Critical Points and Optimisation in Two Variables
Find points where the gradient vanishes and classify them with the second derivative test, including the saddle point, which has no one-variable analogue.
What a learner can do afterwards
- Locate all critical points by solving the gradient equal to zero
- Classify a critical point with the discriminant of the second partials
- Describe a saddle and say why one-variable tests miss it
1 · Read
Critical points are where the gradient vanishes, the flat candidates for peaks, pits, and saddles. Solve fx = 0 and fy = 0 together. For x squared minus 2x + y squared + 6y + 10: 2x minus 2 = 0 gives x = 1, 2y + 6 = 0 gives y = minus 3, and the value there is 0.
The discriminant D = fxx fyy minus fxy squared sorts each candidate. For x squared + y squared: fxx = 2, fyy = 2, fxy = 0, so D = 4 with fxx > 0, a local minimum at the origin. For x squared minus y squared: D = minus 4, a saddle.
Read the table: D > 0 with fxx > 0 is a min, D > 0 with fxx < 0 is a max, D < 0 is a saddle. Jay is right that negative D always means a saddle. Keep this table beside you until the cases feel routine.
The saddle is the new animal here: up in one slice and down in another. Single variable tests only check axis slices, so they can miss the mix entirely. When D goes negative, picture a mountain pass, not a peak or a pit.
Solve gradient zero for candidates, then let D and the sign of fxx deliver the verdict.
2 · Watch
Take it off screen
Where it sits
This opens up
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.