Question 14.5.2: Consider the problem max(min) f (x, y) = x² + y² s.t. g(x, y......
Consider the problem
max(min) f (x, y) = x² + y² s.t. g(x, y) = x² + xy + y² = 3
In Example 14.3.1 we saw that the first-order conditions give the points (1, 1) and (−1,−1) with λ = 2/3, as well as (\sqrt{3},−\sqrt{3}) and (−\sqrt{3},\sqrt{3}) with λ = 2. Check the local second-order conditions of Theorem 14.5.2 in this case.
We find that f_{11}^{\prime\prime}=2,f_{12}^{\prime\prime}=0,f_{22}^{\prime\prime}=2,\,g_{11}^{\prime\prime}=2,\,g_{12}^{\prime\prime}=1,\,\mathrm{and}\,g_{22}^{\prime\prime}=2. So
D(x,y,\lambda)=(2-2\lambda)(x+2y)^{2}+2\lambda(2x+y)(x+2y)+(2-2\lambda)(2x+y)^{2}Hence D(1,1,\textstyle{\frac{2}{3}})=D(-1,-1,\frac{2}{3})=24\ \mathsf{a n d}\ D(\sqrt{3},-\sqrt{3},2)=D(-\sqrt{3},\sqrt{3},2)=-24. From the signs of D at the four points satisfying the first-order conditions, we conclude that (1, 1) and (−1,−1) are local minimum points, whereas (\sqrt{3},−\sqrt{3}) and (−\sqrt{3},\sqrt{3}) are local maximum points.^{15}
^{15} In Example 14.3.1 we proved that these points were actually global extrema.