Optimization in two variables and Lagrange multipliers practice question 10
The function has exactly one critical point, . What does the second derivative test conclude about it?
Find , and at , then . Which case of the test does its value fall in?
, , . Remember that is squared.
Theory UBC Math 100 §16.1.2Key results
Worked solution
Strategy
Work out D at the critical point and read its sign first. D > 0 and D < 0 each give a verdict; D = 0 gives none, and the point has to be settled another way.
Working
f_y = 0 forces y = x, and then f_x = 4x³ = 0 forces x = 0: the one critical point is (0, 0).
f_xx = 12x² − 2 at x = 0; the other two are constants.
D = 0, so the second derivative test draws no conclusion.
More work settles it: f(0, 0) = 0, but arbitrarily close by f is positive along y = x and negative along the x-axis. So (0, 0) is in fact a saddle point, which the test could not detect.
Answer
The answer is The test gives no conclusion, because D = 0. (Further work shows (0, 0) is a saddle point.).
Check
f(0.1, 0.1) = 0.0001 is above f(0, 0) = 0 and f(0.1, 0) = −0.0099 is below it, so (0, 0) is neither a maximum nor a minimum.
Where this goes wrong
Forgetting to square f_xy, which gives D = 4 − 2 = 2 and a false local maximum. With the square, D = 0 and the test is silent; the quadratic part −x² + 2xy − y² = −(x − y)² is flat along y = x, where the x4 term takes over.
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
Saved in this browser only.