0
$\begingroup$

How do I compute the optimum choice of $x$ and $y$ which belong to $[0,1]$ here? I seem to be ending up in an inconsistency?

We wish to maximize $K - [l_x(1-x) + l_y(1-y)] - (\beta_xx + \beta_yy)^2$. And I take the partial derivative with respect to $x$ and $y$ respectively and I obtain that

$\beta_xx^* + \beta_yy^* = \dfrac{l_x}{2\beta_x} = \dfrac{l_y}{2\beta_y}$ but the last equality need not always hold.

What am I missing? Does the inconsistency simply mean that the optimal efforts are corner points?

1 Answers 1

0

What you re missing are multipliers for the $[0,1]$ constraints. Let me first get rid of the annoying constants and rewrite your problem as $\max_{(x,y)\in[0,1]^{2}}l_{x}x+l_{y}y-(\beta_{x}x+\beta_{y}y)^{2}$.

Lagrangian of this problem is $$L(x,y,\mu_{x},\mu_{y},\gamma_{x},\gamma_{y})=l_{x}x+l_{y}y-(\beta_{x}x+\beta_{y}y)^{2}-\mu_{x}(x-1)-\mu_{y}(y-1)+\gamma_{x}x+\gamma_{y}y$$

The Kuhn-Tucker system is: $$\begin{aligned} L_{x}(x^{*},y^{*},\mu_{x}^{*},\mu_{y}^{*},\gamma_{x}^{*},\gamma_{y}^{*})&=0\\ L_{y}(x^{*},y^{*},\mu_{x}^{*},\mu_{y}^{*},\gamma_{x}^{*},\gamma_{y}^{*})&=0\\ \mu_{x}^{*}(x^{*}-1)&=0\\ \mu_{y}^{*}(y^{*}-1)&=0\\ \gamma_{x}^{*}x^{*}&=0\\ \gamma_{y}^{*}y^{*}&=0\\ x^{*}&\in[0,1]\\ y^{*}&\in[0,1]\\ \mu_{x}^{*},\mu_{y}^{*},\gamma_{x}^{*},\gamma_{y}^{*}&\geq0\\ \end{aligned}$$

The Kuhn-Tucker theorem tells you that if $(x^{*},y^{*})$ constitute local maximum in your problem, then there exist $(\mu_{x}^{*},\mu_{y}^{*},\gamma_{x}^{*},\gamma_{y}^{*})$ such that $(x^{*},y^{*},\mu_{x}^{*},\mu_{y}^{*},\gamma_{x}^{*},\gamma_{y}^{*})$ constitutes a solution to the Kuhn-Tucker system (under constraint qualification which holds here since your constraints are cut-out by affine functions).

That is, by finding all solutions to the system, you isolate a set of points that must include your optimum.

Now coming to your inconsistency. It is driven by the fact that the first order conditions (without multipliers) are $$\begin{aligned} l_{x}-2(\beta_{x}x+\beta_{y}y)\beta_{x}&=0\\ l_{y}-2(\beta_{x}x+\beta_{y}y)\beta_{y}&=0\\ \end{aligned}$$ which can holds simultaneously only if $l_{x}/\beta_{x}=l_{y}/\beta_{y}$. If this equality fails, then the multipliers will ensure that the first order conditions (with multipliers) are going to be satisfied. From the complementary slackness conditions (equations 3-6 of the system) this means that either $x$ or $y$, or both, will be at the corner solution , either at zero or unity.

Let me also comment on your question. Your are probably thinking of all the constant as being strictly positive but do not explicitly specify it. The better the question you pose, the better the answers.