12.5 Taylor formulas with remainders 639
In case AC - B2 > 0, the quantity in braces will be nonnegative whenever jhi
and Jkl are small enough to make
(8) (14 + EI)(C + E3) - (B + E2)2 > 0.
It follows that if (4) holds and
(9) Gxx(xo,yo) > 0, Gxx(xo,Yo)Gyy(xo,yo)- [Gx.y(xo,yo)]2 > 0,
then G must have a minimum at (xo,yo) because in this case .4 + El must have the
same sign as .4 and the right member of (7) must be nonnegative when jhj and
iki are sufficiently small. Similarly if (4) holds and
(10) Gxx(xo,Yo) < 0, Gxx(xo,Yo)Gyy(xo,Yo) - [Gxy(xo,Yo)]2 > 0,
then G must have a maximum at (xo,yo) because in this case the right member of
(7) must be nonpositive when jhj and Jki are sufficiently small. In case (4)
holds and 4C - B2 < 0, that is,
(11) G=x(xo,yo)G."(xo,Yo)- [Gxv(x0,Y0)J2 < 0,
the function G cannot have an extremum at (xo,yo). We omit proof of this fact,
and we also omit discussion of the way in which Taylor formulas having more
terms can (when the required derivatives exist) be used to discuss cases in which
.4C-B2=0.
8 Supposing that a 0 and b2 - 4ac 0 0, find the extrema of the function
G for which
(1) G(x,y) = axe + bxy+ cy2.
Remark: When solving problems of this nature, it is usually safe to set
z = axe + bxy + cy2
and use the curly dee notation for partial derivatives. Outline of solution: The
system of equations ax = 0,az= 0 is satisfied only when x = y = 0. The
ay
formulas
(2)
8a2Z
x2 =2a, ax2aye-Cay 8X =4ac - b2
and the italicized statements of Problem 7 show that z has a minimum at (0,0)
if a > 0 and 4ac - b2 > 0, that z has a maximum at (0,0) if a < 0 and 4ac - bo
> 0, and that z has neither a maximum nor a minimum at (0,0) if 4ac - b"- < 0.
Remark: The examples
(3) z=x2+y2, z= -x2-y2, z=x2-Y2
illustrate the three phenomena. The examples
(4) z = 0, z = (2x - y)2
illustrate the cases excluded from this problem by the condition b2 - 4ac s 0.
9 Supposing that a 0 0, determine the points (x,y), if any, at which the
function G defined by
G(x,y) = x3 - 3axy + y3