If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ Can some one help me out on where to go?
If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ ?
 A: We are given that $ a^2 + b^2 = 1$, and $ a, b \geq 0$.
Applying AM-GM or Cauchy-Schwarz, we get that
$$ 20 = 5(a^2 + a^2 + a^2 + a^2 + 4b^2) \geq (a+a+a+a+2b)^2 = (4a+2b)^2$$
Hence $ 2a + b \leq \frac{20}{16} = \frac{5}{4} $, with equality when $ a = 2b, b = \sqrt{\frac{1}{5} }$.
A: Well, you know that $a^2 + b^2 = 1$ (assuming "cathets" means "legs", as suggested by a google search). So $b = \sqrt{1 - a^2}$. Therefore, you're trying to maximize $2a + \sqrt{1 - a^2}$, and your only restriction is that $0 \leq a \leq 1$. (If you don't see why $a$ needs to be in $[0,1]$, convince yourself of this before proceeding.) If you know how to optimize functions using derivatives, the rest is quick. (The maximum of $f(a) = 2a + \sqrt{1-a^2}$ in $[0,1]$ will occur at one of the endpoints or at some $a$ where $f'(a) = 0$.)
A: Let the two sides be $\sin(x)$ and $\cos(x)$.
So we want to maximize $2 \cos(x) + \sin(x)$.
We can write (2,1) in polar form as 
$$
2 = R\cos(y),~ 1 = R\sin(y)~~~~\Rightarrow R = \sqrt{5}$$
Hence
$$
2 \cos(x) + \sin(x) = R (\cos(x) \cos(y) + \sin(x) \sin(y)) =\sqrt{5} \cos(x-y)$$
We really do not need to find $x$ or $y$ since
$$
\text{Maximum value of } \sqrt{5} \cos(x-y) = \sqrt{5}$$
Hence answer is $\sqrt{5}$
A: Your problem is as follows:
\begin{align*}
&\max \,2a+b\\
\\
&\text{s.t. } a^2 + b^2=1
\end{align*}
Using substitution, let $f(b) = 2a + b = 2\sqrt{1 - b^2} +b$. Now, we wish to find the maximum of this function with respect to $b$.
Using simple derivatives, we get
$$
\frac{-2b}{\sqrt{1-b^2}} + 1 = 0\implies 2b=\sqrt{1-b^2} \implies 5b^2=1\implies b=\sqrt{1/5}\implies a = \sqrt{4/5}.
$$
