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Longhand

Before the final

Misapplying the second derivative test in two variables

The habit: Work out D = f_xx f_yy − (f_xy)², with the mixed partial squared, and read its sign first: D < 0 is a saddle whatever f_xx says, and f_xx decides only when D > 0.

3 practice questions are built to catch it, in 1 topic.

Practise against it: 2 questions

What it looks like

A question from optimization in two variables and Lagrange multipliers, with the answer this slip gives beside the right one.

Introductory2 marks · 3 min

The function f(x,y)=x2+4xy+y2−6x−6yf(x, y) = x^{2} + 4xy + y^{2} - 6x - 6y has a critical point at (1,1)(1, 1). What does the second derivative test say about it?

With the slip
It is a local minimum
Worked through
It is a saddle point

f_xx = f_yy = 2 are positive, but the mixed partial f_xy = 4 enters D too: work out D = f_xx f_yy − (f_xy)².

See the worked solution

Where it comes up

Practise against it

Each of these has an answer this slip gives. Work them on paper: if you fall into it, the verdict names it and the question comes back in review.

  1. 01
    How many of the critical points of f(x,y)=x3−3x+y3−3y2f(x, y) = x^{3} - 3x + y^{3} - 3y^{2} are saddle points?

    Optimization in two variables and Lagrange multipliersStandard

  2. 02
    The function f(x,y)=x4−x2+2xy−y2f(x, y) = x^{4} - x^{2} + 2xy - y^{2} has exactly one critical point, (0,0)(0, 0). What does the second derivative test conclude about it?

    Optimization in two variables and Lagrange multipliersStandard

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