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Consider the following maximization problem with constraint: $$ \begin{cases} & \max f(x) + g(y)\\ & x f(x) + y g(y) = B \end{cases} $$

Functions $f(x)$, $g(y)$ and constant $B$ are defined.

I'm interested in getting general solution for this problem.

Stuck on solving it. Lagrange method didn't seem to help me.

Alex Velickiy
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  • What are the definitions of $x$ and $y$? – unseen_rider Aug 14 '16 at 10:54
  • What do you mean by a general solution? Since $f$ and $g$ are not defined explicitly, at most you will be able to get an implicit solution using first order conditions. There may be no solution - you haven't made enough assumptions to guarantee one. – smcc Aug 14 '16 at 10:59
  • @unseen_rider x, y are variables. Say they can get any real value. – Alex Velickiy Aug 14 '16 at 11:21
  • @smcc By general solution I thought of an equation on f, g, B, x, y. And all x,y that apply to this equation are solutions of this optimization problem. – Alex Velickiy Aug 14 '16 at 11:23
  • There may be no solution. Currently we cannot even say there are any $x$ and $y$ that satisfy the constraint. – smcc Aug 14 '16 at 11:26

1 Answers1

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The Lagrangian for the problem is

$$L(x,y,\lambda)=f(x)+g(y)+\lambda[xf(x)+yg(y)-B]$$

Let us assume that there is a solution of the optimization problem and that it is a critical point of the Lagrangian.

The critical point(s) satisfy:

$$L_x=f'(x)+\lambda[f(x)+xf'(x)]=0\tag{1}$$ $$L_y=g'(y)+\lambda[g(y)+yg'(y)]=0\tag{2}$$ $$L_\lambda=xf(x)+yg(y)-B=0\tag{3}$$

Suppose $\lambda=0$. There may be a critical point satisfying $f'(x)=g'(y)=0$ and $xf(x)+yg(y)=B$.

Suppose $\lambda\neq 0$. Then rearranging $(1)$ and $(2)$ and dividing the former by the latter, gives

$$\frac{f'(x)}{g'(y)}=\frac{f(x)+xf'(x)}{g(y)+yg'(y)}.$$

Solving for $y$:

$$y=x+\frac{f(x)}{f'(x)}-\frac{g(y)}{g'(y)}$$

Substitute into $(3)$ and solve to get:

$$x=\frac{Bf'(x)g'(y)+g(y)[f'(x)g(y)-f(x)g'(y)]}{[f(x)+g(y)] f'(x)g'(y)}$$

and

$$y=\frac{Bf'(x)g'(y)+f(x)[f(x)g'(y)-f'(x)g(y)]}{[f(x)+g(y)] f'(x)g'(y)}$$

which may be another critical point (assuming I did not divide by zero somewhere).

smcc
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