What is the maximum of $ab^n$ ?

By Algebra: By AM-GM inequality ,

$\begin{aligned} na + \underbrace{\frac bn + \frac bn + \frac bn + \cdots + \frac bn}_{n \text{ terms}} & \ge (n+1)\sqrt[n+1]{\frac {nab^n}{n^n}} \\ na + b & \ge (n+1)\sqrt[n+1]{\frac {ab^n}{n^{n-1}}} & \small \color{#3D99F6} \text{Raise to the power of }n+1 \text{ on both sides and swap sides.} \\ \frac {(n+1)^{n+1}ab^n}{n^{n-1}} & \le (na+b)^{n+1} & \small \color{#3D99F6} \text{Equality occurs when }na = \frac bn \text{ or } b = an^2 \\ & \le a^{n+1}n^{n+1}(n+1)^{n+1} \\ \implies ab^n & \le \boxed{a^{n+1}n^{2n}} \end{aligned}$

---

By Calculus: Let $na+b= k$ , where $k$ is a constant. Then

$\begin{aligned} b & = k - an \\ \implies f(a) & = ab^n & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }a \\ f'(a) & = b^n + anb^{n-1}\cdot \frac {db}{da} & \small \color{#3D99F6} \text{Note that }\frac {db}{da} = - n \\ & = b^n - an^2b^{n-1} & \small \color{#3D99F6} \text{Putting }f'(a) = 0 \\ \implies b & = an^2 \implies a = \frac b{n^2} & \small \color{#3D99F6} \text{For }b \ne 0 \\ f''(a) & = -n^2b^{n-1} - n^2b^{n-1} + an^3(n-1)b^{n-2} \\ & = n^2b^{n-2}(an - 2b) \\ f'' \left(\frac b{n^2} \right) & = n(1-2n)b^{n-1} < 0 & \small \color{#3D99F6} \text{For }n \in \mathbb N \\ \implies \max (ab^n) & = f \left(\frac b{n^2} \right) = a(an^2)^n = \boxed{a^{n+1}n^{2n}} \end{aligned}$