Q7Advanced Engineering Mathematics

Given function: $f(x_1,x_2)=x_1^3+x_2^3+9x_1^2+18x_2^2+144$.

First, find the partial derivatives and set them to zero: $\frac{\partial f}{\partial x_1}=3x_1^2+18x_1=3x_1(x_1+6)=0⟹x_1=0,-6$

$\frac{\partial f}{\partial x_2}=3x_2^2+36x_2=3x_2(x_2+12)=0⟹x_2=0,-12$

The critical points are therefore $(0,0),(0,-12),(-6,0),(-6,-12)$.

Calculate second-order derivatives: $r=\frac{{\partial }^{2}f}{\partial x_1^2}=6x_1+18$, $t=\frac{{\partial }^{2}f}{\partial x_2^2}=6x_2+36$, $s=\frac{{\partial }^{2}f}{\partial x_1\partial x_2}=0$.

Evaluate $rt-s^2$ at each critical point to classify it. For example, at $(0,0)$, $r=18>0,t=36>0,rt-s^2=(18)(36)>0$, so it is a local minimum.