Lecture 10, October 15
• Critical points. Suppose (x0 , y0 ) is a critical point of f (x, y) in the sense that
fx (x0 , y0 ) = fy (x0 , y0 ) = 0.
To determine the behaviour of f at that point, one looks at the expression
D = fxx (x0 , y0 )fyy (x0 , y0 ) − fxy (x0 , y0 )2 .
(a) If D > 0 and fxx (x0 , y0 ) > 0, then f has a local minimum at (x0 , y0 ).
(b) If D > 0 and fxx (x0 , y0 ) < 0, then f has a local maximum at (x0 , y0 ).
(c) If D < 0, then f has a saddle point at (x0 , y0 ).
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Example 1. Let f (x, y) = x2 − xy + y 2 − 2x − 2y. Then the critical points satisfy
0 = fx (x, y) = 2x − y − 2,
0 = fy (x, y) = −x + 2y − 2.
We multiply the first equation by 2 and then add it to the second equation to get
0 = 3x − 6
=⇒
x=2
=⇒
y = 2x − 2 = 2.
This shows that (2, 2) is the only critical point, while
2
D = fxx fyy − fxy
= 2 · 2 − (−1)2 = 3.
Since D > 0 and fxx = 2 > 0, the critical point (2, 2) is a local minimum.
Example 2. Let f (x, y) = 3xy − x3 − y 3 . To find the critical points, we solve
0 = fx (x, y) = 3y − 3x2 ,
0 = fy (x, y) = 3x − 3y 2 .
These equations give y = x2 and also x = y 2 , so we easily get
x = y 2 = x4
=⇒
x4 − x = 0
=⇒
x = 0, 1.
Since y = x2 , the only critical points are (0, 0) and (1, 1), while
2
D = fxx fyy − fxy
= (−6x)(−6y) − 32 = 36xy − 9.
At the critical point (0, 0), we have D = −9 and we get a saddle point. At the critical
point (1, 1), we have D > 0 and fxx = −6x < 0, so we get a local maximum.