Constrained optimization max $ f(x,y) = x+y$ subject to $x^2+y^2 \leq 4, x \geq0, y \geq0$

Question

max $ f(x,y) = x+y$ subject to $x^2+y^2 \leq 4, x \geq0, y \geq0$

I need to solve this by the Kuhn Tucker conditions without using concavity of the Lagrangian.

Answer 1

The Lagrangian is $$L(x,y,z_1,z_2,z_3)=-x-y-z_1(4-x^2-y^2)-z_2x-z_3y.$$ The KKT conditions are $$\text{Feasibility:}\quad x^2+y^2\leq 4,~x\geq 0,~y\geq 0,~z_1\geq 0,~z_2\geq 0,~z_3\geq 0$$ $$\text{Complementarity:}\quad(4-x^2-y^2)z_1=0 \quad xz_2 = 0 \quad yz_3 = 0$$ $$\text{Lagrangian derivative:}\quad -1+2xz_1-z_2=0, \quad -1+2yz_1-z_3=0$$

If we guess that $x=0$, then the Lagrangian derivative requires $z_2=-1$, contradicting $z_2\geq 0$. Likewise, if $y=0$, then $z_3=-1$, contradicting $z_3\geq 0$. So we can conclude that $x,y>0$.

Given $x,y>0$, then $z_1=1/(2y)=1/(2x)$, which implies $x=y$. Complementarity then requires $x^2+y^2=4$, or $x=y=\sqrt{2}$. So $(x,y,z_1,z_2,z_3)=(\sqrt{2},\sqrt{2},1/(2\sqrt{2}),0,0)$ satisfies all KKT conditions.

Sure, there are easier ways to solve the problem, and it is good to use them to verify this, but if it's the KKT conditions you need, here you are.

Answer 2

You can use the following inequality to get the solution: $$ (a_1b_1+a_2b_2)^2\le(a_1^2+a_2^2)(b_1^2+b_2^2). $$

Answer 3

The constraint $x^2 + y^2 \leq 4$ may be replaced by $x^2 + y^2 = 4$ since if $f$ were maximized when $x^2 + y^2 < 4$ we could simply increase $x$ or $y$ so that $x^2 + y^2 = 4$, contradicting the fact that $x+y$ was maximized.

As such we may write $y = \sqrt{4-x^2}$. Now we need to maximize $f = x + (4-x^2)^{1/2}$. The maximum occurs when $f'(x)=0$, that is when:

$$f'(x) = 1 + -x(4-x^2)^{-1/2} = 0 $$

This is equivalent to the fact that:

$$x = \sqrt{4-x^2}$$

Which after simplifying gives $2x^2 = 4$ and so $ x = \sqrt{2}$. As the equations are symmetric $y = \sqrt{2}$ also.