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Unformatted text preview: Optimization http2/ [www.mantexasedu/users/ gilbert/ asciiinatthptimi zati on. iron; UPTIMEZATIGN The surface to the right is actually the graph of
3 = fix, I) = Sianl sintif), — Tc ‘1 x, y < 7r, a Google map 'street view' of part of Yosemite, perhaps! it has
mountains, Local Maximal, and it has valieys, Locai Miniina,
both of which occured for curves. But it aiso has a pass
through the mountains at which the tenain slopes up in one
direction and down in another direction just iike a saddie. So
we also have to bring in the notion of Saddlepoints,
corresponding to a pass through the mountains. But, just as in
the one variable case, everything centers on an algebraic and
graphical understanding of locai extrema and critical points. j Berni/xiii:
A function 2 x f (x, y) is said to have a Local Extreinnin at (a, b) if
9 Local Maximum: foe, y) S ﬂa, b) for all (x, y) near (a, h), a Local Minimum: ﬂx, y) 2 f (a, b) for all (x, 32) near (a, b). In one variable locating local extrema usually meant ﬁnding where f'(x) m 0. For a ﬁrnction f(x, y) let's see how V f gets
involvad: suppose V f (a, b) at: 0. Then the vector V ﬂat, [2) will point in a direction of positive slope on the graph of f ,
and so we can ﬁnd a larger value of f (x, y) than f(a, 5) near (a, b) by moving in that direction. But then (a, 5) cannot be a locai maximum of f . How would you modiﬁ/ this argument for a locai minimum (remember WV f points in the opposite
direction to V f)? mi N. M ' Deﬁition: A point (a, b) is said to be a Critical Point of f (x, y) when V ﬂat, .6) = O or at Eeast one of fx(a, b), I];(a, 15) does not exist. _ Two important observations: i. if(a, b) is a local extremurn, then (a, b) is a critical point. 2. if j;(a, b), ﬁia, b) both exist, the tangent plane is horizontal at the point
((a, b}, f (a, 13)) on the graph off when (a, b) is a critical point. 1 of4 11/30/2009 4:54 PM Optimization http://www .maartexas.edu/users/gilbere’asciimath/Optimi zationhtnil Critical points, along with the location and classiﬁcation of local extrema, can thus he studied algebraicaliy and
graphically. w. Example: to the right is the earlier graph of 3 = .f(x: 2 3mm) Sing): “7: < x: y < K? now with axes. By the Product Rule=
fi(x, y) 2 0080031110}, ﬁx, :v) = sintx} 003(9 Thus 1;, fy are always deﬁned for 7t < x, y < 7:, so the only critical points occur at solutions of the equations
)2 = O, w 0. These we ﬁnd to be {0, 0) and e a (a a (a a a a The tangent plane is supposed to be horizontal at all these points. Is that clear ﬁom the graph? ﬂmﬁmwmwumwmmmnmmmﬂm The graph off(x, y) = sin(x) sin(y) also helps classify these critical points: 7: 2r 7: 7r
a h (— _)’ w W, n m)
f has alocal maxrmum at 2, 2 ( 2 2 , e f has a local minimum at (— g, g), (g, m Near (O, 0), howevar, f looks like a 'saddle‘, and we formalize this in general with ” 'ijﬁﬁiiiéiﬁ
A critical point (a, b) of a function f is said to be a Saddle Point if,
no matter how ciose (x, y) is to (a, b) there are always points at which f(x, y) < f (a, b) and other points at which f (x, y) > f (a, b). So a critical point (a, b)is either a iocai maximum, at local minimum, or a saddle point. But how can we determine which?
in the case ofﬂx, y) 2 sin(x) sin(y), knowing the graph was enough. There are other ways. 20M 11/30/2009 4:54 PM Optiini and on 30M [E
t
i
i“
l , MWmmmw E
i W ‘ ____ Using gradients: suppose we have a plot of the gradient Wrmhmvmmmﬁrmmmmm A about S and T? mwm  i  WWWMmmmmnwmwm vectors of f as shown to the right for for, y) m 83106) Sinbd with the arrows indicating the direction of(V f)(x, y) and j
the length of the arrow indicating its length KV f)(x, 32)]. This vector points in the direction of maximum slope of
the graph off at (x, y, for, y)) and has length equal to the value of that maximum slope. So, as (x, y) approaches a local maximum point (a, b), the arrows all point inward
to (a, b) and get secessively smaller. Why inward? This ‘
happens when P m (a, b) is the locai maximum in the ﬁrst quadrant. At the iocai minimum Q in the second quadrant, however, the arrows all point outward. Why outward? But at the saddle point R, the arrows get successively
smalier as (x, y) approaches the origin, yet some arrows 3
always point inward and some outward, because the graph siopes both up and down near (0, 0). So what can you say MAW We?sz ' . . hm .fo: m $110”) Sintv): as shown to the right with the contours f (x, y) "e" k shaded so that the darker the color the smaller 16 is. Some contours are labelled with a speciﬁc value ofk. At a local
minimum like Q, the contours encircle Q and the color é
darkens as the contours shrink to Q. Why shrink? And at a local maximum like S, the contours again encircle and shrink to S, except that now the coiors iighten because the height increases as we approach a local maximum. At a _
saddle point, like R, however, the contours won't encircle R because in some directions from R the slope is increasing, while in others it is decreasing. Notice that at R there are two straight contour fines through R having the same height so the tangent plane at R is certainly horizontal! l WmswwwwwwwwmmwmwmmmwmmwyWWWWMWMWWlwiim:Wmmmmvmwmmﬁwmwwwmm a . M if“ m
E Contour map: suppose now we have the contour map for http:f/www.rnautexas.edu/users/gilbeny’asciirnatthptimization.hnnl f ‘F' 2" w («r W ,W‘pcvsrovwm 4w .4, I“? WHM’
"er/WW“ xx; l ‘0‘”
e’ll\‘*xr*l\\" iiilgi Cl}.
*gliél
\‘Y AWWWW/WW mwma wankmem Wmmm These make good visuais to explain what's going on, but for most ﬁinctions the second order partial derivatives turn out to
be more useﬁil just as it was for functions of one variable. Suppose f (x, y) is a function with continuous second order partial derivatives and iet A a few, 5), B = ﬂylaa 5), C a [was a) he the values of these second order derivatives at a point (a, b). To determine whether (a, b) is a Iocai maximum, iocai
minimum or a saddle point, we need to check the sign of the Discriminant D m D{a, 15): ACWBQ. 1 1/30/2009 4:54 PM Optimization http://www.ma.u1:exas.edu/users/gilbert/asciimatlﬂOptimizationhmﬁ ‘ Second Derixlative Test: at a oritioal point (o, b}, a ﬁmc’tion hes. a  l u  9 Local Maximum: ifD > 0 and fn(a, b) < O, @ Local Minimum: if]? > O and fxx(a, b) > 0, @ Saddle Point: ifD < O. So what about D == 0? Well, here the test fails. It tells us nothingll The example of f (x, y) m sin(x) sin(y) illustrates this well. For then ﬂag, 3/) W mama) Slﬂ(y), fxy(x, y) = cos(x)cos(y), I 13,},(x, y) : wsin(x) Sinbl), conﬁrming what was already clear from all the graphs while at (0, 0): 198:0,Dm1>0>§A=C=O’B:E’D=_1<0’§ f has a saddle point. 40f4  11/30/2009 4:54 PM ...
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This note was uploaded on 03/24/2010 for the course M 57500 taught by Professor Gilbert during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Gilbert

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