "non-negative" bit of the question! –  Chris Taylor Jan 16 '13 at 12:21