Maximize the value 2a + b where a and b are the legs of a right triangle with hypotenuse 1

Consider a right triangle with hypotenuse 1. Let $a$ and $b$ denote the lengths of the legs of the triangle. Find the values of $a$ and $b$ such that the value $2a+b$ is maximized.

We have the following right triangle:

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From the Pythagorean theorem we have

$a^2 + b^2 = 1 \quad \implies \quad a = \sqrt{1-b^2}.$

So we want to maximize,

$2a + b = 2\sqrt{1-b^2} + b.$

Calling this function $f(b)$ and taking the derivative,

$f'(b) = -\frac{2b}{\sqrt{1-b^2}} + 1.$

Setting this equal to 0 we find the critical points of the function $f(b)$ ,

$\begin{align*} -\frac{2b}{\sqrt{1-b^2}} + 1 = 0 && \implies && -2b + \sqrt{1-b^2} &= 0 \\ && \implies && 2b &= \sqrt{1-b^2} \\ && \implies && 4b^2 &= 1-b^2 \\ && \implies && b = \frac{\sqrt{5}}{5}. \end{align*}$