Raising the bar

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 $latex f(x_1,x_2)=w_1x_1+w_2x_2$ subject to

$latex h_1(x_1,x_2)=x_1\geq 0$, $latex h_2(x_1,x_2)=x_2\ge q_0$, $latex h_3(x_1,x_2)=x_1^2+x_2^2\geq 1.$

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

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

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

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

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

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

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

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

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

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

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

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

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

and the value of the objective function at this point is

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

Case 5. The only possibilities left are $latex x^{(1)}=(0,1)\in b_1\cap b_3,$ $latex x^{(2)}=(1,0)\in b_2\cap b_3.$ 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 $latex f$ at these points:

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

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

Conclusion: the Kuhn-Tucker does work in this case!