By Seán Dineen (auth.)
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Additional resources for Functions of Two Variables
Strictly speaking we should suppose at this stage that G also depends on () since the c which appears in the mean value theorem may not be unique and thus not fully determined by x and ~x. However, we now put all the above information together and arrive at a formula which gives us a suitable approximation and at the same time shows that G does not depend on (). Our way of writing G was just a little premature. We have f(x + ~x,y + ~y) - f(x,y + ~y) = g(x + ~x) - g(x) = g'(x + ()~x) . ~x of = ox (x + (}~x, y + ~y) .
1)xe zy ) °. °= eO . 1 = 1. Also = (x - 1)xe zy ; x 0 8 8x8y = 8y8x = 8y (e ZY (1 + xy - y)) = xe zY (1 + xy - y) + eZY(x - 1). 82 f 80 88- (O,1)=O·eO(1+0·1-1)+eO(O-1)=-1. x y Functions of two variables 34 We have Hf(O,l) = (_~ -~) and det (Hf(O,l») and (0,1) is a saddle point of the function = -1 < 0 f. Example 29. Let f(x, y) = x 3 + y3 - 3xy. Then Vf = (3x 2 - 3y,3y2 - 3x). For critical points we must solve 3x 2 - 3y = 0 3y2 - 3x = O. e. X4 = x so X4 - X = O. e. x = 1. (X 2 )2 Hence x(x 3 - If x = 1 then y = x 2 = 1 and if x = 0 then y = O.
2 Many directions may lead to an increase in f. Can we choose a direction which gives the maximum rate of increase for f? IT we look at the unit circle vi + v~ = 1 we see that each point gives us a direction and we would ~~ (x, y). The problem can be posed as follows: find v = (VI, V2) satisfying vi + v~ = 1 so like to choose the direction v = as to maximize VI (VI, V2) af ax (x, y) so as to maximize af + V2 ay (x, y) (where (x,y) is some fixed point with V'f(x,y) I:- (0,0)). Having found v what is the maximum value of ~~(x, y)?