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The expression B2 AC also has the property of invariance with respect to plane rotations " " x x cos  y sin  " " y x sin  y cos  It has been discovered that with the identi cations A fxx ; B fxy ; C fyy , we have the partial deri2 vative form fxy fxx fyy that characterizes relative extrema. The demonstration of invariance of this form can be found in analytic geometric books. However, if you would like to put the problem in the context of the second partial derivative, observe that fx fx " fy fx " @x @y fy fx cos  fy sin  " " @x @x
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@x @y fy fx sin  fy cos  " " @y @y
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Then using the chain rule to compute the second partial derivatives and proceeding by straightforward but tedious calculation one shows that
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2 2 fxy fxx fyy fxy fxx fyy : "" "" ""
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The following equivalences are a consequence of this invariant form (independently of direction in the tangent plane at P0 ):
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2 fxy fxx fyy < 0 2 fxy
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fxx fyy > 0 fxx fyy < 0
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fxx fyy > 0
The key relation is (1) because in order that this equivalence hold, both fx fy must have the same sign. We can look to the one variable case (make the same argument for each coordinate direction) and conclude that there is a relative minimum at P0 if both partial derivatives are positive and a relative maximum if both are negative. We can make this argument for any pair of coordinate directions because of the invariance under rotation that was established. If (2) holds, then the point is called a saddle point. If the quadratic form is zero, no information results. Observe that this situation is analogous to the one variable extreme value theory in which the nature of f at x, and with f 0 x 0, is undecided if f 00 x 0.
8.22. Find the relative maxima and minima of f x; y x3 y3 3x 12y 20.
fx 3x2 3 0 when x 1; fy 3y2 12 0 when y 2. Then critical points are P 1; 2 , Q 1; 2 ; R 1; 2 ; S 1; 2 . 2 fxx 6x; fyy 6y; fxy 0. Then fxx fyy fxy 36xy. At P 1; 2 ; > 0 and fxx (or fyy > 0; hence P is a relative minimum point. At Q 1; 2 ; < 0 and Q is neither a relative maximum or minimum point. At R 1; 2 ; < 0 and R is neither a relative maximum or minimum point. At S 1; 2 ; > 0 and fxx (or fyy < 0 so S is a relative maximum point. Thus, the relative minimum value of f x; y occurring at P is 2, while the relative maximum value occurring at S is 38. Points Q and R are saddle points.
8.23. A rectangular box, open at the top, is to have a volume of 32 cubic feet. dimensions so that the total surface is a minimum
If x, y and z are the edges (see Fig. 8-7), then 1 2 Volume of box V xyz 32 Surface area of box S xy 2yz 2xz
What must be the
or, since z 32=xy from (1), S xy 64 64 x y Fig. 8-7
APPLICATIONS OF PARTIAL DERIVATIVES @S 64 y 2 0 when 3 @x x @S 64 x 2 0 when 4 @y y
[CHAP. 8
x2 y 64;
xy2 64
Dividing equations (3) and (4), we nd y x so that x3 64 or x y 4 and z 2.   128 128 128 2 For x y 4, Sxx Syy Sxy 1 > 0 and sxx 3 > 0. Hence, it follows that x3 y3 x the dimensions 4 ft 4 ft 2 ft give the minimum surface.
LAGRANGE MULTIPLIERS FOR MAXIMA AND MINIMA 8.24. Consider F x; y; z subject to the constraint condition G x; y; z 0. Prove that a necessary condition that F x; y; z have an extreme value is that Fx Gy Fy Gx 0.
Since G x; y; z 0, we can consider z as a function of x and y, say z f x; y . A necessary condition that F x; y; f x; y have an extreme value is that the partial derivatives with respect to x and y be zero. This gives 1 Since G x; y; z 0, we also have 3 Gx Gx zx 0 4 Gy Gz zy 0 Fx Fz zx 0 2 Fy Fz Zy 0
From (1) and (3) we have (5) Fx Gx Fx Gx 0, and from (2) and (4) we have (6) Fy Gz Fz Gy 0. Then from (5) and (6) we nd Fx Gy Fy Gx 0: The above results hold only if Fz 6 0; Gz 6 0.
8.25. Referring to the preceding problem, show that the stated condition is equivalent to the conditions x 0; y 0 where  F G and  is a constant.
If x 0; Fx Gx 0. If y 0; Fy Gy 0. Elimination of  between these equations yields Fx Gy Fy Gx 0. The multiplier  is the Lagrange multiplier. If desired we can consider equivalently  F G where x 0; y 0.