Multivariate Calculus - Bruce E. Shapiro

LECTURE 14. UNCONSTRAINED OPTIMIZATION 107
Therefore
d = D(0, 2) = f xx f yy − f 2 xy = (−2e 4 )(−2e 4 ) − (0) 2 = 4e 4 > 0
Since f xx < 0 the point (0, 2) a local maximum.
Example 14.6 Find and classify the stationary points of
Solution. Proceeding as before
f(x, y) = x 2 + 1 − 2x cos y, −π < y < π
f x = 2x − 2 cos y
f y = 2x sin y
Setting these expressions equal to zero gives
x = cos y (14.3)
x sin y = 0 ⇒ x = 0 or sin y = 0 (14.4)
Equation 14.4 can be rewritten (using the fact that −pi < y < pi) as
x = 0 or y = 0 (14.5)
because sin y = 0 implies that y = ±kπ, k = 0, 1, 2, .... Therefore we have two cases
consider. The first case is
x = cos y and x = 0 (14.6)
The second case is
x = cos y and y = 0 (14.7)
From equation 14.6, if x = 0 then cos y = 0, which means y = ±π/2. So our first
two stationary points are
(0, π/2), (0, −π/2)
The second case (equation 14.7 ) has y = 0 and x = cos y = cos 0 = 1. So there is a
third stationary point at
(0, 1)
We now classify the critical points.
derivatives,
To do so we must first calculate the second
f xx = 2
f xy = 2 sin y
f yy = 2x cos y
Hence
D(x, y) = f xx f yy − f 2 xy = 4x cos y − 4 sin 2 y
Math 250, Fall 2006 Revised December 6, 2006.