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.
 A: HINT:
Using AM-GM inequality, 
$$\dfrac{a+b+b+b}4\ge\sqrt[4]{ab^3}$$
A: If $a,b>0$, then $$a+3b=4 \ge 4\sqrt[3]{ab^3}$$From the mean inequality.  Thus $ab^3 \le 1$. 
If $ab<0$, then $ab^3<0$. Thus the maximum of $ab$ cannot be find here. 
If $a,b<0$, then $a+3b<4$. Thus a contradiction.  
A: 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.
A: $4-3b=a\Rightarrow a\cdot b^3=(4-3b)b^3=-3b^4+4b^3$.
Sometimes it is more convenient to set $f(x)=-3x^4+4x^3$ and study the behaviour of this function.
$f'(x)=-12x^3 +12x^2=12x^2(1-x)$ which means that $f'(x)>0$ only when $x<1$.
This means at $x=1$ our function $f(x)$ obtains its greatest value, $f(1)=1$.
Hence $f(x)\leq 1$ which implies also that $a\cdot b^3\leq 1$
A: Even in $\mathbb{R}$ the maximum value is 1. Clearly, as both $a$ and $b$ are nonnegative(otherwise the product will be negative) we can apply A.M G.M inequality.
$$4=a+b+b+b\geq 4(a.b.b.b)^{\frac{1}{4}}$$ we have $ab^3\leq 1$.
A: Hint: You can use A.M.-G.M. inequality or use weighted A.M.-G.M. inequality. Weighted A.M.-G.M. tell us that 
$\frac{ma+nb}{m+n}$ =$[a^m.b^n]^{\frac{1}{m+n}}$.
Use it you can easily do this. It is deducted from A.M-G.M.
