Multivariate Calculus - Bruce E. Shapiro

106 LECTURE 14. UNCONSTRAINED OPTIMIZATION
Example 14.4 . Find and classify the stationary points of f(x, y) = xy
Solution. Differentiating,
f x = y
f y = x
The only stationary point occurs when x = y = 0, i.e., at the origin. Furthermore,
f xx = f yy = 0
f xy = 1
Therefore
D(x, y) = f xx f yy − f 2 xy = (0)(0) − (1) 2 = −1 < 0
for all x, y. We conclude that (0, 0) is a saddle point.
Example 14.5 Find and classify all the stationary points of f(x, y) = e −(x2 +y 2 −4y)
Solution. Differentiating,
f x = −2xe −(x2 +y 2 −4y)
f y = (−2y + 4)e −(x2 +y 2 −4y)
Setting f x = 0 gives
−2xe −(x2 +y 2 −4y) = 0
⇒ x = 0
Setting f y = 0 gives
(−2y + 4)e −(x2 +y 2 −4y) = 0
⇒ −2y + 4 = 0
⇒ y = 2
The only stationary point is at (0, 2). Next, we calculate the second derivatives.
f xx = −2e −(x2 +y 2−4y) + 4x 2 e −(x2 +y 2 −4y)
= (−2 + 4x 2 )e −(x2 +y 2 −4y)
f xy = −2x(−2y + 4)e −(x2 +y 2 −4y)
f yy = −2e −(x2 +y 2−4y) + (−2y + 4) 2 e −(x2 +y 2 −4y)
= (−2 + 4y 2 − 16y + 16)e −(x2 +y 2 −4y)
= (4y 2 − 16y + 14)e −(x2 +y 2 −4y)
At the critical point (0,2), we find that
f xx (0, 2) = (−2 + 4(0) 2 )e −(02 +2 2−4(2)) = −2e 4
f xy (0, 2) = −2(0)(−2(2) + 4)e −(02 +2 2−4(2)) = 0
f yy (0, 2) = (4(2) 2 − 16(2) + 14)e −(02 +2 2−4(2)) = −2e 4
Revised December 6, 2006. Math 250, Fall 2006