How to derive a closed-form optimal solution of the following concave maximization problem?

Question

The following problem

$\max_{x, y} f(x,y) = \log x + \log (x+y) - 2x -3y$

subject to $x \geq 0, y \geq 0$

is a concave maximization problem, and thus can be numerically solved by convex optimization methods. The optimal solution may be $(x^*, y^*) = (1, 0)$.

However, I want to derive the solution analytically. I tried the following approach, for example, by taking partial derivative,

$\frac{\partial f}{\partial x} = \frac{1}{x} + \frac{1}{x+y} - 2$

$\frac{\partial f}{\partial x} = \frac{1}{x+y} - 3$

And there exists no solution that makes both partial derivative to be $0$.

Answer 1

What you need for optimisation under constraints are the Karush-Kuhn-Tucker conditions, which, for your problem can be written as $$ \frac{\partial\mathcal{L}}{\partial x} \left(x,y,\mu\right)=0\\ \frac{\partial\mathcal{L}}{\partial y} \left(x,y,\mu\right)=0\\ x>0, y\ge0, \mu\ge0\\ y\mu=0\ , $$ where $\ \mathcal{L}\ $ is given by $$ \mathcal{L}\left(x,y,\mu\right)=f(x,y)+\mu y\ . $$ With $\ \mathcal{L}\ $ thus defined, we have \begin{align} \frac{\partial\mathcal{L}}{\partial x} \left(x,y,\mu\right)&=\frac{1}{x} +\frac{1}{x+y}-2=0\\ \frac{\partial\mathcal{L}}{\partial x} \left(x,y,\mu\right)&=\frac{1}{x+y}-3+\mu=0\\ .\ \end{align} From this, bit of elementary algebra gives $$ x=\frac{1}{\mu-1}\ , $$ which implies $\ \mu>1\ $, since $\ x>0\ $, and hence $\ y=0\ $, from the condition $\ \mu y=0\ $. Substituting $\ y=0\ $ in the first of the above conditions then gives $\ x=1\ $, and thence, $\ \mu=2\ $.

Answer 2

Considering with $$f(x,y) = \ln x+\ln(x+y)- 2x-3y$$ the Lagrangian

$$ L(x,y,\lambda_1,\lambda_2,s_1,s_2) = f(x,y)+\lambda_1(x-s_1^2)+\lambda_2(y-s_2^2) $$

Here $s_1,s_2$ are slack variables to produce the restrictions $x\ge 0, y\ge 0$

The stationary points are the solutions for

$$ \nabla L = 0 = \cases{\frac 1x+\frac{1}{x+y}+\lambda_1 - 2=0\\ \frac{1}{x+y}+\lambda_2-3=0\\ x-s_1^2=0\\ y-s_2^2\\ \lambda_1s_1=0\\ \lambda_2s_2=0 } $$

giving

$$ \left( \begin{array}{ccccccc} f & x & y & \lambda_1 & \lambda_2 & s_1^2 & s_2^2\\ -2 & 1 & 0 & 0 & 2 & 1 & 0 \end{array} \right) $$

The only real solution is represented by the first numeric row. We can conclude also that the restriction $y \ge 0$ is active because $s_2^2=0$

Answer 3

Elementary solution (no Lagrangian, etc.)

Firstly observe that the objective is concave because it is a sum of concave functions (the concavity of $\log(x+y)$ follows from the concavity of $\log$ and the fact that $x+y$ is affine).

Next, since the derivatives \begin{align*} \frac{\partial f}{\partial x} &= \frac{1}{x} + \frac{1}{x+y} - 2\\ \frac{\partial f}{\partial y} &= \frac{1}{x+y} - 3 \end{align*} do not vanish simultaneously in the first quadrant ($x,y>0$), the maximum must be attained at the boundary of this set. Therefore the optimal point has $x=0$ or $y=0$. Since $x=0$ can be discarded we are left with a point of the form $(x,0)$. For such points the objective becomes $$ g(x) = f(x,0) = 2\log(x) - 2x, $$ which reaches its maximum when $g'(x)=0$, i.e., $$ \frac{2}{x}-2 = 0. $$ Hence $x=1$ and the optimal point is $(1,0)$ with optimal value $g(1)=2\log(1)-2=-2$.