The right solution to Example 6.5

Dec 17

The right solution to Example 6.5

The right solution to Example 6.5

The treatment of Example 6.5 in Baltovic's guide is confusing. The exposition indicates a problem but does not provide the explanation. Before reading this post try to solve the exercise on your own following the economical way.

Example 6.5. Consider the cost-minimisation problem of a consumer:

minimise $f(x_1,x_2)=w_1x_1+w_2x_2$ subject to $h_1(x_1,x_2)=x_1\geq 0$ , $h_2(x_1,x_2)=x_2\geq 0$ , $h_3(x_1,x_2)=x_1^2+x_2^2\geq 1.$

Don't forget that in case of minimization the lambdas in the Lagrangian should be taken with negative signs. It is assumed that $w_1,w_2>0.$

Case 1. Internal solutions are impossible because the first order conditions for $f$ give $w_1=w_2=0.$

Denote the boundaries $b_1=\{x_1=0\},$ $b_2=\{x_2=0\},$ $b_3=\{x_1^2+x_2^2=1\}.$ The Kuhn-Tucker conditions are:

(1) $\lambda_1\geq 0,$ $x_1\geq 0$ , $\lambda_1x_1=0$

(2) $\lambda_2\geq 0$ , $x_2\geq 0$ , $\lambda _2x_2=0$

(3) $\lambda_3\geq 0$ , $1\leq x_1^2+x_2^2,$ $\lambda_3(x_1^2+x_2^2-1)=0$

(4) $w_1-\lambda_1-2\lambda_3x_1=0$

(5) $w_2-\lambda_2-2\lambda_3x_2=0$

The constraint qualification condition is obviously satisfied for $b_1,b_2,b_3$ considered separately.

Case 2. $x$ belongs to $b_{1}$ only. Then $x$ does not belong to $b_{2}$ and $b_{3}$ and from complementary slackness $\lambda_2=\lambda_3=0.$ Then from (5) $w_2=0,$ which is nonsense.

Case 3. Similarly, if $x$ belongs to $b_2$ only, then $w_1=0,$ which is impossible.

Case 4. $x$ belongs to $b_3$ only. Then $x_1,x_2>0$ and from complementary slackness $\lambda_1=\lambda_2=0.$ (3)-(5) simplify to $x_1^2+x_2^2=1$ , $w_1-2\lambda_3x_1=0$ , $w_2-2\lambda_3x_2=0.$ The solution to this system is

$x^{(0)}=\left(\frac{w_1}{\sqrt{w_1^2+w_2^2}},\frac{w_2}{\sqrt{w_1^2+w_2^2}}\right),$ $\lambda=\frac{\sqrt{w_1^2+w_2^2}}{2}$

and the value of the objective function at this point is

$f(x^{(0)})=\sqrt{w_1^2+w_2^2}.$

Case 5. The only possibilities left are $x^{(1)}=(0,1)\in b_1\cap b_3,$ $x^{(2)}=(1,0)\in b_2\cap b_3.$ Don't bother checking the constraint qualification for these points because a) it may fail, in which case the Kuhn-Tucker theorem does not apply, even though any of these points can be a minimum point, and b) even if it holds, none of these points may be a minimum point (the Kuhn-Tucker theorem provides just a necessary condition). Just check directly the values of $f$ at these points:

$f(x^{(1)})=w_2,$ $f(x^{(2)})=w_1.$

Since $w_1,w_2$ are strictly positive, we see that $f(x^{(0)})=\sqrt{w_1^2+w_2^2}>\max\{w_1,w_2\}.$ Thus, $f(x^{(1)})$ is the minimum if $w_2<w_1,$ $f(x^{(2)})$ is the minimum if $w_2>w_1$ and we have two minimum points in case of a tie $w_2=w_1.$