I have to maximize this function : $$ f(x,y) = a\sqrt{x} + b\sqrt{y}$$ $$ a, b \in{R+*} $$ Knowing that $$ 0 ≤ x ≤ 2 − 2y $$ with $$ 0 ≤ y ≤ 1 $$
I said that f is a linear combination of 2 concave functions so it has a maximum (for 0<=x<=2-2y and 0<=y<=1). But because a,b are positive real numbers, I cannot solve $$ \frac{a}{2\sqrt{x}} = 0 $$ And it's the same for y.
How do I deal with such a situation. Thank you very much !!
With the help of some slack variables $s_k$ we transform the inequalities into equations and then with $f(x,y) = a\sqrt x+b\sqrt y$ forming the lagrangian
$$ L = f(x,y)+\lambda_1(x-s_1^2)+\lambda_2(y-s_2^2)+\lambda_3(x-2+2y+s_3^2)+\lambda_4(y-1+s_4^2) $$
The lagrangian stationary points are the solutions for
$$ \nabla L = 0 = \left\{ \begin{array}{l} \frac{a}{2 \sqrt{x}}+\lambda_1+\lambda_3 \\ \frac{b}{2 \sqrt{y}}+\lambda_2+2 \lambda_3+\lambda_4 \\ x-s_1^2 \\ y-s_2^2 \\ s_3^2+x+2 y-2 \\ s_4^2+y-1 \\ -2 s_1 \lambda_1 \\ -2 s_2 \lambda_2 \\ 2 s_3 \lambda_3 \\ 2 s_4 \lambda_4 \\ \end{array} \right. $$
giving the solution
$$ \left[ \begin{array}{ccccccc} f(x,y)& x & y & s_1^2 & s_2^2 & s_3^2 & s_4^2\\ \sqrt{2a^2+b^2} & \frac{4 a^2}{2 a^2+b^2} & \frac{b^2}{2 a^2+b^2} & \frac{4 a^2}{2 a^2+b^2} & \frac{b^2}{2 a^2+b^2} & 0 & \frac{2 a^2}{2 a^2+b^2} \\ \end{array} \right] $$
NOTE
Null $s_k$'s indicates that the corresponding constraint is active.
By cauchy-schwarz-inequality, $$a\sqrt{x} + b\sqrt{y} = (\sqrt2a)(\sqrt{x}/\sqrt2) + b\sqrt{y} \le \sqrt{(2a^2+b^2)(x/2 + y)} \le \sqrt{2a^2+b^2}$$ due to the constraint $x \le 2 − 2y \iff (x+2y)/2 \le 1$, with equality holds iff $$\frac{\sqrt2a}{b} = \frac{\sqrt{x}/\sqrt2}{\sqrt{y}} \iff 2a \sqrt{y} = b \sqrt{x}$$ and $x + 2y = 2$. \begin{align} 2a \sqrt{y} &= b \sqrt{2 - 2y} \\ 2a^2 y &= b^2 (1 - y) \\ y &= \frac{b^2}{2a^2 + b^2} \in (0,1). \tag{$a,b \in \mathbb{R}_+^*$} \\ x &= 2(1-y) = \frac{4a^2}{2a^2 + b^2} \end{align}
