Optimization

• To find a maximum or minimum in some interval, take the derivative, and check all points where the derivative is 0, or undefined, or at the ends of the interval

Multiple Variables

• For multiple variables, find where the Gradient is $\overrightarrow{0}$
  • Note that the Gradient being $\overrightarrow{0}$ does not necessarily imply that that point is a local extrema, since it could be a saddle point
  • This can be difficult since in general, you will get a non-linear system of equations
• To figure out if the critical point is a local max or min or saddle point (these tests are on the formula sheet):
  • If $det(Hf(\overrightarrow{a}))>0$ and $f_{xx}(\overrightarrow{a})>0⟹f(x,y)-f(a,b)>0$ then $\overrightarrow{a}$ is a local min
  • If $det(Hf(\overrightarrow{a}))>0$ and $f_{xx}(\overrightarrow{a})<0⟹f(x,y)-f(a,b)<0$ then $\overrightarrow{a}$ is a local max
  • If $det(Hf(\overrightarrow{a}))<0$ then $\overrightarrow{a}$ is a saddle point
  • If $detHf(\overrightarrow{a})=0$ then you're cooked (degenerate critical point)
• $Hf(\overrightarrow{a})$ is the Hessian Matrix
• When finding absolute min/max, make sure to find the min/max of the entire boundary
  • You might have to use Lagrange Multipliers
  • You could also turn the function into a Parametric Equations

Why This Works

• Taylor Series
• $T_{2,p_0}(x,y)=L_{a,b}+A(x-a{)}^2+B(x-a)(y-b)+C(y-b{)}^2$
• there's some crazy algebra going on here but it turns out if you just extend the stuff from 2-d to 3-d it kinda works