Maximizing The Power

Relevant wiki: Extrema - Problem Solving - Medium

$f(x,y) = x^{y}$
$f(x) = x^{\dfrac{c}{x}} = (x^{\dfrac{1}{x}})^{c}$
This is maximized when $x^{\dfrac{1}{x}}$ is maximized.
$g(x) = x^{\dfrac{1}{x}}$
$g'(x) = x^{\dfrac{1}{x}}\left(\dfrac{1}{x^{2}} - \dfrac{\ln(x)}{x^{2}} \right)$

$g'(x) = 0 \rightarrow \ln(x) = 1, x = e$
$\displaystyle \lim_{x \to e-} g'(x) > 0$
$\displaystyle \lim_{x \to e+} g'(x) < 0$
Thus $x = e$ is a point of local maxima.
$g(x)_{max} = e^{\dfrac{1}{e}}$
$f(x)_{max} = e^{\dfrac{c}{e}}$
$e^{\dfrac{c}{e}} = c$
$e^{c} = c^{e}$
Drawing the graph we can see that there's only one positive solution $c = e$

Relevant wiki: Lagrange Multipliers

We will use Lagrange Multipliers to solve this problem. Namely, we will find the maximum of the general case, and then set this general case to be equal to $c$ , and solve.

The Lagrangian of the given is $\mathcal{L} = x^y - \lambda(xy-c)$ We set the gradient of this to be the zero vector, namely $\nabla \mathcal{L} = \begin{bmatrix} \frac{\partial \mathcal{L}}{\partial x} \\ \frac{\partial \mathcal{L}}{\partial y} \\ \frac{\partial \mathcal{L}}{\partial \lambda} \end{bmatrix} = \begin{bmatrix} yx^{y-1} -\lambda y \\ x^y \ln(x) - \lambda x \\ xy - c \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ Now we have a system of equations, namely $\begin{cases}yx^{y-1} = \lambda y\\ x^y \ln(x) = \lambda x \\ xy = c \end{cases}$ The third equation is simply the constraint, so we focus on the first and second equation. Divide the first equation by $y$ and the second equation by $x$ (we can do this because $x, y \neq 0$ ) to get that $\begin{cases}x^{y-1} = \lambda \\ x^{y-1} \ln(x) = \lambda \end{cases} \implies \ln{x} = 1 \implies x = e$ Thus, the maximal pair is $(x,y) = (e, c/e)$ and $\boxed{\text{max}(x^y) = e^{\frac{c}{e}}}$ We set this to equal $c$ to get that $e^{\frac{c}{e}} = c \implies e^c = c^e$ Obviously, the only positive solution to this is that $c = e \approx \boxed{\textbf{2.718}}$ and we are done.