Multivariate Calculus - Bruce E. Shapiro

LECTURE 14. UNCONSTRAINED OPTIMIZATION 109
This is the function we need to minimize. We define
f(x, y) = xy + 512
y + 512
x
Taking derivatives,
Stationary points occur when
f x = y − 512/x 2
f y = x − 512/y 2
Thus
0 = y − 512/x 2 and 0 = x − 512/y 2
y = 512/x 2 = 512/(512/y 2 ) 2 = y 4 /512
⇒ 512y − y 4 = 0
⇒ y(512 − y 3 ) = 0
⇒ y = 0 or y = 8
Since we also required x = 512/y 2 , we find that at y = 0, x → ∞ hence there is
no stationary point in this case.
At y = 8 we have x = 512/8 2 = 8. Hence the only stationary point is (8, 8). We
now proceed to test this stationary point. The second derivatives are
Thus
and
f xx = 1024/x 3
f xy = 1
f yy = 1024/y 3
f xx (8, 8) = 1024/(8 3 ) = 1024/512 = 2
f xy (8, 8) = 1
f yy (8, 8) = 1024/(8 3 ) = 1024/512 = 2
d = D(8, 8) = f xx (8, 8)f yy (8, 8) − f 2 xy(8, 8) = (2)(2) − (1) 2 = 3 > 0
Since d > 0 and f xx (8, 8) > 0, we conclude that f(x, y) has a local minimum at
(8,8). The third dimension is
z = 256/xy = 256/64 = 4
The dimensions of the minimum area box are 8 by 8 by 4, and its total area is
A = xy + 2xz + 2yz = (8)(8) + 2(8)(4) + 2(8)(4) = 192
Math 250, Fall 2006 Revised December 6, 2006.