Optimization in two variables and Lagrange multipliers practice question 2
The function has a critical point at . What does the second derivative test say about it?
The second derivative test combines all three second partials, the mixed one included, into one number. Which part of the test do you read first?
Work out , and , then . Read the sign of before anything else.
Theory UBC Math 100 §16.1Key results
Worked solution
Strategy
The second derivative test: compute D = f_xx f_yy − (f_xy)² at the critical point. Its sign is read first; only when D > 0 does the sign of f_xx decide between a maximum and a minimum.
Working
Confirms that (1, 1) is a critical point.
The second partials are constants because f is quadratic.
The mixed partial is squared, and here it outweighs the product of the other two.
Along the line y = x, f rises away from (1, 1); along y = 2 − x it falls.
Answer
The answer is a saddle point, since D = −12 is negative.
Check
f(1, 1) = −6, but f(1.1, 1.1) = −5.94 is higher and f(1.1, 0.9) = −6.02 is lower, so (1, 1) is neither a maximum nor a minimum.
Where this goes wrong
Calling it a local minimum because f_xx and f_yy are both positive. The mixed partial can override them: D = 4 − 16 is negative, so the surface curves up in some directions and down in others.
The idea behind it
The optimization in two variables and Lagrange multipliers lesson, a figure to move, in a few steps
Leans on
Second partial derivatives, from Functions of two variables and partial derivatives
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