Exercise 6.2.5

We have with one constraint, $L(x,\lambda )=\frac{1}{2}(x_1^2+x_2^2+x_3^2)+\lambda (x_1+x_2+x_3-3)$

• $\frac{\mathit{\partial L}}{\partial x_1}=x_1+\lambda =0$
• $\frac{\mathit{\partial L}}{\partial x_2}=x_2+\lambda =0$
• $\frac{\mathit{\partial L}}{\partial x_3}=x_3+\lambda =0$
• $\frac{\mathit{\partial L}}{\mathit{\partial \lambda }}=x_1+x_2+x_3-3=0$

Solve it we find $x_1^{\text{∗}}=x_2^{\text{∗}}=x_3^{\text{∗}}=1$ and $\lambda =-1$.

With the second constraint we have $L(x,{\lambda }_1,{\lambda }_2)=\frac{1}{2}(x_1^2+x_2^2+x_3^2)+{\lambda }_1(x_1+x_2+x_3-3)+{\lambda }_2(x_1+2x_2+3x_3-12)$

• $\frac{\mathit{\partial L}}{\partial x_1}=x_1+{\lambda }_1+{\lambda }_2=0$
• $\frac{\mathit{\partial L}}{\partial x_2}=x_2+{\lambda }_1+2{\lambda }_2=0$
• $\frac{\mathit{\partial L}}{\partial x_3}=x_3+{\lambda }_1+3{\lambda }_2=0$
• $\frac{\mathit{\partial L}}{\partial {\lambda }_1}=x_1+x_2+x_3-3=0$
• $\frac{\mathit{\partial L}}{\partial {\lambda }_2}=x_1+2x_2+3x_3-12=0$

Solve it we find $x_1^{\text{∗}}=-2,x_2^{\text{∗}}=1,x_3^{\text{∗}}=4$ and ${\lambda }_1=5,{\lambda }_2=-3$.

The second minimum value should be less than the first minimum value because any solution that satisfies the two constraints must minimize the function with first constraint as well. If it’s larger, that will contradict with whatever the first problem finds.

The first problem has a plane tagent to a sphere in $R^3$, the second problem has a line tagent to a sphere in $R^3$.