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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.

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

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