Raising the bar The right solution to Example 6.5: just follow the Kuhn-Tucker recipe

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

$h_1(x_1,x_2)=x_1\geq 0$ , $h_2(x_1,x_2)=x_2\ge q 0$ , $h_3(x_1,x_2)=x_1^2+x_2^2\geq 1.$

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 $w_2=w_1.$