What is max($ab^3$) if $a+3b=4$? This question was  asked by my professor,i think in the set $\mathbb Z$,its answer is 1.But,he advised me for the further exploration in  $\mathbb  R$.But,what i think(i may be wrong) that in $\mathbb R
$ its answer is not unique,but it should be unique.
Any suggestions are heartly welcome.

Let $f(a,b) = ab^3$ and $g(a,b) = a + 3b = 4$.
Using the method of Lagrange multipliers, we have the following facts:
(1) $ f_a (a,b) = b^3 = \lambda g_a(a,b) = \lambda$
(2) $ f_b (a,b) = 3ab^2 = \lambda g_b(a,b) = 3\lambda$
(3) $g(a,b) = a + 3b = 4$.
We clearly see that since $3\lambda = 3b^3 = 3ab^2$, either $b=0$ or $b=a$ gives us a maximum.
Now appealing to (3), we see that if $b=0$, then $a=4$, and if $b=a, 4a=4 \Rightarrow a = 1 \Rightarrow b=1$.
So, our critical points are (1,1) and (4,0), and we have to test them.
$f(1,1)= 1 * 1^3 = 1$.
$f(4,0) = 4 * 0^3 = 0$. Clearly $f(1,1)$ is our maximum, so $a=1, b= 1$ maximizes $ab^3$, and the maximum is 1.
